CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 3
Liberty University
Question 1
Question
For the reaction
2A →B+C,
the rate constant at 25
°
C is 1.5×10−3s−1. When the temperature is increased
to 50
°
C, the rate constant becomes 5.0×10−3s−1. Calculate the activation
energy for this reaction.
Solution
Step 1: 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)), T1and T2are the
temperatures in Kelvin.
Step 2: Plug in the given values: k1= 1.5×10−3s−1(at 25
°
C = 298 K),
k2= 5.0×10−3s−1(at 50
°
C = 323 K), R= 8.314 J/(mol·K).
Step 3: Substitute the values into the Arrhenius equation:
ln 5.0×10−3
1.5×10−3=−Ea
8.314 1
323 −1
298.
Step 4: Simplify the equation:
ln 5.0
1.5=−Ea
8.314 1
323 −1
298.
Step 5: Solve for the activation energy Ea:
Ea=−8.314 ×ln 5.0
1.5
1
323 −1
298
.
Step 6: Calculate the activation energy:
Ea=−8.314 ×ln 5.0
1.5
1
323 −1
298
≈39.93 kJ/mol.
Therefore, the activation energy for the reaction is approximately 39.93
kJ/mol.
Question 2
Question
For a certain reaction, the rate constant at 25
°
C is 3.0×10−3s−1and the rate
constant at 35
°
C is 2.1×10−2s−1. Calculate the activation energy (in kJ/mol)
for this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at
different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
where - k1and k2are the rate constants at temperatures T1and T2respectively,
-Eais the activation energy, - Ris the ideal gas constant, and - T1and T2are
the temperatures in Kelvin.
Step 2: We can plug in the given values into the equation above:
ln 2.1×10−2
3.0×10−3=−Ea
8.314 1
308 −1
298
Step 3: Solving for the activation energy:
ln (7) = −Ea
8.314 1
308 −1
298
Step 4: Simplifying the equation further:
Ea=−8.314 ×ln(7)
1
308 −1
298
2
Step 5: Finally, calculating the activation energy:
Ea=−8.314 ×ln(7)
1
308 −1
298≈79.8 kJ/mol
Therefore, the activation energy for this reaction is approximately 79.8
kJ/mol.
Question 3
Question
The rate constant for a certain reaction at 25
°
C is 1.20 ×10−3s−1, and the
activation energy for the reaction is 85.0 kJ/mol. Calculate the rate constant at
75
°
C for this reaction.
Solution
Given: - Rate constant at 25
°
C, k1= 1.20 ×10−3s−1- Activation energy,
Ea= 85.0 kJ/mol - Initial temperature, T1= 25C+ 273 = 298 K - Final
temperature, T2= 75C+ 273 = 348 K
We can use the Arrhenius equation to find the rate constant at the final
temperature:
k2=A·e−Ea
R·1
T2
−1
T1
Step 1: Convert the activation energy to joules/mol
Ea= 85.0 kJ/mol ×1000 J/kJ = 85000 J/mol
Step 2: Plug in the given values into the Arrhenius equation
k2= 1.20 ×10−3s−1·e(−85000 J/mol
8.314 J/mol·K·(1
348 K −1
298 K ))
Step 3: Calculate the rate constant at 75
°
C
k2= 1.20 ×10−3s−1·e(−85000 J/mol
8.314 J/mol·K·(1
348 K −1
298 K ))
k2= 1.20 ×10−3s−1·e(−85000
8.314 ·(1
348 −1
298 ))
k2≈8.53 ×10−3s−1
Therefore, the rate constant at 75
°
C for this reaction is approximately 8.53×
10−3s−1.
3
Question 4
Question
The rate constant of a reaction at 25
°
C is 4.22 ×10−3s−1. When the temper-
ature is increased to 45
°
C, the rate constant becomes 0.055 s−1. 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= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol·K), T= temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation: k1= 4.22 ×10−3s−1
(at 25
°
C = 298 K), k2= 0.055 s−1(at 45
°
C = 318 K).
Step 4: Calculate the activation energy, Ea:
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
Ea=8.314
ln 0.055
4.22×10−31
318 −1
298
Ea≈8.314
ln 0.055
4.22×10−31
318 −1
298
Ea≈46.21 kJ/mol
Therefore, the activation energy for this reaction is approximately 46.21
kJ/mol.
4
Question 5
Question
A reaction has an activation energy of 50.0 kJ/mol and a frequency factor of
1.0×1010 s−1. At a certain temperature, the rate constant of the reaction is
5.0×10−4s−1. Calculate the rate constant at a temperature that is 10
°
C higher,
assuming the activation energy does not change.
Solution
Step 1: Calculate the rate constant k1at the initial temperature using the
Arrhenius equation:
k1=A·e−Ea
RT1
where: - k1is the rate constant at the initial temperature, - Ais the frequency
factor, - Eais the activation energy, - Ris the ideal gas constant (8.314 J/(mol
K)), - T1is the initial temperature in Kelvin.
Given that A= 1.0×1010 s−1,Ea= 50.0 kJ/mol = 5.0×104J/mol, and
T1corresponds to the initial temperature, use T1=T1+ 273.15 K.
Step 2: Calculate the rate constant k2at the higher temperature using the
Arrhenius equation with the new temperature T2=T1+ 10
°
C:
k2=A·e−Ea
RT2
Step 3: Substitute the known values into the Arrhenius equation to find the
new rate constant k2at the higher temperature.
Question 6
Question
The rate constant for the reaction
2A+B→C
is found to be 2.2×10−3s−1at 25◦C and 1.5×10−2s−1at 75◦C. Calculate the
activation energy for the reaction.
Solution
Step 1: Let’s first write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J/mol·K), - Tis the
temperature in Kelvin.
5
Step 2: Use the data given at 25◦C and 75◦C to form two equations:
2.2×10−3=A·e−Ea
8.314·(25+273.15)
1.5×10−2=A·e−Ea
8.314·(75+273.15)
Step 3: Take the ratio of the two equations to eliminate A:
2.2×10−3
1.5×10−2=e−Ea
8.314·(25+273.15)
e−Ea
8.314·(75+273.15)
Step 4: Simplify the equation to solve for Ea:
2.2
1.5=e
Ea
8.314 (1
25+273.15 −1
75+273.15 )
Step 5: Calculate and simplify further to find the activation energy Ea.
Question 7
Question
The rate constant for the first-order decomposition of a compound is 6.25 ×
10−4s−1at 350 K and 1.25 ×10−3s−1at 370 K. Calculate the activation energy
for this reaction.
Solution
Step 1: Let’s start by writing 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: Next, we can rearrange the Arrhenius equation to solve for Eausing
two sets of data given:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
k2
k1
=e
Ea
R1
T1
−1
T2
Given that k1= 6.25 ×10−4s−1at 350 K and k2= 1.25 ×10−3s−1at 370 K, we
can plug in these values along with Rto solve for Ea.
Step 3: Substituting the values in, we get:
1.25 ×10−3
6.25 ×10−4=e
Ea
8.314 (1
350 −1
370 )
6
2 = e
Ea
8.314 (1
350 −1
370 )
Step 4: Simplifying further, we find:
ln(2) = Ea
8.314 1
350 −1
370
0.693
8.314 =Ea1
350 −1
370
Step 5: Solve for Ea:
0.693
8.314 =Ea1
350 −1
370
Ea=0.693
8.314∇ · 1
350 −1
370
Ea≈62200 J/mol
Therefore, the activation energy for this reaction is approximately 62200 J/mol.
Question 8
Question
The rate constant for a first-order reaction at 25
°
C is 1.20×10−3s−1. When the
temperature is increased to 45
°
C, the rate constant becomes 6.80 ×10−3s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
k2
k1
= exp Ea
R1
T1
−1
T2
where: - k1and k2are the rate constants at temperatures T1and T2respectively,
-Eais the activation energy, - Ris the gas constant (8.314 J/mol·K), - T1and
T2are the temperatures in Kelvin.
Step 2: Plug in the given values:
6.80 ×10−3
1.20 ×10−3= exp Ea
8.314 1
298 −1
318
Step 3: Simplify the equation by dividing the rate constants and calculating
the temperature differences:
5.67 = exp Ea
8.314 1
298 −1
318
7
Step 4: Calculate the temperature differences:
1
298 −1
318 =318 −298
298 ×318 =20
89484 =5
22371
Step 5: Substitute back into the equation:
5.67 = exp Ea
8.314 ·5
22371
Step 6: Solve for the activation energy Eaby taking the natural logarithm
of both sides:
ln(5.67) = Ea
8.314 ·5
22371
Step 7: Rearrange the equation to solve for Ea:
Ea= 8.314 ·ln(5.67) ·22371
5
Step 8: Calculate Eato find the activation energy of the reaction.
Question 9
Question
The rate constant for a certain reaction is found to be 3.5×10−4s−1at 25 ◦C
and 1.2×10−3s−1at 35 ◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down 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: Rearrange the Arrhenius equation to solve for the activation energy
Ea:
Ea=− ln k
A
1
R·T1
−1
R·T2!
where: - k1= 3.5×10−4s−1at 25 ◦C = 298 K, - k2= 1.2×10−3s−1at
35 ◦C = 308 K.
Step 3: Plug in the values and solve for the activation energy Ea:
Ea=−
ln 1.2×10−3
3.5×10−4
1
8.314·298 −1
8.314·308
8
Ea=−ln (3.43)
1
2485.272 −1
2542.392
Ea=−ln (3.43)
0.000401 −0.000393
Ea=−ln (3.43)
0.000008
Ea=−9.203 ×103K
Therefore, the activation energy for this reaction is 9.203 ×103K.
Question 10
Question
The rate constant of a reaction is observed to double when the temperature
is increased from 25
°
C to 35
°
C. Calculate the activation energy (Ea) for the
reaction. Assume the frequency factor (A) is constant.
Solution
Step 1: Convert temperatures to Kelvin.
The temperatures given are 25
°
C and 35
°
C. To convert to Kelvin, we add 273
to each temperature:
T1= 25C+ 273 = 298 K
T2= 35C+ 273 = 308 K
Step 2: Use the Arrhenius equation.
The Arrhenius equation relates the rate constant of a reaction to the tempera-
ture and the activation energy:
k=Ae−Ea
RT
Step 3: Determine the ratio of rate constants.
