CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 7
Liberty University
Question 1
Question
The rate constant for the reaction 2A + B →C is found to be 0.0034 L/mol·s
at 300 K and 0.012 L/mol·s at 350 K. Calculate the activation energy for this
reaction.
Solution
Step 1: Write the Arrhenius equation: The Arrhenius equation relates the rate
constant of a reaction to the temperature and the activation energy (Ea) of the
reaction. It is given as:
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: Determine the rate constant at each temperature: Given: At 300 K,
k1= 0.0034 L/mol·s, At 350 K, k2= 0.012 L/mol·s.
Step 3: Take the natural logarithm of the Arrhenius equation: Taking the
natural logarithm of both sides of the Arrhenius equation gives:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Plug in the values and solve for the activation energy: Substitute
the known values into the equation and solve for the activation energy Ea:
ln 0.012
0.0034=−Ea
8.314 1
350 −1
300
ln (3.5294) = −Ea
8.314 1
350 −1
300
ln (3.5294) = −Ea
8.314 5
105000
ln (3.5294) = −Ea
8.314 1
21000
ln (3.5294) = −Ea
174.294
Ea=−174.294 ×ln (3.5294)
Ea≈38.45 kJ/mol
Step 5: Finalize the result: Therefore, the activation energy for the reaction
2A + B →C is approximately 38.45 kJ/mol.
Question 2
Question
The rate constant for a reaction is 1.25 ×10−2s−1at 400 K and 4.71 ×10−2
s−1at 600 K. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant of
a reaction to the temperature and activation energy.
k=A·e−Ea
RT
Step 2: We are given two sets of data points:
k1= 1.25 ×10−2s−1at 400 K
k2= 4.71 ×10−2s−1at 600 K
Step 3: Let’s first solve for A, the pre-exponential factor. We can set up a
ratio of the rate constants at the two temperatures.
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
4.71 ×10−2
1.25 ×10−2=e−Ea
R·600 ·eEa
R·400
3.768 = e−Ea
R(1
600 −1
400 )
2
Step 4: Simplify the equation from Step 3 and solve for Ea.
3.768 = e−Ea
R(1
600 −1
400 )
3.768 = e−Ea
R(1
240000 )
ln(3.768) = −Ea
R1
240000
Ea=−ln(3.768) ·(8.314 J/mol·K) ·240000
Step 5: Calculate the activation energy using the values obtained in Step 4.
Ea=−ln(3.768) ·(8.314 J/mol·K) ·240000
Ea≈58600 J/mol ≈58.6 kJ/mol
Therefore, the activation energy for this reaction is approximately 58.6
kJ/mol.
Question 3
Question
The rate constant of a reaction at two different temperatures, T1and T2, are
known to be k1= 1.2×10−4s−1and k2= 8.3×10−2s−1, respectively. If the
activation energy for the reaction is 60 kJ/mol, determine the ratio of the rate
constants (k2/k1) and the temperature at which the rate constant would be
5×10−3s−1.
Solution
Step 1: Calculate the ratio of rate constants (k2/k1) using the Arrhenius equa-
tion: k2
k1
= exp Ea
R1
T1
−1
T2
Step 2: Substitute the given values into the Arrhenius equation:
k2
k1
= exp 60,000 J/mol
8.314 J/mol K 1
T1
−1
T2
Step 3: Substitute the rate constants into the ratio expression:
8.3×10−2
1.2×10−4= exp 60,000 J/mol
8.314 J/mol K 1
T1
−1
T2
Step 4: Simplify the ratio of rate constants:
k2
k1
= 692.5
3
Step 5: Solve for the temperature at which the rate constant would be
5×10−3s−1:
5×10−3=k=Aexp −Ea
RT
Step 6: Rearrange the equation to solve for temperature:
ln k
A=−Ea
R1
T
Step 7: Substitute the values of k,Ea,R, and Ainto the equation:
ln 5×10−3
1.2×10−4=−60,000
8.314 1
T
Step 8: Solve for the temperature, T:
T=1
−60,000
8.314×ln5×10−3
1.2×10−4
= 408 K
Question 4
Question
The rate constant for a certain reaction at 25
°
C is 2.5×10−4s−1, and at 45
°
C
it is 7.5×10−3s−1. Determine the activation energy for this reaction.
Solution
Step 1: First, we can use the Arrhenius equation to relate the rate constants at
two different temperatures to the activation energy. The Arrhenius equation 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: We are given two sets of data: At 25
°
C, k1= 2.5×10−4s−1,
T1= 25C= 298 K, At 45
°
C, k2= 7.5×10−3s−1,T2= 45C= 318 K.
Step 3: Let’s set up the Arrhenius equation for the two temperature points:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 4: Divide the two equations to eliminate Aand solve for Ea:
k2
k1
=e−Ea
R·T2
e−Ea
R·T1
4
k2
k1
=eEa
R·1
T1
−1
T2
Step 5: Now, substitute the given values and solve for Ea:
7.5×10−3
2.5×10−4=e(Ea
8.314 ·(1
298 −1
318 ))
Step 6: Calculate the activation energy Eausing the natural logarithm to
solve for it:
Ea=−8.314 ×ln 7.5×10−3
2.5×10−4×1
298 −1
318
Step 7: Solve for Eato find the activation energy of the reaction.
Question 5
Question
The rate constant of a reaction at two different temperatures is found to be
k1= 1.25 ×10−4s−1at T1= 300 K and k2= 5.00 ×10−3s−1at T2= 350 K.
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: Write down the expression for the rate constant at the two different
temperatures:
k1=A×e−Ea
R×T1
k2=A×e−Ea
R×T2
Step 3: Divide the rate constants to obtain a ratio:
k2
k1
=A×e−Ea
R×T2
A×e−Ea
R×T1
Step 4: Simplify the ratio:
k2
k1
=e−Ea
R×T2
e−Ea
R×T1
5
k2
k1
=e
Ea
R1
T1
−1
T2
Step 5: Plug in the given values and solve for the activation energy Ea:
5.00 ×10−3s−1
1.25 ×10−4s−1=eEa
8.314 (1
300 −1
350 )
Step 6: Calculate the activation energy:
5.00 ×10−3
1.25 ×10−4=eEa
8.314 (1
300 −1
350 )
40 = eEa
8.314 (1
300 −1
350 )
Step 7: Solve for the activation energy, Ea:
ln(40) = Ea
8.314 1
300 −1
350
Ea= 8.314 ×ln(40)
1
300 −1
350
Ea≈50900 J/mol
Question 6
Question
The rate constant for a reaction at 25
°
C is 4.57 ×10−3s−1. When the tempera-
ture is increased to 55
°
C, the rate constant becomes 1.26 ×10−2s−1. Calculate
the activation energy for the reaction.
Solution
Step 1: Convert temperatures to Kelvin.
Given:
T1= 25C
T2= 55C
To convert from Celsius to Kelvin, we use:
T(K) = T(C) + 273.15
Therefore:
T1= 25 + 273.15 = 298.15K
T2= 55 + 273.15 = 328.15K
6
Step 2: Write down 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
Given two sets of data (rate constants and temperatures), we can set up a ratio
and solve for the activation energy.
Step 3: Set up the ratio of rate constants.
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R1
T1
−1
T2
Step 4: Substitute values and solve for activation energy.
Given data:
k1= 4.57 ×10−3s−1
k2= 1.26 ×10−2s−1
T1= 298.15 K
T2= 328.15 K
R= 8.314 J mol−1K−1
Substitute the values into the equation:
1.26 ×10−2
4.57 ×10−3=e−Ea
8.314 (1
298.15 −1
328.15 )
Solve for the activation energy:
ln 1.26 ×10−2
4.57 ×10−3=−Ea
8.314 1
298.15 −1
328.15
Ea=−8.314 ×
ln 1.26×10−2
4.57×10−3
1
298.15 −1
328.15
Calculate the activation energy using the given data.
Question 7
Question
The rate constant for a reaction at 25◦Cis 2.5×10−3s−1and the activation
energy (Ea) for the reaction is 75 kJ/mol. Determine the rate constant at 50◦C
for this reaction.
7
Solution
Step 1: Calculate the ratio of rate constants using the Arrhenius equation:
k2
k1
= exp Ea
R1
T1
−1
T2
where: k1= 2.5×10−3s−1(rate constant at 25◦C), T1= 25◦C= 298 K,
T2= 50◦C= 323 K, Ea= 75 kJ/mol, and R= 8.314
Question
The rate constant for a reaction is found to be 2.5×10−3s−1at 300 K and
1.5×10−2s−1at 350 K. Calculate the activation energy for the reaction. (The
universal gas constant Ris 8.314 J/(mol·K).)
Solution
Step 1: We can use the Arrhenius equation to determine the activation energy
Ea. The Arrhenius equation is given by:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor or frequency factor,
Ea= activation energy, R= universal gas constant (8.314 J/(mol·K)), T=
temperature (in Kelvin).
Step 2: We can rewrite the Arrhenius equation as:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k2and k1are the rate constants at temperatures T2and T1, respec-
tively.
Step 3: Substituting the given values k1= 2.5×10−3s−1at 300 K and
k2= 1.5×10−2s−1at 350 K into the equation, we get:
ln 1.5×10−2
2.5×10−3=−Ea
8.314 1
350 −1
300
Step 4: Simplifying the above equation, we find:
ln (6) = −Ea
8.314 1
350 −1
300
Step 5: Calculate the activation energy Eaby solving the above equation:
Ea
8.314 =−ln (6) ×1
350 −1
300
Ea=−8.314 ×ln (6) ×1
350 −1
300
Calculating the above expression will give us the activation energy Ea.
8
Question 9
Question
The rate constant for a certain reaction is found to be 4.5×10−2s−1at 25◦C and
9.0×10−2s−1at 35◦C. Calculate the activation energy (Ea) for this reaction.
The activation energy is known to be 8.314 J/mol ·K.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation, which is
given by:
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), and - Tis the
temperature in Kelvin.
Let’s denote the rate constants as k1= 4.5×10−2s−1at 25◦C and k2=
9.0×10−2s−1at 35◦C.
Step 2: We can set up two Arrhenius equations for the two sets of tempera-
ture and rate constant given:
k1=A·e
−Ea
R·298
k2=A·e
−Ea
R·308
Step 3: Divide the two equations to eliminate A:
k2
k1
=e
−Ea
8.314 ·308
e
−Ea
8.314 ·298
Step 4: Simplify the equation:
k2
k1
=e 298 −308
8.314 ·308 ·298!·Ea
Step 5: Solve for Ea:
Ea=8.314 ·308 ·298
10 ·ln 9.0×10−2
4.5×10−2
Step 6: Calculate the value of Eato find the activation energy of the reaction.
