CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 8
Liberty University
Question 1
Question
For the reaction
2A →B+C,
the rate constant at 25
°
C is 2.0×10−3s−1. When the temperature is increased
to 35
°
C, the rate constant becomes 6.0×10−3s−1. Calculate the activation
energy for this reaction. Assume the frequency factor remains constant.
Solution
Step 1: Let’s begin by writing the Arrhenius equation, which relates the rate
constant kto the activation energy Ea, temperature T, gas constant R, and
frequency factor A:
ln(k) = ln(A)−Ea
R·1
T.
Step 2: We are given two sets of data points (T1= 25C,k1= 2.0×10−3s−1)
and (T2= 35C,k2= 6.0×10−3s−1). Let’s convert the temperatures to Kelvin:
T1= 25 + 273 = 298 K
T2= 35 + 273 = 308 K
Step 3: Substitute into the Arrhenius equation for the two sets of data
points:
ln(k1) = ln(A)−Ea
R·1
T1
ln2.0×10−3= ln(A)−Ea
8.314 ·1
298
ln(k2) = ln(A)−Ea
R·1
T2
ln6.0×10−3= ln(A)−Ea
8.314 ·1
308
Step 4: Now, we have a system of two equations with two unknowns: Aand
Ea. Let’s solve for Ea. Subtract the first equation from the second equation to
eliminate A:
ln6.0×10−3−ln2.0×10−3=Ea
8.314 1
308 −1
298
Step 5: Simplify and solve for Ea:
ln 6.0×10−3
2.0×10−3=Ea
8.314 1
308 −1
298
Ea= 12080 J/mol ≈12.1 kJ/mol
Step 6: Therefore, the activation energy for the reaction is approximately
12.1 kJ/mol.
Question 2
Question
For a reaction at 25◦C, the rate constant kis found to be 5.75×10−3s−1. When
the temperature is increased to 50◦C, the rate constant increases to 1.15 ×10−2
s−1. Determine the activation energy of the reaction in kJ/mol.
Solution
Step 1: Write down the Arrhenius equation relating the rate constant kto the
temperature T, pre-exponential factor A, and activation energy Ea.
k=A×e−Ea
RT
Step 2: Write down the Arrhenius equation for the two different tempera-
tures given in the question.
At 25◦C (298K):
k1=A×e−Ea
R×298
5.75 ×10−3=A×e−Ea
8.314×298
At 50◦C (323K):
k2=A×e−Ea
R×323
2
1.15 ×10−2=A×e−Ea
8.314×323
Step 3: Divide the equation at 50◦C by the equation at 25◦C to eliminate
A.
k2
k1
=e−Ea
8.314×323
e−Ea
8.314×298
1.15 ×10−2
5.75 ×10−3=e−Ea
8.314 (1
323 −1
298 )
Step 4: Solve for the activation energy Ea.
1.15 ×10−2
5.75 ×10−3=e2692.5Ea
8.314×323×298
2 = e2692.5Ea
8.314×323×298
Step 5: Take the natural logarithm of both sides to solve for Ea.
ln(2) = 2692.5Ea
8.314 ×323 ×298
Ea=−ln(2) ×8.314 ×323 ×298
2692.5≈90.4 kJ/mol
Therefore, the activation energy of the reaction is approximately 90.4 kJ/mol.
Question 3
Question
The rate constant for a certain reaction is found to be 1.20 ×10−3s−1at 25◦C
and 3.85 ×10−2s−1at 55◦C. 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
T2
−1
T1
where: - k1and k2are the rate constants at temperatures T1and T2, - Eais the
activation energy, - Ris the ideal gas constant, - T1and T2are the temperatures
in Kelvin.
Step 2: Substituting the given values into the equation, we get:
3.85 ×10−2
1.20 ×10−3= exp −Ea
R1
328 −1
298
Step 3: Solving for Ea, we have:
3
ln 3.21 ×101
1.20 ×10−3=−Ea
8.314 1
328 −1
298
Step 4: Simplifying the equation further:
ln 3.21 ×101
1.20 ×10−3=−Ea
8.314 1
328 −1
298
Step 5: Finally, solving for Ea, we get:
Ea=−8.314 ×10−3(30.2) ln 3.21 ×101
1.20 ×10−3
Therefore, the activation energy for this reaction is ≈72.5 kJ/mol.
Question 4
Question
The rate constant for a reaction is found to be 5.0×10−3s−1at 25◦C and
3.0×10−2s−1at 100◦C. Calculate the activation energy of the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J K−1mol−1), T= temperature in Kelvin.
Step 2: Use the given data to form two equations: At 25◦C:
5.0×10−3=A·e−Ea
8.314×(25+273)
At 100◦C:
3.0×10−2=A·e−Ea
8.314×(100+273)
Step 3: Take the ratio of the two equations to eliminate A:
5.0×10−3
3.0×10−2=e−Ea
8.314×(25+273)
e−Ea
8.314×(100+273)
Step 4: Simplify the ratio and solve for Ea:
5
30 =e−Ea
8.314×298 −e−Ea
8.314×373
Step 5: Solve for Eato find the activation energy of the reaction.
4
Question 5
Question
The rate constant for the reaction A →B is found to be 1.25×10−3s−1at 25◦C
and 4.68 ×10−3s−1at 35◦C. Calculate the activation energy for this reaction.
Solution
Let’s use the Arrhenius equation to find the activation energy (Ea) for the
reaction:
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.
We are given two sets of data: At 25◦C (298 K): k1= 1.25 ×10−3s−1At
35◦C (308 K): k2= 4.68 ×10−3s−1
Step 1: Write the Arrhenius equation using the given data points.
k1=A·e−Ea
R·298
k2=A·e−Ea
R·308
Step 2: Divide the two equations to eliminate A.
k2
k1
=e−Ea
R·308
e−Ea
R·298
Step 3: Simplify the equation.
k2
k1
=eEa
R·(1
298 −1
308 )
Step 4: Solve for Ea.
ln k2
k1=Ea
R·1
298 −1
308
Step 5: Calculate Ea.
Ea=R·1
298 −1
308−1
·ln k2
k1
Now, plug in the given values for R,k1, and k2to find the activation energy.
5
Question 6
Question
The rate constant for a reaction is found to be 4.63 ×10−3s−1at a temperature
of 320 K, while at a temperature of 360 K the rate constant is 1.42 ×10−2s−1.
Assuming the frequency factor (A) is 5.12 ×1012 s−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 frequency factor - Eais the activation
energy - Ris the gas constant - Tis the temperature in Kelvin
Step 2: Write down the Arrhenius equation at the two temperatures given:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 3: Take the ratio of the two equations:
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 4: Simplify the ratio:
k2
k1
=eEa
R·1
T1
−1
T2
Step 5: Plug in the values given in the question:
1.42 ×10−2
4.63 ×10−3=e(Ea
8.314 ·(1
320 −1
360 ))
Step 6: Solve for Ea:
1.42 ×10−2
4.63 ×10−3=e(Ea
8.314 ·(1
320 −1
360 ))
3.065 = e(Ea
8.314 ·(1
320 −1
360 ))
Step 7: Solve for Eausing natural logarithms:
ln(3.065) = Ea
8.314 ·1
320 −1
360
6
Ea= 8.314 ×ln(3.065) ×1
320 −1
360
Step 8: Calculate Ea:
Ea= 8.314 ×ln(3.065) ×1
320 −1
360
Ea≈46.9 kJ/mol
Therefore, the activation energy for this reaction is approximately 46.9
kJ/mol.
Question 7
Question
The rate constant for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When
the temperature is raised to 45
°
C, the rate constant becomes 7.86 ×10−2s−1.
Calculate the activation energy for this reaction in kJ/mol.
Solution
Step 1: Write the Arrhenius equation:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/mol·K), - Tis the temperature
in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
Ea=−ln k
A
1
R·1
T
Step 3: Convert the given temperatures to Kelvin. - At 25
°
C: T1= 25+273 =
298 K - At 45
°
C: T2= 45 + 273 = 318 K
Step 4: Plug in the values for the rate constants and temperatures into the
Arrhenius equation to find the activation energy:
Ea=−
ln 7.86×10−2
1.25×10−3
1
8.314 ·1
318
Step 5: Calculate the activation energy using a calculator:
Ea≈55.8 kJ/mol
Therefore, the activation energy for this reaction is approximately 55.8
kJ/mol.
7
Question 8
Question
The rate constant of a reaction is found to double when the temperature is
increased from 300 K to 330 K. Calculate the activation energy for this reaction.
(Assume the pre-exponential factor A remains constant.)
Solution
Step 1: Write down the Arrhenius equation The Arrhenius equation relates the
rate constant of a reaction to the temperature and the activation energy:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/(mol
·
K)), and - Tis the
temperature in Kelvin.
Step 2: Given data We are given: - Temperature T1= 300 K, - Temperature
T2= 330 K, - Rate constant k1at T1(initial temperature), - Rate constant k2
at T2(final temperature).
Step 3: Use the given data to find the activation energy Given that the rate
constant doubles when the temperature is increased from 300 K to 330 K, we
can express this relationship using the Arrhenius equation as:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
=e
Ea
R1
T1
−1
T2
Substitute the given values:
2k1
k1
=eEa
R(1
300 −1
330 )
Solve for Ea:
k2
k1
=eEa
R(1
300 −1
330 )
ln 2
1=Ea
R1
300 −1
330
ln 2 = Ea
R1
300 −1
330
Ea=Rln 2
1/300 −1/330
Step 4: Calculate the activation energy Now, we can substitute the values
for Rand calculate the activation energy:
Ea= 8.314 J/(mol
·
K) ln 2
1/300 −1/330
8
Question 9
Question
The rate constant for a certain reaction is known to be 1.08 ×10−3s−1at 25
°
C.