Given that the rate constant doubles when the temperature increases from T1
to T2, we have:
k2
k1
= 2
Step 4: Take the ratio of the Arrhenius equations.
Substitute the Arrhenius equation for k1and k2and simplify:
Ae−Ea
RT2
Ae−Ea
RT1
= 2
9
Step 5: Simplify the equation.
Cancel out the frequency factor Aand rearrange the equation:
e
Ea
R(1
T1
−1
T2)= 2
Step 6: Solve for the activation energy (Ea).
Substitute the values of T1and T2, and solve for Ea:
e
Ea
R(1
298 −1
308 )= 2
e
Ea
R(10
298×308 )= 2
Ea
R(10
298 ×308) = ln 2
Ea= (ln 2) ×R×298 ×308
10
Step 7: Calculate the activation energy.
Now, substitute the value of the gas constant R= 8.314 J/(mol*K) to find the
activation energy Ea.
Question 11
Question
The rate constant of a first-order reaction doubles when the temperature is
increased from 25
°
C to 45
°
C. Calculate the activation energy (in kJ/mol) for
this reaction. (Given: R = 8.314 J/(mol*K))
Solution
Step 1: The Arrhenius equation relates the rate constant of a reaction to tem-
perature and the activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor (frequency
factor), - Eais the activation energy, - Ris the gas constant (8.314 J/(mol*K)),
-Tis the temperature in Kelvin.
Step 2: Given that the rate constant doubles when the temperature is in-
creased from 25
°
C to 45
°
C, we can write:
k2= 2 ·k1
10
Substitute the Arrhenius equation for both temperatures:
A·e−Ea
R·(273+25) = 2 ·A·e−Ea
R·(273+45)
Step 3: Simplify the equation by dividing both sides by Aand cancelling the
factor of 2:
e−Ea
R·(273+25) =e−Ea
R·(273+45)
Step 4: To solve for the activation energy, set the exponents equal to each
other:
−Ea
R·(273 + 25) =−Ea
R·(273 + 45)
Step 5: Solve for the activation energy:
Ea
R·(273 + 25) =Ea
R·(273 + 45)
Step 6: Cross multiply to obtain:
Ea·(273 + 45) = Ea·(273 + 25)
Step 7: The activation energy can be calculated as follows:
Ea=R·273 + 25
45 −25
Step 8: Calculate the activation energy:
Ea= 8.314 J/(mol*K) ·273 + 25
45 −25 = X kJ/mol
Step 9: Calculate the value of Eato determine the activation energy for this
reaction.
Question 12
Question
The rate constant for the reaction A →P is found to be 4.23 ×10−3s−1at
20◦C. When the temperature is increased to 60◦C, the rate constant becomes
0.023 s−1. Calculate the activation energy of the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to tem-
perature and activation energy. The Arrhenius equation is given by:
k=A·e−Ea
RT
11
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J/mol ·K), - Tis the
temperature in Kelvin.
Step 2: Let’s first calculate the Arrhenius parameters at 20◦C. At 20◦C,
T= 20 + 273.15 = 293.15 K. Given k1= 4.23 ×10−3s−1, we have:
4.23 ×10−3=A·e−Ea
8.314·293.15
Step 3: Now, let’s calculate the Arrhenius parameters at 60◦C. At 60◦C,
T= 60 + 273.15 = 333.15 K. Given k2= 0.023 s−1, we have:
0.023 = A·e−Ea
8.314·333.15
Step 4: We now have a system of two equations:
4.23 ×10−3=A·e−Ea
8.314·293.15
0.023 = A·e−Ea
8.314·333.15
Step 5: By dividing the second equation by the first one, we can eliminate
the pre-exponential factor A:
0.023
4.23 ×10−3=e−Ea
8.314·333.15
e−Ea
8.314·293.15
Step 6: Solving the above equation will give us the activation energy Ea.
Let’s go ahead and solve for Ea.
Question 13
Question
The rate constant of a reaction doubles when the temperature is increased from
25
°
C to 35
°
C. Calculate the activation energy of the reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given that T1= 25◦C
and T2= 35◦C, we convert to Kelvin using the formula T(K) = T(C) + 273.15.
Thus, T1= 25 + 273.15 = 298.15 Kand T2= 35 + 273.15 = 308.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants at two
temperatures. The Arrhenius equation is given by 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)), - Tis the temperature in Kelvin.
Since the rate constant doubles when the temperature is increased from 25
°
C
to 35
°
C, we can write:
2k1=k2
12
Substitute the Arrhenius equation into this relationship:
2A·e−Ea
R·298.15 =A·e−Ea
R·308.15
Step 3: Simplify the equation and solve for Ea. Divide both sides by Aand
take the natural logarithm of both sides to get rid of the exponential term:
ln(2) = Ea
R1
298.15 −1
308.15
Now, solve for Ea:
Ea=R·308.15 ·298.15
308.15 −298.15 ·ln(2)
Step 4: Calculate the activation energy. Substitute the values for R=
8.314 J/(mol
·
K) and solve for Ea:
Ea= 8.314 ×308.15 ×298.15
308.15 −298.15 ×ln(2)
Ea≈5.65 ×104J/mol
Therefore, the activation energy of the reaction is approximately 5.65 ×
104J/mol.
Question 14
Question
The rate constant of a reaction at 25
°
C is found to be 1.20 ×10−2s−1. When
the temperature is raised to 45
°
C, the rate constant increases to 5.80 ×10−2
s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the rate constants to their corresponding Arrhenius equations
using the Arrhenius equation:
k=Ae−Ea
RT
where: - k1= 1.20 ×10−2s−1at 25
°
C - k2= 5.80 ×10−2s−1at 45
°
C -
T1= 25 + 273 = 298 K - T2= 45 + 273 = 318 K
For the first temperature:
k1=Ae−Ea
R·298
1.20 ×10−2=Ae−Ea
8.314·298
For the second temperature:
k2=Ae−Ea
8.314·318
13
5.80 ×10−2=Ae−Ea
8.314·318
Step 2: Divide the two Arrhenius equations to eliminate the pre-exponential
factor A:
k2
k1
=e−Ea
8.314·318
e−Ea
8.314·298
5.80 ×10−2
1.20 ×10−2=e−Ea
8.314 (1
318 −1
298 )
Step 3: Solve for the activation energy (Ea) by taking the natural logarithm
of both sides and rearranging the equation:
ln 5.80 ×10−2
1.20 ×10−2=−Ea
8.314 1
318 −1
298
Ea=−8.314 ×
ln 5.80×10−2
1.20×10−2
1
318 −1
298
Step 4: Calculate the activation energy:
Ea=−8.314 ×
ln 5.80×10−2
1.20×10−2
1
318 −1
298
Ea≈43.6 kJ/mol
Therefore, the activation energy for this reaction is approximately 43.6
kJ/mol.
Question 15
Question
The rate constant for a certain reaction is found to be 4.5×10−3s−1at 25◦C
and 1.2×10−2s−1at 35◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula:
T(K) = T(◦C) + 273.15
For 25◦C:
T1= 25 + 273.15 = 298.15 K
For 35◦C:
T2= 35 + 273.15 = 308.15 K
14
Step 2: Calculate the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
We can rearrange the equation to solve for Ea:
1
T1
=−Ea
R·1
T2
1
+ ln(A)
Similarly for the second set of data:
1
T2
=−Ea
R·1
T2
2
+ ln(A)
Step 3: Substitute the known values into the above equations and solve for
Ea:1
298.15 =−Ea
8.314 ·1
298.152+ ln4.5×10−3
1
308.15 =−Ea
8.314 ·1
308.152+ ln1.2×10−2
Solving these two simultaneous equations will give the value of the activation
energy, Ea.
Question 16
Question
The rate constant for a reaction was found to be 6.25 ×10−2s−1at 25
°
C and
8.75 ×10−2s−1at 37
°
C. Calculate the activation energy for the reaction.
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(
°
C) +
273.15. At 25
°
C: T1= 25 + 273.15 = 298.15 K
At 37
°
C: T2= 37 + 273.15 = 310.15 K
Step 2: Write down 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 3: Set up two equations using the given rate constants at the two
temperatures:
k1=A·e
−Ea
R·T1
k2=A·e
−Ea
R·T2
15
Step 4: Solve the system of equations for the activation energy Ea. Divide
the two equations:
k1
k2
=e
−Ea
R·T1
e
−Ea
R·T2
k1
k2
=e
Ea
R1
T2
−1
T1
Step 5: Plug in the known values and solve for Ea.
6.25 ×10−2
8.75 ×10−2=e
Ea
8.314 (1
310.15 −1
298.15 )
5
7=e
Ea
8.314 (1
310.15 −1
298.15 )
Step 6: Solve for Eaby taking the natural logarithm of both sides and
rearranging the equation.
ln 5
7=Ea
8.314 1
310.15 −1
298.15
Ea=−8.314 ×ln 5
7×1
310.15 −1
298.15
Step 7: Calculate the activation energy using the given values and the for-
mula above.
Question 17
Question
The rate constant of a reaction at 25
°
C is 1.8×10−3s−1, while at 50
°
C the rate
constant is 3.2×10−2s−1. Calculate the activation energy for this reaction.
(Hint: Use the Arrhenius equation k=Ae−Ea
RT where kis the rate constant, A
is 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= 25C= 25 + 273.15 = 298.15 K
T2= 50C= 50 + 273.15 = 323.15 K
Step 2: Write the Arrhenius equation for both temperatures.
At T1= 298.15 K:
k1=Ae−Ea
RT1
1.8×10−3=Ae−Ea
R×298.15
16
At T2= 323.15 K:
k2=Ae−Ea
RT2
3.2×10−2=Ae−Ea
R×323.15
Step 3: Divide the two Arrhenius equations to eliminate A and solve for Ea.
k2
k1=e−Ea
R×323.15
e−Ea
R×298.15
3.2×10−2
1.8×10−3=eEa
R(1
298.15 −1
323.15 )
3.2
1.8=eEa
R(1
298.15 −1
323.15 )
ln 3.2
1.8=Ea
R(1
298.15 −1
323.15 )
Step 4: Solve for Ea.
ln 3.2
1.8=Ea
R(1
298.15 −1
323.15 )
Ea=R×ln(3.2
1.8)
1
298.15 −1
323.15
Ea= 8.314 ×ln(3.2
1.8)
1
298.15 −1
323.15
Ea≈48670 J/mol
Ea≈48.67 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.67
kJ/mol.
