CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 9
Liberty University
Question 1
Question
Consider a reaction with an activation energy of 50 kJ/mol. If the rate constant
at 25
°
C is 1.5×10−3s−1, what will be the rate constant at 45
°
C assuming the
frequency factor remains the same?
Solution
Step 1: We can use the Arrhenius equation to calculate the rate constant at
45
°
C. The Arrhenius equation is given by:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the frequency factor, - Eais the activation
energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the temperature in
Kelvin.
Step 2: First, we need to convert the activation energy from kJ/mol to
J/mol:
Ea= 50 kJ/mol ×1000 J/1 kJ = 50000 J/mol
Step 3: Next, we convert the temperature from Celsius to Kelvin for both
25
°
C and 45
°
C:
T1= 25 + 273 = 298 K
T2= 45 + 273 = 318 K
Step 4: Plug in the values into the Arrhenius equation for k1and k2:
k1= 1.5×10−3s−1
k2=A·e−50000
8.314·298
Step 5: Solve for the frequency factor A:
A=k1
e−50000
8.314·298
Step 6: Finally, calculate the rate constant at 45
°
C using the derived fre-
quency factor:
k2=A·e−50000
8.314·318
Question 2
Question
The rate constant for a reaction at 25
°
C is 2.0×10−3s−1, and the activation
energy for the reaction is 70 kJ/mol. Calculate the rate constant at 50
°
C for
this reaction.
Solution
Step 1: Calculate the Arrhenius constant, A, using the Arrhenius equation:
k=A·e−Ea
RT
Given: k1= 2.0×10−3s−1T1= 25 + 273 K Ea= 70 kJ/mol R= 8.314
J/(molK)
First, convert Eato Joules/mol: Ea= 70 ×1000 = 70000 J/mol
Plug in the values into the Arrhenius equation to solve for A: 2.0×10−3=
A·e−70000
8.314×(25+273)
Solving for A:A=2.0×10−3
e−70000
8.314×298
A≈2.0×10−3
e−29921.43
A≈2.0×10−3
2.596×10−13092
A≈7.70 ×1012989 s−1
Step 2: Calculate the rate constant, k2, at 50
°
C using the Arrhenius equation
with the previously calculated A:
k2=A·e−Ea
RT2
Given: T2= 50 + 273 K
Plug in the values and solve for k2:k2= (7.70 ×1012989)·e−70000
8.314×(50+273)
k2= (7.70 ×1012989)·e−70000
8.314×323
k2= (7.70 ×1012989)·e−26.85
k2≈(7.70 ×1012989)·1.30 ×10−12
k2≈1.00 ×1012978 s−1
2
Question 3
Question
The rate constant for the reaction A →B is given by the Arrhenius equation:
k=Ae−Ea
RT , where kis the rate constant, Ais the pre-exponential factor, Ea
is the activation energy, Ris the gas constant (8.31 J/mol·K), and Tis the
temperature in Kelvin.
For a certain reaction, the rate constant at 200 K is 1.5×10−3s−1, and at
250 K is 3.5×10−3s−1. Calculate the activation energy for this reaction.
Solution
Step 1: We are given two sets of temperature and rate constant values. We can
use the Arrhenius equation to write two equations: 1. For T= 200 K:
1.5×10−3=Ae−Ea
8.31×200
2. For T= 250 K:
3.5×10−3=Ae−Ea
8.31×250
Step 2: Divide the second equation by the first one to eliminate A:
3.5×10−3
1.5×10−3=e−Ea
8.31×250
e−Ea
8.31×200
Step 3: Simplify the equation:
2.33 = e−Ea
8.31×250 +Ea
8.31×200
Step 4: Take the natural logarithm of both sides to get rid of the exponential
term:
ln(2.33) = ln e−Ea
8.31×250 +Ea
8.31×200
Step 5: Apply properties of logarithms to simplify the equation:
ln(2.33) = −Ea
8.31 ×250 +Ea
8.31 ×200
Step 6: Solve for the activation energy Ea:
1
8.31 ×200Ea−1
8.31 ×250Ea= ln(2.33)
Ea1
8.31 ×200 −1
8.31 ×250= ln(2.33)
Ea=ln(2.33)
1
8.31×200 −1
8.31×250
Step 7: Calculate the activation energy using the above expression.
3
Question 4
Question
The rate constant for a certain reaction is found to be 4.75 ×10−3s−1at 25◦C
and 1.23 ×10−2s−1at 35◦C. Calculate the activation energy (Ea) for the
reaction.
Solution
Step 1: Write down the Arrhenius equation,
k=Ae−Ea
RT
where: kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J mol−1K−1), and Tis the temperature in
Kelvin.
Step 2: Convert the temperatures to Kelvin, 25◦C to 298 K and 35◦C to
308 K.
Step 3: Use the data provided to form two equations based on the Arrhenius
equation at each given temperature,
4.75 ×10−3=Ae−Ea
8.314×298
1.23 ×10−2=Ae−Ea
8.314×308
Step 4: Take the ratio of the two equations to eliminate A,
4.75 ×10−3
1.23 ×10−2=e−Ea
8.314×298
e−Ea
8.314×308
Step 5: Simplify the equation and solve for Ea,
4.75
1.23 =eEa
8.314 (1
298 −1
308 )
Step 6: Now, solve for Eato find the activation energy of the reaction.
Question 5
Question
The rate constant for a reaction at 30
°
C is 6.2×10−3s−1, while at 60
°
C it is
2.4×10−2s−1. Calculate the activation energy for the reaction.
4
Solution
Step 1: Let’s start by writing the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol ·K)), - Tis the temperature
in Kelvin.
Step 2: We have two sets of data: First set: T1= 30◦C = 30 + 273 = 303 K
k1= 6.2×10−3s−1
Second set: T2= 60◦C = 60 + 273 = 333 K k2= 2.4×10−2s−1
Step 3: We need to set up two equations using the Arrhenius equation:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 4: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R·T1
e−Ea
R·T2
Step 5: Simplify the equation:
k2
k1
=e
Ea
R1
T1
−1
T2
Step 6: Plug in the values and solve for Ea:
2.4×10−2
6.2×10−3=eEa
8.314 (1
303 −1
333 )
3.87 = eEa
8.314 (1
303 −1
333 )
Step 7: Solve for Ea:
ln(3.87) = Ea
8.314 1
303 −1
333
Ea= 8.314 ×ln(3.87)
1
303 −1
333
Calculate Eato find the activation energy of the reaction.
Question 6
Question
The rate constant for the decomposition of a certain compound at 25
°
C is 3.20×
10−3s−1. When the temperature is increased to 50
°
C, the rate constant becomes
2.15 ×10−2s−1. Calculate the activation energy for this reaction.
5
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. At 25
°
C: T1= 25C+ 273.15 = 298.15K
At 50
°
C: T2= 50C+ 273.15 = 323.15K
Step 2: Use the Arrhenius equation: k=A·e−Ea
RT , where: - k1= 3.20×10−3
s−1is the rate constant at T1= 298.15 K, - k2= 2.15 ×10−2s−1is the rate
constant at T2= 323.15 K, - R= 8.3145 J/(mol
·
K) is the ideal gas constant.
Step 3: Take the ratio of the Arrhenius equation for k1and k2to eliminate
the pre-exponential factor A:k2
k1=A·e−Ea
RT2
A·e−Ea
RT1
=e−Ea
R1
T2
−1
T1
Step 4: Plug in the values and solve for Ea:2.15×10−2
3.20×10−3=e−Ea
8.3145 (1
323.15 −1
298.15 )
6.71875 = e−Ea
8.3145 (0.003095−0.003352) 6.71875 = e−Ea
8.3145 (−0.000257) 6.71875 =
e0.000066096Ea
Step 5: Solve for Eaby taking the natural logarithm of both sides and then
multiply by 8.3145
0.000257 : ln(6.71875) = 0.000066096EaEa=ln(6.71875)
0.000066096 ×8.3145
0.000257
Ea= 58990.1 J/mol
Therefore, the activation energy for this reaction is 58.99 kJ/mol.
Question 7
Question
The rate constant of a certain reaction is observed to double when the temper-
ature is increased from 25
°
C to 35
°
C. Calculate the activation energy for this
reaction. Assume that the frequency factor Ais independent of temperature.
Solution
Step 1: Recall the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/mol·K), and T= temperature in Kelvin.
Step 2: Given that the rate constant doubles when the temperature is in-
creased from 25
°
C (298 K) to 35
°
C (308 K), we can write:
2k1=k2
2A·e−Ea
R·298 =A·e−Ea
R·308
Step 3: Simplify the equation by dividing both sides by A:
2e−Ea
R·298 =e−Ea
R·308
6
Step 4: Take the natural logarithm of both sides:
ln2e−Ea
R·298 = lne−Ea
R·308
ln 2 −Ea
R·298 =−Ea
R·308
Step 5: Solve for Ea:
ln 2 = Ea
R1
308 −1
298
Ea=R·ln 2
1
308 −1
298
Step 6: Calculate the activation energy using the gas constant R= 8.314
J/mol·K:
Ea= 8.314 ·ln 2
1
308 −1
298
Ea≈52.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 52.1
kJ/mol.
Question 8
Question
Given a reaction with an activation energy of 50 kJ/mol, calculate the rate
constant at 25
°
C if the activation energy is reduced to 40 kJ/mol. Assume the
pre-exponential factor (A) remains constant.
Solution
Step 1: Write the Arrhenius equation: The Arrhenius equation relates the rate
constant (k) of a reaction to the temperature (T), activation energy (Ea), and
the pre-exponential factor (A). It is given by:
k=A·e−Ea
RT
where: - k= rate constant - A= pre-exponential factor - Ea = activation
energy - R= gas constant (8.314 J/(mol
·
K)) - T= temperature in Kelvin
Step 2: Calculate the rate constant (k1) at 25
°
C (298 K) with an activation
energy of 50 kJ/mol: Given: Ea1= 50 kJ/mol, T= 298 K, and assuming A
remains constant.
Converting activation energy from kJ/mol to J/mol:
Ea1= 50 kJ/mol ×1000 J/kJ = 50000 J/mol
7
Substitute the values into the Arrhenius equation:
k1=A·e−50000
8.314×298
Step 3: Calculate the rate constant (k2) at 25
°
C (298 K) with an activation
energy of 40 kJ/mol: Given: Ea2= 40 kJ/mol, T= 298 K, and assuming A
remains constant.
Converting activation energy from kJ/mol to J/mol:
Ea2= 40 kJ/mol ×1000 J/kJ = 40000 J/mol
Substitute the values into the Arrhenius equation:
k2=A·e−40000
8.314×298
Step 4: Compare the rate constants k1and k2to determine the effect of re-
ducing the activation energy: Calculate k1and k2to compare the rate constants
with different activation energies.
This challenging question tests the understanding of the Arrhenius equation
and its dependency on activation energy and temperature.
Question 9
Question
The rate constant for the decomposition of a certain molecule A is found to be
4.62 ×10−3s−1at 25
°
C and 7.83 ×10−2s−1at 60
°
C. Calculate the activation
energy for this reaction. (Given: gas constant R= 8.31 J/mol ·K)
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
Where: k1= 4.62 ×10−3s−1(at 25
°
C) k2= 7.83 ×10−2s−1(at 60
°
C) T1=
25 + 273 = 298 K T2= 60 + 273 = 333 K R= 8.31 J/mol ·K
Step 2: Substitute the values into the equation and solve for the activation
energy Ea:
ln 7.83 ×10−2
4.62 ×10−3=−Ea
8.31 1
333 −1
298
ln 7.83 ×10−2
4.62 ×10−3=−Ea
8.31 35
10374
ln (16.98) = −Ea
8.31 (0.00337)
8
Step 3: Simplify the equation and solve for Ea:
2.832 = −0.00337Ea
8.31
Ea=−2.832 ×8.31
0.00337
Ea≈ −70000 J/mol
Therefore, the activation energy for the reaction is approximately −70000 J/mol.