9
Question 10
Question
What is the activation energy of a reaction if its rate constant doubles when the
temperature is increased from 25
°
C to 35
°
C? Assume that the pre-exponential
factor remains constant at room temperature.
Solution
Step 1: Recall the Arrhenius equation, which relates the rate constant (k) of a
reaction to the temperature (T) and 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.
Step 2: Let’s denote the rate constant at 25
°
C as k1and the rate constant
at 35
°
C as k2. We know that:
k2= 2 ·k1
Step 3: Let’s substitute these values into the Arrhenius equation:
A·e−Ea
R(25+273.15) = 2 ·A·e−Ea
R(35+273.15)
Step 4: Cancel out the pre-exponential factor A:
e−Ea
R(25+273.15) = 2 ·e−Ea
R(35+273.15)
Step 5: Take the natural logarithm of both sides:
−Ea
R(25 + 273.15) = ln(2) −Ea
R(35 + 273.15)
Step 6: Solve for the activation energy Ea:
Ea
R(1
25 + 273.15 −1
35 + 273.15) = ln(2)
Step 7: Calculate the activation energy Ea:
Ea=R·25 + 273.15
35 + 273.15 ·ln(2)
Ea≈54.32 kcal/mol
Therefore, the activation energy of the reaction is approximately 54.32 kcal/mol.
10
Question 11
Question
The rate constant for the decomposition of a certain compound at 25
°
C is 4.2×
10−3s−1. When the temperature is increased to 50
°
C, the rate constant becomes
7.8×10−2s−1. Determine the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=Ae−Ea
RT )
Solution
Step 1: We can start by writing the Arrhenius equation in its logarithmic form:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1and k2are the rate constants at temperatures T1and T2respec-
tively, Eais the activation energy, Ris the gas constant, and T1and T2are the
temperatures in Kelvin.
Step 2: Plug in the given values: k1= 4.2×10−3s−1,k2= 7.8×10−2s−1,
T1= 298 K, and T2= 323 K. The gas constant Ris 8.314 J/(mol
·
K).
Step 3: Substitute the values into the equation and solve for Ea:
ln 7.8×10−2
4.2×10−3=−Ea
8.314 1
323 −1
298
Step 4: Calculate the natural logarithm:
ln 7.8×10−2
4.2×10−3= ln 18.57 ≈2.92
Step 5: Substitute this back into the equation and solve for Ea:
2.92 = −Ea
8.314 1
323 −1
298
Step 6: Calculate the activation energy:
Ea=−2.92 ×8.314 1
323 −1
298≈58401 J/mol = 58.4 kJ/mol
Therefore, the activation energy for this reaction is approximately 58.4
kJ/mol.
Question 12
Question
The rate constant for a reaction at 25
°
C is 2.50 ×10−4s−1and its activation
energy is 60 kJ/mol. Calculate the rate constant at 35
°
C for this reaction.
11
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol. Given: Activation
energy, Ea= 60 kJ/mol.
Converting kJ to J: 60 kJ = 60,000 J.
Step 2: Use the Arrhenius equation to calculate the rate constant at 35
°
C.
The Arrhenius equation 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.
Given: Rate constant at 25
°
C, k1= 2.50 ×10−4s−1, Temperature at 25
°
C,
T1= 25 + 273.15 = 298.15 K, Temperature at 35
°
C, T2= 35 + 273.15 = 308.15
K, Activation energy, Ea= 60,000 J/mol.
Substitute the given values into the Arrhenius equation:
2.50 ×10−4=A·e−60,000
8.314·298.15
Solving for the pre-exponential factor, A:
A=2.50 ×10−4
e−60,000
8.314·298.15
Step 3: Calculate the rate constant at 35
°
C. Substitute the pre-exponential
factor Aand the new temperature T2into the Arrhenius equation:
k2=A·e−Ea
RT2
Now, calculate the rate constant at 35
°
C:
k2=A·e−60,000
8.314·308.15
Question 13
Question
The rate constant (k) of a reaction at 25◦C is found to be 2.5×10−3s−1. When
the temperature is increased to 35◦C, the rate constant becomes 1.2×10−2s−1.
Calculate the activation energy (Ea) for this reaction.
Solution
Step 1: Find the ratio of rate constants at the two temperatures using the
Arrhenius equation: k2
k1
=eEa
R(1
T1
−1
T2)
12
Step 2: Plug in the given values:
1.2×10−2
2.5×10−3=eEa
8.314 (1
298 −1
308 )
Step 3: Calculate the ratio of rate constants:
4.8
1=eEa
8.314 (0.0034)
Step 4: Solve for Ea:
ln 4.8 = Ea
8.314(0.0034)
Step 5: Rearrange the equation to solve for Ea:
Ea= 8.314 ×0.0034 ×ln 4.8
Step 6: Calculate Ea:
Ea≈54.73 kJ/mol
Therefore, the activation energy for this reaction is approximately 54.73
kJ/mol.
Question 14
Question
The rate constant for a reaction at 25
°
C is 4.50 ×10−4s−1. When the tempera-
ture is increased to 50
°
C, the rate constant becomes 1.80 ×10−3s−1. Calculate
the activation energy of the reaction.
Solution
Step 1: Convert the given temperatures into Kelvin. 25C= 25 + 273.15 =
298.15 K
50C= 50 + 273.15 = 323.15 K
Step 2: Calculate the activation energy using the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
Substitute the given values:
ln 1.80 ×10−3
4.50 ×10−4=−Ea
8.314 1
323.15 −1
298.15
Step 3: Solve for the activation energy Ea.
ln 4 = −Ea
8.314 1
323.15 −1
298.15
13
ln 4 = −Ea
8.314 298.15 −323.15
298.15 ×323.15
Step 4: Calculate the activation energy.
Ea=−8.314 ×ln 4 ×298.15 ×323.15
25
Ea≈69.31 kJ/mol
Question 15
Question
The rate constant of a reaction at 40
°
C is 4.28 ×10−3s−1, and at 80
°
C it is 1.97
s−1. Calculate the activation energy (in kJ/mol) for this reaction. Assume the
frequency factor (A) is 1.21 ×1014 s−1.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 40C+ 273.15 = 313.15 K
T2= 80C+ 273.15 = 353.15 K
Step 2: Calculate the value of the gas constant (R) in J/mol K. Since R=
8.314 J/mol K,
R= 8.314 J/mol K
Step 3: Use the Arrhenius equation to find the activation energy (Ea) in
Joules.
ln k2
k1=−Ea
R1
T2
−1
T1
Plugging in the given values,
ln 1.97
4.28 ×10−3=−Ea
8.314 1
353.15 −1
313.15
Solving for Eain Joules, we get
Ea=−8.314 ×ln 1.97
4.28 ×10−3 1
353.15 −1
313.15
Step 4: Convert the activation energy from Joules to kilojoules.
Ea=
−8.314 ×10−3×ln 1.97
4.28×10−31
353.15 −1
313.15
1000
Therefore, the activation energy for this reaction is approximately 48.3 kJ/mol .
14
Question 16
Question
The rate constant (k) for a certain reaction was found to be 6.3×10−3s−1at a
temperature of 350 K and 2.1×10−2s−1at a temperature of 400 K. Calculate
the activation energy (Ea) for this reaction.
Solution
Step 1: Recall 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), and T= temperature (in Kelvin).
Step 2: We can write two Arrhenius equations for the given data:
k1=A·e−Ea
R·T1
k2=A·e−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 expression in Step 3:
k1
k2
=eEa
R·1
T2
−1
T1
Step 5: Plug in the given values and solve for Ea:
6.3×10−3
2.1×10−2=e(Ea
8.314 ·(1
400 −1
350 ))
Step 6: Solve for Ea:3
10 =e(Ea
8.314 ·(7
140000 ))
Step 7: Take the natural logarithm of both sides to solve for Ea:
ln 3
10=Ea
8.314 ·7
140000
Step 8: Calculate Ea:
Ea= 8.314 ·140000
7·ln 3
10≈61.0 kJ/mol
Therefore, the activation energy for this reaction is approximately 61.0
kJ/mol.
15
Question 17
Question
The rate constant (k) for a first-order reaction at 25
°
C is 5.0×10−3s−1. When
the temperature is increased to 45
°
C, the rate constant becomes 1.4×10−2s−1.
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 (frequency
factor), - Eais the activation energy, - Ris the gas constant (8.314 J/(mol·K)),
- and Tis the temperature in Kelvin.
Step 2: Set up the equation for the rate constant at the two different tem-
peratures:
k1=A·e−Ea
RT1
k2=A·e−Ea
RT2
Given:
k1= 5.0×10−3s−1
k2= 1.4×10−2s−1
T1= 25 + 273 = 298 K
T2= 45 + 273 = 318 K
Step 3: Take the ratio of the two rate constants to eliminate A:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
k2
k1
=e
Ea
R1
T1
−1
T2
Substitute the values: 1.4×10−2
5.0×10−3=eEa
8.314 (1
298 −1
318 )
Step 4: Solve for the activation energy (Ea):
ln 1.4×10−2
5.0×10−3=Ea
8.314 1
298 −1
318
Ea= 8.314 ×ln 1.4×10−2
5.0×10−3×1
298 −1
318
16
Step 5: Calculate the activation energy:
Ea= 8.314 ×ln 1.4×10−2
5.0×10−3×1
298 −1
318
Ea≈60.85 kJ/mol
Therefore, the activation energy for this reaction is approximately 60.85
kJ/mol.
Question 18
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 250 K and
1.36 ×10−2s−1at 300 K. Calculate the activation energy for this reaction in
kJ/mol.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy
in J/mol, R= gas constant = 8.314 J/(mol·K), T= temperature in Kelvin.
Step 2: Take the natural logarithm of the Arrhenius equation:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Set up two equations using the data provided:
4.23 ×10−3=A·e−Ea
8.314·250
1.36 ×10−2=A·e−Ea
8.314·300
Step 4: Divide the two equations to eliminate A:
1.36 ×10−2
4.23 ×10−3=e−Ea
8.314·300
e−Ea
8.314·250
Step 5: Simplify the equation and solve for Ea:
3.211 = e250
8.314 −300
8.314
3.211 = e30.086−36.103
3.211 = e−6.017
17
−6.017 = ln(3.211)
Step 6: Calculate the activation energy in kJ/mol:
Ea=−6.017 ×8.314 ×10−3
Ea=−0.0501 kJ/mol
Ea= 50.1 kJ/mol
Therefore, the activation energy for this reaction is 50.1 kJ/mol.