Question 18
Question
The rate constant (k) for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When
the temperature is increased to 50
°
C, the rate constant becomes 8.75×10−3s−1.
Calculate the activation energy (Ea) for this reaction. (R = 8.314 J/mol
·
K)
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. The temperature in
Kelvin is given by T(K) = T(C) + 273.15. For 25
°
C: T1= 25 + 273.15 = 298.15
K. For 50
°
C: T2= 50 + 273.15 = 323.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures. The Arrhenius equation is given by:
k=A×e−Ea
RT
where: k1= 1.25 ×10−3s−1,T1= 298.15 K, k2= 8.75 ×10−3s−1,T2= 323.15
K, R= 8.314 J/mol
·
K.
Step 3: Take the ratio of the two Arrhenius equations to eliminate the pre-
exponential factor.
k2
k1
=A×e−Ea
RT2
A×e−Ea
RT1
17
k2
k1
=e−Ea
R1
T1
−1
T2
Step 4: Solve for the activation energy (Ea) using the known values. Sub-
stitute the given values:
8.75 ×10−3
1.25 ×10−3=e−Ea
8.314 (1
298.15 −1
323.15 )
Step 5: Calculate Ea.
8.75 ×10−3
1.25 ×10−3=e
−Ea
8.314 (1
323.15 −1
298.15 )
7 = e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 6: Solve for Ea.
ln(7) = −Ea
8.314 1
323.15 −1
298.15
Ea=−8.314 ×ln(7)
1
323.15 −1
298.15
Question 19
Question
The rate constant for a certain reaction is known to double when the tempera-
ture is increased from 25◦C to 35◦C. Calculate the activation energy (in kJ/mol)
for this reaction.
Given: Activation Energy at 25◦C = 50 kJ/mol Gas constant, R= 8.314
J/(mol·K)
Solution
Step 1: Convert the given activation energy to joules: Let Eabe the activation
energy in joules.
Activation Energy (J) = 50 kJ/mol ×1000 J/kJ = 50000 J/mol
Step 2: Use the Arrhenius equation to relate the rate constants at two
different temperatures: The Arrhenius equation is given by:
k=A·e−Ea
RT
Given that the rate constant doubles when the temperature is increased from
25◦C to 35◦C, we have:
k2
k1
= 2
18
A·e−Ea
R(25+273.15)
A·e−Ea
R(35+273.15)
= 2
e−50000
8.314×(25+273.15)
e−50000
8.314×(35+273.15)
= 2
Step 3: Solve for the activation energy, Ea:
−50000
8.314 ×(25 + 273.15) +50000
8.314 ×(35 + 273.15) = ln(2)
−1931.79 + 2447.85 = 0.69315
516.06 = 0.69315
Therefore, the activation energy for this reaction is approximately 516.06
kJ/mol.
Question 20
Question
The rate constant (k) for a certain reaction is known to be 4.72 ×10−3s−1at
25
°
C. When the temperature is increased to 45
°
C, the rate constant becomes
7.61 ×10−2s−1. Calculate the activation energy (Ea) for this reaction. The
activation energy is expressed in units of kJ/mol.
Solution
Step 1: Determine the value of the gas constant R. Given: Temperature (T1)
= 25
°
C = 25 + 273 = 298 K Temperature (T2) = 45
°
C = 45 + 273 = 318 K
We can use the Arrhenius equation to relate the rate constants at different
temperatures:
k2
k1
= exp Ea
R1
T1
−1
T2
where: k1= 4.72 ×10−3s−1,k2= 7.61 ×10−2s−1,T1= 298 K, T2= 318 K,
and Ris the gas constant.
Step 2: Calculate R. We can rearrange the Arrhenius equation to solve for
R:
R=Ea
ln k2
k11
T2
−1
T1
Substitute the given values into the equation to find R.
Step 3: Calculate Ea. Once Ris determined, we can solve for the activation
energy Eausing the formula:
Ea=Rln k2
k1 1
T2
−1
T1
Substitute the values of R,k1,k2,T1, and T2into the equation to find the
activation energy in kJ/mol.
19
Question 21
Question
The rate constant for a certain reaction is 4.2×10−4s−1at 25◦C and 1.5×10−2
s−1at 50◦C. Calculate the activation energy (Ea) of the reaction.
Solution
Step 1: Let’s start by writing the Arrhenius equation:
k=Ae−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: We are given two sets of data: At 25◦C (298 K): k1= 4.2×10−4
s−1At 50◦C (323 K): k2= 1.5×10−2s−1
Step 3: Divide the two Arrhenius equations to eliminate A:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R(1
T1
−1
T2)
Step 4: Plug in the known values:
1.5×10−2
4.2×10−4=e−Ea
8.314 (1
298 −1
323 )
Step 5: Solve for Ea:
1.5×10−2
4.2×10−4=e−Ea
8.314 (0.00336)
35.71 = e−3.1256×10−3Ea
Step 6: Take the natural logarithm of both sides:
ln(35.71) = lne−3.1256×10−3Ea
ln(35.71) = −3.1256 ×10−3Ea
Step 7: Solve for Ea:
Ea=−ln(35.71)
3.1256 ×10−3≈53.57 kJ/mol
Therefore, the activation energy of the reaction is approximately 53.57 kJ/mol.
20
Question 22
Question
The rate constant for a reaction at 25◦C is 1.20 ×10−3s−1. When the tempera-
ture is increased to 55◦C, the rate constant becomes 2.75 ×10−2s−1. Calculate
the activation energy in kJ/mol for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given: T1= 25◦C =
298 K T2= 55◦C = 328 K
Step 2: Use the Arrhenius equation to find the activation energy. The
Arrhenius equation relates the rate constant kto the temperature Tand the
activation energy Ea:
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
Taking the ratio of the rate constants at the two temperatures and solving
for Ea:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
k2
k1
=e
Ea
R1
T1
−1
T2
ln k2
k1=Ea
R1
T1
−1
T2
Ea=R·
ln k2
k1
1
T1
−1
T2
Step 3: Substitute the given values into the equation to find the activation
energy.
Ea= 8.314 J/mol·K·
ln 2.75×10−2
1.20×10−3
1
298 −1
328
Calculating the activation energy gives:
Ea≈66.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 66.5
kJ/mol.
21
Question 23
Question
The rate constant for a reaction is 4.17 ×10−3s−1at 25
°
C and 2.50 ×10−2s−1
at 55
°
C. Calculate the activation energy (in kJ/mol) for this reaction. (Hint:
Use 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, and
Tis the temperature in Kelvin.)
Solution
Step 1: Convert temperatures to Kelvin
Given: - Temperature at 25
°
C: T1= 25C+ 273.15 = 298.15 K - Temperature
at 55
°
C: T2= 55C+ 273.15 = 328.15 K
Step 2: Use the Arrhenius equation
Given rate constants: - At 25
°
C: k1= 4.17×10−3s−1- At 55
°
C: k2= 2.50×10−2
s−1
The Arrhenius equation can be written as:
k=Ae−Ea
RT
Substitute the rate constants and temperatures into the Arrhenius equation:
k1=Ae−Ea
R·298.15
k2=Ae−Ea
R·328.15
Step 3: Take the ratio of the two rate constants
Divide the second equation by the first to eliminate A:
k2
k1
=e−Ea
R·328.15
e−Ea
R·298.15
k2
k1
=e−Ea
R(1
328.15 −1
298.15 )
Step 4: Solve for activation energy
Take natural logarithms of both sides of the equation to solve for the activation
energy, Ea:
ln k2
k1=−Ea
R1
328.15 −1
298.15
Finally, calculate the activation energy Eain kJ/mol by using the gas con-
stant R= 8.314 J/(mol
·
K).
22
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is 2.35 ×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.23 ×10−2s−1.
Calculate the activation energy for this reaction. (Given: R= 8.314 J mol−1K−1)
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25C+ 273.15 = 298.15 K
T2= 50C+ 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at different
temperatures.
k2=k1·e
−Ea
R1
T2
−1
T1
Step 3: Plug in the given values and the rate constants.
1.23 ×10−2= 2.35 ×10−3·e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 4: Solve for the activation energy Ea.
1.23 ×10−2
2.35 ×10−3=e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 5: Simplify and solve for Ea.
5.23 = e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 6: Take the natural logarithm of both sides and solve for Ea.
ln(5.23) = −Ea
8.314 1
323.15 −1
298.15
Step 7: Calculate Ea.
Ea=−8.314 ×(ln(5.23)) ×1
323.15 −1
298.15
Step 8: Calculate the final answer for the activation energy Ea. Remember
to include the appropriate units.
Ea≈67.8 kJ/mol
23
Question 25
Question
The rate constant (k) for a certain reaction is found to be 1.25 ×10−2s−1at
25◦C and 4.18 ×10−2s−1at 45◦C. Calculate the activation energy (in kJ/mol)
for the reaction.
(Assume the activation energy is independent of temperature and use the
universal gas constant R= 8.314 J/mol ·K for calculations.)
Solution
Step 1: Convert temperatures to Kelvin using T(K) = T(C) + 273.15.
T1= 25 + 273.15 = 298.15 K
T2= 45 + 273.15 = 318.15 K
Step 2: Write out the Arrhenius equation:
k=A·e−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.
Step 3: Take the ratio of the rate constants at the two temperatures and
solve for the activation energy (Ea):
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
k2
k1
=e−Ea
R1
T1
−1
T2
ln k2
k1=−Ea
R1
T1
−1
T2
Ea=−R·ln k2
k1 1
T1
−1
T2
Step 4: Plug in the given values:
Ea=−8.314 J/mol ·K×ln 4.18 ×10−2
1.25 ×10−2×1
298.15 −1
318.15
Step 5: Calculate the activation energy (Ea) in kJ/mol.
Ea=−8.314 ×ln (3.344) ×(0.003354 −0.003144)
Ea≈ −8.314 ×ln (3.344) ×0.000210
Ea≈ −8.314 ×1.204 ×0.000210
Ea≈ −8.314 ×0.00025284
Ea≈ −0.0021 kJ/mol ≈21 kJ/mol
Therefore, the activation energy for the reaction is 21 kJ/mol.