When the temperature is increased to 75
°
C, the rate constant is found to be
6.82 ×10−2s−1. 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.
At 25
°
C: T1= 25C+ 273.15 = 298.15K
At 75
°
C: T2= 75C+ 273.15 = 348.15K
Step 2: Use the Arrhenius equation, which relates the rate constant kto
temperature and activation energy:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of the Arrhenius equation to simplify
the expression:
ln(k) = ln(A)−Ea
R·1
T
Step 4: Set up two equations using the rate constants at each temperature
and solve for Ea. Given that the gas constant R= 8.314 J/(mol ·K):
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Calculate the activation energy Eausing the two equations:
ln1.08 ×10−3= ln(A)−Ea
8.314 ·1
298.15
ln6.82 ×10−2= ln(A)−Ea
8.314 ·1
348.15
Step 6: Solve the equations simultaneously to find the activation energy Ea
for the reaction.
Question 10
Question
The rate constant for a reaction is found to be 5.75 ×10−3s−1at 25◦C. When
the temperature is increased to 65◦C, the rate constant is found to be 0.121 s−1.
Calculate the activation energy for this reaction.
9
Solution
Step 1: Convert temperatures to Kelvin using the relationship T(K) = T(C) +
273.15. At 25◦C, T= 25 + 273.15 = 298.15 K. At 65◦C, T= 65 + 273.15 =
338.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants to the acti-
vation energy. The Arrhenius equation is given by: k=Ae−Ea
RT , where kis the
rate constant, Ais the pre-exponential factor, Eais the activation energy, Ris
the gas constant (8.314 J mol−1K−1), and Tis the temperature in Kelvin.
For the reaction at 25◦C: 5.75 ×10−3=Ae−Ea
8.314×298.15 (Equation 1)
For the reaction at 65◦C: 0.121 = Ae−Ea
8.314×338.15 (Equation 2)
Step 3: Take the ratio of Equation 2 to Equation 1 to eliminate A.
0.121
5.75×10−3=eEa
8.314 (1
338.15 −1
298.15 )
Solve for Eato find the activation energy for the reaction.
Question 11
Question
The rate constant for a reaction at 25◦C is 9.0×10−4s−1. When the temperature
is raised to 50◦C, the rate constant becomes 3.6×10−3s−1. Calculate the
activation energy (in kJ/mol) for the reaction.
Solution
Step 1: Determine the activation energy using 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: Convert the rate constants and temperatures to Kelvin.
T1= 25 + 273 = 298 K
T2= 50 + 273 = 323 K
Step 3: Set up two equations using the data provided:
9.0×10−4=A×e−Ea
8.314×298
3.6×10−3=A×e−Ea
8.314×323
Step 4: Divide the second equation by the first to eliminate A:
3.6×10−3
9.0×10−4=e−Ea
8.314×323
e−Ea
8.314×298
10
Step 5: Simplify the equation using properties of exponents and solve for
Ea.
4 = e−Ea
8.314×323 +Ea
8.314×298
4 = eEa(1
8.314×298 −1
8.314×323 )
Step 6: Solve for Ea:
Ea=−8.314 ×298 ×323 ×ln 1
4
Step 7: Calculate the activation energy in kJ/mol.
Ea=−8.314 ×298 ×323 ×ln(0.25) ≈64.3 kJ/mol
Question 12
Question
The rate constant for a reaction at 25
°
C is 1.23 ×10−2s−1, and the activation
energy for the reaction is 75 kJ/mol. Calculate the rate constant at 35
°
C.
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol. Given: Activation
energy (Ea) = 75 kJ/mol Conversion factor: 1 kJ = 103J
Activation energy = 75 ×103J/mol = 7.5×104J/mol
Step 2: Calculate the rate constant at 35
°
C using the Arrhenius equation:
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 kat 25
°
C (k25)=1.23 ×10−2s−1Ea= 7.5 ×104J/mol T25C= 25
°
C
= 25 + 273 = 298 K T35C= 35
°
C = 35 + 273 = 308 K
Let’s first find the pre-exponential factor A:
k25 =A×e−Ea
R×T25
1.23 ×10−2=A×e−7.5×104
8.314×298
1.23 ×10−2=A×e−31.73
A=1.23 ×10−2
e−31.73
11
A≈1.07 ×1033
Step 3: Calculate the rate constant at 35
°
C using the Arrhenius equation:
k35 =A×e−Ea
R×T35
k35 = 1.07 ×1033 ×e−7.5×104
8.314×308
k35 ≈2.78 ×10−2s−1
Therefore, the rate constant at 35
°
C is approximately 2.78 ×10−2s−1.
Question 13
Question
For the reaction
2H2O2(aq)→2H2O(l)+O2(g)
the rate constant at 25
°
C is 1.20 ×10−4s−1. At what temperature will the rate
constant be twice as large if the activation energy is 75 kJ/mol?
Solution
Step 1: We can use the Arrhenius equation to determine the relationship be-
tween the rate constant (k2) at a different temperature (T2) compared to the
rate constant (k1) at the initial temperature (T1). 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.
Step 2: Let T1= 25C= 25 + 273 = 298 K, k1= 1.20 ×10−4s−1, and
T2be the temperature we need to find where k2= 2k1= 2 ×1.20 ×10−4=
2.40 ×10−4s−1.
Step 3: To find T2, we set up the Arrhenius equation for both temperatures:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 4: We can divide the two equations to eliminate A:
k2
k1
=e−Ea
R·T2
e−Ea
R·T1
12
Step 5: Substitute the known values into the equation:
2.40 ×10−4
1.20 ×10−4=e−75000
8.314 1
T2
−1
298
Step 6: Solve for T2:
2 = e−75000
8.314 1
T2
−1
298
Step 7: Taking the natural logarithm of both sides:
ln(2) = −75000
8.314 1
T2
−1
298
Step 8: Solve for T2:
T2=75000
8.314 ·(ln(2) + 75000
8.314·298 )≈446 K
Therefore, the rate constant will be twice as large at approximately 446 K.
Question 14
Question
The rate constant for a reaction is found to be 2.5×10−3s−1at 298 Kand
8.0×10−3s−1at 333 K. Calculate the activation energy for this reaction.
Solution
Step 1: Understand the Arrhenius equation and how it relates to the rate con-
stants at different temperatures.
The Arrhenius equation relates the rate constant of a reaction to the tem-
perature and activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/(molK)), and - Tis the
temperature in Kelvin.
Step 2: Write down the given rate constants and temperatures.
Given: - k1= 2.5×10−3s−1at T1= 298 K, - k2= 8.0×10−3s−1at
T2= 333 K.
Step 3: Set up the Arrhenius equation with the two sets of data.
k1
k2
=A·e−Ea
R·T1
A·e−Ea
R·T2
Step 4: Simplify the equation and solve for the activation energy (Ea).
13
k1
k2
=eEa
R·1
T2
−1
T1
ln k1
k2=Ea
R·1
T2
−1
T1
Ea=R·
ln k1
k2
1
T2
−1
T1
Step 5: Substitute the given values and solve for the activation energy.
Ea= 8.314
ln 2.5×10−3
8.0×10−3
1
333 −1
298
Ea= 8.314 ln (0.3125)
0.0030
Ea= 8.314 −1.1568
0.0030
Ea≈ −321,741 J/mol
Question 15
Question
For a certain reaction, the rate constant at 300 K is 2.5×10−3s−1, and the
rate constant at 350 K is 8.0×10−3s−1. Calculate the activation energy (in
kJ/mol) for the reaction.
Solution
Step 1: 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 can set up two equations using the Arrhenius equation for the
rate constants at 300 K and 350 K:
k1=A·e−Ea
R·300
k2=A·e−Ea
R·350
14
where k1= 2.5×10−3s−1and k2= 8.0×10−3s−1.
Step 3: Taking the ratio of the two equations, we have:
k2
k1
=A·e−Ea
R·350
A·e−Ea
R·300
8.0×10−3
2.5×10−3=e−Ea
R(1
350 −1
300 )
Step 4: Simplifying the ratio on the left side gives:
3.2 = eEa
8.314 (1
300 −1
350 )
Step 5: Solving for the activation energy, we find:
ln(3.2) = Ea
8.314 1
300 −1
350
Ea= 8.314 ×10−3×1
300 −1
350−1
ln(3.2)
Step 6: Calculating the activation energy:
Ea≈47.9 kJ/mol
Therefore, the activation energy for the reaction is approximately 47.9 kJ/mol.
Question 16
Question
For a certain reaction, the rate constant at 35
°
C is 0.0023 s−1and at 50
°
C is
0.157 s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write the Arrhenius equation:
The Arrhenius equation is given by:
k=Ae−Ea
RT
Where: - kis the rate constant - Ais the pre-exponential factor - Eais the
activation energy (in J/mol) - Ris the gas constant (8.314 J/mol*K) - Tis the
temperature in Kelvin
Step 2: Calculate the activation energy:
Given data: - k1= 0.0023 s−1at T1= 35 ◦C = 308 K - k2= 0.157 s−1at
T2= 50 ◦C = 323 K
15
Using the Arrhenius equation for both cases:
k1=Ae−Ea
R·T1
k2=Ae−Ea
R·T2
Step 3: Take the ratio of the two equations:
Taking the ratio of the two Arrhenius equations:
k2
k1
=Ae−Ea
R·T2
Ae−Ea
R·T1
k2
k1
=e−Ea
R1
T2
−1
T1
Step 4: Solve for the activation energy:
Plugging in the values:
0.157
0.0023 =e−Ea
8.314 (1
323 −1
308 )
Solving for Eagives:
Ea≈97679 J/mol
Question 17
Question
The rate constant for a certain reaction at 25
°
C is 6.20 ×10−4s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.86 ×10−3s−1.
Calculate the activation energy for this 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 gas constant (8.31 J/mol-K), and Tis the temperature in Kelvin.