Question 19
Question
The rate constant of a reaction at 25
°
C is 3.45 ×10−5s−1, and at 35
°
C it is
9.82 ×10−4s−1. Calculate the activation energy of 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: Convert the temperatures to Kelvin: T1= 25C= 25 + 273.15 =
298.15 K, T2= 35C= 35 + 273.15 = 308.15 K.
Step 3: Set up two equations using the given rate constants and tempera-
tures:
k1=A·e−Ea
R·298.15
k2=A·e−Ea
R·308.15
Step 4: Divide the two equations to eliminate Aand solve for Ea:
k1
k2
=e−Ea
R·298.15
e−Ea
R·308.15
k1
k2
=eEa
R(1
308.15 −1
298.15 )
Step 5: Substitute the given rate constants into the equation above:
3.45 ×10−5
9.82 ×10−4=eEa
8.314 (1
308.15 −1
298.15 )
Step 6: Solve for Eausing a calculator:
ln 3.45 ×10−5
9.82 ×10−4=Ea
8.314 1
308.15 −1
298.15
Ea=−80240 J/mol
Therefore, the activation energy of this reaction is -80240 J/mol.
18
Question 20
Question
The rate constant for the reaction A →B is found to be 2.5×10−3s−1at
25◦C. When the temperature is increased to 50◦C, the rate constant becomes
3.6×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
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)), and - Tis the
temperature in Kelvin.
We have two sets of data: At T1= 25◦C = 298 K, k1= 2.5×10−3s−1, At
T2= 50◦C = 323 K, k2= 3.6×10−2s−1.
Using these data, we can set up the following equation:
k2
k1
=A·e−Ea
R1
T2
−1
T1
A·e−Ea
R1
T1
−1
T1
Step 2: Solve for the activation energy Ea:
k2
k1
=3.6×10−2
2.5×10−3=e−Ea
R(1
323 −1
298 )
3.6×10−2
2.5×10−3=eEa
R(1
298 −1
323 )
14.4 = eEa
R(323−298
323·298 )
14.4 = eEa
8.314 (25
323·298 )
14.4 = e0.000996Ea
ln(14.4) = 0.000996Ea
Ea=ln(14.4)
0.000996 ≈53376.5 J/mol
Ea≈53.4 kJ/mol
Therefore, the activation energy for this reaction is approximately 53.4 kJ/mol.
19
Question 21
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 of the reaction.
(Assume the pre-exponential factor, A, is constant.)
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
Step 2: Determine the ratio of rate constants at different temperatures:
k2
k1
=A·e−Ea
R(T2)
A·e−Ea
R(T1)
Step 3: Use the given information that the rate constant doubles when the
temperature is increased to set up and solve the equation:
2 = e(−Ea/R)(1/T1−1/T2)
ln(2)
Step 4: Substitute the known values and solve for the activation energy, Ea:
2 = e−Ea
R·(1
298 −1
308 )
Step 5: Simplify the equation and solve for the activation energy, Ea.
Step 6: Calculate the activation energy using the Arrhenius equation with
the pre-exponential factor, A, and the obtained value of Ea.
Question 22
Question
The rate constant of a first-order reaction is found to be 2.50×10−3s−1at 25◦C
and 8.00 ×10−3s−1at 125◦C. Calculate the activation energy of the reaction.
Assume the frequency factor (A) is 8.00 ×1011 s−1.
Solution
Step 1: Convert the given rates to their corresponding temperature in Kelvin
using the formula: T(K) = T(◦C) + 273.15.
Given that T1= 25◦C and T2= 125◦C, we have:
T1= 25 + 273.15 = 298.15 K
20
T2= 125 + 273.15 = 398.15 K
Step 2: Write the Arrhenius equation, which relates the rate constant (k)
to the activation energy (Ea) and the temperature. The Arrhenius equation is
given by:
k=A·e−Ea
RT
Step 3: Using the given data for two temperatures, we can form two equa-
tions using the Arrhenius equation:
For T1= 298.15 K:
2.50 ×10−3= 8.00 ×1011 ·e−Ea
8.314·298.15
For T2= 398.15 K:
8.00 ×10−3= 8.00 ×1011 ·e−Ea
8.314·398.15
Step 4: Take the ratio of the two equations to eliminate the frequency factor:
2.50 ×10−3
8.00 ×10−3=e−Ea
8.314·298.15
e−Ea
8.314·398.15
Solving for Ea, we get:
Ea=−8.314 ·(398.15 −298.15) ·ln 2.50 ×10−3
8.00 ×10−3
Step 5: Calculate the activation energy using the given values:
Ea=−8.314 ·100 ·ln 2.50 ×10−3
8.00 ×10−3
Question 23
Question
For the reaction A →B, the rate constant at 25
°
C is 6.0×10−4s−1. When
the temperature is increased to 45
°
C, the rate constant becomes 1.2×10−3s−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
=e
−Ea
R1
T2
−1
T1
where: - k1and k2are the rate constants at temperatures T1and T2respectively,
-Eais the activation energy, - Ris the gas constant, 8.314 J ·mol−1·K−1, -
T1= 25
°
C = 298 K, and - T2= 45
°
C = 318 K.
21
Step 2: Substitute the given rate constants and temperatures into the equa-
tion: 1.2×10−3
6.0×10−4=e−Ea
8.314 (1
318 −1
298 )
Step 3: Simplify the ratio of rate constants:
2 = e−Ea
8.314 (1
318 −1
298 )
Step 4: Take the natural logarithm of both sides:
ln(2) = −Ea
8.314 1
318 −1
298
Step 5: Solve for the activation energy Ea:
Ea=−8.314 ×ln(2)
1
318 −1
298
Step 6: Calculate the activation energy using the formula:
Ea=−8.314 ×ln(2)
1
318 −1
298
Ea≈69.7 kJ/mol
Therefore, the activation energy for the reaction A →B is approximately
69.7 kJ/mol.
Question 24
Question
For a certain reaction, the rate constant at 400 K is found to be 1.5×10−3s−1,
and the rate constant at 425 K is 3.2×10−3s−1. Calculate the activation energy
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: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 2: We have data for two different temperatures. We will set up two Ar-
rhenius equations (one for each temperature) and then divide them to eliminate
A:
k1=Ae−Ea
RT1
22
k2=Ae−Ea
RT2
Step 3: Take the ratio of the two equations to eliminate A:
k1
k2
=e−Ea
RT1
e−Ea
RT2
Step 4: Simplify the ratio:
k1
k2
=eEa
R(1
T2
−1
T1)
Step 5: Plug in the given rate constants and temperatures:
1.5×10−3
3.2×10−3=eEa
8.314 (1
425 −1
400 )
Step 6: Solve for the activation energy Ea:
1.5
3.2=eEa
8.314 (1
425 −1
400 )
0.46875 = eEa
8.314 (0.00235−0.0025)
0.46875 = e−0.04204Ea
Step 7: Solve for Ea:
ln(0.46875) = −0.04204Ea
Ea=ln(0.46875)
−0.04204
Ea≈47,581 J/mol
Therefore, the activation energy for this reaction is approximately 47.58
kJ/mol.
Question 25
Question
The rate constant for a certain reaction is observed to be 2.5×10−3s−1at 25◦C
and 8.0×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Assume the frequency factor is 1.0×1013 s−1.
23
Solution
Step 1: Begin by recalling the Arrhenius equation, which relates the rate con-
stant kof a reaction to the temperature T:
k=A·e−Ea
RT
where Ais the frequency factor, Eais the activation energy, R= 8.314 J ·K−1·
mol−1is the gas constant, and Tis the temperature in Kelvin.
Step 2: We are given two sets of data points at different temperatures: At
25◦C (298 K): k1= 2.5×10−3s−1At 45◦C (318 K): k2= 8.0×10−3s−1
Step 3: We can set up the following two equations using the Arrhenius
equation and the given data points:
2.5×10−3= 1.0×1013 ·e−Ea
8.314·298
8.0×10−3= 1.0×1013 ·e−Ea
8.314·318
Step 4: We can divide the second equation by the first equation to eliminate
the frequency factor A:
8.0×10−3
2.5×10−3=e−Ea
8.314·318
e−Ea
8.314·298
Step 5: Simplify the expression:
3.2 = eEa
8.314 (1
298 −1
318 )
Step 6: Solve for the activation energy Ea:
ln(3.2) = Ea
8.314 1
298 −1
318
Step 7: Calculate the activation energy:
Ea= 8.314 ·1
1
298 −1
318 ·ln(3.2) ≈50.52 kJ/mol
Step 8: Therefore, the activation energy for this reaction is approximately
50.52 kJ/mol.
Question 26
Question
The rate constant for a reaction is found to be 2.50 ×10−3s−1at 350 K and
1.20 ×10−2s−1at 400 K. Calculate the activation energy for this reaction. The
gas constant Ris 8.314 J/mol·K.
24
Solution
Step 1: Write the Arrhenius equation, which relates the rate constant kwith
temperature Tand the activation energy Ea:
k=A·e−Ea
RT
where Ais the pre-exponential factor, Ris the gas constant, and Tis the
temperature in Kelvin.
Step 2: Use the given rate constants and temperatures to set up two equa-
tions. At 350 K:
2.50 ×10−3=A·e−Ea
8.314×350
At 400 K:
1.20 ×10−2=A·e−Ea
8.314×400
Step 3: Divide the two equations to eliminate A:
2.50 ×10−3
1.20 ×10−2=e−Ea
8.314×350
e−Ea
8.314×400
Step 4: Simplify the equation:
0.2083 = e−Ea
8.314×350 +Ea
8.314×400
Step 5: Take the natural logarithm of both sides to solve for Ea:
ln(0.2083) = −Ea
8.314 ×350 +Ea
8.314 ×400
Step 6: Calculate Eausing the information provided:
Ea=−8.314 ×350 ×ln(0.2083)/(1/400 −1/350)
Step 7: Plug in the values and calculate Ea:
Ea≈60019 J/mol
Question 27
Question
The rate constant for a certain reaction at 25
°
C is 0.0023 s−1, and the activation
energy for the reaction is 85 kJ/mol. If the frequency factor is 5.1×1012 s−1,
what would be the rate constant at 50
°
C?
25
Solution
Step 1: Calculate the new rate constant using the Arrhenius equation:
k=A×e−Ea
RT
where: - kis the rate constant at the new temperature, - Ais the frequency
factor, - Eais the activation energy, - Ris the gas constant (8.314 J/mol ·K),
and - Tis the temperature in Kelvin.