Question 10
Question
The rate constant for the reaction 2A→Bis found to be 4.50 ×10−3s−1at
25◦C. When the temperature is increased to 45◦C, the rate constant becomes
1.20 ×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Let’s first write down the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
where: - k1and k2are the rate constants at temperatures T1and T2, - Eais
the activation energy, - Ris the gas constant (8.314 J/mol ·K).
Step 2: Substituting the given values into the equation:
ln 1.20 ×10−2s−1
4.50 ×10−3s−1=−Ea
8.314 1
318 K −1
298 K
Step 3: Simplifying the equation:
ln 1.20 ×10−2
4.50 ×10−3=−Ea
8.314 1
318 −1
298
Step 4: Calculating the natural logarithm:
ln 2.67
1=−Ea
8.314 1
318 −1
298
Step 5: Simplifying the equation further:
ln(2.67) = −Ea
8.314 1
318 −1
298
9
Step 6: Solving for the activation energy, Ea:
Ea=−8.314 ×ln(2.67)
1
318 −1
298
Step 7: Finally, plug in the given values to find the activation energy:
Ea=−8.314 ×ln(2.67)
1
318 −1
298≈62.52 kJ/mol
Therefore, the activation energy for the reaction 2A→Bis approximately
62.52 kJ/mol.
Question 11
Question
The rate constant of a reaction is found to be 0.0052 s−1at 25◦C and 0.053 s−1
at 55◦C. Calculate the activation energy of the reaction.
Solution
Step 1: Identify the given information.
Given: - Rate constant at 25◦C (k1) = 0.0052 s−1- Rate constant at 55◦C
(k2) = 0.053 s−1- Temperature at 25◦C (T1) = 25◦C + 273.15 = 298.15 K -
Temperature at 55◦C (T2) = 55◦C + 273.15 = 328.15 K
Step 2: Recall the Arrhenius equation.
The Arrhenius equation relates the rate constant of a reaction, temperature,
and the activation energy. It is given by:
k=A·e−Ea
RT
where: - k= rate constant - A= pre-exponential factor - Ea= activation energy
-R= gas constant (8.314 J/(mol.K)) - T= temperature in Kelvin
Step 3: Write the Arrhenius equation for k1and k2.
For k1at 25◦C:
k1=A·e−Ea
R·298.15
For k2at 55◦C:
k2=A·e−Ea
R·328.15
Step 4: Take the ratio of k2to k1.
k2
k1
=A·e−Ea
R·328.15
A·e−Ea
R·298.15
10
k2
k1
=eEa
R(1
298.15 −1
328.15 )
k2
k1
=eEa
R(328.15−298.15
298.15·328.15 )
k2
k1
=eEa
R·0.03506
Step 5: Solve for the activation energy (Ea).
ln k2
k1=Ea
R·0.03506
Ea=−R·0.03506 ·ln k2
k1
Step 6: Substitute the given values to calculate the activation energy.
Ea=−8.314 ×0.03506 ×ln 0.053
0.0052
Ea≈77.3 kJ/mol
Therefore, the activation energy of the reaction is approximately 77.3 kJ/mol.
Question 12
Question
The rate constant (k) for a reaction at two different temperatures (298 K and 350
K) are known to be 3.21×10−3s−1and 7.85×10−2s−1, respectively. Determine
the activation energy for this reaction. (Hint: Use the Arrhenius equation:
k=A·e−Ea
RT where Ais the pre-exponential factor, Eais the activation energy,
Ris the ideal gas constant, and Tis the temperature in Kelvin.)
Solution
Step 1: Convert the rate constants to the natural logarithm of the rate constants.
Let’s denote the rate constants at 298 K and 350 K as k1and k2, respectively.
Then, we have:
ln(k1) = ln3.21 ×10−3and ln(k2) = ln7.85 ×10−2
Step 2: Use the Arrhenius equation to set up and solve the system of equations.
The Arrhenius equation can be expressed as:
ln(k) = ln(A)−Ea
RT
11
Subtracting one equation from the other gives:
ln(k2)−ln(k1) = −Ea
R1
T2
−1
T1
Substitute the known values:
ln7.85 ×10−2−ln3.21 ×10−3=−Ea
R1
350 −1
298
Step 3: Solve for the activation energy (Ea). Now, we can calculate Eaby
rearranging the equation:
Ea=−R ln7.85 ×10−2−ln3.21 ×10−3
1
350 −1
298 !
Substitute R= 8.314 J/(mol·K) to find the activation energy in Joules per mole.
Question 13
Question
The rate constant for the reaction
2A→B+C
is found to be 4.63 ×10−3s−1at 25◦C. When the temperature is increased
to 45◦C, the rate constant becomes 1.38 ×10−2s−1. Calculate the activation
energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin.
T1= 25◦C + 273 = 298 K
T2= 45◦C + 273 = 318 K
Step 2: Use the Arrhenius equation to relate the rate constants to the tem-
peratures and activation energy.
k2
k1
=eEa
R1
T1
−1
T2
Step 3: Plug in the given values and solve for the activation energy Ea.
1.38 ×10−2
4.63 ×10−3=e(Ea
8.314 (1
298 −1
318 ))
1.38 ×10−2
4.63 ×10−3=e(Ea
8.314 (0.00335−0.00314))
12
1.38 ×10−2
4.63 ×10−3=e(Ea
8.314 ×0.00021)
2.98 = e0.00021Ea
Step 4: Take the natural logarithm of both sides to solve for the activation
energy.
ln(2.98) = lne0.00021Ea
ln(2.98) = 0.00021Ea
Ea=ln(2.98)
0.00021
Step 5: Calculate the activation energy Ea.
Ea=ln(2.98)
0.00021 ≈26289.26 J/mol
Question 14
Question
For a certain reaction, the rate constant at 25
°
C is 1.2×10−3s−1, and the rate
constant at 35
°
C is 5.6×10−2s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: We can utilize the Arrhenius equation to determine the activation energy
(Ea) for the reaction. The Arrhenius equation is expressed as:
k=A·e−Ea
RT
Where: k= rate constant
A= pre-exponential factor
Ea= activation energy
R= gas constant (8.314 J·mol−1·K−1)
T= temperature in Kelvin
Step 2: We are given two sets of conditions (25
°
C and 35
°
C) with their
respective rate constants. We can set up two equations using the Arrhenius
equation and solve for Ea.
At 25
°
C (298 K): 1.2×10−3=A·e−Ea
8.314·298
At 35
°
C (308 K): 5.6×10−2=A·e−Ea
8.314·308
Step 3: Taking the ratio of the two equations will eliminate the pre-exponential
factor A:5.6×10−2
1.2×10−3=eEa
8.314 (1
308 −1
298 )
13
Step 4: Simplifying the ratio gives:
46.67 = eEa
8.314 ·10
298·308
Step 5: Taking the natural logarithm of both sides to solve for Ea:
ln(46.67) = Ea
8.314 ·10
298 ·308
Step 6: Finally, solve for Ea:
Ea= 8.314 ×10
298 ·308 ×ln(46.67)
Step 7: Calculating the value:
Ea≈58.6 kJ/mol
Question 15
Question
The rate constant for a reaction at 25
°
C is 6.21 ×10−3s−1, and at 35
°
C it is
2.04 ×10−2s−1. Calculate the activation energy (in kJ/mol) for the reaction.
Assume the pre-exponential factor is 1.00 ×1012 s−1.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation. The
Arrhenius equation relates the rate constant of a reaction to temperature and
activation energy:
k=A×e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/mol·K), and - Tis the
temperature in Kelvin.
We can rewrite the equation as:
ln(k) = ln(A)−Ea
RT
Step 2: Calculate the activation energy using the rate constant values pro-
vided.
At 25
°
C (298K), we have:
ln6.21 ×10−3= ln1.00 ×1012−Ea
8.314 ×298
14
Solving for Eagives:
Ea=−8.314 ×298 ×(ln6.21 ×10−3−ln1.00 ×1012)
Ea≈55.2 kJ/mol
Step 3: Repeat the calculation using the rate constant at 35
°
C.
At 35
°
C (308K), we have:
ln2.04 ×10−2= ln1.00 ×1012−Ea
8.314 ×308
Solving for Eagives:
Ea=−8.314 ×308 ×(ln2.04 ×10−2−ln1.00 ×1012)
Ea≈54.8 kJ/mol
Step 4: Calculate the average activation energy. Average activation energy
=55.2 kJ/mol+54.8 kJ/mol
2
Therefore, the activation energy for the reaction is approximately 55.0 kJ/mol.
Question 16
Question
The rate constant for a certain reaction at 25
°
C is 1.80 ×10−3s−1and the
activation energy is 75 kJ/mol. Calculate the rate constant at 50
°
C for this
reaction.
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol.
Given: Activation energy, Ea= 75 kJ/mol.
To convert kJ to J, we multiply by 1000:
Ea= 75 ×1000 = 75000 J/mol
Step 2: Calculate the new rate constant at 50
°
C using the Arrhenius equa-
tion:
The Arrhenius equation is given by:
k=A×e−Ea
RT
Where: - kis the rate constant at a certain temperature, - Ais the pre-
exponential factor, - Eais the activation energy, - Ris the gas constant (8.314
J/(mol K)), - Tis the temperature in Kelvin.
15
Given: - Rate constant at 25
°
C, k1= 1.80 ×10−3s−1, - Activation energy,
Ea= 75000 J/mol, - Temperature at 25
°
C, T1= 25 + 273 = 298 K.
We need to find the rate constant at 50
°
C, k2, with temperature T2= 50 +
273 = 323 K.
Step 3: Use the Arrhenius equation to calculate the new rate constant.
Substitute the known values into the Arrhenius equation:
k1=A×e−Ea
R·T1
Solving for A:
A=k1
e−Ea
R·T1
Substitute R= 8.314 J/(mol K), T1= 298 K, and Ea= 75000 J/mol:
A=1.80 ×10−3
e−75000
8.314·298
A=1.80 ×10−3
e−90.08
Step 4: Calculate the new rate constant, k2, at 50
°
C.
Using the Arrhenius equation:
k2=A×e−Ea
R·T2
Substitute Afrom Step 3, R= 8.314 J/(mol K), T2= 323 K, and Ea= 75000
J/mol:
k2=1.80 ×10−3
e−90.08 ×e−75000
8.314·323
k2=1.80 ×10−3
e−90.08 ×e−28.89
k2≈1.80 ×10−3
e−90.08 ×1.416 ×10−13
k2≈5.08 ×10−15 s−1
Thus, the rate constant at 50
°
C for this reaction is approximately 5.08×10−15
s−1.
16
Question 17
Question
The rate constant for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When
the temperature is raised to 45
°
C, the rate constant becomes 3.75 ×10−2s−1.
Calculate the activation energy for this reaction. The activation energy, Ea, can
be found using the Arrhenius equation:
k=A×e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J/(mol*K)), and Tis the temperature in
Kelvin.