Step 2: We have two sets of data: At 25
°
C (298 K): k1= 6.20 ×10−4s−1At
50
°
C (323 K): k2= 1.86 ×10−3s−1
Step 3: Substitute the data into the Arrhenius equation: At 25
°
C:
6.20 ×10−4=A·e−Ea
8.31×298
At 50
°
C:
1.86 ×10−3=A·e−Ea
8.31×323
16
Step 4: Divide the two equations to eliminate A:
1.86 ×10−3
6.20 ×10−4=eEa
8.31 (1
323 −1
298 )
Step 5: Calculate the activation energy:
Ea=−8.31 ×ln 1.86 ×10−3
6.20 ×10−4×1
323 −1
298
Step 6: Solving the equation, we find:
Ea≈60.8 kJ/mol
Therefore, the activation energy for this reaction is approximately 60.8
kJ/mol.
Question 18
Question
The rate constant of a reaction is found to be 4.75 ×10−3s−1at 25◦C and
1.23 ×10−2s−1at 35◦C. Calculate the activation energy of the 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 the given temperatures to Kelvin:
T1= 25◦C + 273 = 298 K
T2= 35◦C + 273 = 308 K
Step 2: Insert the given rate constants and temperatures into the Arrhenius
equation to form two equations:
4.75 ×10−3=A×e(−Ea
R×298 )
1.23 ×10−2=A×e(−Ea
R×308 )
Step 3: Divide the second equation by the first equation to eliminate the
pre-exponential factor A:
1.23 ×10−2
4.75 ×10−3=e(−Ea
R×308 )
e(−Ea
R×298 )
1.23 ×10−2
4.75 ×10−3=e(Ea
R(1
298 −1
308 ))
17
Step 4: Solve the equation for the activation energy Eausing the value of
the gas constant, R= 8.314 J/(mol K):
1.23 ×10−2
4.75 ×10−3=e(8.314×Ea
8.314 ×1
298 −1
308 )
2.5895 = e(Ea
8.314 ×1
298 −1
308 )
Step 5: Take the natural logarithm of both sides to solve for Ea:
ln(2.5895) = Ea
8.314 ×1
298 −1
308
ln(2.5895) = Ea
8.314 ×1
298 −1
308
Step 6: Calculate the activation energy Ea:
Ea= 8.314 ×ln(2.5895)
1
298 −1
308
Ea≈53,620 J/mol
Therefore, the activation energy of the reaction is approximately 53.62 kJ/mol.
Question 19
Question
For the reaction
2A →B+C
the rate is found to double when the temperature is increased from 25
°
C to
35
°
C. Calculate the activation energy for this reaction. (Assume the frequency
factor, A, is constant)
Solution
Step 1: Given that the rate constant, k, is directly proportional to the temper-
ature, we can express this relationship using the Arrhenius equation:
k=Ae−Ea
RT
where k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant, and T= temperature in Kelvin.
Step 2: We are told that the rate doubles when the temperature is increased
from 25
°
C to 35
°
C, which means:
k2
k1
= 2
18
Using the Arrhenius equation, we can express the rates in terms of their corre-
sponding temperatures:
Ae−Ea
R(25+273.15)
Ae−Ea
R(35+273.15)
= 2
Simplifying this equation gives:
e−Ea
2984.75
e−Ea
4112.05
= 2
Step 3: To solve for the activation energy, we need to isolate Eain the above
equation. We can do this by taking the natural logarithm of both sides:
ln e−Ea
2984.75
e−Ea
4112.05 != ln(2)
Using the properties of logarithms, we simplify the left side to:
−Ea
2984.75 +Ea
4112.05 = ln(2)
Step 4: Further simplification yields the equation:
1127.3
2984.75Ea= ln(2)
From which we can solve for the activation energy, Ea:
Ea=2984.75 ·ln(2)
1127.3≈10.16 kJ/mol
Therefore, the activation energy for the reaction is approximately 10.16
kJ/mol.
Question 20
Question
The rate constant for a certain reaction at 25
°
C is 3.2×10−3s−1, and at 35
°
C
it is 1.3×10−2s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·exp −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.
19
Step 2: Rearrange the Arrhenius equation to solve for the activation energy
(Ea):
Ea=−RT ln(k/A)
ln(k/A)
Step 3: Convert the given temperatures to Kelvin (25
°
C = 298 K and 35
°
C
= 308 K).
Step 4: Plug in the values for the rate constants and temperatures into the
equation to solve for the activation energy:
Ea=−(8.314 J/(mol
·
K))(298 K) ln3.2×10−3/A
ln(3.2×10−3/A)
Step 5: Repeat the calculation for the other temperature:
Ea=−(8.314 J/(mol
·
K))(308 K) ln1.3×10−2/A
ln(1.3×10−2/A)
Step 6: Subtract the activation energy calculated at 298 K from the activa-
tion energy calculated at 308 K to find the overall activation energy.
Question 21
Question
The rate constant at 25
°
C for the reaction
2A →B
is 5.0×10−3s−1. When the temperature is increased to 35
°
C, the rate constant
is found to be 2.0×10−2s−1. Calculate the activation energy for this reaction.
Given: R= 8.314 J/(mol·K)
Solution
Step 1: Convert the temperatures to Kelvin.
T1= 25 + 273 = 298 K
T2= 35 + 273 = 308 K
Step 2: Use the Arrhenius equation to find the activation energy (Ea).
k1=A×e−Ea
RT1
k2=A×e−Ea
RT2
20
Step 3: Divide the second equation by the first to eliminate A.
k2
k1
=A×e−Ea
RT2
A×e−Ea
RT1
k2
k1
=e−Ea
R1
T1
−1
T2
Step 4: Substitute the given values and solve for Ea.
2.0×10−2
5.0×10−3=e−Ea
8.314 (1
298 −1
308 )
4 = e−Ea
8.314 ×0.0034
ln(4) = −Ea
8.314 ×0.0034
Ea=−8.314 ×0.0034
ln(4)
Ea= 48.97 kJ/mol
Therefore, the activation energy for the reaction is 48.97 kJ/mol.
Question 22
Question
The rate constant, k, for a certain reaction is found to be 6.0×10−3s−1at
30
°
C and 4.0×10−2s−1at 50
°
C. Determine the activation energy, Ea, for this
reaction.
Solution
Step 1: Write the Arrhenius equation relating the rate constant, temperature,
and activation energy.
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·K)), and - Tis the
absolute temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin. - 30◦C = 30 + 273 = 303 K -
50◦C = 50 + 273 = 323 K
Step 3: Substitute the given data points into the Arrhenius equation to form
two equations.
6.0×10−3=A·exp −Ea
8.314×303
4.0×10−2=A·exp −Ea
8.314×323
21
Step 4: Divide the two equations to eliminate A.
6.0×10−3
4.0×10−2=
exp −Ea
8.314×303
exp −Ea
8.314×323
Step 5: Simplify the equation.
0.15 = exp Ea
8.314 1
303 −1
323
Step 6: Solve for the activation energy, Ea.
Ea=−8.314 ·ln(0.15) ·1
303 −1
323
Step 7: Calculate the activation energy.
Ea≈64.4 kJ/mol
Question 23
Question
The rate constant of a certain reaction has been measured at two different
temperatures, T1and T2, with values of k1= 2.5×10−3s−1at T1= 300 K and
k2= 3.6×10−2s−1at T2= 350 K. Calculate the activation energy for this
reaction in kJ/mol.
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), T= temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
Ea=−R
ln k
A·1
T
Step 3: Calculate the activation energy using the given data: At T1= 300 K,
k1= 2.5×10−3s−1. At T2= 350 K, k2= 3.6×10−2s−1.
Step 4: Calculate Ea:
Ea=−8.314
ln 3.6×10−2
2.5×10−3·1
350 −1
300
22
Ea≈ − 8.314
ln(14.4) ·1
350 −1
300
Step 5: Perform the calculations to find the activation energy Eain J/mol.
Convert the result to kJ/mol for convenience.
Step 6: Therefore, the activation energy for this reaction is approximately
95.7 kJ/mol .
Question 24
Question
A certain reaction has an activation energy of 75 kJ/mol. When the temperature
is increased from 25
°
C to 45
°
C, the rate constant increases by a factor of 5.
Calculate the rate constant at 25
°
C and the rate constant at 45
°
C for this
reaction.
Solution
Step 1: We know that the Arrhenius equation is given by:
k=Ae−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: Let’s denote the rate constant at 25
°
C as k1and at 45
°
C as k2. The
rate constant increases by a factor of 5 when the temperature increases from
25
°
C to 45
°
C. Therefore, we can write:
k2= 5k1
Step 3: Converting the temperatures to Kelvin:
T1= 25C= 25 + 273 = 298K
T2= 45C= 45 + 273 = 318K
Step 4: Substituting the values into the Arrhenius equation for k1and k2:
For k1at 25
°
C:
k1=Ae−75000
8.314×298
For k2at 45
°
C:
k2= 5k1= 5Ae−75000
8.314×318
Step 5: Now we can solve for k1and k2. From the given information, we
know that k2= 5k1. Therefore, substitute the expressions for k1and k2into
this equation and solve for A.
5Ae−75000
8.314×318 = 5Ae−75000
8.314×298
Step 6: After solving for A, substitute the value back into the expressions
for k1and k2to find the rate constants.
23
Question 25
Question
The rate constant for the decomposition of compound A is found to be 6.32 ×
10−4s−1at 65◦C and 2.54 ×10−3s−1at 75◦C. Calculate the activation energy
for the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. For 65◦C:
T= 65 + 273.15 = 338.15 K
For 75◦C:
T= 75 + 273.15 = 348.15 K
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation:
k=A·e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the
activation energy, Ris the gas constant (8.314 J ·mol−1·K−1), and Tis the
temperature in Kelvin.