Step 2: Convert the activation energy to joules:
Ea= 85 kJ/mol ×1000 J
1 kJ = 85000 J/mol
Step 3: Convert the temperature to Kelvin:
25C+ 273.15 = 298.15 K
Step 4: Calculate the rate constant at 25
°
C:
k= 5.1×1012 s−1×e−85000 J/mol
8.314 J/mol·K×298.15 K
Calculating this expression gives:
k≈0.0023 s−1
Step 5: Convert the new temperature to Kelvin:
50C+ 273.15 = 323.15 K
Step 6: Calculate the rate constant at 50
°
C:
k= 5.1×1012 s−1×e−85000 J/mol
8.314 J/mol·K×323.15 K
Calculating this expression gives:
k≈0.0105 s−1
Therefore, the rate constant at 50
°
C would be approximately 0.0105 s−1.
Question 28
Question
The rate constant for a reaction at 45
°
C is 2.75 ×10−3s−1. When the tempera-
ture is increased to 65
°
C, the rate constant becomes 1.20 ×10−2s−1. Calculate
the activation energy of the reaction. (Assume the frequency factor is 3.5×1012
s−1).
26
Solution
Step 1: Write the Arrhenius equation:
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), and - Tis the temperature in
Kelvin.
Step 2: Convert the temperatures to Kelvin. T1= 45C= 45 + 273 = 318 K
and T2= 65C= 65 + 273 = 338 K.
Step 3: Calculate the rate constant at each temperature using the given
values. For the first temperature: k1= 2.75 ×10−3s−1and T= 318 K. For the
second temperature: k2= 1.20 ×10−2s−1and T= 338 K.
Step 4: Substituting the values into the Arrhenius equation for both tem-
peratures, we get the following equations:
k1=A·e−Ea
R·318
k2=A·e−Ea
R·338
Step 5: Take the ratio of the two equations to eliminate the frequency factor
A:
k2
k1
=e−Ea
R·338
e−Ea
R·318
=eEa
R(1
318 −1
338 )
Step 6: Solve the equation for the activation energy Ea:
k2
k1
=eEa
R(1
318 −1
338 )
ln k2
k1=Ea
R1
318 −1
338
Ea=−R·ln k2
k1 1
318 −1
338
Step 7: Substitute the given values to calculate the activation energy.
Ea=−8.314 ·ln 1.20 ×10−2
2.75 ×10−3 1
318 −1
338
Question 29
Question
The rate constant for a reaction is 4.25 ×10−3s−1at 20◦C and 8.63 ×10−3s−1
at 50◦C. Calculate the activation energy for this reaction.
27
Solution
Step 1: Let’s write down the Arrhenius equation, which relates the rate constant
to temperature and the activation energy:
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: We can write two Arrhenius equations for the given temperatures:
For 20◦C (293 K):
4.25 ×10−3=A·e−Ea
8.314·293
For 50◦C (323 K):
8.63 ×10−3=A·e−Ea
8.314·323
Step 3: Let’s divide the second equation by the first equation to eliminate
A:
8.63 ×10−3
4.25 ×10−3=e−Ea
8.314·323
e−Ea
8.314·293
Step 4: Simplifying the left side:
2.0282 = eEa
8.314 (1
293 −1
323 )
Step 5: Further simplifying:
2.0282 = eEa
8.314 (30
293·323 )
Step 6: Taking the natural logarithm of both sides:
ln(2.0282) = Ea
8.314 30
293 ·323
Step 7: Solving for the activation energy (Ea):
Ea=−8.314 ·ln(2.0282) ·293 ·323
30
Step 8: Calculating the activation energy:
Ea≈48.6 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.6
kJ/mol.
Question 30
Question
The rate constant for the reaction 2NO2(g)→2NO(g) + O2(g) is found to be
3.21 ×10−3s−1at 25◦Cand 1.52 ×10−2s−1at 75◦C. Determine the activation
energy for this reaction.
28
Solution
Step 1: Write 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−1·K−1), - Tis the
temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for the activation energy
Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Plug in the given values:
k1= 3.21 ×10−3s−1
T1= 25 + 273 = 298 K
k2= 1.52 ×10−2s−1
T2= 75 + 273 = 348 K
Step 4: Calculate the activation energy:
ln 1.52 ×10−2
3.21 ×10−3=−Ea
8.314 1
348 −1
298
ln 4.74
1=−Ea
8.314 1
348 −1
298
ln(4.74) = −Ea
8.314 1
348 −1
298
Step 5: Calculate the activation energy Ea:
−ln(4.74) = Ea
8.314 1
298 −1
348
Ea=−8.314 ×ln(4.74) 1
298 −1
348
Ea≈9.52 kJ/mol
Therefore, the activation energy for the reaction is approximately 9.52 kJ/mol.
Question 31
Question
The rate constant (k) for a certain reaction is found to be 2.5×10−3s−1at 300
K and 1.5×10−2s−1at 350 K. Calculate the activation energy (Ea) for this
reaction.
29
Solution
Step 1: First, we can use the Arrhenius equation to relate the rate constants at
two different temperatures:
k2
k1
= e−Ea
R(1
T2
−1
T1)
where: k1= 2.5×10−3s−1(rate constant at 300 K),
k2= 1.5×10−2s−1(rate constant at 350 K),
T1= 300 K,
T2= 350 K,
Eais the activation energy, and
Ris the ideal gas constant (8.314 J mol−1K−1).
Step 2: Plug in the values and solve for Ea:
1.5×10−2
2.5×10−3= e−Ea
8.314 (1
350 −1
300 )
6=e−Ea
8.314 (1
350 −1
300 )
Step 3: Simplify the equation and solve for Ea:
6=e−Ea
8.314 (5
1050 )
6=e−Ea
4986
4986 ln(6) = −Ea
Ea=−4986 ln(6) ≈9.02 ×103J/mol
Step 4: Therefore, the activation energy for this reaction is approximately
9.02 ×103J/mol.
Question 32
Question
The rate constant for a certain reaction is found to be 2.45 ×10−3s−1at 25◦C
and 7.80 ×10−3s−1at 50◦C. Calculate the activation energy for this reaction.
(Given: R= 8.314 J mol−1K−1)
Solution
Step 1: Let’s first write 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−1K−1), and - Tis the
temperature in Kelvin.
30
Step 2: We are given two sets of data points: At 25◦C = 298 K, k1=
2.45 ×10−3s−1
At 50◦C = 323 K, k2= 7.80 ×10−3s−1
Step 3: We can set up two Arrhenius equations using the given data points:
k1=A e−Ea
R×298
k2=A e−Ea
R×323
Step 4: Divide the two equations to eliminate the pre-exponential factor A:
k2
k1
=e−Ea
R×323
e−Ea
R×298
Step 5: Simplify the expression:
k2
k1
=eEa
R(1
298 −1
323 )
Step 6: Solve for Ea:
ln k2
k1=Ea
R1
298 −1
323
Step 7: Calculate Ea:
Ea=R×1
298 −1
323−1
×ln k2
k1
Substitute the given values:
Ea= 8.314 ×1
298 −1
323−1
×ln 7.80 ×10−3
2.45 ×10−3
After calculation, we find:
Ea≈48.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.1
kJ/mol.
Question 33
Question
The rate constant (k) for a certain reaction is found to be 4.28 ×10−3s−1at
25◦C and 8.75 ×10−2s−1at 45◦C. Calculate the activation energy (Ea) for this
reaction. Assume the frequency factor (A) is 8.30 ×1011 s−1.
31
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: Rearrange the Arrhenius equation to solve for Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values and solve for Ea: Given: k1= 4.28×10−3
s−1,k2= 8.75 ×10−2s−1,A= 8.30 ×1011 s−1,R= 8.314 J/(mol
·
K), T1=
25 + 273 = 298 K, T2= 45 + 273 = 318 K.
ln 8.75 ×10−2
4.28 ×10−3=−Ea
8.314 1
318 −1
298
ln(20.45) = −Ea
8.314 1
318 −1
298
Ea=−8.314 ×ln(20.45) ×1
318 −1
298
Ea≈74260 J/mol
Therefore, the activation energy (Ea) for this reaction is approximately 74260
J/mol.
Question 34
Question
The rate constant for a certain reaction is found to be 3.2×10−4s−1at 40◦C
and 1.2×10−2s−1at 60◦C. Calculate the activation energy for the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to tem-
perature and the activation energy. The Arrhenius equation is given by:
k=A·exp −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.
32
Step 2: Let’s start by calculating the activation energy using the information
provided at 40◦Cand 60◦C. Converting these temperatures to Kelvin:
T1= 40 + 273 = 313K
T2= 60 + 273 = 333K
Step 3: Now we can set up two equations using the rate constants and
temperatures given:
3.2×10−4=A×exp −Ea
8.314 ×313
1.2×10−2=A×exp −Ea
8.314 ×333
Step 4: Let’s divide the second equation by the first equation to eliminate
A:1.2×10−2
3.2×10−4= exp Ea
8.314(1
333 −1
313)
Step 5: Simplify the left side of the equation:
1.2×10−2
3.2×10−4= 37.5
Step 6: Solve for Eaby taking the natural logarithm of both sides:
ln(37.5) = Ea
8.314(1
333 −1
313)
Step 7: Calculate the activation energy Ea:
Ea= 8.314 ×(ln(37.5)) ×(1
333 −1
313)
Step 8: After performing the calculation, we find:
Ea≈56.8kJ/mol
Therefore, the activation energy for the reaction is approximately 56.8kJ/mol.
Question 35
Question
The rate constant for a certain reaction at 25
°
C is 1.20 ×10−3s−1. When the
temperature is increased to 55
°
C, the rate constant is found to be 4.80 ×10−3
s−1. Calculate the activation energy for this reaction. (Hint: Use the Arrhenius
equation: k=Ae−Ea
RT )
33
Solution
Step 1: Convert the temperatures to Kelvin. Given: Initial temperature, T1=
25CFinal temperature, T2= 55C
Converting to Kelvin: T1= 25+273.15 = 298.15 K T2= 55+273.15 = 328.15
K
Step 2: Use the Arrhenius equation to find the activation energy (Ea). The
Arrhenius equation is given by: k=Ae−Ea
RT
Given: k1= 1.20 ×10−3s−1(at T1= 298.15 K) k2= 4.80 ×10−3s−1(at
T2= 328.15 K)
Taking the ratio of the two rate constants: k2
k1=Ae−Ea
RT2
Ae−Ea
RT1
Solving for Ea:Ea=−R
lnk2
k1×1
1
T2
−1
T1
Step 3: Calculate the activation energy. Given: Gas constant, R= 8.314
J/mol·K
Plugging in the values: Ea=−8.314
ln4.80×10−3
1.20×10−3×1
(1
328.15 −1
298.15 )
Ea=−8.314
ln(4) ×1
(0.003046−0.003356)
Ea=−8.314
1.3863 ×1
−0.000309
Ea= 56,767 J/mol or 56.77 kJ/mol
Therefore, the activation energy for this reaction is 56.77 kJ/mol.