Solution
Step 1: Convert the given temperatures to Kelvin. Given: T1= 25◦C and
T2= 45◦C. Converting to Kelvin: T1= 25+273 = 298 K T2= 45+273 = 318 K
Step 2: Calculate the activation energy using the two rate constants and
temperatures. Using the Arrhenius equation, we have two equations: For T1:
1.25 ×10−3=A×e−Ea
8.314×298
For T2:
3.75 ×10−2=A×e−Ea
8.314×318
Dividing the two equations eliminates A:
3.75 ×10−2
1.25 ×10−3=eEa
8.314 ×(1
318 −1
298 )
30 = eEa
8.314 ×(1
318 −1
298 )
Step 3: Solve for the activation energy, Ea. Taking the natural logarithm of
both sides:
ln(30) = Ea
8.314 ×1
318 −1
298
Ea= 8.314 ×ln(30)
1
318 −1
298
Ea≈1.48 kJ/mol
Therefore, the activation energy for this reaction is approximately 1.48
kJ/mol.
17
Question 18
Question
The rate constant for the reaction A →B is found to be 2.00 ×10−2s−1at
T1= 300 K and 4.00 ×10−2s−1at T2= 320 K. Calculate the activation energy
for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol
·
K)), T= temperature in Kelvin.
Step 2: Calculate the ratio of rate constants at T1and T2:
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
4.00 ×10−2
2.00 ×10−2=e−Ea
8.314·320
e−Ea
8.314·300
Step 3: Simplify the equation:
2 = eEa
8.314 (1
300 −1
320 )
Step 4: Solve for Eaby taking natural logarithms of both sides:
ln(2) = Ea
8.314 1
300 −1
320
Ea= 8.314 1
300 −1
320ln(2)
Step 5: Calculate the activation energy:
Ea= 8.314 1
300 −1
320ln(2)
Ea≈48.9 kJ/mol
Question 19
Question
The rate constant for a reaction is found to be 1.25 ×10−3s−1at 25◦Cand
2.50 ×10−3s−1at 45◦C. Calculate the activation energy (Ea) for this reaction.
18
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C+ 273.15 = 298.15K
T2= 45◦C+ 273.15 = 318.15K
Step 2: Write down the Arrhenius equation which relates rate constants at
different temperatures. k2
k1
=e
−Ea
R1
T2
−1
T1
Step 3: Plug in the given rate constants, temperatures in Kelvin, and the
gas constant (R= 8.314J/mol ·K) into the Arrhenius equation.
2.50 ×10−3
1.25 ×10−3=e−Ea
8.314 (1
318.15 −1
298.15 )
Step 4: Solve for the activation energy (Ea).
2.5
1.25 =e−Ea
8.314 (1
318.15 −1
298.15 )
2 = e−Ea
8.314 (1
318.15 −1
298.15 )
ln(2) = −Ea
8.314 1
318.15 −1
298.15
Ea=−8.314 ·ln(2) 1
318.15 −1
298.15
Ea≈45.84 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.84
kJ/mol.
Question 20
Question
The rate constant for a certain reaction is k= 2.5×10−2s−1at a temperature
of 25◦C. If the activation energy for the reaction is 50 kJ/mol, calculate the
rate constant at 75◦C. The activation energy can be expressed in both joules
per mole and electron volts (eV) per molecule. Assume the temperature is in
Kelvin.
Solution
Step 1: Convert the activation energy from kilojoules per mole to joules per
mole.
Activation energy (in joules per mole) = 50 kJ/mol ×1000 J/kJ = 50000 J/mol
19
Step 2: Calculate the rate constant at 25◦C using the Arrhenius equation:
k1=A×exp −Ea
RT1
where Ais the pre-exponential factor, Eais the activation energy, Ris the gas
constant (8.314 J/(mol
·
K)), and T1is the temperature in Kelvin.
Given that k1= 2.5×10−2s−1and T1= 25◦C = 298 K,
2.5×10−2=A×exp −50000
8.314 ×298
Step 3: Solve for A:
A=2.5×10−2
exp −50000
8.314×298 ≈2.5×10−2
0.0432 ≈0.5787 s−1
Step 4: Use the Arrhenius equation to calculate the rate constant at 75◦C
(T2= 75◦C = 348 K):
k2=A×exp −Ea
RT2= 0.5787 ×exp −50000
8.314 ×348
Step 5: Calculate k2:
k2≈0.5787 ×exp −50000
8.314 ×348≈0.5787 ×exp (−17.41)
k2≈0.5787 ×1.629 ×10−8≈9.43 ×10−9s−1
Therefore, the rate constant at 75◦C is approximately 9.43 ×10−9s−1.
Question 21
Question
For a certain chemical reaction, the rate constant at 25
°
C is 2.0×10−3mol−1L s−1
and the activation energy is 80 kJ/mol. Calculate the rate constant at 50
°
C for
this reaction.
Solution
Step 1: Convert the activation energy to the proper units in Joules. Given that
the activation energy is 80 kJ/mol, we convert it to Joules using the conversion
factor 1 kJ = 1000 J:
80 kJ/mol ×1000 J/kJ = 80,000 J/mol
20
Step 2: Apply the Arrhenius equation to find the rate constant at 50
°
C. The
Arrhenius equation relates the rate constant (k) to the activation energy (Ea),
the gas constant (R), and the temperature in Kelvin (T):
k=A×e−Ea
RT
where: k= rate constant at 50
°
CA= pre-exponential factor (assumed to remain
constant) Ea= activation energy in Joules (80,000 J/mol) R= gas constant
(8.314 J/molK) T= temperature in Kelvin
Step 3: Convert the temperature to Kelvin. To convert 50
°
C to Kelvin, we
use the formula:
T(K) = T(C) + 273.15
T(K) = 50 + 273.15 = 323.15 K
Step 4: Plug in the values and solve for the rate constant at 50
°
C.
k= (2.0×10−3mol−1L s−1)×e−80000
8.314×323.15
k≈0.0052 mol−1L s−1
Therefore, the rate constant at 50
°
C for this reaction is approximately 5.2×
10−3mol−1L s−1.
Question 22
Question
The rate constant for a reaction at 25
°
C is 5.08 ×10−4s−1. When the tempera-
ture is increased to 45
°
C, the rate constant becomes 3.24 ×10−3s−1. Calculate
the activation energy (Ea) for this reaction. (Hint: Use the Arrhenius equation:
k=Ae−Ea
RT , where kis the rate constant, Ais the pre-exponential factor, Ea
is the activation energy, Ris the gas constant, and Tis the temperature in
Kelvin.)
Solution
Step 1: Convert temperatures from Celsius to Kelvin. Given T1= 25C=
25 + 273 = 298Kand T2= 45C= 45 + 273 = 318K.
Step 2: Calculate the ratio of rate constants.
k2
k1
=3.24 ×10−3s−1
5.08 ×10−4s−1= 6.38
Step 3: Substitute values into the Arrhenius equation. Since k2=Ae−Ea
RT2
and k1=Ae−Ea
RT1, we have:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
21
6.38 = e−Ea
318R
e−Ea
298R
Step 4: Simplify the equation.
6.38 = eEa
298R−Ea
318R=eEa(1
298R−1
318R)
6.38 = eEa(318−298
318×298 )=eEa(20
95064 )
Step 5: Solve for the activation energy (Ea).
ln(6.38) = Ea×20
95064
Ea=95064 ×ln(6.38)
20
Ea≈47760 J/mol
Question 23
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1, and the
activation energy is 50 kJ/mol. Calculate the rate constant at 50
°
C for this
reaction.
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol. Given that
the activation energy is 50 kJ/mol, we need to convert it to joules per mole:
50 kJ/mol ×1000 J/kJ = 50000 J/mol.
Step 2: Use the Arrhenius equation to find the rate constant at 50
°
C. The
Arrhenius equation relates the rate constant kto temperature Tand activation
energy Ea:
k=A·e−Ea
RT
where: - kis the rate constant - Ais the pre-exponential factor (a constant for
a given reaction) - Eais the activation energy - Ris the gas constant (8.314
J/mol·K) - Tis the temperature in Kelvin.
We are given: - k1= 2.5×10−3s−1-Ea= 50000 J/mol - T1= 25C= 298
K - T2= 50C= 323 K.
Step 3: Calculate the rate constant at 50
°
C. Substitute the given values into
the Arrhenius equation:
k1=A·e−Ea
RT1
2.5×10−3=A·e−50000
8.314·298
A= 2.5×10−3·e50000
8.314·298
22
Now, calculate the rate constant at 50
°
C:
k2=A·e−Ea
RT2=A·e−50000
8.314·323
Now, plug in the calculated value of Aand solve for k2.
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is found to be 1.5×10−3s−1
and the rate constant at 55
°
C is 0.024 s−1. Calculate the activation energy for
this reaction given that the universal gas constant is 8.314 J/(mol·K).
Solution
Step 1: Convert the given rate constants to Arrhenius equation form.
At T1= 25◦C = 298 K,the rate constant k1= 1.5×10−3s−1.
At T2= 55◦C = 328 K,the rate constant k2= 0.024 s−1.
The Arrhenius equation is given by:
k=Ae−Ea
RT ,
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the universal gas constant, and Tis the temperature in Kelvin.
At T1:
1.5×10−3=Ae −Ea
8.314×298 .
At T2:
0.024 = Ae −Ea
8.314×328 .
Step 2: Solve the simultaneous equations to find Ea. Divide the two Arrhe-
nius equations:
1.5×10−3
0.024 =Ae −Ea
8.314×298
Ae −Ea
8.314×328
,
62.5 = e−Ea
8.314 (1
298 −1
328 ).
Step 3: Solve for Ea.
ln 62.5 = −Ea
8.314(1
298 −1
328),
−2.07 = −Ea
8.314 ×0.0034,
Ea= 56.83 kJ/mol.
Therefore, the activation energy for this reaction is 56.83 kJ/mol.
23
Question 25
Question
The rate constant for a certain reaction at 25
°
C is 5.0×10−2s−1. When the
temperature is increased to 45
°
C, the rate constant becomes 3.0 s−1. Calculate
the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
Given: - Initial temperature (T1) = 25
°
C - Final temperature (T2) = 45
°
C
Converting to Kelvin:
T1= 25 + 273 = 298 K
T2= 45 + 273 = 318 K
Step 2: Write the Arrhenius equation.
The Arrhenius equation relates the rate constant kto the activation energy Ea
and the temperature T:
k=A·e−Ea
RT
where: - k= rate constant - A= pre-exponential factor - Ea= activation energy
-R= gas constant (8.314 J/(mol
·
K)) - T= temperature in Kelvin
Step 3: Set up the equations using the rate constants and temperatures.
At T1:
k1=A·e−Ea
RT1
5.0×10−2=A·e−Ea
8.314×298
At T2:
k2=A·e−Ea
RT2
3.0 = A·e−Ea
8.314×318
Step 4: Divide the second equation by the first equation to eliminate A.
3.0
5.0×10−2=e−Ea
8.314×318
e−Ea
8.314×298
Step 5: Simplify the equation by dividing and taking the natural logarithm.
60 = eEa
8.314 (1
298 −1
318 )
ln(60) = Ea
8.314 1
298 −1
318
Step 6: Solve for Ea.
Ea= 8.314 ×318 ·298
318 −298 ln(60) ≈112.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 112.5
kJ/mol.
24
Question 26
Question
The rate constant for a certain reaction is found to be 1.20 ×10−3s−1at 25◦C
and 3.98 ×10−3s−1at 35◦C. Calculate the activation energy (in kJ/mol) for
this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant with
temperature:
k=A·e−Ea
RT
where: k= rate constant, T= temperature (in Kelvin), A= pre-exponential
factor, Ea= activation energy, and R= gas constant (8.314 J/mol·K).
Step 2: Let’s first convert the given temperatures to Kelvin: 25◦C = 25 +
273.15 = 298.15 K 35◦C = 35 + 273.15 = 308.15 K
Step 3: Next, we can write the Arrhenius equation for each temperature:
k1=A·e−Ea
R·298.15
k2=A·e−Ea
R·308.15
Step 4: We are given that k1= 1.20 ×10−3s−1and k2= 3.98 ×10−3s−1.