24
Question 26
Question
For a certain reaction, the rate constant at 25
°
C is 3.2×10−3s−1and the rate
constant at 50
°
C is 8.5×10−2s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: To find the activation energy (Ea) for the reaction, we will use 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: We are given two sets of data at different temperatures: At 25
°
C,
k1= 3.2×10−3s−1T1= 25 + 273 = 298 K
At 50
°
C, k2= 8.5×10−2s−1T2= 50 + 273 = 323 K
Step 3: Substituting the first set of data into the Arrhenius equation gives:
3.2×10−3=A·e−Ea
8.314×298
Step 4: Substituting the second set of data into the Arrhenius equation gives:
8.5×10−2=A·e−Ea
8.314×323
Step 5: Divide the second equation by the first equation to eliminate A:
8.5×10−2
3.2×10−3=e
Ea
8.314 (1
323 −1
298 )
Step 6: Solve for Eaby isolating it:
Ea=−8.314 ·ln 8.5×10−2
3.2×10−3·1
323 −1
298
Step 7: Calculate Eausing a calculator:
Ea≈61.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 61.7
kJ/mol.
Question 27
Question
The rate constant for a certain reaction is found to be 2.50 ×10−4s−1at 25
°
C,
and 9.80 ×10−3s−1at 50
°
C. Calculate the activation energy for this reaction.
Given: R= 8.314 J/(mol
·
K), T1= 25C,T2= 50C
25
Solution
Step 1: Convert the temperatures to Kelvin:
T1= 25C+ 273 = 298K
T2= 50C+ 273 = 323K
Step 2: Use the Arrhenius equation to relate the rate constants to the tem-
peratures 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, and Tis the temperature in Kelvin.
Step 3: Set up two equations for the given data:
2.50 ×10−4=A·e−Ea
8.314·298
9.80 ×10−3=A·e−Ea
8.314·323
Step 4: Divide the two equations to eliminate A:
2.50 ×10−4
9.80 ×10−3=e−Ea
8.314·298
e−Ea
8.314·323
Step 5: Simplify the equation and solve for Ea:
2.50
9.80 =e
Ea
8.314 (1
323 −1
298 )
2.50
9.80 =e−Ea
8.314·38354
0.255 = e−Ea
254.81
ln(0.255) = lne−Ea
254.81
−1.366 = −Ea
254.81
Ea= 1.366 ×254.81 = 348.04 kJ/mol
Therefore, the activation energy for this reaction is 348.04 kJ/mol.
Question 28
Question
The rate constant for a certain reaction was measured at two different tempera-
tures, 25◦C and 50◦C. At 25◦C, the rate constant was found to be 1.5×10−3s−1,
while at 50◦C, the rate constant was 7.5×10−3s−1. Calculate the activation
energy (in kJ/mol) for this reaction.
26
Solution
Step 1: Convert the given temperatures to Kelvin using the equation T(K) =
T(◦C) + 273.15.
T1= 25 + 273.15 = 298.15 K
T2= 50 + 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures to the activation energy.
k2
k1
=e
Ea
R(1
T1
−1
T2)
Step 3: Plug in the given values for k1,k2,T1, and T2, and the universal gas
constant R= 8.314 J/(mol K).
7.5×10−3
1.5×10−3=e
Ea
8.314 (1
298.15 −1
323.15 )
Step 4: Solve for the activation energy Ea.
7.5
1.5=e
Ea
8.314 (1
298.15 −1
323.15 )
5 = e
Ea
8.314 (−0.001356)
Step 5: Take the natural logarithm of both sides to solve for the activation
energy.
ln(5) = Ea
8.314(−0.001356)
Step 6: Solve for the activation energy Eain kJ/mol.
Ea=−8.314 ×0.001356 ×ln(5) ≈61.96 kJ/mol
Therefore, the activation energy for this reaction is approximately 61.96
kJ/mol.
Question 29
Question
For a certain chemical reaction, the rate constant at 25
°
C is 1.2×10−3s−1, and
the rate constant at 45
°
C is 2.5×10−2s−1. Calculate the activation energy of
the reaction in kJ/mol.
27
Solution
Step 1: Convert the given temperatures from Celsius to Kelvin.
T1= 25C+ 273 = 298 K
T2= 45C+ 273 = 318 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures.
k2
k1
=e(−Ea
R)1
T2
−1
T1
Step 3: Substitute the given rate constants, temperatures, and gas constant
R= 8.314 J/mol ·K into the Arrhenius equation.
2.5×10−2
1.2×10−3=e(−Ea
8.314 )( 1
318 −1
298 )
Step 4: Solve for the activation energy Ea.
2.5×10−2
1.2×10−3=e(−Ea
8.314 )( 1
318 −1
298 )
20.83 = e(−Ea
8.314 )( 1
318 −1
298 )
ln(20.83) = −Ea
8.314 1
318 −1
298
Step 5: Calculate the activation energy Eain J/mol.
Ea=−8.314 × 1
318 −1
298×ln(20.83)
≈92736 J/mol
≈92.7 kJ/mol
Therefore, the activation energy of the reaction is approximately 92.7 kJ/mol.
Question 30
Question
The rate constant for the decomposition of a certain compound at 298 K is
5.32 ×10−3s−1, and the activation energy for the reaction is 78.5 kJ/mol.
Calculate the rate constant at 313 K if the activation energy remains constant.
28
Solution
Step 1: Calculate the rate constant at 313 K using the Arrhenius equation:
k2=k1·e−Ea
R1
T2
−1
T1
where: k2= rate constant at 313 K k1= rate constant at 298 K Ea= acti-
vation energy = 78.5 kJ/mol R= gas constant = 8.314 J/(mol*K) T2= new
temperature = 313 K T1= initial temperature = 298 K
Step 2: Substitute the given values into the equation and solve for k2:
k2= 5.32 ×10−3s−1·e−78.5 kJ/mol
8.314 J/(mol*K) (1
313 K −1
298 K )
Step 3: Calculate the value of k2:
k2= 5.32 ×10−3s−1·e−78.5×103J/mol
8.314 J/(mol*K) (1
313 K −1
298 K )
Step 4: Simplify the equation and calculate k2:
k2≈0.0305 s−1
Therefore, the rate constant at 313 K is approximately 0.0305 s−1.
Question 31
Question
The rate constant for the reaction 2A→Bis 0.02 s−1at 25◦C and 0.12 s−1at
45◦C. Calculate the activation energy for the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants to the
temperature and activation energy. The Arrhenius equation is given by:
k=Ae−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 can write two equations based on the rate constants given at
two different temperatures:
k1=Ae−Ea
R·(25+273.15)
k2=Ae−Ea
R·(45+273.15)
29
Step 3: We can solve these two equations simultaneously to find the activa-
tion energy, Ea. We first take the ratio of the two equations:
k2
k1
=Ae−Ea
R·(45+273.15)
Ae−Ea
R·(25+273.15)
k2
k1
=e−Ea
R(1
45+273.15 −1
25+273.15 )
Step 4: Given that k1= 0.02 s−1and k2= 0.12 s−1, we can plug in the
values and solve for the activation energy, Ea.
Step 5: Calculate the activation energy, Ea, using the values obtained in
Step 4 and the gas constant R= 8.314 J ·mol−1·K−1.
Thus, the activation energy for the reaction is
Ea= 40.2 kJ ·mol−1
.
Question 32
Question
The rate constant for a certain reaction is 2.83 ×10−3s−1at 35◦C, while at
45◦C it is 1.37 ×10−2s−1. Calculate 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, R= 8.314 J K−1mol−1is the gas constant,
and Tis temperature in Kelvin)
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 35◦C + 273.15 = 308.15 K
T2= 45◦C + 273.15 = 318.15 K
Step 2: Write down the Arrhenius equation for the two temperatures and
the given rate constants.
k1=A·e−Ea
RT1
k2=A·e−Ea
RT2
Step 3: Take the ratio of the two rate constants to eliminate A.
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
30
k2
k1
=e−Ea
R1
T2
−1
T1
Step 4: Take the natural logarithm of both sides to solve for Ea.
ln k2
k1=−Ea
R1
T2
−1
T1
Step 5: Substitute the given values and solve for Ea.
ln 1.37 ×10−2
2.83 ×10−3=−Ea
8.314 1
318.15 −1
308.15
ln (4.82) = −Ea
8.314 1
318.15 −1
308.15
Question 33
Question
The rate constant for the reaction 2A →B is given by the Arrhenius equation:
k=Ae−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the
activation energy, R= 8.314
Step 1: Convert the temperatures given to Kelvin.
T1= 25C= 25 + 273 = 298 K
T2= 50C= 50 + 273 = 323 K
Step 2: Set up two Arrhenius equations using the rate constants and tem-
peratures given:
k1=Ae−Ea
R·T1
k2=Ae−Ea
R·T2
Step 3: Divide the two equations to eliminate A:
k1
k2
=e−Ea
R·T1
e−Ea
R·T2
Step 4: Simplify the equation by canceling out A:
k1
k2
=e(Ea
R)1
T2
−1
T1
Step 5: Substitute the given rate constants and temperatures:
5.0×10−3
1.0×10−2=e(Ea
8.314 )( 1
323 −1
298 )
31
Step 6: Solve for the activation energy Ea:
ln 5.0×10−3
1.0×10−2=Ea
8.314 1
323 −1
298
⇒
8.314 ·ln 5.0×10−3
1.0×10−2
1
323 −1
298
=Ea
Step 7: Calculate the activation energy Ea:
Ea≈35.8 kJ/mol
Therefore, the activation energy for the reaction 2A →B is approximately
35.8 kJ/mol.
Question 34
Question
For a certain reaction, the rate constant at 25
°
C is 1.5×10−4s−1and the
activation energy is 50 kJ/mol. Determine the rate constant at 50
°
C for this
reaction.
Solution
Step 1: Given the activation energy and the rate constant at 25
°
C, we can
use the Arrhenius equation to find the rate constant at 50
°
C. The Arrhenius
equation is given by:
k=Ae−Ea
RT
where: k= rate constant at temperature T (in Kelvin), A= pre-exponential
factor, Ea= activation energy, R= gas constant (8.314 J/mol-K), T= tem-
perature in Kelvin.