34
ln (3.5294) = −Ea
8.314 1
350 −1
300
ln (3.5294) = −Ea
8.314 5
105000
ln (3.5294) = −Ea
8.314 1
21000
ln (3.5294) = −Ea
174.294
Ea=−174.294 ×ln (3.5294)
Ea≈38.45 kJ/mol
Step 5: Finalize the result: Therefore, the activation energy for the reaction
2A + B →C is approximately 38.45 kJ/mol.
Question 2
Question
The rate constant for a reaction is 1.25 ×10−2s−1at 400 K and 4.71 ×10−2
s−1at 600 K. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant of
a reaction to the temperature and activation energy.
k=A·e−Ea
RT
Step 2: We are given two sets of data points:
k1= 1.25 ×10−2s−1at 400 K
k2= 4.71 ×10−2s−1at 600 K
Step 3: Let’s first solve for A, the pre-exponential factor. We can set up a
ratio of the rate constants at the two temperatures.
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
4.71 ×10−2
1.25 ×10−2=e−Ea
R·600 ·eEa
R·400
3.768 = e−Ea
R(1
600 −1
400 )
2
Step 4: Simplify the equation from Step 3 and solve for Ea.
3.768 = e−Ea
R(1
600 −1
400 )
3.768 = e−Ea
R(1
240000 )
ln(3.768) = −Ea
R1
240000
Ea=−ln(3.768) ·(8.314 J/mol·K) ·240000
Step 5: Calculate the activation energy using the values obtained in Step 4.
Ea=−ln(3.768) ·(8.314 J/mol·K) ·240000
Ea≈58600 J/mol ≈58.6 kJ/mol
Therefore, the activation energy for this reaction is approximately 58.6
kJ/mol.
Question 3
Question
The rate constant of a reaction at two different temperatures, T1and T2, are
known to be k1= 1.2×10−4s−1and k2= 8.3×10−2s−1, respectively. If the
activation energy for the reaction is 60 kJ/mol, determine the ratio of the rate
constants (k2/k1) and the temperature at which the rate constant would be
5×10−3s−1.
Solution
Step 1: Calculate the ratio of rate constants (k2/k1) using the Arrhenius equa-
tion: k2
k1
= exp Ea
R1
T1
−1
T2
Step 2: Substitute the given values into the Arrhenius equation:
k2
k1
= exp 60,000 J/mol
8.314 J/mol K 1
T1
−1
T2
Step 3: Substitute the rate constants into the ratio expression:
8.3×10−2
1.2×10−4= exp 60,000 J/mol
8.314 J/mol K 1
T1
−1
T2
Step 4: Simplify the ratio of rate constants:
k2
k1
= 692.5
3
Step 5: Solve for the temperature at which the rate constant would be
5×10−3s−1:
5×10−3=k=Aexp −Ea
RT
Step 6: Rearrange the equation to solve for temperature:
ln k
A=−Ea
R1
T
Step 7: Substitute the values of k,Ea,R, and Ainto the equation:
ln 5×10−3
1.2×10−4=−60,000
8.314 1
T
Step 8: Solve for the temperature, T:
T=1
−60,000
8.314×ln5×10−3
1.2×10−4
= 408 K
Question 4
Question
The rate constant for a certain reaction at 25
°
C is 2.5×10−4s−1, and at 45
°
C
it is 7.5×10−3s−1. Determine the activation energy for this reaction.
Solution
Step 1: First, we can use the Arrhenius equation to relate the rate constants at
two different temperatures to the activation energy. The Arrhenius equation 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: We are given two sets of data: At 25
°
C, k1= 2.5×10−4s−1,
T1= 25C= 298 K, At 45
°
C, k2= 7.5×10−3s−1,T2= 45C= 318 K.
Step 3: Let’s set up the Arrhenius equation for the two temperature points:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 4: Divide the two equations to eliminate Aand solve for Ea:
k2
k1
=e−Ea
R·T2
e−Ea
R·T1
4
k2
k1
=eEa
R·1
T1
−1
T2
Step 5: Now, substitute the given values and solve for Ea:
7.5×10−3
2.5×10−4=e(Ea
8.314 ·(1
298 −1
318 ))
Step 6: Calculate the activation energy Eausing the natural logarithm to
solve for it:
Ea=−8.314 ×ln 7.5×10−3
2.5×10−4×1
298 −1
318
Step 7: Solve for Eato find the activation energy of the reaction.
Question 5
Question
The rate constant of a reaction at two different temperatures is found to be
k1= 1.25 ×10−4s−1at T1= 300 K and k2= 5.00 ×10−3s−1at T2= 350 K.
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: Write down the expression for the rate constant at the two different
temperatures:
k1=A×e−Ea
R×T1
k2=A×e−Ea
R×T2
Step 3: Divide the rate constants to obtain a ratio:
k2
k1
=A×e−Ea
R×T2
A×e−Ea
R×T1
Step 4: Simplify the ratio:
k2
k1
=e−Ea
R×T2
e−Ea
R×T1
5
k2
k1
=e
Ea
R1
T1
−1
T2
Step 5: Plug in the given values and solve for the activation energy Ea:
5.00 ×10−3s−1
1.25 ×10−4s−1=eEa
8.314 (1
300 −1
350 )
Step 6: Calculate the activation energy:
5.00 ×10−3
1.25 ×10−4=eEa
8.314 (1
300 −1
350 )
40 = eEa
8.314 (1
300 −1
350 )
Step 7: Solve for the activation energy, Ea:
ln(40) = Ea
8.314 1
300 −1
350
Ea= 8.314 ×ln(40)
1
300 −1
350
Ea≈50900 J/mol
Question 6
Question
The rate constant for a reaction at 25
°
C is 4.57 ×10−3s−1. When the tempera-
ture is increased to 55
°
C, the rate constant becomes 1.26 ×10−2s−1. Calculate
the activation energy for the reaction.
Solution
Step 1: Convert temperatures to Kelvin.
Given:
T1= 25C
T2= 55C
To convert from Celsius to Kelvin, we use:
T(K) = T(C) + 273.15
Therefore:
T1= 25 + 273.15 = 298.15K
T2= 55 + 273.15 = 328.15K
6
Step 2: Write down 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
Given two sets of data (rate constants and temperatures), we can set up a ratio
and solve for the activation energy.
Step 3: Set up the ratio of rate constants.
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R1
T1
−1
T2
Step 4: Substitute values and solve for activation energy.
Given data:
k1= 4.57 ×10−3s−1
k2= 1.26 ×10−2s−1
T1= 298.15 K
T2= 328.15 K
R= 8.314 J mol−1K−1
Substitute the values into the equation:
1.26 ×10−2
4.57 ×10−3=e−Ea
8.314 (1
298.15 −1
328.15 )
Solve for the activation energy:
ln 1.26 ×10−2
4.57 ×10−3=−Ea
8.314 1
298.15 −1
328.15
Ea=−8.314 ×
ln 1.26×10−2
4.57×10−3
1
298.15 −1
328.15
Calculate the activation energy using the given data.
Question 7
Question
The rate constant for a reaction at 25◦Cis 2.5×10−3s−1and the activation
energy (Ea) for the reaction is 75 kJ/mol. Determine the rate constant at 50◦C
for this reaction.
7
Solution
Step 1: Calculate the ratio of rate constants using the Arrhenius equation:
k2
k1
= exp Ea
R1
T1
−1
T2
where: k1= 2.5×10−3s−1(rate constant at 25◦C), T1= 25◦C= 298 K,
T2= 50◦C= 323 K, Ea= 75 kJ/mol, and R= 8.314
Question
The rate constant for a reaction is found to be 2.5×10−3s−1at 300 K and
1.5×10−2s−1at 350 K. Calculate the activation energy for the reaction. (The
universal gas constant Ris 8.314 J/(mol·K).)
Solution
Step 1: We can use the Arrhenius equation to determine the activation energy
Ea. The Arrhenius equation is given by:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor or frequency factor,
Ea= activation energy, R= universal gas constant (8.314 J/(mol·K)), T=
temperature (in Kelvin).
Step 2: We can rewrite the Arrhenius equation as:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k2and k1are the rate constants at temperatures T2and T1, respec-
tively.
Step 3: Substituting the given values k1= 2.5×10−3s−1at 300 K and
k2= 1.5×10−2s−1at 350 K into the equation, we get:
ln 1.5×10−2
2.5×10−3=−Ea
8.314 1
350 −1
300
Step 4: Simplifying the above equation, we find:
ln (6) = −Ea
8.314 1
350 −1
300
Step 5: Calculate the activation energy Eaby solving the above equation:
Ea
8.314 =−ln (6) ×1
350 −1
300
Ea=−8.314 ×ln (6) ×1
350 −1
300
Calculating the above expression will give us the activation energy Ea.
8
Question 9
Question
The rate constant for a certain reaction is found to be 4.5×10−2s−1at 25◦C and
9.0×10−2s−1at 35◦C. Calculate the activation energy (Ea) for this reaction.
The activation energy is known to be 8.314 J/mol ·K.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation, which is
given by:
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), and - Tis the
temperature in Kelvin.
Let’s denote the rate constants as k1= 4.5×10−2s−1at 25◦C and k2=
9.0×10−2s−1at 35◦C.
Step 2: We can set up two Arrhenius equations for the two sets of tempera-
ture and rate constant given:
k1=A·e
−Ea
R·298
k2=A·e
−Ea
R·308
Step 3: Divide the two equations to eliminate A:
k2
k1
=e
−Ea
8.314 ·308
e
−Ea
8.314 ·298
Step 4: Simplify the equation:
k2
k1
=e 298 −308
8.314 ·308 ·298!·Ea
Step 5: Solve for Ea:
Ea=8.314 ·308 ·298
10 ·ln 9.0×10−2
4.5×10−2
Step 6: Calculate the value of Eato find the activation energy of the reaction.
9
Question 10
Question
What is the activation energy of a reaction if its rate constant doubles when the
temperature is increased from 25
°
C to 35
°
C? Assume that the pre-exponential
factor remains constant at room temperature.
Solution
Step 1: Recall the Arrhenius equation, which relates the rate constant (k) of a
reaction to the temperature (T) and 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.