We can set up a system of equations:
1.20 ×10−3=A·e−Ea
R·298.15
3.98 ×10−3=A·e−Ea
R·308.15
Step 5: Divide the second equation by the first to eliminate the pre-exponential
factor:
3.98 ×10−3
1.20 ×10−3=e−Ea
R·308.15
e−Ea
R·298.15
Step 6: Simplify the ratio of rate constants:
3.32 = e−Ea
8.314·308.15 +Ea
8.314·298.15
Step 7: Combine the exponents using the properties of exponents:
3.32 = e−Ea(1/308.15−1/298.15)
8.314
Step 8: Solve for the activation energy Ea:
Ea=−8.314 ·ln(3.32) ·1
(1/308.15 −1/298.15)
Step 9: Calculate Eato find the activation energy in kJ/mol.
25
Question 27
Question
The rate constant for a certain reaction is 1.25 ×10−2s−1at 25◦C and 5.25 ×
10−2s−1at 50◦C. Calculate the activation energy (Ea) for the reaction. (Hint:
Use the Arrhenius equation: k=A·e−Ea
RT where kis the rate constant, Ais the
pre-exponential factor, Eais the activation energy, Ris the gas constant, and
Tis the temperature in Kelvin.)
Solution
Step 1: Convert the temperatures to Kelvin.
Given: T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 50◦C = 50 + 273.15 = 323.15 K
Step 2: Write the Arrhenius equation for each temperature.
At T= 298.15 K : k1= 1.25 ×10−2s−1
At T= 323.15 K : k2= 5.25 ×10−2s−1
The Arrhenius equation can be written as:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of both sides of the Arrhenius equation
for each temperature.
ln k= ln A−Ea
RT
Step 4: Set up the equations for T1and T2. For T= 298.15 K : ln k1=
ln A−Ea
R·298.15
For T= 323.15 K : ln k2= ln A−Ea
R·323.15
Step 5: Subtract the equation for T1from the equation for T2to eliminate
ln A.
ln k2−ln k1=−Ea
R1
323.15 −1
298.15
Step 6: Solve for Eausing the gas constant R= 8.314 J ·mol−1·K−1.
Ea=−R·1
323.15 −1
298.15−1
·(ln k2−ln k1)
Now, calculate Eausing the given values of k1and k2.
Question 28
Question
The rate constant for a reaction is given by the Arrhenius equation: k=
Ae−Ea/RT , where Ais the pre-exponential factor, Eais the activation energy,
26
Ris the gas constant (8.314 J/mol·K), Tis the temperature in Kelvin, and kis
the rate constant.
For a certain reaction, the rate constant is found to be 2.5×10−4s−1at 37◦C
and 1.2×10−3s−1at 57◦C. Determine the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin. Given: T1= 37◦C and
T2= 57◦C.
To convert to Kelvin, we use the formula T(K) = T(◦C) + 273.15. Thus,
T1= 37 + 273.15 = 310.15 K and T2= 57 + 273.15 = 330.15 K.
Step 2: Use the Arrhenius equation to set up two equations. Using the
Arrhenius equation k=Ae−Ea/RT , we can write:
k1=Ae−Ea/RT1and k2=Ae−Ea/RT2
where k1= 2.5×10−4s−1,T1= 310.15 K, k2= 1.2×10−3s−1, and T2= 330.15
K.
Step 3: Take the ratio of the two equations. Dividing the second equation
by the first gives:
k2
k1
=Ae−Ea/RT2
Ae−Ea/RT1
k2
k1
=e−Ea/R·(1/T2−1/T1)
1.2×10−3
2.5×10−4=e−Ea/(8.314)·(1/330.15−1/310.15)
Step 4: Solve for the activation energy, Ea.
4.8×10−3
2.5×10−4=e−Ea/(8.314)·(0.003034−0.003225)
19.2×10−3=e−Ea/(8.314)·(−0.000191)
19.2 = e0.001588Ea
Taking the natural logarithm of both sides:
ln(19.2) = lne0.001588Ea
ln(19.2) = 0.001588Ea
Ea≈ln(19.2)
0.001588 ≈45105 J/mol
Therefore, the activation energy for this reaction is approximately 45.1
kJ/mol.
27
Question 29
Question
For a certain reaction, the rate constant kat 25
°
C is 1.5×10−3s−1. If the
activation energy for the reaction is 50 kJ/mol, calculate the rate constant at
50
°
C.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
k2=k1×e
Ea
R× 1
T1
−
1
T2!
,
where: - k1= 1.5×10−3s−1is the rate constant at 25
°
C, - Ea= 50 kJ/mol is the
activation energy, - R= 8.314 J/mol·K is the gas constant, - T1= 25C= 298 K,
and - T2= 50C= 323 K.
Step 2: Now, we can substitute the given values into the equation:
k2= 1.5×10−3×e
50 ×103
8.314 × 1
298−
1
323!
,
Step 3: Calculate the value of k2:
k2= 1.5×10−3×e(6024.26×(0.0034)),
k2= 1.5×10−3×e20.4982,
k21.5×10−3×5.82 ×108,
k28.73 ×105s−1.
Therefore, the rate constant at 50
°
C is approximately 8.73 ×105s−1.
Question 30
Question
The rate constant for a reaction is found to be 4.26 ×10−3s−1at 25◦C and
3.22 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
28
Solution
Step 1: Determine the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
Step 2: Take the natural logarithm of both sides of the Arrhenius equation
to linearize the equation:
ln k= ln A−Ea
R·1
T
Step 3: Create two linear equations using the data provided for T1= 25◦C
and T2= 45◦C:
ln k1= ln A−Ea
R·1
T1
ln k2= ln A−Ea
R·1
T2
Step 4: Subtract the two linear equations to eliminate ln A:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 5: Solve for the activation energy (Ea):
Ea=−R·1
T2
−1
T1−1
·ln k2
k1
Step 6: Plug in the values for R,T1,T2,k1, and k2to calculate the activation
energy. Remember to convert the temperatures to Kelvin by adding 273.15.
Ea=−8.314 J mol−1K−1·1
318.15 −1
318.15−1
·ln 3.22 ×10−2
4.26 ×10−3
Calculate the final answer to find the activation energy.
Question 31
Question
The rate constant (k) for a certain reaction was found to be 5.2×10−3s−1at
T= 25◦C and 1.6×10−2s−1at T= 50◦C. Calculate the activation energy
(Ea) for this reaction. (Given: R= 8.314 J K−1mol−1)
29
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(C) +
273.15.
At T= 25◦C:
T= 25 + 273.15 = 298.15 K
At T= 50◦C:
T= 50 + 273.15 = 323.15 K
Step 2: Calculate the ratio of rate constants using the Arrhenius equation:
k2
k1
=eEa
R1
T1
−1
T2
Substitute the given values:
1.6×10−2
5.2×10−3=e(Ea
8.314 (1
298.15 −1
323.15 ))
Step 3: Solve for the activation energy (Ea).
e(Ea
8.314 (1
298.15 −1
323.15 )) =1.6×10−2
5.2×10−3
1.6×10−2
5.2×10−3=e(Ea
8.314 (0.00335))
Step 4: Taking the natural logarithm of both sides to solve for Ea.
ln 1.6×10−2
5.2×10−3=Ea
8.314 ×0.00335
Step 5: Calculate the activation energy (Ea).
Ea= 8.314 ×0.00335 ×ln 1.6×10−2
5.2×10−3
Ea≈54.8 kJ/mol
Therefore, the activation energy for this reaction is approximately 54.8
kJ/mol.
Question 32
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 400 K and
1.06 ×10−2s−1at 450 K. Calculate the activation energy for this reaction.
30
Solution
Step 1: Write the Arrhenius equation relating rate constant (k), activation
energy (Ea), gas constant (R), temperature (T), and pre-exponential factor
(A):
k=A·e−Ea
RT
Step 2: We are given two sets of data points for kand T:
k1= 4.23 ×10−3s−1, T1= 400 K
k2= 1.06 ×10−2s−1, T2= 450 K
Step 3: Take the natural logarithm of the Arrhenius equation to simplify
the equation:
ln k= ln A−Ea
RT
Step 4: Subtract the second equation from the first equation to eliminate A:
ln k2
k1
=−Ea
R1
T2
−1
T1
Step 5: Plug in the values for k1,k2,T1, and T2:
ln 1.06 ×10−2
4.23 ×10−3=−Ea
R1
450 −1
400
Step 6: Solve for activation energy:
ln 1.06
0.423 =−Ea
R1
450 −1
400
ln 2.51
0.423 =Ea
8.314 1
400 −1
450
Ea= 8.314 ×ln 2.51
0.423
1
400 −1
450
Step 7: Calculate the activation energy:
Ea= 8.314 ×ln 2.51
0.423
1
400 −1
450
≈45.76 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.76
kJ/mol.
31
Question 33
Question
For a certain reaction, the rate constant at 25
°
C is 4.60 ×10−4s−1and the
rate constant at 60
°
C is 2.90 ×10−2s−1. Calculate the activation energy for
this reaction. The activation energy is given by the Arrhenius equation: k=
A·e−Ea/RT , where kis the rate constant, Ais the pre-exponential factor, Ea
is the activation energy, Ris the gas constant (8.31 J/mol
·
K), and Tis the
temperature in Kelvin.
Solution
Step 1: Convert the given temperatures to Kelvin. Given: Temperature at 25
°
C:
T1= 25C+273.15 = 298.15 K Temperature at 60
°
C: T2= 60C+273.15 = 333.15
K
Step 2: Calculate the activation energy using the two rate constants and
temperatures. From the Arrhenius equation, we have: k1=A·e−Ea/(R·T1)
k2=A·e−Ea/(R·T2)
Step 3: Take the ratio of the two rate constants. Taking the ratio of the two
rate constants gives: k2
k1=A·e−Ea/(R·T2)
A·e−Ea/(R·T1)Simplify to find: k2
k1=e−Ea/R·(1/T1−1/T2)
Step 4: Plug in the given rate constants and temperatures to solve for the
activation energy. Substitute the given rate constants and temperatures into
the equation: 2.90×10−2
4.60×10−4=e−Ea/8.31·(1/298.15−1/333.15)
Step 5: Simplify and solve for the activation energy. Solve for Eain the above
equation: e−Ea/8.31·(1/298.15−1/333.15) =2.90×10−2
4.60×10−4eEa/8.31·(0.00336−0.002999) =
2.90×10−2
4.60×10−4eEa/8.31·0.000361 = 63.0435
Step 6: Finally, solve for Ea. Taking the natural logarithm of both sides:
Ea/8.31 ·0.000361 = ln(63.0435) Ea= 8.31 ·0.000361 ·ln(63.0435) Ea≈60.7
kJ/mol
Therefore, the activation energy for this reaction is approximately 60.7
kJ/mol.
Question 34
Question
The rate constant for a certain reaction is 4.80 ×10−3s−1at 20◦C and 1.14 ×
10−2s−1at 40◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Look up the value of the universal gas constant R.
From tables, R= 8.314 J ·mol−1·K−1
32
Step 2: Write down the Arrhenius equation, which relates the rate con-
stant (k), the pre-exponential factor (A), the activation energy (Ea), and the
temperature (T).
k=A·e−Ea
RT
Step 3: Rearrange the Arrhenius equation to solve for the activation energy
(Ea).
Ea=−R·1
T2
−1
T1ln k2
k1
Step 4: Convert temperatures to Kelvin.