Step 2: Let’s first convert the activation energy to joules:
Ea= 50 kJ/mol ×1000 J/kJ = 50000 J/mol
Step 3: Calculate the rate constant at 25
°
C:
k25 = 1.5×10−4s−1
Step 4: Convert the temperatures to Kelvin:
T25 = 25C+ 273 = 298 K
T50 = 50C+ 273 = 323 K
32
Step 5: Plug in the values into the Arrhenius equation to find the pre-
exponential factor A:
1.5×10−4=Ae−50000
8.314×298
Step 6: Solve for A:
A= 1.5×10−4×e50000
8.314×298
Step 7: Finally, calculate the rate constant at 50
°
C using the pre-exponential
factor Aand the activation energy Ea:
k50 =Ae−50000
8.314×323
Question 35
Question
Consider a reaction with an activation energy of 75 kJ/mol. If the rate constant
at 25
°
C is 1.8×10−2s−1and the activation energy is reduced to 50 kJ/mol,
what will be the new rate constant at 25
°
C? (Assume the pre-exponential factor
A remains constant.)
Solution
Step 1: Calculate the rate constant using the Arrhenius equation for the original
activation energy.
k1=A·e−Ea
RT
Given: Original activation energy Ea1= 75 kJ/mol, Rate constant k1= 1.8×
10−2s−1, Temperature T= 25
°
C = 298 K.
Plugging in the values:
k1=A·e−75000
8.314×298
k1=A·e−31.77 = 1.8×10−2
Step 2: Calculate the new rate constant using the modified activation energy.
k2=A·e−
Ea2
RT =A·e−50000
8.314×298
k2=A·e−21.18
Step 3: Determine the ratio between the new and original rate constants.
k2
k1
=A·e−21.18
A·e−31.77
e−21.18+31.77 =e10.59
k2=k1·e10.59
k2= 1.8×10−2·e10.59
33
Question 4
Question
The rate constant of a reaction at 25
°
C is 4.22 ×10−3s−1. When the temper-
ature is increased to 45
°
C, the rate constant becomes 0.055 s−1. 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= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol·K), T= temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation: k1= 4.22 ×10−3s−1
(at 25
°
C = 298 K), k2= 0.055 s−1(at 45
°
C = 318 K).
Step 4: Calculate the activation energy, Ea:
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
ln 0.055
4.22 ×10−3=−Ea
8.314 1
318 −1
298
Ea=8.314
ln 0.055
4.22×10−31
318 −1
298
Ea≈8.314
ln 0.055
4.22×10−31
318 −1
298
Ea≈46.21 kJ/mol
Therefore, the activation energy for this reaction is approximately 46.21
kJ/mol.
4
Question 5
Question
A reaction has an activation energy of 50.0 kJ/mol and a frequency factor of
1.0×1010 s−1. At a certain temperature, the rate constant of the reaction is
5.0×10−4s−1. Calculate the rate constant at a temperature that is 10
°
C higher,
assuming the activation energy does not change.
Solution
Step 1: Calculate the rate constant k1at the initial temperature using the
Arrhenius equation:
k1=A·e−Ea
RT1
where: - k1is the rate constant at the initial temperature, - Ais the frequency
factor, - Eais the activation energy, - Ris the ideal gas constant (8.314 J/(mol
K)), - T1is the initial temperature in Kelvin.
Given that A= 1.0×1010 s−1,Ea= 50.0 kJ/mol = 5.0×104J/mol, and
T1corresponds to the initial temperature, use T1=T1+ 273.15 K.
Step 2: Calculate the rate constant k2at the higher temperature using the
Arrhenius equation with the new temperature T2=T1+ 10
°
C:
k2=A·e−Ea
RT2
Step 3: Substitute the known values into the Arrhenius equation to find the
new rate constant k2at the higher temperature.
Question 6
Question
The rate constant for the reaction
2A+B→C
is found to be 2.2×10−3s−1at 25◦C and 1.5×10−2s−1at 75◦C. Calculate the
activation energy for the reaction.
Solution
Step 1: Let’s first write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J/mol·K), - Tis the
temperature in Kelvin.
5
Step 2: Use the data given at 25◦C and 75◦C to form two equations:
2.2×10−3=A·e−Ea
8.314·(25+273.15)
1.5×10−2=A·e−Ea
8.314·(75+273.15)
Step 3: Take the ratio of the two equations to eliminate A:
2.2×10−3
1.5×10−2=e−Ea
8.314·(25+273.15)
e−Ea
8.314·(75+273.15)
Step 4: Simplify the equation to solve for Ea:
2.2
1.5=e
Ea
8.314 (1
25+273.15 −1
75+273.15 )
Step 5: Calculate and simplify further to find the activation energy Ea.
Question 7
Question
The rate constant for the first-order decomposition of a compound is 6.25 ×
10−4s−1at 350 K and 1.25 ×10−3s−1at 370 K. Calculate the activation energy
for this reaction.
Solution
Step 1: Let’s start by writing 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: Next, we can rearrange the Arrhenius equation to solve for Eausing
two sets of data given:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
k2
k1
=e
Ea
R1
T1
−1
T2
Given that k1= 6.25 ×10−4s−1at 350 K and k2= 1.25 ×10−3s−1at 370 K, we
can plug in these values along with Rto solve for Ea.
Step 3: Substituting the values in, we get:
1.25 ×10−3
6.25 ×10−4=e
Ea
8.314 (1
350 −1
370 )
6
2 = e
Ea
8.314 (1
350 −1
370 )
Step 4: Simplifying further, we find:
ln(2) = Ea
8.314 1
350 −1
370
0.693
8.314 =Ea1
350 −1
370
Step 5: Solve for Ea:
0.693
8.314 =Ea1
350 −1
370
Ea=0.693
8.314∇ · 1
350 −1
370
Ea≈62200 J/mol
Therefore, the activation energy for this reaction is approximately 62200 J/mol.
Question 8
Question
The rate constant for a first-order reaction at 25
°
C is 1.20×10−3s−1. When the
temperature is increased to 45
°
C, the rate constant becomes 6.80 ×10−3s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
k2
k1
= exp Ea
R1
T1
−1
T2
where: - k1and k2are the rate constants at temperatures T1and T2respectively,
-Eais the activation energy, - Ris the gas constant (8.314 J/mol·K), - T1and
T2are the temperatures in Kelvin.
Step 2: Plug in the given values:
6.80 ×10−3
1.20 ×10−3= exp Ea
8.314 1
298 −1
318
Step 3: Simplify the equation by dividing the rate constants and calculating
the temperature differences:
5.67 = exp Ea
8.314 1
298 −1
318
7
Step 4: Calculate the temperature differences:
1
298 −1
318 =318 −298
298 ×318 =20
89484 =5
22371
Step 5: Substitute back into the equation:
5.67 = exp Ea
8.314 ·5
22371
Step 6: Solve for the activation energy Eaby taking the natural logarithm
of both sides:
ln(5.67) = Ea
8.314 ·5
22371
Step 7: Rearrange the equation to solve for Ea:
Ea= 8.314 ·ln(5.67) ·22371
5
Step 8: Calculate Eato find the activation energy of the reaction.
Question 9
Question
The rate constant for a certain reaction is found to be 3.5×10−4s−1at 25 ◦C
and 1.2×10−3s−1at 35 ◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down 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: Rearrange the Arrhenius equation to solve for the activation energy
Ea:
Ea=− ln k
A
1
R·T1
−1
R·T2!
where: - k1= 3.5×10−4s−1at 25 ◦C = 298 K, - k2= 1.2×10−3s−1at
35 ◦C = 308 K.
Step 3: Plug in the values and solve for the activation energy Ea:
Ea=−
ln 1.2×10−3
3.5×10−4
1
8.314·298 −1
8.314·308
8
Ea=−ln (3.43)
1
2485.272 −1
2542.392
Ea=−ln (3.43)
0.000401 −0.000393
Ea=−ln (3.43)
0.000008
Ea=−9.203 ×103K
Therefore, the activation energy for this reaction is 9.203 ×103K.
Question 10
Question
The rate constant of a reaction is observed to double when the temperature
is increased from 25
°
C to 35
°
C. Calculate the activation energy (Ea) for the
reaction. Assume the frequency factor (A) is constant.
Solution
Step 1: Convert temperatures to Kelvin.
The temperatures given are 25
°
C and 35
°
C. To convert to Kelvin, we add 273
to each temperature:
T1= 25C+ 273 = 298 K
T2= 35C+ 273 = 308 K
Step 2: Use the Arrhenius equation.
The Arrhenius equation relates the rate constant of a reaction to the tempera-
ture and the activation energy:
k=Ae−Ea
RT
Step 3: Determine the ratio of rate constants.
Given that the rate constant doubles when the temperature increases from T1
to T2, we have:
k2
k1
= 2
Step 4: Take the ratio of the Arrhenius equations.
Substitute the Arrhenius equation for k1and k2and simplify:
Ae−Ea
RT2
Ae−Ea
RT1
= 2
9
Step 5: Simplify the equation.
Cancel out the frequency factor Aand rearrange the equation:
e
Ea
R(1
T1
−1
T2)= 2
Step 6: Solve for the activation energy (Ea).
Substitute the values of T1and T2, and solve for Ea:
e
Ea
R(1
298 −1
308 )= 2
e
Ea
R(10
298×308 )= 2
Ea
R(10
298 ×308) = ln 2
Ea= (ln 2) ×R×298 ×308
10
Step 7: Calculate the activation energy.
Now, substitute the value of the gas constant R= 8.314 J/(mol*K) to find the
activation energy Ea.
Question 11
Question
The rate constant of a first-order reaction doubles when the temperature is
increased from 25
°
C to 45
°
C. Calculate the activation energy (in kJ/mol) for
this reaction. (Given: R = 8.314 J/(mol*K))
Solution
Step 1: The Arrhenius equation relates the rate constant of a reaction to tem-
perature and the activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor (frequency
factor), - Eais the activation energy, - Ris the gas constant (8.314 J/(mol*K)),
-Tis the temperature in Kelvin.