Step 2: Let’s denote the rate constant at 25
°
C as k1and the rate constant
at 35
°
C as k2. We know that:
k2= 2 ·k1
Step 3: Let’s substitute these values into the Arrhenius equation:
A·e−Ea
R(25+273.15) = 2 ·A·e−Ea
R(35+273.15)
Step 4: Cancel out the pre-exponential factor A:
e−Ea
R(25+273.15) = 2 ·e−Ea
R(35+273.15)
Step 5: Take the natural logarithm of both sides:
−Ea
R(25 + 273.15) = ln(2) −Ea
R(35 + 273.15)
Step 6: Solve for the activation energy Ea:
Ea
R(1
25 + 273.15 −1
35 + 273.15) = ln(2)
Step 7: Calculate the activation energy Ea:
Ea=R·25 + 273.15
35 + 273.15 ·ln(2)
Ea≈54.32 kcal/mol
Therefore, the activation energy of the reaction is approximately 54.32 kcal/mol.
10
Question 11
Question
The rate constant for the decomposition of a certain compound at 25
°
C is 4.2×
10−3s−1. When the temperature is increased to 50
°
C, the rate constant becomes
7.8×10−2s−1. Determine the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=Ae−Ea
RT )
Solution
Step 1: We can start by writing the Arrhenius equation in its logarithmic form:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1and k2are the rate constants at temperatures T1and T2respec-
tively, Eais the activation energy, Ris the gas constant, and T1and T2are the
temperatures in Kelvin.
Step 2: Plug in the given values: k1= 4.2×10−3s−1,k2= 7.8×10−2s−1,
T1= 298 K, and T2= 323 K. The gas constant Ris 8.314 J/(mol
·
K).
Step 3: Substitute the values into the equation and solve for Ea:
ln 7.8×10−2
4.2×10−3=−Ea
8.314 1
323 −1
298
Step 4: Calculate the natural logarithm:
ln 7.8×10−2
4.2×10−3= ln 18.57 ≈2.92
Step 5: Substitute this back into the equation and solve for Ea:
2.92 = −Ea
8.314 1
323 −1
298
Step 6: Calculate the activation energy:
Ea=−2.92 ×8.314 1
323 −1
298≈58401 J/mol = 58.4 kJ/mol
Therefore, the activation energy for this reaction is approximately 58.4
kJ/mol.
Question 12
Question
The rate constant for a reaction at 25
°
C is 2.50 ×10−4s−1and its activation
energy is 60 kJ/mol. Calculate the rate constant at 35
°
C for this reaction.
11
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol. Given: Activation
energy, Ea= 60 kJ/mol.
Converting kJ to J: 60 kJ = 60,000 J.
Step 2: Use the Arrhenius equation to calculate the rate constant at 35
°
C.
The Arrhenius equation 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.
Given: Rate constant at 25
°
C, k1= 2.50 ×10−4s−1, Temperature at 25
°
C,
T1= 25 + 273.15 = 298.15 K, Temperature at 35
°
C, T2= 35 + 273.15 = 308.15
K, Activation energy, Ea= 60,000 J/mol.
Substitute the given values into the Arrhenius equation:
2.50 ×10−4=A·e−60,000
8.314·298.15
Solving for the pre-exponential factor, A:
A=2.50 ×10−4
e−60,000
8.314·298.15
Step 3: Calculate the rate constant at 35
°
C. Substitute the pre-exponential
factor Aand the new temperature T2into the Arrhenius equation:
k2=A·e−Ea
RT2
Now, calculate the rate constant at 35
°
C:
k2=A·e−60,000
8.314·308.15
Question 13
Question
The rate constant (k) of a reaction at 25◦C is found to be 2.5×10−3s−1. When
the temperature is increased to 35◦C, the rate constant becomes 1.2×10−2s−1.
Calculate the activation energy (Ea) for this reaction.
Solution
Step 1: Find the ratio of rate constants at the two temperatures using the
Arrhenius equation: k2
k1
=eEa
R(1
T1
−1
T2)
12
Step 2: Plug in the given values:
1.2×10−2
2.5×10−3=eEa
8.314 (1
298 −1
308 )
Step 3: Calculate the ratio of rate constants:
4.8
1=eEa
8.314 (0.0034)
Step 4: Solve for Ea:
ln 4.8 = Ea
8.314(0.0034)
Step 5: Rearrange the equation to solve for Ea:
Ea= 8.314 ×0.0034 ×ln 4.8
Step 6: Calculate Ea:
Ea≈54.73 kJ/mol
Therefore, the activation energy for this reaction is approximately 54.73
kJ/mol.
Question 14
Question
The rate constant for a reaction at 25
°
C is 4.50 ×10−4s−1. When the tempera-
ture is increased to 50
°
C, the rate constant becomes 1.80 ×10−3s−1. Calculate
the activation energy of the reaction.
Solution
Step 1: Convert the given temperatures into Kelvin. 25C= 25 + 273.15 =
298.15 K
50C= 50 + 273.15 = 323.15 K
Step 2: Calculate the activation energy using the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
Substitute the given values:
ln 1.80 ×10−3
4.50 ×10−4=−Ea
8.314 1
323.15 −1
298.15
Step 3: Solve for the activation energy Ea.
ln 4 = −Ea
8.314 1
323.15 −1
298.15
13
ln 4 = −Ea
8.314 298.15 −323.15
298.15 ×323.15
Step 4: Calculate the activation energy.
Ea=−8.314 ×ln 4 ×298.15 ×323.15
25
Ea≈69.31 kJ/mol
Question 15
Question
The rate constant of a reaction at 40
°
C is 4.28 ×10−3s−1, and at 80
°
C it is 1.97
s−1. Calculate the activation energy (in kJ/mol) for this reaction. Assume the
frequency factor (A) is 1.21 ×1014 s−1.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 40C+ 273.15 = 313.15 K
T2= 80C+ 273.15 = 353.15 K
Step 2: Calculate the value of the gas constant (R) in J/mol K. Since R=
8.314 J/mol K,
R= 8.314 J/mol K
Step 3: Use the Arrhenius equation to find the activation energy (Ea) in
Joules.
ln k2
k1=−Ea
R1
T2
−1
T1
Plugging in the given values,
ln 1.97
4.28 ×10−3=−Ea
8.314 1
353.15 −1
313.15
Solving for Eain Joules, we get
Ea=−8.314 ×ln 1.97
4.28 ×10−3 1
353.15 −1
313.15
Step 4: Convert the activation energy from Joules to kilojoules.
Ea=
−8.314 ×10−3×ln 1.97
4.28×10−31
353.15 −1
313.15
1000
Therefore, the activation energy for this reaction is approximately 48.3 kJ/mol .
14
Question 16
Question
The rate constant (k) for a certain reaction was found to be 6.3×10−3s−1at a
temperature of 350 K and 2.1×10−2s−1at a temperature of 400 K. Calculate
the activation energy (Ea) for this reaction.
Solution
Step 1: Recall 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), and T= temperature (in Kelvin).
Step 2: We can write two Arrhenius equations for the given data:
k1=A·e−Ea
R·T1
k2=A·e−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 expression in Step 3:
k1
k2
=eEa
R·1
T2
−1
T1
Step 5: Plug in the given values and solve for Ea:
6.3×10−3
2.1×10−2=e(Ea
8.314 ·(1
400 −1
350 ))
Step 6: Solve for Ea:3
10 =e(Ea
8.314 ·(7
140000 ))
Step 7: Take the natural logarithm of both sides to solve for Ea:
ln 3
10=Ea
8.314 ·7
140000
Step 8: Calculate Ea:
Ea= 8.314 ·140000
7·ln 3
10≈61.0 kJ/mol
Therefore, the activation energy for this reaction is approximately 61.0
kJ/mol.
15
Question 17
Question
The rate constant (k) for a first-order reaction at 25
°
C is 5.0×10−3s−1. When
the temperature is increased to 45
°
C, the rate constant becomes 1.4×10−2s−1.
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 (frequency
factor), - Eais the activation energy, - Ris the gas constant (8.314 J/(mol·K)),
- and Tis the temperature in Kelvin.
Step 2: Set up the equation for the rate constant at the two different tem-
peratures:
k1=A·e−Ea
RT1
k2=A·e−Ea
RT2
Given:
k1= 5.0×10−3s−1
k2= 1.4×10−2s−1
T1= 25 + 273 = 298 K
T2= 45 + 273 = 318 K
Step 3: Take the ratio of the two rate constants to eliminate A:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
k2
k1
=e
Ea
R1
T1
−1
T2
Substitute the values: 1.4×10−2
5.0×10−3=eEa
8.314 (1
298 −1
318 )
Step 4: Solve for the activation energy (Ea):
ln 1.4×10−2
5.0×10−3=Ea
8.314 1
298 −1
318
Ea= 8.314 ×ln 1.4×10−2
5.0×10−3×1
298 −1
318
16
Step 5: Calculate the activation energy:
Ea= 8.314 ×ln 1.4×10−2
5.0×10−3×1
298 −1
318
Ea≈60.85 kJ/mol
Therefore, the activation energy for this reaction is approximately 60.85
kJ/mol.
Question 18
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 250 K and
1.36 ×10−2s−1at 300 K. Calculate the activation energy for this reaction in
kJ/mol.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy
in J/mol, R= gas constant = 8.314 J/(mol·K), T= temperature in Kelvin.
Step 2: Take the natural logarithm of the Arrhenius equation:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Set up two equations using the data provided:
4.23 ×10−3=A·e−Ea
8.314·250
1.36 ×10−2=A·e−Ea
8.314·300
Step 4: Divide the two equations to eliminate A:
1.36 ×10−2
4.23 ×10−3=e−Ea
8.314·300
e−Ea
8.314·250
Step 5: Simplify the equation and solve for Ea:
3.211 = e250
8.314 −300
8.314
3.211 = e30.086−36.103
3.211 = e−6.017
17
−6.017 = ln(3.211)
Step 6: Calculate the activation energy in kJ/mol:
Ea=−6.017 ×8.314 ×10−3
Ea=−0.0501 kJ/mol
Ea= 50.1 kJ/mol
Therefore, the activation energy for this reaction is 50.1 kJ/mol.