We can rearrange the equation in the following form:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the values into the equation:
ln 2.54 ×10−3
6.32 ×10−4=−Ea
8.314 1
348.15 −1
338.15
Step 4: Solve for Ea:
Ea=−8.314 ×
ln 2.54×10−3
6.32×10−4
1
348.15 −1
338.15
Ea≈53320 J/mol
Therefore, the activation energy for the reaction is approximately 53.32 kJ/mol.
Question 26
Question
The rate constant for the reaction H2O2→H2O+O2at 25◦Cis 1.20×10−4s−1.
When the temperature is increased to 35◦C, the rate constant becomes 5.80 ×
10−3s−1. Calculate the activation energy for this reaction.
24
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at
different temperatures:
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: First, let’s convert the temperatures to Kelvin: - 25◦C= 25 +273 =
298 K - 35◦C= 35 + 273 = 308 K
Step 3: We can set up two equations using the rate constants provided:
1.20 ×10−4=A·e−Ea
8.314·298
5.80 ×10−3=A·e−Ea
8.314·308
Step 4: Now we can divide the two equations to eliminate A:
5.80 ×10−3
1.20 ×10−4=e−Ea
8.314·308
e−Ea
8.314·298
Step 5: Simplifying, we get:
48.33 = eEa
8.314 (1
298 −1
308 )
Step 6: Further simplifying, we find:
ln(48.33) = Ea
8.314 1
298 −1
308
Step 7: Solving for Ea, we have:
Ea= 8.314 ×1
298 −1
308×ln(48.33)
After calculations, we can find the activation energy for this reaction.
Question 27
Question
The rate constant for a certain reaction is found to be 2.45 ×10−3s−1at 25
°
C
and 1.53 ×10−2s−1at 35
°
C. Determine the activation energy for the reaction
in kJ/mol.
25
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−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 K).
Step 2: Take the logarithm of the Arrhenius equation to simplify the calcu-
lation:
ln(k) = ln(A)−Ea
RT
Step 3: Rearrange the equation to solve for Ea:
Ea
R=1
T1
−1
T2
Step 4: Convert the given temperatures to Kelvin:
T1= 25C+ 273.15 = 298.15K
T2= 35C+ 273.15 = 308.15K
Step 5: Substitute the values into the equation and solve for Ea:
Ea
8.314 =1
298.15 −1
308.15
Ea= 8.314 ×1
298.15 −1
308.15
Ea≈56184.15 J/mol
Step 6: Convert the activation energy to kilojoules per mole:
Ea≈56184.15 J
1000 kJ/mol
Ea≈56.18 kJ/mol
Therefore, the activation energy for the reaction is approximately 56.18
kJ/mol.
Question 28
Question
The rate constant of a certain reaction doubles when the temperature is raised
from 25
°
C to 35
°
C. Calculate the activation energy (in kJ/mol) for this reaction.
Assume the frequency factor is 1.2×1013 s−1.
26
Solution
Step 1: Calculate the temperature values in Kelvin. Given: T1= 25◦C =
25 + 273 = 298 K T2= 35◦C = 35 + 273 = 308 K
Step 2: Using the Arrhenius equation, we have:
k2
k1
= exp −Ea
R1
T2
−1
T1
Given that k2
k1= 2 and R= 8.314 J/mol K, we can rewrite the equation as:
2 = exp −Ea
8.314 1
308 −1
298
Step 3: Simplify the equation:
2 = exp −Ea
8.314 10
298 ×308
Step 4: Solve for the activation energy Ea:
ln(2) = −Ea
8.314 ×10
298 ×308
Ea=−8.314 ×10
298 ×308 ×ln(2)
Step 5: Calculate the activation energy Ea:
Ea≈ − 8.314 ×10
298 ×308 ×ln(2)
Ea≈56.31 kJ/mol
Therefore, the activation energy for this reaction is approximately 56.31
kJ/mol.
Question 29
Question
The rate constant for a reaction at 25
°
C is 3.42 ×10−4s−1. When the tempera-
ture is increased to 55
°
C, the rate constant becomes 1.26 ×10−2s−1. Calculate
the activation energy for this reaction.
27
Solution
Step 1: 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=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 can set up two equations based on the given information:
3.42 ×10−4=Ae−Ea
(8.314)(298)
1.26 ×10−2=Ae−Ea
(8.314)(328)
Step 3: Divide the two equations to eliminate A:
3.42 ×10−4
1.26 ×10−2=e−Ea
(8.314)(298)
e−Ea
(8.314)(328)
Step 4: Simplify the above expression:
e
Ea
(8.314)(328) −Ea
(8.314)(298) =3.42 ×10−4
1.26 ×10−2
Step 5: Solve for Eaby taking natural logarithm of both sides:
Ea
(8.314) 1
328 −1
298= ln 3.42 ×10−4
1.26 ×10−2
Step 6: Calculate the activation energy, Ea:
Ea= 8.314 ×
ln 3.42×10−4
1.26×10−2
1
328 −1
298
Therefore, the activation energy for this reaction is approximately 47.8
kJ/mol.
Question 30
Question
The rate constant (k) for a certain reaction is found to be 1.5×10−4s−1at 25
°
C
and 3.2×10−3s−1at 35
°
C. Calculate the activation energy for this reaction.
28
Solution
Step 1: Write down the Arrhenius equation. The Arrhenius equation relates
the rate constant (k) of a reaction to the temperature (T) and the activation
energy (Ea).
k=Ae−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.
Step 2: Identify the two sets of data given. We are given two sets of data:
At 25
°
C: k1= 1.5×10−4s−1T1= 25 + 273 = 298 K
At 35
°
C: k2= 3.2×10−3s−1T2= 35 + 273 = 308 K
Step 3: Plug the data into the Arrhenius equation. We can set up two
equations using the Arrhenius equation:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 4: Take the ratio of the two equations. Dividing the two equations will
eliminate A:
k1
k2
=e−Ea
RT1
e−Ea
RT2
Step 5: Simplify the expression.
k1
k2
=e
Ea
R1
T2
−1
T1
Step 6: Solve for Ea. Using the known values k1,k2,T1,T2, and R, we can
now solve for Ea:
1.5×10−4
3.2×10−3=eEa
8.314 (1
308 −1
298 )
0.046875 = eEa
8.314 (1
308 −1
298 )
ln(0.046875) = Ea
8.314 1
308 −1
298
Ea= 2.83 kJ/mol
Therefore, the activation energy for this reaction is 2.83 kJ/mol.
Question 31
Question
The rate constant of a certain reaction is found to be 5.00 ×10−3s−1at 25
°
C
and 1.00 ×10−2s−1at 35
°
C. Calculate the activation energy of the reaction.
Given: R= 8.314 J/(mol·K), T1= 25
°
C, T2= 35
°
C.
29
Solution
Step 1: Convert the given temperatures from Celsius to Kelvin.
T1= 25 + 273 = 298 K
T2= 35 + 273 = 308 K
Step 2: Use the Arrhenius equation to relate the rate constants at different
temperatures:
k2=k1·e(−Ea
R(1
T2
−1
T1))
Step 3: Substitute the given rate constants and temperatures into the equa-
tion and solve for the activation energy Ea.
1.00 ×10−2= 5.00 ×10−3·e(−Ea
8.314 (1
308 −1
298 ))
Step 4: Solve for Eaby taking the natural logarithm of both sides and
rearranging the terms.
ln1.00 ×10−2
5.00 ×10−3=−Ea
8.314(1
308 −1
298)
Step 5: Calculate the activation energy Eaby plugging in the known values
and solving for it.
Ea=−8.314 ·ln1.00 ×10−2
5.00 ×10−3·1
308 −1
298
Ea≈30.4 kJ/mol
Therefore, the activation energy of the reaction is approximately 30.4 kJ/mol.
Question 32
Question
The rate constant, k, for a certain reaction is found to be 8.21 ×10−4s−1at
300 K and 3.12 ×10−3s−1at 325 K. Calculate the activation energy for the
reaction.
Solution
Step 1: We can use the Arrhenius equation to relate 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.
30
Step 2: Taking the natural logarithm of both sides of the Arrhenius equation
gives:
ln(k) = −Ea
R·1
T+ ln(A)
Step 3: We will use the given data at two temperatures (300 K and 325 K)
to set up a system of equations:
(ln8.21 ×10−4=−Ea
8.314 ·1
300 + ln(A)
ln3.12 ×10−3=−Ea
8.314 ·1
325 + ln(A)
Step 4: Solve the system of equations and find the values of Eaand ln(A).
Step 5: With the values of Eaand ln(A), calculate the activation energy Ea
using one of the equations.
Step 6: Calculate the activation energy Eafor the reaction.
Question 33
Question
The rate constant (k) for a certain reaction is found to double when the tem-
perature is increased from 25
°
C to 35
°
C. Calculate the activation energy (Ea)
for this reaction. (R = 8.314 J/mol·K)
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant (k) to
temperature and activation energy:
k=A·e−Ea
RT
where Ais the pre-exponential factor, Eais the activation energy, Ris the gas
constant, and Tis the temperature in Kelvin.
Step 2: We are given that the rate constant doubles when the temperature
is increased from 25
°
C (298 K) to 35
°
C (308 K). This means that:
k2= 2k1
A·e−Ea
R·308 = 2(A·e−Ea
R·298 )
Step 3: Divide the second equation by the first equation to eliminate A:
e−Ea
R·308
e−Ea
R·298
= 2
e−Ea
R·308 +Ea
R·298 = 2
eEa
R(1
298 −1
308 )= 2
31
Step 4: Solve for Ea:
Ea
R1
298 −1
308= ln(2)
Ea
8.314 1
298 −1
308= ln(2)
Ea= 8.314 ×ln(2) ×1
1
298 −1
308
Ea≈36590.56 J/mol
Therefore, the activation energy for this reaction is approximately 36.59
kJ/mol.