Step 2: Given that the rate constant doubles when the temperature is in-
creased from 25
°
C to 45
°
C, we can write:
k2= 2 ·k1
10
Substitute the Arrhenius equation for both temperatures:
A·e−Ea
R·(273+25) = 2 ·A·e−Ea
R·(273+45)
Step 3: Simplify the equation by dividing both sides by Aand cancelling the
factor of 2:
e−Ea
R·(273+25) =e−Ea
R·(273+45)
Step 4: To solve for the activation energy, set the exponents equal to each
other:
−Ea
R·(273 + 25) =−Ea
R·(273 + 45)
Step 5: Solve for the activation energy:
Ea
R·(273 + 25) =Ea
R·(273 + 45)
Step 6: Cross multiply to obtain:
Ea·(273 + 45) = Ea·(273 + 25)
Step 7: The activation energy can be calculated as follows:
Ea=R·273 + 25
45 −25
Step 8: Calculate the activation energy:
Ea= 8.314 J/(mol*K) ·273 + 25
45 −25 = X kJ/mol
Step 9: Calculate the value of Eato determine the activation energy for this
reaction.
Question 12
Question
The rate constant for the reaction A →P is found to be 4.23 ×10−3s−1at
20◦C. When the temperature is increased to 60◦C, the rate constant becomes
0.023 s−1. Calculate the activation energy of the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to tem-
perature and activation energy. The Arrhenius equation is given by:
k=A·e−Ea
RT
11
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J/mol ·K), - Tis the
temperature in Kelvin.
Step 2: Let’s first calculate the Arrhenius parameters at 20◦C. At 20◦C,
T= 20 + 273.15 = 293.15 K. Given k1= 4.23 ×10−3s−1, we have:
4.23 ×10−3=A·e−Ea
8.314·293.15
Step 3: Now, let’s calculate the Arrhenius parameters at 60◦C. At 60◦C,
T= 60 + 273.15 = 333.15 K. Given k2= 0.023 s−1, we have:
0.023 = A·e−Ea
8.314·333.15
Step 4: We now have a system of two equations:
4.23 ×10−3=A·e−Ea
8.314·293.15
0.023 = A·e−Ea
8.314·333.15
Step 5: By dividing the second equation by the first one, we can eliminate
the pre-exponential factor A:
0.023
4.23 ×10−3=e−Ea
8.314·333.15
e−Ea
8.314·293.15
Step 6: Solving the above equation will give us the activation energy Ea.
Let’s go ahead and solve for Ea.
Question 13
Question
The rate constant of a reaction doubles when the temperature is increased from
25
°
C to 35
°
C. Calculate the activation energy of the reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given that T1= 25◦C
and T2= 35◦C, we convert to Kelvin using the formula T(K) = T(C) + 273.15.
Thus, T1= 25 + 273.15 = 298.15 Kand T2= 35 + 273.15 = 308.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants at two
temperatures. The Arrhenius equation is given by 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)), - Tis the temperature in Kelvin.
Since the rate constant doubles when the temperature is increased from 25
°
C
to 35
°
C, we can write:
2k1=k2
12
Substitute the Arrhenius equation into this relationship:
2A·e−Ea
R·298.15 =A·e−Ea
R·308.15
Step 3: Simplify the equation and solve for Ea. Divide both sides by Aand
take the natural logarithm of both sides to get rid of the exponential term:
ln(2) = Ea
R1
298.15 −1
308.15
Now, solve for Ea:
Ea=R·308.15 ·298.15
308.15 −298.15 ·ln(2)
Step 4: Calculate the activation energy. Substitute the values for R=
8.314 J/(mol
·
K) and solve for Ea:
Ea= 8.314 ×308.15 ×298.15
308.15 −298.15 ×ln(2)
Ea≈5.65 ×104J/mol
Therefore, the activation energy of the reaction is approximately 5.65 ×
104J/mol.
Question 14
Question
The rate constant of a reaction at 25
°
C is found to be 1.20 ×10−2s−1. When
the temperature is raised to 45
°
C, the rate constant increases to 5.80 ×10−2
s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the rate constants to their corresponding Arrhenius equations
using the Arrhenius equation:
k=Ae−Ea
RT
where: - k1= 1.20 ×10−2s−1at 25
°
C - k2= 5.80 ×10−2s−1at 45
°
C -
T1= 25 + 273 = 298 K - T2= 45 + 273 = 318 K
For the first temperature:
k1=Ae−Ea
R·298
1.20 ×10−2=Ae−Ea
8.314·298
For the second temperature:
k2=Ae−Ea
8.314·318
13
5.80 ×10−2=Ae−Ea
8.314·318
Step 2: Divide the two Arrhenius equations to eliminate the pre-exponential
factor A:
k2
k1
=e−Ea
8.314·318
e−Ea
8.314·298
5.80 ×10−2
1.20 ×10−2=e−Ea
8.314 (1
318 −1
298 )
Step 3: Solve for the activation energy (Ea) by taking the natural logarithm
of both sides and rearranging the equation:
ln 5.80 ×10−2
1.20 ×10−2=−Ea
8.314 1
318 −1
298
Ea=−8.314 ×
ln 5.80×10−2
1.20×10−2
1
318 −1
298
Step 4: Calculate the activation energy:
Ea=−8.314 ×
ln 5.80×10−2
1.20×10−2
1
318 −1
298
Ea≈43.6 kJ/mol
Therefore, the activation energy for this reaction is approximately 43.6
kJ/mol.
Question 15
Question
The rate constant for a certain reaction is found to be 4.5×10−3s−1at 25◦C
and 1.2×10−2s−1at 35◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula:
T(K) = T(◦C) + 273.15
For 25◦C:
T1= 25 + 273.15 = 298.15 K
For 35◦C:
T2= 35 + 273.15 = 308.15 K
14
Step 2: Calculate the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
We can rearrange the equation to solve for Ea:
1
T1
=−Ea
R·1
T2
1
+ ln(A)
Similarly for the second set of data:
1
T2
=−Ea
R·1
T2
2
+ ln(A)
Step 3: Substitute the known values into the above equations and solve for
Ea:1
298.15 =−Ea
8.314 ·1
298.152+ ln4.5×10−3
1
308.15 =−Ea
8.314 ·1
308.152+ ln1.2×10−2
Solving these two simultaneous equations will give the value of the activation
energy, Ea.
Question 16
Question
The rate constant for a reaction was found to be 6.25 ×10−2s−1at 25
°
C and
8.75 ×10−2s−1at 37
°
C. Calculate the activation energy for the reaction.
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(
°
C) +
273.15. At 25
°
C: T1= 25 + 273.15 = 298.15 K
At 37
°
C: T2= 37 + 273.15 = 310.15 K
Step 2: Write down 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 3: Set up two equations using the given rate constants at the two
temperatures:
k1=A·e
−Ea
R·T1
k2=A·e
−Ea
R·T2
15
Step 4: Solve the system of equations for the activation energy Ea. Divide
the two equations:
k1
k2
=e
−Ea
R·T1
e
−Ea
R·T2
k1
k2
=e
Ea
R1
T2
−1
T1
Step 5: Plug in the known values and solve for Ea.
6.25 ×10−2
8.75 ×10−2=e
Ea
8.314 (1
310.15 −1
298.15 )
5
7=e
Ea
8.314 (1
310.15 −1
298.15 )
Step 6: Solve for Eaby taking the natural logarithm of both sides and
rearranging the equation.
ln 5
7=Ea
8.314 1
310.15 −1
298.15
Ea=−8.314 ×ln 5
7×1
310.15 −1
298.15
Step 7: Calculate the activation energy using the given values and the for-
mula above.
Question 17
Question
The rate constant of a reaction at 25
°
C is 1.8×10−3s−1, while at 50
°
C the rate
constant is 3.2×10−2s−1. Calculate the activation energy for this reaction.
(Hint: Use the Arrhenius equation k=Ae−Ea
RT where kis the rate constant, A
is 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= 25C= 25 + 273.15 = 298.15 K
T2= 50C= 50 + 273.15 = 323.15 K
Step 2: Write the Arrhenius equation for both temperatures.
At T1= 298.15 K:
k1=Ae−Ea
RT1
1.8×10−3=Ae−Ea
R×298.15
16
At T2= 323.15 K:
k2=Ae−Ea
RT2
3.2×10−2=Ae−Ea
R×323.15
Step 3: Divide the two Arrhenius equations to eliminate A and solve for Ea.
k2
k1=e−Ea
R×323.15
e−Ea
R×298.15
3.2×10−2
1.8×10−3=eEa
R(1
298.15 −1
323.15 )
3.2
1.8=eEa
R(1
298.15 −1
323.15 )
ln 3.2
1.8=Ea
R(1
298.15 −1
323.15 )
Step 4: Solve for Ea.
ln 3.2
1.8=Ea
R(1
298.15 −1
323.15 )
Ea=R×ln(3.2
1.8)
1
298.15 −1
323.15
Ea= 8.314 ×ln(3.2
1.8)
1
298.15 −1
323.15
Ea≈48670 J/mol
Ea≈48.67 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.67
kJ/mol.
Question 18
Question
The rate constant (k) for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When
the temperature is increased to 50
°
C, the rate constant becomes 8.75×10−3s−1.
Calculate the activation energy (Ea) for this reaction. (R = 8.314 J/mol
·
K)
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. The temperature in
Kelvin is given by T(K) = T(C) + 273.15. For 25
°
C: T1= 25 + 273.15 = 298.15
K. For 50
°
C: T2= 50 + 273.15 = 323.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures. The Arrhenius equation is given by:
k=A×e−Ea
RT
where: k1= 1.25 ×10−3s−1,T1= 298.15 K, k2= 8.75 ×10−3s−1,T2= 323.15
K, R= 8.314 J/mol
·
K.
Step 3: Take the ratio of the two Arrhenius equations to eliminate the pre-
exponential factor.
k2
k1
=A×e−Ea
RT2
A×e−Ea
RT1
17
k2
k1
=e−Ea
R1
T1
−1
T2
Step 4: Solve for the activation energy (Ea) using the known values. Sub-
stitute the given values:
8.75 ×10−3
1.25 ×10−3=e−Ea
8.314 (1
298.15 −1
323.15 )
Step 5: Calculate Ea.
8.75 ×10−3
1.25 ×10−3=e
−Ea
8.314 (1
323.15 −1
298.15 )
7 = e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 6: Solve for Ea.
ln(7) = −Ea
8.314 1
323.15 −1
298.15
Ea=−8.314 ×ln(7)
1
323.15 −1
298.15
Question 19
Question
The rate constant for a certain reaction is known to double when the tempera-
ture is increased from 25◦C to 35◦C. Calculate the activation energy (in kJ/mol)
for this reaction.