Question 19
Question
The rate constant of a reaction at 25
°
C is 3.45 ×10−5s−1, and at 35
°
C it is
9.82 ×10−4s−1. Calculate the activation energy of 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: Convert the temperatures to Kelvin: T1= 25C= 25 + 273.15 =
298.15 K, T2= 35C= 35 + 273.15 = 308.15 K.
Step 3: Set up two equations using the given rate constants and tempera-
tures:
k1=A·e−Ea
R·298.15
k2=A·e−Ea
R·308.15
Step 4: Divide the two equations to eliminate Aand solve for Ea:
k1
k2
=e−Ea
R·298.15
e−Ea
R·308.15
k1
k2
=eEa
R(1
308.15 −1
298.15 )
Step 5: Substitute the given rate constants into the equation above:
3.45 ×10−5
9.82 ×10−4=eEa
8.314 (1
308.15 −1
298.15 )
Step 6: Solve for Eausing a calculator:
ln 3.45 ×10−5
9.82 ×10−4=Ea
8.314 1
308.15 −1
298.15
Ea=−80240 J/mol
Therefore, the activation energy of this reaction is -80240 J/mol.
18
Question 20
Question
The rate constant for the reaction A →B is found to be 2.5×10−3s−1at
25◦C. When the temperature is increased to 50◦C, the rate constant becomes
3.6×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
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)), and - Tis the
temperature in Kelvin.
We have two sets of data: At T1= 25◦C = 298 K, k1= 2.5×10−3s−1, At
T2= 50◦C = 323 K, k2= 3.6×10−2s−1.
Using these data, we can set up the following equation:
k2
k1
=A·e−Ea
R1
T2
−1
T1
A·e−Ea
R1
T1
−1
T1
Step 2: Solve for the activation energy Ea:
k2
k1
=3.6×10−2
2.5×10−3=e−Ea
R(1
323 −1
298 )
3.6×10−2
2.5×10−3=eEa
R(1
298 −1
323 )
14.4 = eEa
R(323−298
323·298 )
14.4 = eEa
8.314 (25
323·298 )
14.4 = e0.000996Ea
ln(14.4) = 0.000996Ea
Ea=ln(14.4)
0.000996 ≈53376.5 J/mol
Ea≈53.4 kJ/mol
Therefore, the activation energy for this reaction is approximately 53.4 kJ/mol.
19
Question 21
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 of the reaction.
(Assume the pre-exponential factor, A, is constant.)
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
Step 2: Determine the ratio of rate constants at different temperatures:
k2
k1
=A·e−Ea
R(T2)
A·e−Ea
R(T1)
Step 3: Use the given information that the rate constant doubles when the
temperature is increased to set up and solve the equation:
2 = e(−Ea/R)(1/T1−1/T2)
ln(2)
Step 4: Substitute the known values and solve for the activation energy, Ea:
2 = e−Ea
R·(1
298 −1
308 )
Step 5: Simplify the equation and solve for the activation energy, Ea.
Step 6: Calculate the activation energy using the Arrhenius equation with
the pre-exponential factor, A, and the obtained value of Ea.
Question 22
Question
The rate constant of a first-order reaction is found to be 2.50×10−3s−1at 25◦C
and 8.00 ×10−3s−1at 125◦C. Calculate the activation energy of the reaction.
Assume the frequency factor (A) is 8.00 ×1011 s−1.
Solution
Step 1: Convert the given rates to their corresponding temperature in Kelvin
using the formula: T(K) = T(◦C) + 273.15.
Given that T1= 25◦C and T2= 125◦C, we have:
T1= 25 + 273.15 = 298.15 K
20
T2= 125 + 273.15 = 398.15 K
Step 2: Write the Arrhenius equation, which relates the rate constant (k)
to the activation energy (Ea) and the temperature. The Arrhenius equation is
given by:
k=A·e−Ea
RT
Step 3: Using the given data for two temperatures, we can form two equa-
tions using the Arrhenius equation:
For T1= 298.15 K:
2.50 ×10−3= 8.00 ×1011 ·e−Ea
8.314·298.15
For T2= 398.15 K:
8.00 ×10−3= 8.00 ×1011 ·e−Ea
8.314·398.15
Step 4: Take the ratio of the two equations to eliminate the frequency factor:
2.50 ×10−3
8.00 ×10−3=e−Ea
8.314·298.15
e−Ea
8.314·398.15
Solving for Ea, we get:
Ea=−8.314 ·(398.15 −298.15) ·ln 2.50 ×10−3
8.00 ×10−3
Step 5: Calculate the activation energy using the given values:
Ea=−8.314 ·100 ·ln 2.50 ×10−3
8.00 ×10−3
Question 23
Question
For the reaction A →B, the rate constant at 25
°
C is 6.0×10−4s−1. When
the temperature is increased to 45
°
C, the rate constant becomes 1.2×10−3s−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
=e
−Ea
R1
T2
−1
T1
where: - k1and k2are the rate constants at temperatures T1and T2respectively,
-Eais the activation energy, - Ris the gas constant, 8.314 J ·mol−1·K−1, -
T1= 25
°
C = 298 K, and - T2= 45
°
C = 318 K.
21
Step 2: Substitute the given rate constants and temperatures into the equa-
tion: 1.2×10−3
6.0×10−4=e−Ea
8.314 (1
318 −1
298 )
Step 3: Simplify the ratio of rate constants:
2 = e−Ea
8.314 (1
318 −1
298 )
Step 4: Take the natural logarithm of both sides:
ln(2) = −Ea
8.314 1
318 −1
298
Step 5: Solve for the activation energy Ea:
Ea=−8.314 ×ln(2)
1
318 −1
298
Step 6: Calculate the activation energy using the formula:
Ea=−8.314 ×ln(2)
1
318 −1
298
Ea≈69.7 kJ/mol
Therefore, the activation energy for the reaction A →B is approximately
69.7 kJ/mol.
Question 24
Question
For a certain reaction, the rate constant at 400 K is found to be 1.5×10−3s−1,
and the rate constant at 425 K is 3.2×10−3s−1. Calculate the activation energy
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: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 2: We have data for two different temperatures. We will set up two Ar-
rhenius equations (one for each temperature) and then divide them to eliminate
A:
k1=Ae−Ea
RT1
22
k2=Ae−Ea
RT2
Step 3: Take the ratio of the two equations to eliminate A:
k1
k2
=e−Ea
RT1
e−Ea
RT2
Step 4: Simplify the ratio:
k1
k2
=eEa
R(1
T2
−1
T1)
Step 5: Plug in the given rate constants and temperatures:
1.5×10−3
3.2×10−3=eEa
8.314 (1
425 −1
400 )
Step 6: Solve for the activation energy Ea:
1.5
3.2=eEa
8.314 (1
425 −1
400 )
0.46875 = eEa
8.314 (0.00235−0.0025)
0.46875 = e−0.04204Ea
Step 7: Solve for Ea:
ln(0.46875) = −0.04204Ea
Ea=ln(0.46875)
−0.04204
Ea≈47,581 J/mol
Therefore, the activation energy for this reaction is approximately 47.58
kJ/mol.
Question 25
Question
The rate constant for a certain reaction is observed to be 2.5×10−3s−1at 25◦C
and 8.0×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Assume the frequency factor is 1.0×1013 s−1.
23
Solution
Step 1: Begin by recalling the Arrhenius equation, which relates the rate con-
stant kof a reaction to the temperature T:
k=A·e−Ea
RT
where Ais the frequency factor, Eais the activation energy, R= 8.314 J ·K−1·
mol−1is the gas constant, and Tis the temperature in Kelvin.
Step 2: We are given two sets of data points at different temperatures: At
25◦C (298 K): k1= 2.5×10−3s−1At 45◦C (318 K): k2= 8.0×10−3s−1
Step 3: We can set up the following two equations using the Arrhenius
equation and the given data points:
2.5×10−3= 1.0×1013 ·e−Ea
8.314·298
8.0×10−3= 1.0×1013 ·e−Ea
8.314·318
Step 4: We can divide the second equation by the first equation to eliminate
the frequency factor A:
8.0×10−3
2.5×10−3=e−Ea
8.314·318
e−Ea
8.314·298
Step 5: Simplify the expression:
3.2 = eEa
8.314 (1
298 −1
318 )
Step 6: Solve for the activation energy Ea:
ln(3.2) = Ea
8.314 1
298 −1
318
Step 7: Calculate the activation energy:
Ea= 8.314 ·1
1
298 −1
318 ·ln(3.2) ≈50.52 kJ/mol
Step 8: Therefore, the activation energy for this reaction is approximately
50.52 kJ/mol.
Question 26
Question
The rate constant for a reaction is found to be 2.50 ×10−3s−1at 350 K and
1.20 ×10−2s−1at 400 K. Calculate the activation energy for this reaction. The
gas constant Ris 8.314 J/mol·K.
24
Solution
Step 1: Write the Arrhenius equation, which relates the rate constant kwith
temperature Tand the activation energy Ea:
k=A·e−Ea
RT
where Ais the pre-exponential factor, Ris the gas constant, and Tis the
temperature in Kelvin.
Step 2: Use the given rate constants and temperatures to set up two equa-
tions. At 350 K:
2.50 ×10−3=A·e−Ea
8.314×350
At 400 K:
1.20 ×10−2=A·e−Ea
8.314×400
Step 3: Divide the two equations to eliminate A:
2.50 ×10−3
1.20 ×10−2=e−Ea
8.314×350
e−Ea
8.314×400
Step 4: Simplify the equation:
0.2083 = e−Ea
8.314×350 +Ea
8.314×400
Step 5: Take the natural logarithm of both sides to solve for Ea:
ln(0.2083) = −Ea
8.314 ×350 +Ea
8.314 ×400
Step 6: Calculate Eausing the information provided:
Ea=−8.314 ×350 ×ln(0.2083)/(1/400 −1/350)
Step 7: Plug in the values and calculate Ea:
Ea≈60019 J/mol
Question 27
Question
The rate constant for a certain reaction at 25
°
C is 0.0023 s−1, and the activation
energy for the reaction is 85 kJ/mol. If the frequency factor is 5.1×1012 s−1,
what would be the rate constant at 50
°
C?
25
Solution
Step 1: Calculate the new rate constant using the Arrhenius equation:
k=A×e−Ea
RT
where: - kis the rate constant at the new temperature, - Ais the frequency
factor, - Eais the activation energy, - Ris the gas constant (8.314 J/mol ·K),
and - Tis the temperature in Kelvin.
Step 2: Convert the activation energy to joules:
Ea= 85 kJ/mol ×1000 J
1 kJ = 85000 J/mol
Step 3: Convert the temperature to Kelvin:
25C+ 273.15 = 298.15 K
Step 4: Calculate the rate constant at 25
°
C:
k= 5.1×1012 s−1×e−85000 J/mol
8.314 J/mol·K×298.15 K
Calculating this expression gives:
k≈0.0023 s−1
Step 5: Convert the new temperature to Kelvin:
50C+ 273.15 = 323.15 K
Step 6: Calculate the rate constant at 50
°
C:
k= 5.1×1012 s−1×e−85000 J/mol
8.314 J/mol·K×323.15 K
Calculating this expression gives:
k≈0.0105 s−1
Therefore, the rate constant at 50
°
C would be approximately 0.0105 s−1.