T1= 20◦C = 20 + 273.15 = 293.15 K
T2= 40◦C = 40 + 273.15 = 313.15 K
Step 5: Substitute the given values for k1,k2,T1, and T2into the equation
obtained in Step 3.
Ea=−8.314 J ·mol−1·K−11
313.15 K −1
293.15 Kln 1.14 ×10−2s−1
4.80 ×10−3s−1
Step 6: Perform the calculations to find the activation energy (Ea).
Ea≈5.29 ×104J/mol
Question 35
Question
The rate constant for a certain reaction is found to be 4.20 ×10−3s−1at 25◦C
and 0.295 s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Let’s first write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J mol−1K−1), T= temperature in Kelvin.
Step 2: We are given two sets of data points: At T1= 298 K, k1= 4.20 ×
10−3s−1, At T2= 323 K, k2= 0.295 s−1.
Step 3: We can rewrite the Arrhenius equation for each temperature: At T1:
k1=A·e−Ea
R·298
At T2:
k2=A·e−Ea
R·323
33
Question 3
Question
The rate constant for the reaction A →B is given by the Arrhenius equation:
k=Ae−Ea
RT , where kis the rate constant, Ais the pre-exponential factor, Ea
is the activation energy, Ris the gas constant (8.31 J/mol·K), and Tis the
temperature in Kelvin.
For a certain reaction, the rate constant at 200 K is 1.5×10−3s−1, and at
250 K is 3.5×10−3s−1. Calculate the activation energy for this reaction.
Solution
Step 1: We are given two sets of temperature and rate constant values. We can
use the Arrhenius equation to write two equations: 1. For T= 200 K:
1.5×10−3=Ae−Ea
8.31×200
2. For T= 250 K:
3.5×10−3=Ae−Ea
8.31×250
Step 2: Divide the second equation by the first one to eliminate A:
3.5×10−3
1.5×10−3=e−Ea
8.31×250
e−Ea
8.31×200
Step 3: Simplify the equation:
2.33 = e−Ea
8.31×250 +Ea
8.31×200
Step 4: Take the natural logarithm of both sides to get rid of the exponential
term:
ln(2.33) = ln e−Ea
8.31×250 +Ea
8.31×200
Step 5: Apply properties of logarithms to simplify the equation:
ln(2.33) = −Ea
8.31 ×250 +Ea
8.31 ×200
Step 6: Solve for the activation energy Ea:
1
8.31 ×200Ea−1
8.31 ×250Ea= ln(2.33)
Ea1
8.31 ×200 −1
8.31 ×250= ln(2.33)
Ea=ln(2.33)
1
8.31×200 −1
8.31×250
Step 7: Calculate the activation energy using the above expression.
3
Question 4
Question
The rate constant for a certain reaction is found to be 4.75 ×10−3s−1at 25◦C
and 1.23 ×10−2s−1at 35◦C. Calculate the activation energy (Ea) for the
reaction.
Solution
Step 1: Write down the Arrhenius equation,
k=Ae−Ea
RT
where: kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J mol−1K−1), and Tis the temperature in
Kelvin.
Step 2: Convert the temperatures to Kelvin, 25◦C to 298 K and 35◦C to
308 K.
Step 3: Use the data provided to form two equations based on the Arrhenius
equation at each given temperature,
4.75 ×10−3=Ae−Ea
8.314×298
1.23 ×10−2=Ae−Ea
8.314×308
Step 4: Take the ratio of the two equations to eliminate A,
4.75 ×10−3
1.23 ×10−2=e−Ea
8.314×298
e−Ea
8.314×308
Step 5: Simplify the equation and solve for Ea,
4.75
1.23 =eEa
8.314 (1
298 −1
308 )
Step 6: Now, solve for Eato find the activation energy of the reaction.
Question 5
Question
The rate constant for a reaction at 30
°
C is 6.2×10−3s−1, while at 60
°
C it is
2.4×10−2s−1. Calculate the activation energy for the reaction.
4
Solution
Step 1: Let’s start by writing the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol ·K)), - Tis the temperature
in Kelvin.
Step 2: We have two sets of data: First set: T1= 30◦C = 30 + 273 = 303 K
k1= 6.2×10−3s−1
Second set: T2= 60◦C = 60 + 273 = 333 K k2= 2.4×10−2s−1
Step 3: We need to set up two equations using the Arrhenius equation:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 4: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R·T1
e−Ea
R·T2
Step 5: Simplify the equation:
k2
k1
=e
Ea
R1
T1
−1
T2
Step 6: Plug in the values and solve for Ea:
2.4×10−2
6.2×10−3=eEa
8.314 (1
303 −1
333 )
3.87 = eEa
8.314 (1
303 −1
333 )
Step 7: Solve for Ea:
ln(3.87) = Ea
8.314 1
303 −1
333
Ea= 8.314 ×ln(3.87)
1
303 −1
333
Calculate Eato find the activation energy of the reaction.
Question 6
Question
The rate constant for the decomposition of a certain compound at 25
°
C is 3.20×
10−3s−1. When the temperature is increased to 50
°
C, the rate constant becomes
2.15 ×10−2s−1. Calculate the activation energy for this reaction.
5
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. At 25
°
C: T1= 25C+ 273.15 = 298.15K
At 50
°
C: T2= 50C+ 273.15 = 323.15K
Step 2: Use the Arrhenius equation: k=A·e−Ea
RT , where: - k1= 3.20×10−3
s−1is the rate constant at T1= 298.15 K, - k2= 2.15 ×10−2s−1is the rate
constant at T2= 323.15 K, - R= 8.3145 J/(mol
·
K) is the ideal gas constant.
Step 3: Take the ratio of the Arrhenius equation for k1and k2to eliminate
the pre-exponential factor A:k2
k1=A·e−Ea
RT2
A·e−Ea
RT1
=e−Ea
R1
T2
−1
T1
Step 4: Plug in the values and solve for Ea:2.15×10−2
3.20×10−3=e−Ea
8.3145 (1
323.15 −1
298.15 )
6.71875 = e−Ea
8.3145 (0.003095−0.003352) 6.71875 = e−Ea
8.3145 (−0.000257) 6.71875 =
e0.000066096Ea
Step 5: Solve for Eaby taking the natural logarithm of both sides and then
multiply by 8.3145
0.000257 : ln(6.71875) = 0.000066096EaEa=ln(6.71875)
0.000066096 ×8.3145
0.000257
Ea= 58990.1 J/mol
Therefore, the activation energy for this reaction is 58.99 kJ/mol.
Question 7
Question
The rate constant of a certain reaction is observed to double when the temper-
ature is increased from 25
°
C to 35
°
C. Calculate the activation energy for this
reaction. Assume that the frequency factor Ais independent of temperature.
Solution
Step 1: Recall the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/mol·K), and T= temperature in Kelvin.
Step 2: Given that the rate constant doubles when the temperature is in-
creased from 25
°
C (298 K) to 35
°
C (308 K), we can write:
2k1=k2
2A·e−Ea
R·298 =A·e−Ea
R·308
Step 3: Simplify the equation by dividing both sides by A:
2e−Ea
R·298 =e−Ea
R·308
6
Step 4: Take the natural logarithm of both sides:
ln2e−Ea
R·298 = lne−Ea
R·308
ln 2 −Ea
R·298 =−Ea
R·308
Step 5: Solve for Ea:
ln 2 = Ea
R1
308 −1
298
Ea=R·ln 2
1
308 −1
298
Step 6: Calculate the activation energy using the gas constant R= 8.314
J/mol·K:
Ea= 8.314 ·ln 2
1
308 −1
298
Ea≈52.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 52.1
kJ/mol.
Question 8
Question
Given a reaction with an activation energy of 50 kJ/mol, calculate the rate
constant at 25
°
C if the activation energy is reduced to 40 kJ/mol. Assume the
pre-exponential factor (A) remains constant.
Solution
Step 1: Write the Arrhenius equation: The Arrhenius equation relates the rate
constant (k) of a reaction to the temperature (T), activation energy (Ea), and
the pre-exponential factor (A). It is given by:
k=A·e−Ea
RT
where: - k= rate constant - A= pre-exponential factor - Ea = activation
energy - R= gas constant (8.314 J/(mol
·
K)) - T= temperature in Kelvin
Step 2: Calculate the rate constant (k1) at 25
°
C (298 K) with an activation
energy of 50 kJ/mol: Given: Ea1= 50 kJ/mol, T= 298 K, and assuming A
remains constant.
Converting activation energy from kJ/mol to J/mol:
Ea1= 50 kJ/mol ×1000 J/kJ = 50000 J/mol
7
Substitute the values into the Arrhenius equation:
k1=A·e−50000
8.314×298
Step 3: Calculate the rate constant (k2) at 25
°
C (298 K) with an activation
energy of 40 kJ/mol: Given: Ea2= 40 kJ/mol, T= 298 K, and assuming A
remains constant.
Converting activation energy from kJ/mol to J/mol:
Ea2= 40 kJ/mol ×1000 J/kJ = 40000 J/mol
Substitute the values into the Arrhenius equation:
k2=A·e−40000
8.314×298
Step 4: Compare the rate constants k1and k2to determine the effect of re-
ducing the activation energy: Calculate k1and k2to compare the rate constants
with different activation energies.
This challenging question tests the understanding of the Arrhenius equation
and its dependency on activation energy and temperature.
Question 9
Question
The rate constant for the decomposition of a certain molecule A is found to be
4.62 ×10−3s−1at 25
°
C and 7.83 ×10−2s−1at 60
°
C. Calculate the activation
energy for this reaction. (Given: gas constant R= 8.31 J/mol ·K)
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
Where: k1= 4.62 ×10−3s−1(at 25
°
C) k2= 7.83 ×10−2s−1(at 60
°
C) T1=
25 + 273 = 298 K T2= 60 + 273 = 333 K R= 8.31 J/mol ·K
Step 2: Substitute the values into the equation and solve for the activation
energy Ea:
ln 7.83 ×10−2
4.62 ×10−3=−Ea
8.31 1
333 −1
298
ln 7.83 ×10−2
4.62 ×10−3=−Ea
8.31 35
10374
ln (16.98) = −Ea
8.31 (0.00337)
8
Step 3: Simplify the equation and solve for Ea:
2.832 = −0.00337Ea
8.31
Ea=−2.832 ×8.31
0.00337
Ea≈ −70000 J/mol
Therefore, the activation energy for the reaction is approximately −70000 J/mol.
Question 10
Question
The rate constant for the reaction 2A→Bis found to be 4.50 ×10−3s−1at
25◦C. When the temperature is increased to 45◦C, the rate constant becomes
1.20 ×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Let’s first write down the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
where: - k1and k2are the rate constants at temperatures T1and T2, - Eais
the activation energy, - Ris the gas constant (8.314 J/mol ·K).
Step 2: Substituting the given values into the equation:
ln 1.20 ×10−2s−1
4.50 ×10−3s−1=−Ea
8.314 1
318 K −1
298 K
Step 3: Simplifying the equation:
ln 1.20 ×10−2
4.50 ×10−3=−Ea
8.314 1
318 −1
298
Step 4: Calculating the natural logarithm:
ln 2.67
1=−Ea
8.314 1
318 −1
298
Step 5: Simplifying the equation further:
ln(2.67) = −Ea
8.314 1
318 −1
298
9
Step 6: Solving for the activation energy, Ea:
Ea=−8.314 ×ln(2.67)
1
318 −1
298
Step 7: Finally, plug in the given values to find the activation energy:
Ea=−8.314 ×ln(2.67)
1
318 −1
298≈62.52 kJ/mol
Therefore, the activation energy for the reaction 2A→Bis approximately
62.52 kJ/mol.