Question 34
Question
The reaction between nitrogen dioxide gas (NO2) and carbon monoxide gas
(CO) is believed to follow the following rate law at a certain temperature:
Rate = k[NO2][CO]2
Given that the activation energy for this reaction is 120 kJ/mol and the
pre-exponential factor is 1.0×1010 s−1, calculate the rate constant (in s−1) at a
temperature where the rate of reaction is observed to be 3.5×10−3mol L−1s−1
when the concentrations of NO2 and CO are both 0.10 mol L−1.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to the
rate of reaction and the temperature. The Arrhenius equation is given by:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J mol−1K−1), - Tis the
temperature in Kelvin.
Given that A= 1.0×1010 s−1and Ea= 120 kJ mol−1, we can substitute
these values into the Arrhenius equation.
Step 2: Rearranging the Arrhenius equation to solve for T, we have:
T=−Ea
Rln k
A
Step 3: Given that the rate of reaction is observed to be 3.5×10−3mol L−1s−1
and the concentrations of NO2 and CO are both 0.10 mol L−1, we can substitute
32
these values along with the rate constant into the rate law equation to solve for
k.
3.5×10−3=k(0.10)(0.10)2
Step 4: Solve for kin the rate law equation:
k=3.5×10−3
0.001 = 3.5 s−1
Step 5: Finally, substitute the calculated value of kinto the Arrhenius equa-
tion to solve for the temperature T.
Question 35
Question
The rate constant of a reaction is measured at two different temperatures, T1and
T2, and found to be k1= 5.0×10−3s−1at T1= 300 K and k2= 2.0×10−2s−1
at T2= 325 K. Calculate the activation energy of the reaction.
Solution
Step 1: The Arrhenius equation relates the rate constant of a reaction to tem-
perature 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 2: We can take the ratio of the Arrhenius equation at two different
temperatures:
k2
k1
=A e−Ea
R T2
A e−Ea
R T1
Step 3: Substitute the given values:
2.0×10−2
5.0×10−3=e−Ea
8.314×325
e−Ea
8.314×300
Step 4: Simplify the equation:
4 = e−Ea
2651.1×eEa
2494.2
Step 5: Combine the exponents:
4 = e−Ea
2651.1+Ea
2494.2
Step 6: Solve for the activation energy Ea:
33
Question 5
Question
The rate constant for the reaction A →B is found to be 1.25×10−3s−1at 25◦C
and 4.68 ×10−3s−1at 35◦C. Calculate the activation energy for this reaction.
Solution
Let’s use the Arrhenius equation to find the activation energy (Ea) for the
reaction:
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.
We are given two sets of data: At 25◦C (298 K): k1= 1.25 ×10−3s−1At
35◦C (308 K): k2= 4.68 ×10−3s−1
Step 1: Write the Arrhenius equation using the given data points.
k1=A·e−Ea
R·298
k2=A·e−Ea
R·308
Step 2: Divide the two equations to eliminate A.
k2
k1
=e−Ea
R·308
e−Ea
R·298
Step 3: Simplify the equation.
k2
k1
=eEa
R·(1
298 −1
308 )
Step 4: Solve for Ea.
ln k2
k1=Ea
R·1
298 −1
308
Step 5: Calculate Ea.
Ea=R·1
298 −1
308−1
·ln k2
k1
Now, plug in the given values for R,k1, and k2to find the activation energy.
5
Question 6
Question
The rate constant for a reaction is found to be 4.63 ×10−3s−1at a temperature
of 320 K, while at a temperature of 360 K the rate constant is 1.42 ×10−2s−1.
Assuming the frequency factor (A) is 5.12 ×1012 s−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 frequency factor - Eais the activation
energy - Ris the gas constant - Tis the temperature in Kelvin
Step 2: Write down the Arrhenius equation at the two temperatures given:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 3: Take the ratio of the two equations:
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 4: Simplify the ratio:
k2
k1
=eEa
R·1
T1
−1
T2
Step 5: Plug in the values given in the question:
1.42 ×10−2
4.63 ×10−3=e(Ea
8.314 ·(1
320 −1
360 ))
Step 6: Solve for Ea:
1.42 ×10−2
4.63 ×10−3=e(Ea
8.314 ·(1
320 −1
360 ))
3.065 = e(Ea
8.314 ·(1
320 −1
360 ))
Step 7: Solve for Eausing natural logarithms:
ln(3.065) = Ea
8.314 ·1
320 −1
360
6
Ea= 8.314 ×ln(3.065) ×1
320 −1
360
Step 8: Calculate Ea:
Ea= 8.314 ×ln(3.065) ×1
320 −1
360
Ea≈46.9 kJ/mol
Therefore, the activation energy for this reaction is approximately 46.9
kJ/mol.
Question 7
Question
The rate constant for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When
the temperature is raised to 45
°
C, the rate constant becomes 7.86 ×10−2s−1.
Calculate the activation energy for this reaction in kJ/mol.
Solution
Step 1: Write the Arrhenius equation:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/mol·K), - Tis the temperature
in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
Ea=−ln k
A
1
R·1
T
Step 3: Convert the given temperatures to Kelvin. - At 25
°
C: T1= 25+273 =
298 K - At 45
°
C: T2= 45 + 273 = 318 K
Step 4: Plug in the values for the rate constants and temperatures into the
Arrhenius equation to find the activation energy:
Ea=−
ln 7.86×10−2
1.25×10−3
1
8.314 ·1
318
Step 5: Calculate the activation energy using a calculator:
Ea≈55.8 kJ/mol
Therefore, the activation energy for this reaction is approximately 55.8
kJ/mol.
7
Question 8
Question
The rate constant of a reaction is found to double when the temperature is
increased from 300 K to 330 K. Calculate the activation energy for this reaction.
(Assume the pre-exponential factor A remains constant.)
Solution
Step 1: Write down the Arrhenius equation The Arrhenius equation relates the
rate constant of a reaction to the temperature and the activation energy:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/(mol
·
K)), and - Tis the
temperature in Kelvin.
Step 2: Given data We are given: - Temperature T1= 300 K, - Temperature
T2= 330 K, - Rate constant k1at T1(initial temperature), - Rate constant k2
at T2(final temperature).
Step 3: Use the given data to find the activation energy Given that the rate
constant doubles when the temperature is increased from 300 K to 330 K, we
can express this relationship using the Arrhenius equation as:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
=e
Ea
R1
T1
−1
T2
Substitute the given values:
2k1
k1
=eEa
R(1
300 −1
330 )
Solve for Ea:
k2
k1
=eEa
R(1
300 −1
330 )
ln 2
1=Ea
R1
300 −1
330
ln 2 = Ea
R1
300 −1
330
Ea=Rln 2
1/300 −1/330
Step 4: Calculate the activation energy Now, we can substitute the values
for Rand calculate the activation energy:
Ea= 8.314 J/(mol
·
K) ln 2
1/300 −1/330
8
Question 9
Question
The rate constant for a certain reaction is known to be 1.08 ×10−3s−1at 25
°
C.
When the temperature is increased to 75
°
C, the rate constant is found to be
6.82 ×10−2s−1. 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.
At 25
°
C: T1= 25C+ 273.15 = 298.15K
At 75
°
C: T2= 75C+ 273.15 = 348.15K
Step 2: Use the Arrhenius equation, which relates the rate constant kto
temperature and activation energy:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of the Arrhenius equation to simplify
the expression:
ln(k) = ln(A)−Ea
R·1
T
Step 4: Set up two equations using the rate constants at each temperature
and solve for Ea. Given that the gas constant R= 8.314 J/(mol ·K):
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Calculate the activation energy Eausing the two equations:
ln1.08 ×10−3= ln(A)−Ea
8.314 ·1
298.15
ln6.82 ×10−2= ln(A)−Ea
8.314 ·1
348.15
Step 6: Solve the equations simultaneously to find the activation energy Ea
for the reaction.
Question 10
Question
The rate constant for a reaction is found to be 5.75 ×10−3s−1at 25◦C. When
the temperature is increased to 65◦C, the rate constant is found to be 0.121 s−1.
Calculate the activation energy for this reaction.
9
Solution
Step 1: Convert temperatures to Kelvin using the relationship T(K) = T(C) +
273.15. At 25◦C, T= 25 + 273.15 = 298.15 K. At 65◦C, T= 65 + 273.15 =
338.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants to the acti-
vation energy. The Arrhenius equation is given by: k=Ae−Ea
RT , where kis the
rate constant, Ais the pre-exponential factor, Eais the activation energy, Ris
the gas constant (8.314 J mol−1K−1), and Tis the temperature in Kelvin.
For the reaction at 25◦C: 5.75 ×10−3=Ae−Ea
8.314×298.15 (Equation 1)
For the reaction at 65◦C: 0.121 = Ae−Ea
8.314×338.15 (Equation 2)
Step 3: Take the ratio of Equation 2 to Equation 1 to eliminate A.
0.121
5.75×10−3=eEa
8.314 (1
338.15 −1
298.15 )
Solve for Eato find the activation energy for the reaction.
Question 11
Question
The rate constant for a reaction at 25◦C is 9.0×10−4s−1. When the temperature
is raised to 50◦C, the rate constant becomes 3.6×10−3s−1. Calculate the
activation energy (in kJ/mol) for the reaction.
Solution
Step 1: Determine the activation energy using 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: Convert the rate constants and temperatures to Kelvin.
T1= 25 + 273 = 298 K
T2= 50 + 273 = 323 K
Step 3: Set up two equations using the data provided:
9.0×10−4=A×e−Ea
8.314×298
3.6×10−3=A×e−Ea
8.314×323
Step 4: Divide the second equation by the first to eliminate A:
3.6×10−3
9.0×10−4=e−Ea
8.314×323
e−Ea
8.314×298
10
Step 5: Simplify the equation using properties of exponents and solve for
Ea.