Question 28
Question
The rate constant for a reaction at 45
°
C is 2.75 ×10−3s−1. When the tempera-
ture is increased to 65
°
C, the rate constant becomes 1.20 ×10−2s−1. Calculate
the activation energy of the reaction. (Assume the frequency factor is 3.5×1012
s−1).
26
Solution
Step 1: Write the Arrhenius equation:
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), and - Tis the temperature in
Kelvin.
Step 2: Convert the temperatures to Kelvin. T1= 45C= 45 + 273 = 318 K
and T2= 65C= 65 + 273 = 338 K.
Step 3: Calculate the rate constant at each temperature using the given
values. For the first temperature: k1= 2.75 ×10−3s−1and T= 318 K. For the
second temperature: k2= 1.20 ×10−2s−1and T= 338 K.
Step 4: Substituting the values into the Arrhenius equation for both tem-
peratures, we get the following equations:
k1=A·e−Ea
R·318
k2=A·e−Ea
R·338
Step 5: Take the ratio of the two equations to eliminate the frequency factor
A:
k2
k1
=e−Ea
R·338
e−Ea
R·318
=eEa
R(1
318 −1
338 )
Step 6: Solve the equation for the activation energy Ea:
k2
k1
=eEa
R(1
318 −1
338 )
ln k2
k1=Ea
R1
318 −1
338
Ea=−R·ln k2
k1 1
318 −1
338
Step 7: Substitute the given values to calculate the activation energy.
Ea=−8.314 ·ln 1.20 ×10−2
2.75 ×10−3 1
318 −1
338
Question 29
Question
The rate constant for a reaction is 4.25 ×10−3s−1at 20◦C and 8.63 ×10−3s−1
at 50◦C. Calculate the activation energy for this reaction.
27
Solution
Step 1: Let’s write down the Arrhenius equation, which relates the rate constant
to temperature and the activation energy:
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: We can write two Arrhenius equations for the given temperatures:
For 20◦C (293 K):
4.25 ×10−3=A·e−Ea
8.314·293
For 50◦C (323 K):
8.63 ×10−3=A·e−Ea
8.314·323
Step 3: Let’s divide the second equation by the first equation to eliminate
A:
8.63 ×10−3
4.25 ×10−3=e−Ea
8.314·323
e−Ea
8.314·293
Step 4: Simplifying the left side:
2.0282 = eEa
8.314 (1
293 −1
323 )
Step 5: Further simplifying:
2.0282 = eEa
8.314 (30
293·323 )
Step 6: Taking the natural logarithm of both sides:
ln(2.0282) = Ea
8.314 30
293 ·323
Step 7: Solving for the activation energy (Ea):
Ea=−8.314 ·ln(2.0282) ·293 ·323
30
Step 8: Calculating the activation energy:
Ea≈48.6 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.6
kJ/mol.
Question 30
Question
The rate constant for the reaction 2NO2(g)→2NO(g) + O2(g) is found to be
3.21 ×10−3s−1at 25◦Cand 1.52 ×10−2s−1at 75◦C. Determine the activation
energy for this reaction.
28
Solution
Step 1: Write 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−1·K−1), - Tis the
temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for the activation energy
Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Plug in the given values:
k1= 3.21 ×10−3s−1
T1= 25 + 273 = 298 K
k2= 1.52 ×10−2s−1
T2= 75 + 273 = 348 K
Step 4: Calculate the activation energy:
ln 1.52 ×10−2
3.21 ×10−3=−Ea
8.314 1
348 −1
298
ln 4.74
1=−Ea
8.314 1
348 −1
298
ln(4.74) = −Ea
8.314 1
348 −1
298
Step 5: Calculate the activation energy Ea:
−ln(4.74) = Ea
8.314 1
298 −1
348
Ea=−8.314 ×ln(4.74) 1
298 −1
348
Ea≈9.52 kJ/mol
Therefore, the activation energy for the reaction is approximately 9.52 kJ/mol.
Question 31
Question
The rate constant (k) for a certain reaction is found to be 2.5×10−3s−1at 300
K and 1.5×10−2s−1at 350 K. Calculate the activation energy (Ea) for this
reaction.
29
Solution
Step 1: First, we can use the Arrhenius equation to relate the rate constants at
two different temperatures:
k2
k1
= e−Ea
R(1
T2
−1
T1)
where: k1= 2.5×10−3s−1(rate constant at 300 K),
k2= 1.5×10−2s−1(rate constant at 350 K),
T1= 300 K,
T2= 350 K,
Eais the activation energy, and
Ris the ideal gas constant (8.314 J mol−1K−1).
Step 2: Plug in the values and solve for Ea:
1.5×10−2
2.5×10−3= e−Ea
8.314 (1
350 −1
300 )
6=e−Ea
8.314 (1
350 −1
300 )
Step 3: Simplify the equation and solve for Ea:
6=e−Ea
8.314 (5
1050 )
6=e−Ea
4986
4986 ln(6) = −Ea
Ea=−4986 ln(6) ≈9.02 ×103J/mol
Step 4: Therefore, the activation energy for this reaction is approximately
9.02 ×103J/mol.
Question 32
Question
The rate constant for a certain reaction is found to be 2.45 ×10−3s−1at 25◦C
and 7.80 ×10−3s−1at 50◦C. Calculate the activation energy for this reaction.
(Given: R= 8.314 J mol−1K−1)
Solution
Step 1: Let’s first write 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−1K−1), and - Tis the
temperature in Kelvin.
30
Step 2: We are given two sets of data points: At 25◦C = 298 K, k1=
2.45 ×10−3s−1
At 50◦C = 323 K, k2= 7.80 ×10−3s−1
Step 3: We can set up two Arrhenius equations using the given data points:
k1=A e−Ea
R×298
k2=A e−Ea
R×323
Step 4: Divide the two equations to eliminate the pre-exponential factor A:
k2
k1
=e−Ea
R×323
e−Ea
R×298
Step 5: Simplify the expression:
k2
k1
=eEa
R(1
298 −1
323 )
Step 6: Solve for Ea:
ln k2
k1=Ea
R1
298 −1
323
Step 7: Calculate Ea:
Ea=R×1
298 −1
323−1
×ln k2
k1
Substitute the given values:
Ea= 8.314 ×1
298 −1
323−1
×ln 7.80 ×10−3
2.45 ×10−3
After calculation, we find:
Ea≈48.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.1
kJ/mol.
Question 33
Question
The rate constant (k) for a certain reaction is found to be 4.28 ×10−3s−1at
25◦C and 8.75 ×10−2s−1at 45◦C. Calculate the activation energy (Ea) for this
reaction. Assume the frequency factor (A) is 8.30 ×1011 s−1.
31
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: Rearrange the Arrhenius equation to solve for Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values and solve for Ea: Given: k1= 4.28×10−3
s−1,k2= 8.75 ×10−2s−1,A= 8.30 ×1011 s−1,R= 8.314 J/(mol
·
K), T1=
25 + 273 = 298 K, T2= 45 + 273 = 318 K.
ln 8.75 ×10−2
4.28 ×10−3=−Ea
8.314 1
318 −1
298
ln(20.45) = −Ea
8.314 1
318 −1
298
Ea=−8.314 ×ln(20.45) ×1
318 −1
298
Ea≈74260 J/mol
Therefore, the activation energy (Ea) for this reaction is approximately 74260
J/mol.
Question 34
Question
The rate constant for a certain reaction is found to be 3.2×10−4s−1at 40◦C
and 1.2×10−2s−1at 60◦C. Calculate the activation energy for the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to tem-
perature and the activation energy. The Arrhenius equation is given by:
k=A·exp −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.
32
Step 2: Let’s start by calculating the activation energy using the information
provided at 40◦Cand 60◦C. Converting these temperatures to Kelvin:
T1= 40 + 273 = 313K
T2= 60 + 273 = 333K
Step 3: Now we can set up two equations using the rate constants and
temperatures given:
3.2×10−4=A×exp −Ea
8.314 ×313
1.2×10−2=A×exp −Ea
8.314 ×333
Step 4: Let’s divide the second equation by the first equation to eliminate
A:1.2×10−2
3.2×10−4= exp Ea
8.314(1
333 −1
313)
Step 5: Simplify the left side of the equation:
1.2×10−2
3.2×10−4= 37.5
Step 6: Solve for Eaby taking the natural logarithm of both sides:
ln(37.5) = Ea
8.314(1
333 −1
313)
Step 7: Calculate the activation energy Ea:
Ea= 8.314 ×(ln(37.5)) ×(1
333 −1
313)
Step 8: After performing the calculation, we find:
Ea≈56.8kJ/mol
Therefore, the activation energy for the reaction is approximately 56.8kJ/mol.
Question 35
Question
The rate constant for a certain reaction at 25
°
C is 1.20 ×10−3s−1. When the
temperature is increased to 55
°
C, the rate constant is found to be 4.80 ×10−3
s−1. Calculate the activation energy for this reaction. (Hint: Use the Arrhenius
equation: k=Ae−Ea
RT )
33
Solution
Step 1: Convert the temperatures to Kelvin. Given: Initial temperature, T1=
25CFinal temperature, T2= 55C
Converting to Kelvin: T1= 25+273.15 = 298.15 K T2= 55+273.15 = 328.15
K
Step 2: Use the Arrhenius equation to find the activation energy (Ea). The
Arrhenius equation is given by: k=Ae−Ea
RT
Given: k1= 1.20 ×10−3s−1(at T1= 298.15 K) k2= 4.80 ×10−3s−1(at
T2= 328.15 K)
Taking the ratio of the two rate constants: k2
k1=Ae−Ea
RT2
Ae−Ea
RT1
Solving for Ea:Ea=−R
lnk2
k1×1
1
T2
−1
T1
Step 3: Calculate the activation energy. Given: Gas constant, R= 8.314
J/mol·K
Plugging in the values: Ea=−8.314
ln4.80×10−3
1.20×10−3×1
(1
328.15 −1
298.15 )
Ea=−8.314
ln(4) ×1
(0.003046−0.003356)
Ea=−8.314
1.3863 ×1
−0.000309
Ea= 56,767 J/mol or 56.77 kJ/mol
Therefore, the activation energy for this reaction is 56.77 kJ/mol.
34