Question 11
Question
The rate constant of a reaction is found to be 0.0052 s−1at 25◦C and 0.053 s−1
at 55◦C. Calculate the activation energy of the reaction.
Solution
Step 1: Identify the given information.
Given: - Rate constant at 25◦C (k1) = 0.0052 s−1- Rate constant at 55◦C
(k2) = 0.053 s−1- Temperature at 25◦C (T1) = 25◦C + 273.15 = 298.15 K -
Temperature at 55◦C (T2) = 55◦C + 273.15 = 328.15 K
Step 2: Recall the Arrhenius equation.
The Arrhenius equation relates the rate constant of a reaction, temperature,
and the activation energy. It is given by:
k=A·e−Ea
RT
where: - k= rate constant - A= pre-exponential factor - Ea= activation energy
-R= gas constant (8.314 J/(mol.K)) - T= temperature in Kelvin
Step 3: Write the Arrhenius equation for k1and k2.
For k1at 25◦C:
k1=A·e−Ea
R·298.15
For k2at 55◦C:
k2=A·e−Ea
R·328.15
Step 4: Take the ratio of k2to k1.
k2
k1
=A·e−Ea
R·328.15
A·e−Ea
R·298.15
10
k2
k1
=eEa
R(1
298.15 −1
328.15 )
k2
k1
=eEa
R(328.15−298.15
298.15·328.15 )
k2
k1
=eEa
R·0.03506
Step 5: Solve for the activation energy (Ea).
ln k2
k1=Ea
R·0.03506
Ea=−R·0.03506 ·ln k2
k1
Step 6: Substitute the given values to calculate the activation energy.
Ea=−8.314 ×0.03506 ×ln 0.053
0.0052
Ea≈77.3 kJ/mol
Therefore, the activation energy of the reaction is approximately 77.3 kJ/mol.
Question 12
Question
The rate constant (k) for a reaction at two different temperatures (298 K and 350
K) are known to be 3.21×10−3s−1and 7.85×10−2s−1, respectively. Determine
the activation energy for this reaction. (Hint: Use the Arrhenius equation:
k=A·e−Ea
RT where Ais the pre-exponential factor, Eais the activation energy,
Ris the ideal gas constant, and Tis the temperature in Kelvin.)
Solution
Step 1: Convert the rate constants to the natural logarithm of the rate constants.
Let’s denote the rate constants at 298 K and 350 K as k1and k2, respectively.
Then, we have:
ln(k1) = ln3.21 ×10−3and ln(k2) = ln7.85 ×10−2
Step 2: Use the Arrhenius equation to set up and solve the system of equations.
The Arrhenius equation can be expressed as:
ln(k) = ln(A)−Ea
RT
11
Subtracting one equation from the other gives:
ln(k2)−ln(k1) = −Ea
R1
T2
−1
T1
Substitute the known values:
ln7.85 ×10−2−ln3.21 ×10−3=−Ea
R1
350 −1
298
Step 3: Solve for the activation energy (Ea). Now, we can calculate Eaby
rearranging the equation:
Ea=−R ln7.85 ×10−2−ln3.21 ×10−3
1
350 −1
298 !
Substitute R= 8.314 J/(mol·K) to find the activation energy in Joules per mole.
Question 13
Question
The rate constant for the reaction
2A→B+C
is found to be 4.63 ×10−3s−1at 25◦C. When the temperature is increased
to 45◦C, the rate constant becomes 1.38 ×10−2s−1. Calculate the activation
energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin.
T1= 25◦C + 273 = 298 K
T2= 45◦C + 273 = 318 K
Step 2: Use the Arrhenius equation to relate the rate constants to the tem-
peratures and activation energy.
k2
k1
=eEa
R1
T1
−1
T2
Step 3: Plug in the given values and solve for the activation energy Ea.
1.38 ×10−2
4.63 ×10−3=e(Ea
8.314 (1
298 −1
318 ))
1.38 ×10−2
4.63 ×10−3=e(Ea
8.314 (0.00335−0.00314))
12
1.38 ×10−2
4.63 ×10−3=e(Ea
8.314 ×0.00021)
2.98 = e0.00021Ea
Step 4: Take the natural logarithm of both sides to solve for the activation
energy.
ln(2.98) = lne0.00021Ea
ln(2.98) = 0.00021Ea
Ea=ln(2.98)
0.00021
Step 5: Calculate the activation energy Ea.
Ea=ln(2.98)
0.00021 ≈26289.26 J/mol
Question 14
Question
For a certain reaction, the rate constant at 25
°
C is 1.2×10−3s−1, and the rate
constant at 35
°
C is 5.6×10−2s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: We can utilize the Arrhenius equation to determine the activation energy
(Ea) for the reaction. The Arrhenius equation is expressed as:
k=A·e−Ea
RT
Where: k= rate constant
A= pre-exponential factor
Ea= activation energy
R= gas constant (8.314 J·mol−1·K−1)
T= temperature in Kelvin
Step 2: We are given two sets of conditions (25
°
C and 35
°
C) with their
respective rate constants. We can set up two equations using the Arrhenius
equation and solve for Ea.
At 25
°
C (298 K): 1.2×10−3=A·e−Ea
8.314·298
At 35
°
C (308 K): 5.6×10−2=A·e−Ea
8.314·308
Step 3: Taking the ratio of the two equations will eliminate the pre-exponential
factor A:5.6×10−2
1.2×10−3=eEa
8.314 (1
308 −1
298 )
13
Step 4: Simplifying the ratio gives:
46.67 = eEa
8.314 ·10
298·308
Step 5: Taking the natural logarithm of both sides to solve for Ea:
ln(46.67) = Ea
8.314 ·10
298 ·308
Step 6: Finally, solve for Ea:
Ea= 8.314 ×10
298 ·308 ×ln(46.67)
Step 7: Calculating the value:
Ea≈58.6 kJ/mol
Question 15
Question
The rate constant for a reaction at 25
°
C is 6.21 ×10−3s−1, and at 35
°
C it is
2.04 ×10−2s−1. Calculate the activation energy (in kJ/mol) for the reaction.
Assume the pre-exponential factor is 1.00 ×1012 s−1.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation. The
Arrhenius equation relates the rate constant of a reaction to temperature and
activation energy:
k=A×e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/mol·K), and - Tis the
temperature in Kelvin.
We can rewrite the equation as:
ln(k) = ln(A)−Ea
RT
Step 2: Calculate the activation energy using the rate constant values pro-
vided.
At 25
°
C (298K), we have:
ln6.21 ×10−3= ln1.00 ×1012−Ea
8.314 ×298
14
Solving for Eagives:
Ea=−8.314 ×298 ×(ln6.21 ×10−3−ln1.00 ×1012)
Ea≈55.2 kJ/mol
Step 3: Repeat the calculation using the rate constant at 35
°
C.
At 35
°
C (308K), we have:
ln2.04 ×10−2= ln1.00 ×1012−Ea
8.314 ×308
Solving for Eagives:
Ea=−8.314 ×308 ×(ln2.04 ×10−2−ln1.00 ×1012)
Ea≈54.8 kJ/mol
Step 4: Calculate the average activation energy. Average activation energy
=55.2 kJ/mol+54.8 kJ/mol
2
Therefore, the activation energy for the reaction is approximately 55.0 kJ/mol.
Question 16
Question
The rate constant for a certain reaction at 25
°
C is 1.80 ×10−3s−1and the
activation energy is 75 kJ/mol. Calculate the rate constant at 50
°
C for this
reaction.
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol.
Given: Activation energy, Ea= 75 kJ/mol.
To convert kJ to J, we multiply by 1000:
Ea= 75 ×1000 = 75000 J/mol
Step 2: Calculate the new rate constant at 50
°
C using the Arrhenius equa-
tion:
The Arrhenius equation is given by:
k=A×e−Ea
RT
Where: - kis the rate constant at a certain temperature, - Ais the pre-
exponential factor, - Eais the activation energy, - Ris the gas constant (8.314
J/(mol K)), - Tis the temperature in Kelvin.
15
Given: - Rate constant at 25
°
C, k1= 1.80 ×10−3s−1, - Activation energy,
Ea= 75000 J/mol, - Temperature at 25
°
C, T1= 25 + 273 = 298 K.
We need to find the rate constant at 50
°
C, k2, with temperature T2= 50 +
273 = 323 K.
Step 3: Use the Arrhenius equation to calculate the new rate constant.
Substitute the known values into the Arrhenius equation:
k1=A×e−Ea
R·T1
Solving for A:
A=k1
e−Ea
R·T1
Substitute R= 8.314 J/(mol K), T1= 298 K, and Ea= 75000 J/mol:
A=1.80 ×10−3
e−75000
8.314·298
A=1.80 ×10−3
e−90.08
Step 4: Calculate the new rate constant, k2, at 50
°
C.
Using the Arrhenius equation:
k2=A×e−Ea
R·T2
Substitute Afrom Step 3, R= 8.314 J/(mol K), T2= 323 K, and Ea= 75000
J/mol:
k2=1.80 ×10−3
e−90.08 ×e−75000
8.314·323
k2=1.80 ×10−3
e−90.08 ×e−28.89
k2≈1.80 ×10−3
e−90.08 ×1.416 ×10−13
k2≈5.08 ×10−15 s−1
Thus, the rate constant at 50
°
C for this reaction is approximately 5.08×10−15
s−1.
16
Question 17
Question
The rate constant for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When
the temperature is raised to 45
°
C, the rate constant becomes 3.75 ×10−2s−1.
Calculate the activation energy for this reaction. The activation energy, Ea, can
be found using the Arrhenius equation:
k=A×e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J/(mol*K)), and Tis the temperature in
Kelvin.
Solution
Step 1: Convert the given temperatures to Kelvin. Given: T1= 25◦C and
T2= 45◦C. Converting to Kelvin: T1= 25+273 = 298 K T2= 45+273 = 318 K
Step 2: Calculate the activation energy using the two rate constants and
temperatures. Using the Arrhenius equation, we have two equations: For T1:
1.25 ×10−3=A×e−Ea
8.314×298
For T2:
3.75 ×10−2=A×e−Ea
8.314×318
Dividing the two equations eliminates A:
3.75 ×10−2
1.25 ×10−3=eEa
8.314 ×(1
318 −1
298 )
30 = eEa
8.314 ×(1
318 −1
298 )
Step 3: Solve for the activation energy, Ea. Taking the natural logarithm of
both sides:
ln(30) = Ea
8.314 ×1
318 −1
298
Ea= 8.314 ×ln(30)
1
318 −1
298
Ea≈1.48 kJ/mol
Therefore, the activation energy for this reaction is approximately 1.48
kJ/mol.
17
Question 18
Question
The rate constant for the reaction A →B is found to be 2.00 ×10−2s−1at
T1= 300 K and 4.00 ×10−2s−1at T2= 320 K. Calculate the activation energy
for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol
·
K)), T= temperature in Kelvin.
Step 2: Calculate the ratio of rate constants at T1and T2:
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
4.00 ×10−2
2.00 ×10−2=e−Ea
8.314·320
e−Ea
8.314·300
Step 3: Simplify the equation:
2 = eEa
8.314 (1
300 −1
320 )
Step 4: Solve for Eaby taking natural logarithms of both sides:
ln(2) = Ea
8.314 1
300 −1
320
Ea= 8.314 1
300 −1
320ln(2)
Step 5: Calculate the activation energy:
Ea= 8.314 1
300 −1
320ln(2)
Ea≈48.9 kJ/mol
Question 19
Question
The rate constant for a reaction is found to be 1.25 ×10−3s−1at 25◦Cand
2.50 ×10−3s−1at 45◦C. Calculate the activation energy (Ea) for this reaction.