4 = e−Ea
8.314×323 +Ea
8.314×298
4 = eEa(1
8.314×298 −1
8.314×323 )
Step 6: Solve for Ea:
Ea=−8.314 ×298 ×323 ×ln 1
4
Step 7: Calculate the activation energy in kJ/mol.
Ea=−8.314 ×298 ×323 ×ln(0.25) ≈64.3 kJ/mol
Question 12
Question
The rate constant for a reaction at 25
°
C is 1.23 ×10−2s−1, and the activation
energy for the reaction is 75 kJ/mol. Calculate the rate constant at 35
°
C.
Solution
Step 1: Convert the activation energy from kJ/mol to J/mol. Given: Activation
energy (Ea) = 75 kJ/mol Conversion factor: 1 kJ = 103J
Activation energy = 75 ×103J/mol = 7.5×104J/mol
Step 2: Calculate the rate constant at 35
°
C using the Arrhenius equation:
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 kat 25
°
C (k25)=1.23 ×10−2s−1Ea= 7.5 ×104J/mol T25C= 25
°
C
= 25 + 273 = 298 K T35C= 35
°
C = 35 + 273 = 308 K
Let’s first find the pre-exponential factor A:
k25 =A×e−Ea
R×T25
1.23 ×10−2=A×e−7.5×104
8.314×298
1.23 ×10−2=A×e−31.73
A=1.23 ×10−2
e−31.73
11
A≈1.07 ×1033
Step 3: Calculate the rate constant at 35
°
C using the Arrhenius equation:
k35 =A×e−Ea
R×T35
k35 = 1.07 ×1033 ×e−7.5×104
8.314×308
k35 ≈2.78 ×10−2s−1
Therefore, the rate constant at 35
°
C is approximately 2.78 ×10−2s−1.
Question 13
Question
For the reaction
2H2O2(aq)→2H2O(l)+O2(g)
the rate constant at 25
°
C is 1.20 ×10−4s−1. At what temperature will the rate
constant be twice as large if the activation energy is 75 kJ/mol?
Solution
Step 1: We can use the Arrhenius equation to determine the relationship be-
tween the rate constant (k2) at a different temperature (T2) compared to the
rate constant (k1) at the initial temperature (T1). 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.
Step 2: Let T1= 25C= 25 + 273 = 298 K, k1= 1.20 ×10−4s−1, and
T2be the temperature we need to find where k2= 2k1= 2 ×1.20 ×10−4=
2.40 ×10−4s−1.
Step 3: To find T2, we set up the Arrhenius equation for both temperatures:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 4: We can divide the two equations to eliminate A:
k2
k1
=e−Ea
R·T2
e−Ea
R·T1
12
Step 5: Substitute the known values into the equation:
2.40 ×10−4
1.20 ×10−4=e−75000
8.314 1
T2
−1
298
Step 6: Solve for T2:
2 = e−75000
8.314 1
T2
−1
298
Step 7: Taking the natural logarithm of both sides:
ln(2) = −75000
8.314 1
T2
−1
298
Step 8: Solve for T2:
T2=75000
8.314 ·(ln(2) + 75000
8.314·298 )≈446 K
Therefore, the rate constant will be twice as large at approximately 446 K.
Question 14
Question
The rate constant for a reaction is found to be 2.5×10−3s−1at 298 Kand
8.0×10−3s−1at 333 K. Calculate the activation energy for this reaction.
Solution
Step 1: Understand the Arrhenius equation and how it relates to the rate con-
stants at different temperatures.
The Arrhenius equation relates the rate constant of a reaction to the tem-
perature and activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/(molK)), and - Tis the
temperature in Kelvin.
Step 2: Write down the given rate constants and temperatures.
Given: - k1= 2.5×10−3s−1at T1= 298 K, - k2= 8.0×10−3s−1at
T2= 333 K.
Step 3: Set up the Arrhenius equation with the two sets of data.
k1
k2
=A·e−Ea
R·T1
A·e−Ea
R·T2
Step 4: Simplify the equation and solve for the activation energy (Ea).
13
k1
k2
=eEa
R·1
T2
−1
T1
ln k1
k2=Ea
R·1
T2
−1
T1
Ea=R·
ln k1
k2
1
T2
−1
T1
Step 5: Substitute the given values and solve for the activation energy.
Ea= 8.314
ln 2.5×10−3
8.0×10−3
1
333 −1
298
Ea= 8.314 ln (0.3125)
0.0030
Ea= 8.314 −1.1568
0.0030
Ea≈ −321,741 J/mol
Question 15
Question
For a certain reaction, the rate constant at 300 K is 2.5×10−3s−1, and the
rate constant at 350 K is 8.0×10−3s−1. Calculate the activation energy (in
kJ/mol) for the reaction.
Solution
Step 1: 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 can set up two equations using the Arrhenius equation for the
rate constants at 300 K and 350 K:
k1=A·e−Ea
R·300
k2=A·e−Ea
R·350
14
where k1= 2.5×10−3s−1and k2= 8.0×10−3s−1.
Step 3: Taking the ratio of the two equations, we have:
k2
k1
=A·e−Ea
R·350
A·e−Ea
R·300
8.0×10−3
2.5×10−3=e−Ea
R(1
350 −1
300 )
Step 4: Simplifying the ratio on the left side gives:
3.2 = eEa
8.314 (1
300 −1
350 )
Step 5: Solving for the activation energy, we find:
ln(3.2) = Ea
8.314 1
300 −1
350
Ea= 8.314 ×10−3×1
300 −1
350−1
ln(3.2)
Step 6: Calculating the activation energy:
Ea≈47.9 kJ/mol
Therefore, the activation energy for the reaction is approximately 47.9 kJ/mol.
Question 16
Question
For a certain reaction, the rate constant at 35
°
C is 0.0023 s−1and at 50
°
C is
0.157 s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write the Arrhenius equation:
The Arrhenius equation is given by:
k=Ae−Ea
RT
Where: - kis the rate constant - Ais the pre-exponential factor - Eais the
activation energy (in J/mol) - Ris the gas constant (8.314 J/mol*K) - Tis the
temperature in Kelvin
Step 2: Calculate the activation energy:
Given data: - k1= 0.0023 s−1at T1= 35 ◦C = 308 K - k2= 0.157 s−1at
T2= 50 ◦C = 323 K
15
Using the Arrhenius equation for both cases:
k1=Ae−Ea
R·T1
k2=Ae−Ea
R·T2
Step 3: Take the ratio of the two equations:
Taking the ratio of the two Arrhenius equations:
k2
k1
=Ae−Ea
R·T2
Ae−Ea
R·T1
k2
k1
=e−Ea
R1
T2
−1
T1
Step 4: Solve for the activation energy:
Plugging in the values:
0.157
0.0023 =e−Ea
8.314 (1
323 −1
308 )
Solving for Eagives:
Ea≈97679 J/mol
Question 17
Question
The rate constant for a certain reaction at 25
°
C is 6.20 ×10−4s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.86 ×10−3s−1.
Calculate the activation energy for this 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 gas constant (8.31 J/mol-K), and Tis the temperature in Kelvin.
Step 2: We have two sets of data: At 25
°
C (298 K): k1= 6.20 ×10−4s−1At
50
°
C (323 K): k2= 1.86 ×10−3s−1
Step 3: Substitute the data into the Arrhenius equation: At 25
°
C:
6.20 ×10−4=A·e−Ea
8.31×298
At 50
°
C:
1.86 ×10−3=A·e−Ea
8.31×323
16
Step 4: Divide the two equations to eliminate A:
1.86 ×10−3
6.20 ×10−4=eEa
8.31 (1
323 −1
298 )
Step 5: Calculate the activation energy:
Ea=−8.31 ×ln 1.86 ×10−3
6.20 ×10−4×1
323 −1
298
Step 6: Solving the equation, we find:
Ea≈60.8 kJ/mol
Therefore, the activation energy for this reaction is approximately 60.8
kJ/mol.
Question 18
Question
The rate constant of a reaction is found to be 4.75 ×10−3s−1at 25◦C and
1.23 ×10−2s−1at 35◦C. Calculate the activation energy of the 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 the given temperatures to Kelvin:
T1= 25◦C + 273 = 298 K
T2= 35◦C + 273 = 308 K
Step 2: Insert the given rate constants and temperatures into the Arrhenius
equation to form two equations:
4.75 ×10−3=A×e(−Ea
R×298 )
1.23 ×10−2=A×e(−Ea
R×308 )
Step 3: Divide the second equation by the first equation to eliminate the
pre-exponential factor A:
1.23 ×10−2
4.75 ×10−3=e(−Ea
R×308 )
e(−Ea
R×298 )
1.23 ×10−2
4.75 ×10−3=e(Ea
R(1
298 −1
308 ))
17
Step 4: Solve the equation for the activation energy Eausing the value of
the gas constant, R= 8.314 J/(mol K):
1.23 ×10−2
4.75 ×10−3=e(8.314×Ea
8.314 ×1
298 −1
308 )
2.5895 = e(Ea
8.314 ×1
298 −1
308 )
Step 5: Take the natural logarithm of both sides to solve for Ea:
ln(2.5895) = Ea
8.314 ×1
298 −1
308
ln(2.5895) = Ea
8.314 ×1
298 −1
308
Step 6: Calculate the activation energy Ea:
Ea= 8.314 ×ln(2.5895)
1
298 −1
308
Ea≈53,620 J/mol
Therefore, the activation energy of the reaction is approximately 53.62 kJ/mol.
Question 19
Question
For the reaction
2A →B+C
the rate is found to double when the temperature is increased from 25
°
C to
35
°
C. Calculate the activation energy for this reaction. (Assume the frequency
factor, A, is constant)
Solution
Step 1: Given that the rate constant, k, is directly proportional to the temper-
ature, we can express this relationship using the Arrhenius equation:
k=Ae−Ea
RT
where k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant, and T= temperature in Kelvin.