Given: Activation Energy at 25◦C = 50 kJ/mol Gas constant, R= 8.314
J/(mol·K)
Solution
Step 1: Convert the given activation energy to joules: Let Eabe the activation
energy in joules.
Activation Energy (J) = 50 kJ/mol ×1000 J/kJ = 50000 J/mol
Step 2: Use the Arrhenius equation to relate the rate constants at two
different temperatures: The Arrhenius equation is given by:
k=A·e−Ea
RT
Given that the rate constant doubles when the temperature is increased from
25◦C to 35◦C, we have:
k2
k1
= 2
18
A·e−Ea
R(25+273.15)
A·e−Ea
R(35+273.15)
= 2
e−50000
8.314×(25+273.15)
e−50000
8.314×(35+273.15)
= 2
Step 3: Solve for the activation energy, Ea:
−50000
8.314 ×(25 + 273.15) +50000
8.314 ×(35 + 273.15) = ln(2)
−1931.79 + 2447.85 = 0.69315
516.06 = 0.69315
Therefore, the activation energy for this reaction is approximately 516.06
kJ/mol.
Question 20
Question
The rate constant (k) for a certain reaction is known to be 4.72 ×10−3s−1at
25
°
C. When the temperature is increased to 45
°
C, the rate constant becomes
7.61 ×10−2s−1. Calculate the activation energy (Ea) for this reaction. The
activation energy is expressed in units of kJ/mol.
Solution
Step 1: Determine the value of the gas constant R. Given: Temperature (T1)
= 25
°
C = 25 + 273 = 298 K Temperature (T2) = 45
°
C = 45 + 273 = 318 K
We can use the Arrhenius equation to relate the rate constants at different
temperatures:
k2
k1
= exp Ea
R1
T1
−1
T2
where: k1= 4.72 ×10−3s−1,k2= 7.61 ×10−2s−1,T1= 298 K, T2= 318 K,
and Ris the gas constant.
Step 2: Calculate R. We can rearrange the Arrhenius equation to solve for
R:
R=Ea
ln k2
k11
T2
−1
T1
Substitute the given values into the equation to find R.
Step 3: Calculate Ea. Once Ris determined, we can solve for the activation
energy Eausing the formula:
Ea=Rln k2
k1 1
T2
−1
T1
Substitute the values of R,k1,k2,T1, and T2into the equation to find the
activation energy in kJ/mol.
19
Question 21
Question
The rate constant for a certain reaction is 4.2×10−4s−1at 25◦C and 1.5×10−2
s−1at 50◦C. Calculate the activation energy (Ea) of the reaction.
Solution
Step 1: Let’s start by writing the Arrhenius equation:
k=Ae−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: We are given two sets of data: At 25◦C (298 K): k1= 4.2×10−4
s−1At 50◦C (323 K): k2= 1.5×10−2s−1
Step 3: Divide the two Arrhenius equations to eliminate A:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R(1
T1
−1
T2)
Step 4: Plug in the known values:
1.5×10−2
4.2×10−4=e−Ea
8.314 (1
298 −1
323 )
Step 5: Solve for Ea:
1.5×10−2
4.2×10−4=e−Ea
8.314 (0.00336)
35.71 = e−3.1256×10−3Ea
Step 6: Take the natural logarithm of both sides:
ln(35.71) = lne−3.1256×10−3Ea
ln(35.71) = −3.1256 ×10−3Ea
Step 7: Solve for Ea:
Ea=−ln(35.71)
3.1256 ×10−3≈53.57 kJ/mol
Therefore, the activation energy of the reaction is approximately 53.57 kJ/mol.
20
Question 22
Question
The rate constant for a reaction at 25◦C is 1.20 ×10−3s−1. When the tempera-
ture is increased to 55◦C, the rate constant becomes 2.75 ×10−2s−1. Calculate
the activation energy in kJ/mol for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given: T1= 25◦C =
298 K T2= 55◦C = 328 K
Step 2: Use the Arrhenius equation to find the activation energy. The
Arrhenius equation relates the rate constant kto the temperature Tand the
activation energy Ea:
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
Taking the ratio of the rate constants at the two temperatures and solving
for Ea:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
k2
k1
=e
Ea
R1
T1
−1
T2
ln k2
k1=Ea
R1
T1
−1
T2
Ea=R·
ln k2
k1
1
T1
−1
T2
Step 3: Substitute the given values into the equation to find the activation
energy.
Ea= 8.314 J/mol·K·
ln 2.75×10−2
1.20×10−3
1
298 −1
328
Calculating the activation energy gives:
Ea≈66.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 66.5
kJ/mol.
21
Question 23
Question
The rate constant for a reaction is 4.17 ×10−3s−1at 25
°
C and 2.50 ×10−2s−1
at 55
°
C. Calculate the activation energy (in kJ/mol) for this reaction. (Hint:
Use 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, and
Tis the temperature in Kelvin.)
Solution
Step 1: Convert temperatures to Kelvin
Given: - Temperature at 25
°
C: T1= 25C+ 273.15 = 298.15 K - Temperature
at 55
°
C: T2= 55C+ 273.15 = 328.15 K
Step 2: Use the Arrhenius equation
Given rate constants: - At 25
°
C: k1= 4.17×10−3s−1- At 55
°
C: k2= 2.50×10−2
s−1
The Arrhenius equation can be written as:
k=Ae−Ea
RT
Substitute the rate constants and temperatures into the Arrhenius equation:
k1=Ae−Ea
R·298.15
k2=Ae−Ea
R·328.15
Step 3: Take the ratio of the two rate constants
Divide the second equation by the first to eliminate A:
k2
k1
=e−Ea
R·328.15
e−Ea
R·298.15
k2
k1
=e−Ea
R(1
328.15 −1
298.15 )
Step 4: Solve for activation energy
Take natural logarithms of both sides of the equation to solve for the activation
energy, Ea:
ln k2
k1=−Ea
R1
328.15 −1
298.15
Finally, calculate the activation energy Eain kJ/mol by using the gas con-
stant R= 8.314 J/(mol
·
K).
22
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is 2.35 ×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.23 ×10−2s−1.
Calculate the activation energy for this reaction. (Given: R= 8.314 J mol−1K−1)
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25C+ 273.15 = 298.15 K
T2= 50C+ 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at different
temperatures.
k2=k1·e
−Ea
R1
T2
−1
T1
Step 3: Plug in the given values and the rate constants.
1.23 ×10−2= 2.35 ×10−3·e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 4: Solve for the activation energy Ea.
1.23 ×10−2
2.35 ×10−3=e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 5: Simplify and solve for Ea.
5.23 = e
−Ea
8.314 (1
323.15 −1
298.15 )
Step 6: Take the natural logarithm of both sides and solve for Ea.
ln(5.23) = −Ea
8.314 1
323.15 −1
298.15
Step 7: Calculate Ea.
Ea=−8.314 ×(ln(5.23)) ×1
323.15 −1
298.15
Step 8: Calculate the final answer for the activation energy Ea. Remember
to include the appropriate units.
Ea≈67.8 kJ/mol
23
Question 25
Question
The rate constant (k) for a certain reaction is found to be 1.25 ×10−2s−1at
25◦C and 4.18 ×10−2s−1at 45◦C. Calculate the activation energy (in kJ/mol)
for the reaction.
(Assume the activation energy is independent of temperature and use the
universal gas constant R= 8.314 J/mol ·K for calculations.)
Solution
Step 1: Convert temperatures to Kelvin using T(K) = T(C) + 273.15.
T1= 25 + 273.15 = 298.15 K
T2= 45 + 273.15 = 318.15 K
Step 2: Write out the Arrhenius equation:
k=A·e−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.
Step 3: Take the ratio of the rate constants at the two temperatures and
solve for the activation energy (Ea):
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
k2
k1
=e−Ea
R1
T1
−1
T2
ln k2
k1=−Ea
R1
T1
−1
T2
Ea=−R·ln k2
k1 1
T1
−1
T2
Step 4: Plug in the given values:
Ea=−8.314 J/mol ·K×ln 4.18 ×10−2
1.25 ×10−2×1
298.15 −1
318.15
Step 5: Calculate the activation energy (Ea) in kJ/mol.
Ea=−8.314 ×ln (3.344) ×(0.003354 −0.003144)
Ea≈ −8.314 ×ln (3.344) ×0.000210
Ea≈ −8.314 ×1.204 ×0.000210
Ea≈ −8.314 ×0.00025284
Ea≈ −0.0021 kJ/mol ≈21 kJ/mol
Therefore, the activation energy for the reaction is 21 kJ/mol.
24
Question 26
Question
For a certain reaction, the rate constant at 25
°
C is 3.2×10−3s−1and the rate
constant at 50
°
C is 8.5×10−2s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: To find the activation energy (Ea) for the reaction, we will use 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: We are given two sets of data at different temperatures: At 25
°
C,
k1= 3.2×10−3s−1T1= 25 + 273 = 298 K
At 50
°
C, k2= 8.5×10−2s−1T2= 50 + 273 = 323 K
Step 3: Substituting the first set of data into the Arrhenius equation gives:
3.2×10−3=A·e−Ea
8.314×298
Step 4: Substituting the second set of data into the Arrhenius equation gives:
8.5×10−2=A·e−Ea
8.314×323
Step 5: Divide the second equation by the first equation to eliminate A:
8.5×10−2
3.2×10−3=e
Ea
8.314 (1
323 −1
298 )
Step 6: Solve for Eaby isolating it:
Ea=−8.314 ·ln 8.5×10−2
3.2×10−3·1
323 −1
298
Step 7: Calculate Eausing a calculator:
Ea≈61.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 61.7
kJ/mol.
Question 27
Question
The rate constant for a certain reaction is found to be 2.50 ×10−4s−1at 25
°
C,
and 9.80 ×10−3s−1at 50
°
C. Calculate the activation energy for this reaction.
Given: R= 8.314 J/(mol
·
K), T1= 25C,T2= 50C
25
Solution
Step 1: Convert the temperatures to Kelvin:
T1= 25C+ 273 = 298K
T2= 50C+ 273 = 323K
Step 2: Use the Arrhenius equation to relate the rate constants to the tem-
peratures 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, and Tis the temperature in Kelvin.