18
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C+ 273.15 = 298.15K
T2= 45◦C+ 273.15 = 318.15K
Step 2: Write down the Arrhenius equation which relates rate constants at
different temperatures. k2
k1
=e
−Ea
R1
T2
−1
T1
Step 3: Plug in the given rate constants, temperatures in Kelvin, and the
gas constant (R= 8.314J/mol ·K) into the Arrhenius equation.
2.50 ×10−3
1.25 ×10−3=e−Ea
8.314 (1
318.15 −1
298.15 )
Step 4: Solve for the activation energy (Ea).
2.5
1.25 =e−Ea
8.314 (1
318.15 −1
298.15 )
2 = e−Ea
8.314 (1
318.15 −1
298.15 )
ln(2) = −Ea
8.314 1
318.15 −1
298.15
Ea=−8.314 ·ln(2) 1
318.15 −1
298.15
Ea≈45.84 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.84
kJ/mol.
Question 20
Question
The rate constant for a certain reaction is k= 2.5×10−2s−1at a temperature
of 25◦C. If the activation energy for the reaction is 50 kJ/mol, calculate the
rate constant at 75◦C. The activation energy can be expressed in both joules
per mole and electron volts (eV) per molecule. Assume the temperature is in
Kelvin.
Solution
Step 1: Convert the activation energy from kilojoules per mole to joules per
mole.
Activation energy (in joules per mole) = 50 kJ/mol ×1000 J/kJ = 50000 J/mol
19
Step 2: Calculate the rate constant at 25◦C using the Arrhenius equation:
k1=A×exp −Ea
RT1
where Ais the pre-exponential factor, Eais the activation energy, Ris the gas
constant (8.314 J/(mol
·
K)), and T1is the temperature in Kelvin.
Given that k1= 2.5×10−2s−1and T1= 25◦C = 298 K,
2.5×10−2=A×exp −50000
8.314 ×298
Step 3: Solve for A:
A=2.5×10−2
exp −50000
8.314×298 ≈2.5×10−2
0.0432 ≈0.5787 s−1
Step 4: Use the Arrhenius equation to calculate the rate constant at 75◦C
(T2= 75◦C = 348 K):
k2=A×exp −Ea
RT2= 0.5787 ×exp −50000
8.314 ×348
Step 5: Calculate k2:
k2≈0.5787 ×exp −50000
8.314 ×348≈0.5787 ×exp (−17.41)
k2≈0.5787 ×1.629 ×10−8≈9.43 ×10−9s−1
Therefore, the rate constant at 75◦C is approximately 9.43 ×10−9s−1.
Question 21
Question
For a certain chemical reaction, the rate constant at 25
°
C is 2.0×10−3mol−1L s−1
and the activation energy is 80 kJ/mol. Calculate the rate constant at 50
°
C for
this reaction.
Solution
Step 1: Convert the activation energy to the proper units in Joules. Given that
the activation energy is 80 kJ/mol, we convert it to Joules using the conversion
factor 1 kJ = 1000 J:
80 kJ/mol ×1000 J/kJ = 80,000 J/mol
20
Step 2: Apply the Arrhenius equation to find the rate constant at 50
°
C. The
Arrhenius equation relates the rate constant (k) to the activation energy (Ea),
the gas constant (R), and the temperature in Kelvin (T):
k=A×e−Ea
RT
where: k= rate constant at 50
°
CA= pre-exponential factor (assumed to remain
constant) Ea= activation energy in Joules (80,000 J/mol) R= gas constant
(8.314 J/molK) T= temperature in Kelvin
Step 3: Convert the temperature to Kelvin. To convert 50
°
C to Kelvin, we
use the formula:
T(K) = T(C) + 273.15
T(K) = 50 + 273.15 = 323.15 K
Step 4: Plug in the values and solve for the rate constant at 50
°
C.
k= (2.0×10−3mol−1L s−1)×e−80000
8.314×323.15
k≈0.0052 mol−1L s−1
Therefore, the rate constant at 50
°
C for this reaction is approximately 5.2×
10−3mol−1L s−1.
Question 22
Question
The rate constant for a reaction at 25
°
C is 5.08 ×10−4s−1. When the tempera-
ture is increased to 45
°
C, the rate constant becomes 3.24 ×10−3s−1. Calculate
the activation energy (Ea) for this reaction. (Hint: Use the Arrhenius equation:
k=Ae−Ea
RT , where kis the rate constant, Ais the pre-exponential factor, Ea
is the activation energy, Ris the gas constant, and Tis the temperature in
Kelvin.)
Solution
Step 1: Convert temperatures from Celsius to Kelvin. Given T1= 25C=
25 + 273 = 298Kand T2= 45C= 45 + 273 = 318K.
Step 2: Calculate the ratio of rate constants.
k2
k1
=3.24 ×10−3s−1
5.08 ×10−4s−1= 6.38
Step 3: Substitute values into the Arrhenius equation. Since k2=Ae−Ea
RT2
and k1=Ae−Ea
RT1, we have:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
21
6.38 = e−Ea
318R
e−Ea
298R
Step 4: Simplify the equation.
6.38 = eEa
298R−Ea
318R=eEa(1
298R−1
318R)
6.38 = eEa(318−298
318×298 )=eEa(20
95064 )
Step 5: Solve for the activation energy (Ea).
ln(6.38) = Ea×20
95064
Ea=95064 ×ln(6.38)
20
Ea≈47760 J/mol
Question 23
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1, and the
activation energy is 50 kJ/mol. Calculate the rate constant at 50
°
C for this
reaction.
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol. Given that
the activation energy is 50 kJ/mol, we need to convert it to joules per mole:
50 kJ/mol ×1000 J/kJ = 50000 J/mol.
Step 2: Use the Arrhenius equation to find the rate constant at 50
°
C. The
Arrhenius equation relates the rate constant kto temperature Tand activation
energy Ea:
k=A·e−Ea
RT
where: - kis the rate constant - Ais the pre-exponential factor (a constant for
a given reaction) - Eais the activation energy - Ris the gas constant (8.314
J/mol·K) - Tis the temperature in Kelvin.
We are given: - k1= 2.5×10−3s−1-Ea= 50000 J/mol - T1= 25C= 298
K - T2= 50C= 323 K.
Step 3: Calculate the rate constant at 50
°
C. Substitute the given values into
the Arrhenius equation:
k1=A·e−Ea
RT1
2.5×10−3=A·e−50000
8.314·298
A= 2.5×10−3·e50000
8.314·298
22
Now, calculate the rate constant at 50
°
C:
k2=A·e−Ea
RT2=A·e−50000
8.314·323
Now, plug in the calculated value of Aand solve for k2.
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is found to be 1.5×10−3s−1
and the rate constant at 55
°
C is 0.024 s−1. Calculate the activation energy for
this reaction given that the universal gas constant is 8.314 J/(mol·K).
Solution
Step 1: Convert the given rate constants to Arrhenius equation form.
At T1= 25◦C = 298 K,the rate constant k1= 1.5×10−3s−1.
At T2= 55◦C = 328 K,the rate constant k2= 0.024 s−1.
The Arrhenius equation is given by:
k=Ae−Ea
RT ,
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the universal gas constant, and Tis the temperature in Kelvin.
At T1:
1.5×10−3=Ae −Ea
8.314×298 .
At T2:
0.024 = Ae −Ea
8.314×328 .
Step 2: Solve the simultaneous equations to find Ea. Divide the two Arrhe-
nius equations:
1.5×10−3
0.024 =Ae −Ea
8.314×298
Ae −Ea
8.314×328
,
62.5 = e−Ea
8.314 (1
298 −1
328 ).
Step 3: Solve for Ea.
ln 62.5 = −Ea
8.314(1
298 −1
328),
−2.07 = −Ea
8.314 ×0.0034,
Ea= 56.83 kJ/mol.
Therefore, the activation energy for this reaction is 56.83 kJ/mol.
23
Question 25
Question
The rate constant for a certain reaction at 25
°
C is 5.0×10−2s−1. When the
temperature is increased to 45
°
C, the rate constant becomes 3.0 s−1. Calculate
the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
Given: - Initial temperature (T1) = 25
°
C - Final temperature (T2) = 45
°
C
Converting to Kelvin:
T1= 25 + 273 = 298 K
T2= 45 + 273 = 318 K
Step 2: Write the Arrhenius equation.
The Arrhenius equation relates the rate constant kto the activation energy Ea
and the temperature T:
k=A·e−Ea
RT
where: - k= rate constant - A= pre-exponential factor - Ea= activation energy
-R= gas constant (8.314 J/(mol
·
K)) - T= temperature in Kelvin
Step 3: Set up the equations using the rate constants and temperatures.
At T1:
k1=A·e−Ea
RT1
5.0×10−2=A·e−Ea
8.314×298
At T2:
k2=A·e−Ea
RT2
3.0 = A·e−Ea
8.314×318
Step 4: Divide the second equation by the first equation to eliminate A.
3.0
5.0×10−2=e−Ea
8.314×318
e−Ea
8.314×298
Step 5: Simplify the equation by dividing and taking the natural logarithm.
60 = eEa
8.314 (1
298 −1
318 )
ln(60) = Ea
8.314 1
298 −1
318
Step 6: Solve for Ea.
Ea= 8.314 ×318 ·298
318 −298 ln(60) ≈112.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 112.5
kJ/mol.
24
Question 26
Question
The rate constant for a certain reaction is found to be 1.20 ×10−3s−1at 25◦C
and 3.98 ×10−3s−1at 35◦C. Calculate the activation energy (in kJ/mol) for
this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant with
temperature:
k=A·e−Ea
RT
where: k= rate constant, T= temperature (in Kelvin), A= pre-exponential
factor, Ea= activation energy, and R= gas constant (8.314 J/mol·K).
Step 2: Let’s first convert the given temperatures to Kelvin: 25◦C = 25 +
273.15 = 298.15 K 35◦C = 35 + 273.15 = 308.15 K
Step 3: Next, we can write the Arrhenius equation for each temperature:
k1=A·e−Ea
R·298.15
k2=A·e−Ea
R·308.15
Step 4: We are given that k1= 1.20 ×10−3s−1and k2= 3.98 ×10−3s−1.
We can set up a system of equations:
1.20 ×10−3=A·e−Ea
R·298.15
3.98 ×10−3=A·e−Ea
R·308.15
Step 5: Divide the second equation by the first to eliminate the pre-exponential
factor:
3.98 ×10−3
1.20 ×10−3=e−Ea
R·308.15
e−Ea
R·298.15
Step 6: Simplify the ratio of rate constants:
3.32 = e−Ea
8.314·308.15 +Ea
8.314·298.15
Step 7: Combine the exponents using the properties of exponents:
3.32 = e−Ea(1/308.15−1/298.15)
8.314
Step 8: Solve for the activation energy Ea:
Ea=−8.314 ·ln(3.32) ·1
(1/308.15 −1/298.15)
Step 9: Calculate Eato find the activation energy in kJ/mol.
25
Question 27
Question
The rate constant for a certain reaction is 1.25 ×10−2s−1at 25◦C and 5.25 ×
10−2s−1at 50◦C. Calculate the activation energy (Ea) for the reaction. (Hint:
Use the Arrhenius equation: k=A·e−Ea
RT where kis the rate constant, Ais the
pre-exponential factor, Eais the activation energy, Ris the gas constant, and
Tis the temperature in Kelvin.)
Solution
Step 1: Convert the temperatures to Kelvin.
Given: T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 50◦C = 50 + 273.15 = 323.15 K
Step 2: Write the Arrhenius equation for each temperature.