Step 2: We are told that the rate doubles when the temperature is increased
from 25
°
C to 35
°
C, which means:
k2
k1
= 2
18
Using the Arrhenius equation, we can express the rates in terms of their corre-
sponding temperatures:
Ae−Ea
R(25+273.15)
Ae−Ea
R(35+273.15)
= 2
Simplifying this equation gives:
e−Ea
2984.75
e−Ea
4112.05
= 2
Step 3: To solve for the activation energy, we need to isolate Eain the above
equation. We can do this by taking the natural logarithm of both sides:
ln e−Ea
2984.75
e−Ea
4112.05 != ln(2)
Using the properties of logarithms, we simplify the left side to:
−Ea
2984.75 +Ea
4112.05 = ln(2)
Step 4: Further simplification yields the equation:
1127.3
2984.75Ea= ln(2)
From which we can solve for the activation energy, Ea:
Ea=2984.75 ·ln(2)
1127.3≈10.16 kJ/mol
Therefore, the activation energy for the reaction is approximately 10.16
kJ/mol.
Question 20
Question
The rate constant for a certain reaction at 25
°
C is 3.2×10−3s−1, and at 35
°
C
it is 1.3×10−2s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·exp −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.
19
Step 2: Rearrange the Arrhenius equation to solve for the activation energy
(Ea):
Ea=−RT ln(k/A)
ln(k/A)
Step 3: Convert the given temperatures to Kelvin (25
°
C = 298 K and 35
°
C
= 308 K).
Step 4: Plug in the values for the rate constants and temperatures into the
equation to solve for the activation energy:
Ea=−(8.314 J/(mol
·
K))(298 K) ln3.2×10−3/A
ln(3.2×10−3/A)
Step 5: Repeat the calculation for the other temperature:
Ea=−(8.314 J/(mol
·
K))(308 K) ln1.3×10−2/A
ln(1.3×10−2/A)
Step 6: Subtract the activation energy calculated at 298 K from the activa-
tion energy calculated at 308 K to find the overall activation energy.
Question 21
Question
The rate constant at 25
°
C for the reaction
2A →B
is 5.0×10−3s−1. When the temperature is increased to 35
°
C, the rate constant
is found to be 2.0×10−2s−1. Calculate the activation energy for this reaction.
Given: R= 8.314 J/(mol·K)
Solution
Step 1: Convert the temperatures to Kelvin.
T1= 25 + 273 = 298 K
T2= 35 + 273 = 308 K
Step 2: Use the Arrhenius equation to find the activation energy (Ea).
k1=A×e−Ea
RT1
k2=A×e−Ea
RT2
20
Step 3: Divide the second equation by the first to eliminate A.
k2
k1
=A×e−Ea
RT2
A×e−Ea
RT1
k2
k1
=e−Ea
R1
T1
−1
T2
Step 4: Substitute the given values and solve for Ea.
2.0×10−2
5.0×10−3=e−Ea
8.314 (1
298 −1
308 )
4 = e−Ea
8.314 ×0.0034
ln(4) = −Ea
8.314 ×0.0034
Ea=−8.314 ×0.0034
ln(4)
Ea= 48.97 kJ/mol
Therefore, the activation energy for the reaction is 48.97 kJ/mol.
Question 22
Question
The rate constant, k, for a certain reaction is found to be 6.0×10−3s−1at
30
°
C and 4.0×10−2s−1at 50
°
C. Determine the activation energy, Ea, for this
reaction.
Solution
Step 1: Write the Arrhenius equation relating the rate constant, temperature,
and activation energy.
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·K)), and - Tis the
absolute temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin. - 30◦C = 30 + 273 = 303 K -
50◦C = 50 + 273 = 323 K
Step 3: Substitute the given data points into the Arrhenius equation to form
two equations.
6.0×10−3=A·exp −Ea
8.314×303
4.0×10−2=A·exp −Ea
8.314×323
21
Step 4: Divide the two equations to eliminate A.
6.0×10−3
4.0×10−2=
exp −Ea
8.314×303
exp −Ea
8.314×323
Step 5: Simplify the equation.
0.15 = exp Ea
8.314 1
303 −1
323
Step 6: Solve for the activation energy, Ea.
Ea=−8.314 ·ln(0.15) ·1
303 −1
323
Step 7: Calculate the activation energy.
Ea≈64.4 kJ/mol
Question 23
Question
The rate constant of a certain reaction has been measured at two different
temperatures, T1and T2, with values of k1= 2.5×10−3s−1at T1= 300 K and
k2= 3.6×10−2s−1at T2= 350 K. Calculate the activation energy for this
reaction in kJ/mol.
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), T= temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
Ea=−R
ln k
A·1
T
Step 3: Calculate the activation energy using the given data: At T1= 300 K,
k1= 2.5×10−3s−1. At T2= 350 K, k2= 3.6×10−2s−1.
Step 4: Calculate Ea:
Ea=−8.314
ln 3.6×10−2
2.5×10−3·1
350 −1
300
22
Ea≈ − 8.314
ln(14.4) ·1
350 −1
300
Step 5: Perform the calculations to find the activation energy Eain J/mol.
Convert the result to kJ/mol for convenience.
Step 6: Therefore, the activation energy for this reaction is approximately
95.7 kJ/mol .
Question 24
Question
A certain reaction has an activation energy of 75 kJ/mol. When the temperature
is increased from 25
°
C to 45
°
C, the rate constant increases by a factor of 5.
Calculate the rate constant at 25
°
C and the rate constant at 45
°
C for this
reaction.
Solution
Step 1: We know that the Arrhenius equation is given by:
k=Ae−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: Let’s denote the rate constant at 25
°
C as k1and at 45
°
C as k2. The
rate constant increases by a factor of 5 when the temperature increases from
25
°
C to 45
°
C. Therefore, we can write:
k2= 5k1
Step 3: Converting the temperatures to Kelvin:
T1= 25C= 25 + 273 = 298K
T2= 45C= 45 + 273 = 318K
Step 4: Substituting the values into the Arrhenius equation for k1and k2:
For k1at 25
°
C:
k1=Ae−75000
8.314×298
For k2at 45
°
C:
k2= 5k1= 5Ae−75000
8.314×318
Step 5: Now we can solve for k1and k2. From the given information, we
know that k2= 5k1. Therefore, substitute the expressions for k1and k2into
this equation and solve for A.
5Ae−75000
8.314×318 = 5Ae−75000
8.314×298
Step 6: After solving for A, substitute the value back into the expressions
for k1and k2to find the rate constants.
23
Question 25
Question
The rate constant for the decomposition of compound A is found to be 6.32 ×
10−4s−1at 65◦C and 2.54 ×10−3s−1at 75◦C. Calculate the activation energy
for the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. For 65◦C:
T= 65 + 273.15 = 338.15 K
For 75◦C:
T= 75 + 273.15 = 348.15 K
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation:
k=A·e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the
activation energy, Ris the gas constant (8.314 J ·mol−1·K−1), and Tis the
temperature in Kelvin.
We can rearrange the equation in the following form:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the values into the equation:
ln 2.54 ×10−3
6.32 ×10−4=−Ea
8.314 1
348.15 −1
338.15
Step 4: Solve for Ea:
Ea=−8.314 ×
ln 2.54×10−3
6.32×10−4
1
348.15 −1
338.15
Ea≈53320 J/mol
Therefore, the activation energy for the reaction is approximately 53.32 kJ/mol.
Question 26
Question
The rate constant for the reaction H2O2→H2O+O2at 25◦Cis 1.20×10−4s−1.
When the temperature is increased to 35◦C, the rate constant becomes 5.80 ×
10−3s−1. Calculate the activation energy for this reaction.
24
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at
different temperatures:
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: First, let’s convert the temperatures to Kelvin: - 25◦C= 25 +273 =
298 K - 35◦C= 35 + 273 = 308 K
Step 3: We can set up two equations using the rate constants provided:
1.20 ×10−4=A·e−Ea
8.314·298
5.80 ×10−3=A·e−Ea
8.314·308
Step 4: Now we can divide the two equations to eliminate A:
5.80 ×10−3
1.20 ×10−4=e−Ea
8.314·308
e−Ea
8.314·298
Step 5: Simplifying, we get:
48.33 = eEa
8.314 (1
298 −1
308 )
Step 6: Further simplifying, we find:
ln(48.33) = Ea
8.314 1
298 −1
308
Step 7: Solving for Ea, we have:
Ea= 8.314 ×1
298 −1
308×ln(48.33)
After calculations, we can find the activation energy for this reaction.
Question 27
Question
The rate constant for a certain reaction is found to be 2.45 ×10−3s−1at 25
°
C
and 1.53 ×10−2s−1at 35
°
C. Determine the activation energy for the reaction
in kJ/mol.
25
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−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 K).
Step 2: Take the logarithm of the Arrhenius equation to simplify the calcu-
lation:
ln(k) = ln(A)−Ea
RT
Step 3: Rearrange the equation to solve for Ea:
Ea
R=1
T1
−1
T2
Step 4: Convert the given temperatures to Kelvin:
T1= 25C+ 273.15 = 298.15K
T2= 35C+ 273.15 = 308.15K
Step 5: Substitute the values into the equation and solve for Ea:
Ea
8.314 =1
298.15 −1
308.15
Ea= 8.314 ×1
298.15 −1
308.15
Ea≈56184.15 J/mol
Step 6: Convert the activation energy to kilojoules per mole:
Ea≈56184.15 J
1000 kJ/mol
Ea≈56.18 kJ/mol
Therefore, the activation energy for the reaction is approximately 56.18
kJ/mol.