Step 3: Set up two equations for the given data:
2.50 ×10−4=A·e−Ea
8.314·298
9.80 ×10−3=A·e−Ea
8.314·323
Step 4: Divide the two equations to eliminate A:
2.50 ×10−4
9.80 ×10−3=e−Ea
8.314·298
e−Ea
8.314·323
Step 5: Simplify the equation and solve for Ea:
2.50
9.80 =e
Ea
8.314 (1
323 −1
298 )
2.50
9.80 =e−Ea
8.314·38354
0.255 = e−Ea
254.81
ln(0.255) = lne−Ea
254.81
−1.366 = −Ea
254.81
Ea= 1.366 ×254.81 = 348.04 kJ/mol
Therefore, the activation energy for this reaction is 348.04 kJ/mol.
Question 28
Question
The rate constant for a certain reaction was measured at two different tempera-
tures, 25◦C and 50◦C. At 25◦C, the rate constant was found to be 1.5×10−3s−1,
while at 50◦C, the rate constant was 7.5×10−3s−1. Calculate the activation
energy (in kJ/mol) for this reaction.
26
Solution
Step 1: Convert the given temperatures to Kelvin using the equation T(K) =
T(◦C) + 273.15.
T1= 25 + 273.15 = 298.15 K
T2= 50 + 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures to the activation energy.
k2
k1
=e
Ea
R(1
T1
−1
T2)
Step 3: Plug in the given values for k1,k2,T1, and T2, and the universal gas
constant R= 8.314 J/(mol K).
7.5×10−3
1.5×10−3=e
Ea
8.314 (1
298.15 −1
323.15 )
Step 4: Solve for the activation energy Ea.
7.5
1.5=e
Ea
8.314 (1
298.15 −1
323.15 )
5 = e
Ea
8.314 (−0.001356)
Step 5: Take the natural logarithm of both sides to solve for the activation
energy.
ln(5) = Ea
8.314(−0.001356)
Step 6: Solve for the activation energy Eain kJ/mol.
Ea=−8.314 ×0.001356 ×ln(5) ≈61.96 kJ/mol
Therefore, the activation energy for this reaction is approximately 61.96
kJ/mol.
Question 29
Question
For a certain chemical reaction, the rate constant at 25
°
C is 1.2×10−3s−1, and
the rate constant at 45
°
C is 2.5×10−2s−1. Calculate the activation energy of
the reaction in kJ/mol.
27
Solution
Step 1: Convert the given temperatures from Celsius to Kelvin.
T1= 25C+ 273 = 298 K
T2= 45C+ 273 = 318 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures.
k2
k1
=e(−Ea
R)1
T2
−1
T1
Step 3: Substitute the given rate constants, temperatures, and gas constant
R= 8.314 J/mol ·K into the Arrhenius equation.
2.5×10−2
1.2×10−3=e(−Ea
8.314 )( 1
318 −1
298 )
Step 4: Solve for the activation energy Ea.
2.5×10−2
1.2×10−3=e(−Ea
8.314 )( 1
318 −1
298 )
20.83 = e(−Ea
8.314 )( 1
318 −1
298 )
ln(20.83) = −Ea
8.314 1
318 −1
298
Step 5: Calculate the activation energy Eain J/mol.
Ea=−8.314 × 1
318 −1
298×ln(20.83)
≈92736 J/mol
≈92.7 kJ/mol
Therefore, the activation energy of the reaction is approximately 92.7 kJ/mol.
Question 30
Question
The rate constant for the decomposition of a certain compound at 298 K is
5.32 ×10−3s−1, and the activation energy for the reaction is 78.5 kJ/mol.
Calculate the rate constant at 313 K if the activation energy remains constant.
28
Solution
Step 1: Calculate the rate constant at 313 K using the Arrhenius equation:
k2=k1·e−Ea
R1
T2
−1
T1
where: k2= rate constant at 313 K k1= rate constant at 298 K Ea= acti-
vation energy = 78.5 kJ/mol R= gas constant = 8.314 J/(mol*K) T2= new
temperature = 313 K T1= initial temperature = 298 K
Step 2: Substitute the given values into the equation and solve for k2:
k2= 5.32 ×10−3s−1·e−78.5 kJ/mol
8.314 J/(mol*K) (1
313 K −1
298 K )
Step 3: Calculate the value of k2:
k2= 5.32 ×10−3s−1·e−78.5×103J/mol
8.314 J/(mol*K) (1
313 K −1
298 K )
Step 4: Simplify the equation and calculate k2:
k2≈0.0305 s−1
Therefore, the rate constant at 313 K is approximately 0.0305 s−1.
Question 31
Question
The rate constant for the reaction 2A→Bis 0.02 s−1at 25◦C and 0.12 s−1at
45◦C. Calculate the activation energy for the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants to the
temperature and activation energy. The Arrhenius equation is given by:
k=Ae−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 can write two equations based on the rate constants given at
two different temperatures:
k1=Ae−Ea
R·(25+273.15)
k2=Ae−Ea
R·(45+273.15)
29
Step 3: We can solve these two equations simultaneously to find the activa-
tion energy, Ea. We first take the ratio of the two equations:
k2
k1
=Ae−Ea
R·(45+273.15)
Ae−Ea
R·(25+273.15)
k2
k1
=e−Ea
R(1
45+273.15 −1
25+273.15 )
Step 4: Given that k1= 0.02 s−1and k2= 0.12 s−1, we can plug in the
values and solve for the activation energy, Ea.
Step 5: Calculate the activation energy, Ea, using the values obtained in
Step 4 and the gas constant R= 8.314 J ·mol−1·K−1.
Thus, the activation energy for the reaction is
Ea= 40.2 kJ ·mol−1
.
Question 32
Question
The rate constant for a certain reaction is 2.83 ×10−3s−1at 35◦C, while at
45◦C it is 1.37 ×10−2s−1. Calculate 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, R= 8.314 J K−1mol−1is the gas constant,
and Tis temperature in Kelvin)
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 35◦C + 273.15 = 308.15 K
T2= 45◦C + 273.15 = 318.15 K
Step 2: Write down the Arrhenius equation for the two temperatures and
the given rate constants.
k1=A·e−Ea
RT1
k2=A·e−Ea
RT2
Step 3: Take the ratio of the two rate constants to eliminate A.
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
30
k2
k1
=e−Ea
R1
T2
−1
T1
Step 4: Take the natural logarithm of both sides to solve for Ea.
ln k2
k1=−Ea
R1
T2
−1
T1
Step 5: Substitute the given values and solve for Ea.
ln 1.37 ×10−2
2.83 ×10−3=−Ea
8.314 1
318.15 −1
308.15
ln (4.82) = −Ea
8.314 1
318.15 −1
308.15
Question 33
Question
The rate constant for the reaction 2A →B is given by the Arrhenius equation:
k=Ae−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the
activation energy, R= 8.314
Step 1: Convert the temperatures given to Kelvin.
T1= 25C= 25 + 273 = 298 K
T2= 50C= 50 + 273 = 323 K
Step 2: Set up two Arrhenius equations using the rate constants and tem-
peratures given:
k1=Ae−Ea
R·T1
k2=Ae−Ea
R·T2
Step 3: Divide the two equations to eliminate A:
k1
k2
=e−Ea
R·T1
e−Ea
R·T2
Step 4: Simplify the equation by canceling out A:
k1
k2
=e(Ea
R)1
T2
−1
T1
Step 5: Substitute the given rate constants and temperatures:
5.0×10−3
1.0×10−2=e(Ea
8.314 )( 1
323 −1
298 )
31
Step 6: Solve for the activation energy Ea:
ln 5.0×10−3
1.0×10−2=Ea
8.314 1
323 −1
298
⇒
8.314 ·ln 5.0×10−3
1.0×10−2
1
323 −1
298
=Ea
Step 7: Calculate the activation energy Ea:
Ea≈35.8 kJ/mol
Therefore, the activation energy for the reaction 2A →B is approximately
35.8 kJ/mol.
Question 34
Question
For a certain reaction, the rate constant at 25
°
C is 1.5×10−4s−1and the
activation energy is 50 kJ/mol. Determine the rate constant at 50
°
C for this
reaction.
Solution
Step 1: Given the activation energy and the rate constant at 25
°
C, we can
use the Arrhenius equation to find the rate constant at 50
°
C. The Arrhenius
equation is given by:
k=Ae−Ea
RT
where: k= rate constant at temperature T (in Kelvin), A= pre-exponential
factor, Ea= activation energy, R= gas constant (8.314 J/mol-K), T= tem-
perature in Kelvin.
Step 2: Let’s first convert the activation energy to joules:
Ea= 50 kJ/mol ×1000 J/kJ = 50000 J/mol
Step 3: Calculate the rate constant at 25
°
C:
k25 = 1.5×10−4s−1
Step 4: Convert the temperatures to Kelvin:
T25 = 25C+ 273 = 298 K
T50 = 50C+ 273 = 323 K
32
Step 5: Plug in the values into the Arrhenius equation to find the pre-
exponential factor A:
1.5×10−4=Ae−50000
8.314×298
Step 6: Solve for A:
A= 1.5×10−4×e50000
8.314×298
Step 7: Finally, calculate the rate constant at 50
°
C using the pre-exponential
factor Aand the activation energy Ea:
k50 =Ae−50000
8.314×323
Question 35
Question
Consider a reaction with an activation energy of 75 kJ/mol. If the rate constant
at 25
°
C is 1.8×10−2s−1and the activation energy is reduced to 50 kJ/mol,
what will be the new rate constant at 25
°
C? (Assume the pre-exponential factor
A remains constant.)
Solution
Step 1: Calculate the rate constant using the Arrhenius equation for the original
activation energy.
k1=A·e−Ea
RT
Given: Original activation energy Ea1= 75 kJ/mol, Rate constant k1= 1.8×
10−2s−1, Temperature T= 25
°
C = 298 K.
Plugging in the values:
k1=A·e−75000
8.314×298
k1=A·e−31.77 = 1.8×10−2
Step 2: Calculate the new rate constant using the modified activation energy.
k2=A·e−
Ea2
RT =A·e−50000
8.314×298
k2=A·e−21.18
Step 3: Determine the ratio between the new and original rate constants.
k2
k1
=A·e−21.18
A·e−31.77
e−21.18+31.77 =e10.59
k2=k1·e10.59
k2= 1.8×10−2·e10.59
33