At T= 298.15 K : k1= 1.25 ×10−2s−1
At T= 323.15 K : k2= 5.25 ×10−2s−1
The Arrhenius equation can be written as:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of both sides of the Arrhenius equation
for each temperature.
ln k= ln A−Ea
RT
Step 4: Set up the equations for T1and T2. For T= 298.15 K : ln k1=
ln A−Ea
R·298.15
For T= 323.15 K : ln k2= ln A−Ea
R·323.15
Step 5: Subtract the equation for T1from the equation for T2to eliminate
ln A.
ln k2−ln k1=−Ea
R1
323.15 −1
298.15
Step 6: Solve for Eausing the gas constant R= 8.314 J ·mol−1·K−1.
Ea=−R·1
323.15 −1
298.15−1
·(ln k2−ln k1)
Now, calculate Eausing the given values of k1and k2.
Question 28
Question
The rate constant for a reaction is given by the Arrhenius equation: k=
Ae−Ea/RT , where Ais the pre-exponential factor, Eais the activation energy,
26
Ris the gas constant (8.314 J/mol·K), Tis the temperature in Kelvin, and kis
the rate constant.
For a certain reaction, the rate constant is found to be 2.5×10−4s−1at 37◦C
and 1.2×10−3s−1at 57◦C. Determine the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin. Given: T1= 37◦C and
T2= 57◦C.
To convert to Kelvin, we use the formula T(K) = T(◦C) + 273.15. Thus,
T1= 37 + 273.15 = 310.15 K and T2= 57 + 273.15 = 330.15 K.
Step 2: Use the Arrhenius equation to set up two equations. Using the
Arrhenius equation k=Ae−Ea/RT , we can write:
k1=Ae−Ea/RT1and k2=Ae−Ea/RT2
where k1= 2.5×10−4s−1,T1= 310.15 K, k2= 1.2×10−3s−1, and T2= 330.15
K.
Step 3: Take the ratio of the two equations. Dividing the second equation
by the first gives:
k2
k1
=Ae−Ea/RT2
Ae−Ea/RT1
k2
k1
=e−Ea/R·(1/T2−1/T1)
1.2×10−3
2.5×10−4=e−Ea/(8.314)·(1/330.15−1/310.15)
Step 4: Solve for the activation energy, Ea.
4.8×10−3
2.5×10−4=e−Ea/(8.314)·(0.003034−0.003225)
19.2×10−3=e−Ea/(8.314)·(−0.000191)
19.2 = e0.001588Ea
Taking the natural logarithm of both sides:
ln(19.2) = lne0.001588Ea
ln(19.2) = 0.001588Ea
Ea≈ln(19.2)
0.001588 ≈45105 J/mol
Therefore, the activation energy for this reaction is approximately 45.1
kJ/mol.
27
Question 29
Question
For a certain reaction, the rate constant kat 25
°
C is 1.5×10−3s−1. If the
activation energy for the reaction is 50 kJ/mol, calculate the rate constant at
50
°
C.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
k2=k1×e
Ea
R× 1
T1
−
1
T2!
,
where: - k1= 1.5×10−3s−1is the rate constant at 25
°
C, - Ea= 50 kJ/mol is the
activation energy, - R= 8.314 J/mol·K is the gas constant, - T1= 25C= 298 K,
and - T2= 50C= 323 K.
Step 2: Now, we can substitute the given values into the equation:
k2= 1.5×10−3×e
50 ×103
8.314 × 1
298−
1
323!
,
Step 3: Calculate the value of k2:
k2= 1.5×10−3×e(6024.26×(0.0034)),
k2= 1.5×10−3×e20.4982,
k21.5×10−3×5.82 ×108,
k28.73 ×105s−1.
Therefore, the rate constant at 50
°
C is approximately 8.73 ×105s−1.
Question 30
Question
The rate constant for a reaction is found to be 4.26 ×10−3s−1at 25◦C and
3.22 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
28
Solution
Step 1: Determine the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
Step 2: Take the natural logarithm of both sides of the Arrhenius equation
to linearize the equation:
ln k= ln A−Ea
R·1
T
Step 3: Create two linear equations using the data provided for T1= 25◦C
and T2= 45◦C:
ln k1= ln A−Ea
R·1
T1
ln k2= ln A−Ea
R·1
T2
Step 4: Subtract the two linear equations to eliminate ln A:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 5: Solve for the activation energy (Ea):
Ea=−R·1
T2
−1
T1−1
·ln k2
k1
Step 6: Plug in the values for R,T1,T2,k1, and k2to calculate the activation
energy. Remember to convert the temperatures to Kelvin by adding 273.15.
Ea=−8.314 J mol−1K−1·1
318.15 −1
318.15−1
·ln 3.22 ×10−2
4.26 ×10−3
Calculate the final answer to find the activation energy.
Question 31
Question
The rate constant (k) for a certain reaction was found to be 5.2×10−3s−1at
T= 25◦C and 1.6×10−2s−1at T= 50◦C. Calculate the activation energy
(Ea) for this reaction. (Given: R= 8.314 J K−1mol−1)
29
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(C) +
273.15.
At T= 25◦C:
T= 25 + 273.15 = 298.15 K
At T= 50◦C:
T= 50 + 273.15 = 323.15 K
Step 2: Calculate the ratio of rate constants using the Arrhenius equation:
k2
k1
=eEa
R1
T1
−1
T2
Substitute the given values:
1.6×10−2
5.2×10−3=e(Ea
8.314 (1
298.15 −1
323.15 ))
Step 3: Solve for the activation energy (Ea).
e(Ea
8.314 (1
298.15 −1
323.15 )) =1.6×10−2
5.2×10−3
1.6×10−2
5.2×10−3=e(Ea
8.314 (0.00335))
Step 4: Taking the natural logarithm of both sides to solve for Ea.
ln 1.6×10−2
5.2×10−3=Ea
8.314 ×0.00335
Step 5: Calculate the activation energy (Ea).
Ea= 8.314 ×0.00335 ×ln 1.6×10−2
5.2×10−3
Ea≈54.8 kJ/mol
Therefore, the activation energy for this reaction is approximately 54.8
kJ/mol.
Question 32
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 400 K and
1.06 ×10−2s−1at 450 K. Calculate the activation energy for this reaction.
30
Solution
Step 1: Write the Arrhenius equation relating rate constant (k), activation
energy (Ea), gas constant (R), temperature (T), and pre-exponential factor
(A):
k=A·e−Ea
RT
Step 2: We are given two sets of data points for kand T:
k1= 4.23 ×10−3s−1, T1= 400 K
k2= 1.06 ×10−2s−1, T2= 450 K
Step 3: Take the natural logarithm of the Arrhenius equation to simplify
the equation:
ln k= ln A−Ea
RT
Step 4: Subtract the second equation from the first equation to eliminate A:
ln k2
k1
=−Ea
R1
T2
−1
T1
Step 5: Plug in the values for k1,k2,T1, and T2:
ln 1.06 ×10−2
4.23 ×10−3=−Ea
R1
450 −1
400
Step 6: Solve for activation energy:
ln 1.06
0.423 =−Ea
R1
450 −1
400
ln 2.51
0.423 =Ea
8.314 1
400 −1
450
Ea= 8.314 ×ln 2.51
0.423
1
400 −1
450
Step 7: Calculate the activation energy:
Ea= 8.314 ×ln 2.51
0.423
1
400 −1
450
≈45.76 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.76
kJ/mol.
31
Question 33
Question
For a certain reaction, the rate constant at 25
°
C is 4.60 ×10−4s−1and the
rate constant at 60
°
C is 2.90 ×10−2s−1. Calculate the activation energy for
this reaction. The activation energy is given by the Arrhenius equation: k=
A·e−Ea/RT , where kis the rate constant, Ais the pre-exponential factor, Ea
is the activation energy, Ris the gas constant (8.31 J/mol
·
K), and Tis the
temperature in Kelvin.
Solution
Step 1: Convert the given temperatures to Kelvin. Given: Temperature at 25
°
C:
T1= 25C+273.15 = 298.15 K Temperature at 60
°
C: T2= 60C+273.15 = 333.15
K
Step 2: Calculate the activation energy using the two rate constants and
temperatures. From the Arrhenius equation, we have: k1=A·e−Ea/(R·T1)
k2=A·e−Ea/(R·T2)
Step 3: Take the ratio of the two rate constants. Taking the ratio of the two
rate constants gives: k2
k1=A·e−Ea/(R·T2)
A·e−Ea/(R·T1)Simplify to find: k2
k1=e−Ea/R·(1/T1−1/T2)
Step 4: Plug in the given rate constants and temperatures to solve for the
activation energy. Substitute the given rate constants and temperatures into
the equation: 2.90×10−2
4.60×10−4=e−Ea/8.31·(1/298.15−1/333.15)
Step 5: Simplify and solve for the activation energy. Solve for Eain the above
equation: e−Ea/8.31·(1/298.15−1/333.15) =2.90×10−2
4.60×10−4eEa/8.31·(0.00336−0.002999) =
2.90×10−2
4.60×10−4eEa/8.31·0.000361 = 63.0435
Step 6: Finally, solve for Ea. Taking the natural logarithm of both sides:
Ea/8.31 ·0.000361 = ln(63.0435) Ea= 8.31 ·0.000361 ·ln(63.0435) Ea≈60.7
kJ/mol
Therefore, the activation energy for this reaction is approximately 60.7
kJ/mol.
Question 34
Question
The rate constant for a certain reaction is 4.80 ×10−3s−1at 20◦C and 1.14 ×
10−2s−1at 40◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Look up the value of the universal gas constant R.
From tables, R= 8.314 J ·mol−1·K−1
32
Step 2: Write down the Arrhenius equation, which relates the rate con-
stant (k), the pre-exponential factor (A), the activation energy (Ea), and the
temperature (T).
k=A·e−Ea
RT
Step 3: Rearrange the Arrhenius equation to solve for the activation energy
(Ea).
Ea=−R·1
T2
−1
T1ln k2
k1
Step 4: Convert temperatures to Kelvin.
T1= 20◦C = 20 + 273.15 = 293.15 K
T2= 40◦C = 40 + 273.15 = 313.15 K
Step 5: Substitute the given values for k1,k2,T1, and T2into the equation
obtained in Step 3.
Ea=−8.314 J ·mol−1·K−11
313.15 K −1
293.15 Kln 1.14 ×10−2s−1
4.80 ×10−3s−1
Step 6: Perform the calculations to find the activation energy (Ea).
Ea≈5.29 ×104J/mol
Question 35
Question
The rate constant for a certain reaction is found to be 4.20 ×10−3s−1at 25◦C
and 0.295 s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Let’s first write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J mol−1K−1), T= temperature in Kelvin.
Step 2: We are given two sets of data points: At T1= 298 K, k1= 4.20 ×
10−3s−1, At T2= 323 K, k2= 0.295 s−1.
Step 3: We can rewrite the Arrhenius equation for each temperature: At T1:
k1=A·e−Ea
R·298
At T2:
k2=A·e−Ea
R·323
33
Step 4: Divide the two equations:
k2
k1
=A·e−Ea
R·323
A·e−Ea
R·298
Step 5: Simplifying the above expression gives:
k2
k1
=e−Ea
R(1
323 −1
298 )
Step 6: Solve for Ea:k2
k1
=e−Ea
R(1
323 −1
298 )
ln k2
k1=−Ea
R1
323 −1
298
Step 7: Plug in the values and solve for Ea:
ln 0.295
4.20 ×10−3=−Ea
8.314 1
323 −1
298
Ea= (−8.314) ·
ln 0.295
4.20×10−3
1
323 −1
298
Step 8: Calculate the activation energy using the above formula.
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