Question 28
Question
The rate constant of a certain reaction doubles when the temperature is raised
from 25
°
C to 35
°
C. Calculate the activation energy (in kJ/mol) for this reaction.
Assume the frequency factor is 1.2×1013 s−1.
26
Solution
Step 1: Calculate the temperature values in Kelvin. Given: T1= 25◦C =
25 + 273 = 298 K T2= 35◦C = 35 + 273 = 308 K
Step 2: Using the Arrhenius equation, we have:
k2
k1
= exp −Ea
R1
T2
−1
T1
Given that k2
k1= 2 and R= 8.314 J/mol K, we can rewrite the equation as:
2 = exp −Ea
8.314 1
308 −1
298
Step 3: Simplify the equation:
2 = exp −Ea
8.314 10
298 ×308
Step 4: Solve for the activation energy Ea:
ln(2) = −Ea
8.314 ×10
298 ×308
Ea=−8.314 ×10
298 ×308 ×ln(2)
Step 5: Calculate the activation energy Ea:
Ea≈ − 8.314 ×10
298 ×308 ×ln(2)
Ea≈56.31 kJ/mol
Therefore, the activation energy for this reaction is approximately 56.31
kJ/mol.
Question 29
Question
The rate constant for a reaction at 25
°
C is 3.42 ×10−4s−1. When the tempera-
ture is increased to 55
°
C, the rate constant becomes 1.26 ×10−2s−1. Calculate
the activation energy for this reaction.
27
Solution
Step 1: 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=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 can set up two equations based on the given information:
3.42 ×10−4=Ae−Ea
(8.314)(298)
1.26 ×10−2=Ae−Ea
(8.314)(328)
Step 3: Divide the two equations to eliminate A:
3.42 ×10−4
1.26 ×10−2=e−Ea
(8.314)(298)
e−Ea
(8.314)(328)
Step 4: Simplify the above expression:
e
Ea
(8.314)(328) −Ea
(8.314)(298) =3.42 ×10−4
1.26 ×10−2
Step 5: Solve for Eaby taking natural logarithm of both sides:
Ea
(8.314) 1
328 −1
298= ln 3.42 ×10−4
1.26 ×10−2
Step 6: Calculate the activation energy, Ea:
Ea= 8.314 ×
ln 3.42×10−4
1.26×10−2
1
328 −1
298
Therefore, the activation energy for this reaction is approximately 47.8
kJ/mol.
Question 30
Question
The rate constant (k) for a certain reaction is found to be 1.5×10−4s−1at 25
°
C
and 3.2×10−3s−1at 35
°
C. Calculate the activation energy for this reaction.
28
Solution
Step 1: Write down the Arrhenius equation. The Arrhenius equation relates
the rate constant (k) of a reaction to the temperature (T) and the activation
energy (Ea).
k=Ae−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.
Step 2: Identify the two sets of data given. We are given two sets of data:
At 25
°
C: k1= 1.5×10−4s−1T1= 25 + 273 = 298 K
At 35
°
C: k2= 3.2×10−3s−1T2= 35 + 273 = 308 K
Step 3: Plug the data into the Arrhenius equation. We can set up two
equations using the Arrhenius equation:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 4: Take the ratio of the two equations. Dividing the two equations will
eliminate A:
k1
k2
=e−Ea
RT1
e−Ea
RT2
Step 5: Simplify the expression.
k1
k2
=e
Ea
R1
T2
−1
T1
Step 6: Solve for Ea. Using the known values k1,k2,T1,T2, and R, we can
now solve for Ea:
1.5×10−4
3.2×10−3=eEa
8.314 (1
308 −1
298 )
0.046875 = eEa
8.314 (1
308 −1
298 )
ln(0.046875) = Ea
8.314 1
308 −1
298
Ea= 2.83 kJ/mol
Therefore, the activation energy for this reaction is 2.83 kJ/mol.
Question 31
Question
The rate constant of a certain reaction is found to be 5.00 ×10−3s−1at 25
°
C
and 1.00 ×10−2s−1at 35
°
C. Calculate the activation energy of the reaction.
Given: R= 8.314 J/(mol·K), T1= 25
°
C, T2= 35
°
C.
29
Solution
Step 1: Convert the given temperatures from Celsius to Kelvin.
T1= 25 + 273 = 298 K
T2= 35 + 273 = 308 K
Step 2: Use the Arrhenius equation to relate the rate constants at different
temperatures:
k2=k1·e(−Ea
R(1
T2
−1
T1))
Step 3: Substitute the given rate constants and temperatures into the equa-
tion and solve for the activation energy Ea.
1.00 ×10−2= 5.00 ×10−3·e(−Ea
8.314 (1
308 −1
298 ))
Step 4: Solve for Eaby taking the natural logarithm of both sides and
rearranging the terms.
ln1.00 ×10−2
5.00 ×10−3=−Ea
8.314(1
308 −1
298)
Step 5: Calculate the activation energy Eaby plugging in the known values
and solving for it.
Ea=−8.314 ·ln1.00 ×10−2
5.00 ×10−3·1
308 −1
298
Ea≈30.4 kJ/mol
Therefore, the activation energy of the reaction is approximately 30.4 kJ/mol.
Question 32
Question
The rate constant, k, for a certain reaction is found to be 8.21 ×10−4s−1at
300 K and 3.12 ×10−3s−1at 325 K. Calculate the activation energy for the
reaction.
Solution
Step 1: We can use the Arrhenius equation to relate 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.
30
Step 2: Taking the natural logarithm of both sides of the Arrhenius equation
gives:
ln(k) = −Ea
R·1
T+ ln(A)
Step 3: We will use the given data at two temperatures (300 K and 325 K)
to set up a system of equations:
(ln8.21 ×10−4=−Ea
8.314 ·1
300 + ln(A)
ln3.12 ×10−3=−Ea
8.314 ·1
325 + ln(A)
Step 4: Solve the system of equations and find the values of Eaand ln(A).
Step 5: With the values of Eaand ln(A), calculate the activation energy Ea
using one of the equations.
Step 6: Calculate the activation energy Eafor the reaction.
Question 33
Question
The rate constant (k) for a certain reaction is found to double when the tem-
perature is increased from 25
°
C to 35
°
C. Calculate the activation energy (Ea)
for this reaction. (R = 8.314 J/mol·K)
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant (k) to
temperature and activation energy:
k=A·e−Ea
RT
where Ais the pre-exponential factor, Eais the activation energy, Ris the gas
constant, and Tis the temperature in Kelvin.
Step 2: We are given that the rate constant doubles when the temperature
is increased from 25
°
C (298 K) to 35
°
C (308 K). This means that:
k2= 2k1
A·e−Ea
R·308 = 2(A·e−Ea
R·298 )
Step 3: Divide the second equation by the first equation to eliminate A:
e−Ea
R·308
e−Ea
R·298
= 2
e−Ea
R·308 +Ea
R·298 = 2
eEa
R(1
298 −1
308 )= 2
31
Step 4: Solve for Ea:
Ea
R1
298 −1
308= ln(2)
Ea
8.314 1
298 −1
308= ln(2)
Ea= 8.314 ×ln(2) ×1
1
298 −1
308
Ea≈36590.56 J/mol
Therefore, the activation energy for this reaction is approximately 36.59
kJ/mol.
Question 34
Question
The reaction between nitrogen dioxide gas (NO2) and carbon monoxide gas
(CO) is believed to follow the following rate law at a certain temperature:
Rate = k[NO2][CO]2
Given that the activation energy for this reaction is 120 kJ/mol and the
pre-exponential factor is 1.0×1010 s−1, calculate the rate constant (in s−1) at a
temperature where the rate of reaction is observed to be 3.5×10−3mol L−1s−1
when the concentrations of NO2 and CO are both 0.10 mol L−1.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to the
rate of reaction and the temperature. The Arrhenius equation is given by:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J mol−1K−1), - Tis the
temperature in Kelvin.
Given that A= 1.0×1010 s−1and Ea= 120 kJ mol−1, we can substitute
these values into the Arrhenius equation.
Step 2: Rearranging the Arrhenius equation to solve for T, we have:
T=−Ea
Rln k
A
Step 3: Given that the rate of reaction is observed to be 3.5×10−3mol L−1s−1
and the concentrations of NO2 and CO are both 0.10 mol L−1, we can substitute
32
these values along with the rate constant into the rate law equation to solve for
k.
3.5×10−3=k(0.10)(0.10)2
Step 4: Solve for kin the rate law equation:
k=3.5×10−3
0.001 = 3.5 s−1
Step 5: Finally, substitute the calculated value of kinto the Arrhenius equa-
tion to solve for the temperature T.
Question 35
Question
The rate constant of a reaction is measured at two different temperatures, T1and
T2, and found to be k1= 5.0×10−3s−1at T1= 300 K and k2= 2.0×10−2s−1
at T2= 325 K. Calculate the activation energy of the reaction.
Solution
Step 1: The Arrhenius equation relates the rate constant of a reaction to tem-
perature 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 2: We can take the ratio of the Arrhenius equation at two different
temperatures:
k2
k1
=A e−Ea
R T2
A e−Ea
R T1
Step 3: Substitute the given values:
2.0×10−2
5.0×10−3=e−Ea
8.314×325
e−Ea
8.314×300
Step 4: Simplify the equation:
4 = e−Ea
2651.1×eEa
2494.2
Step 5: Combine the exponents:
4 = e−Ea
2651.1+Ea
2494.2
Step 6: Solve for the activation energy Ea:
33
4 = e2494.2−2651.1
2651.1×2494.2Ea
Step 7: Taking the natural logarithm of both sides, we have:
ln(4) = 2494.2−2651.1
2651.1×2494.2Ea
Ea=2651.1×2494.2
2494.2−2651.1ln(4)
Step 8: Calculate the activation energy using the given values:
Ea=2651.1×2494.2
2494.2−2651.1ln(4) ≈72.5 kJ/mol
34
