CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 4
Liberty University
Question 1
Question
The rate constant (k) for a certain reaction at 25
°
C is 6.0×10−4s−1, and the
activation energy (Ea) is 80 kJ/mol. Calculate the value of kif the temperature
is increased to 50
°
C.
Solution
Step 1: Convert the activation energy (Ea) from kilojoules per mole to joules
per molecule:
Ea= 80 kJ/mol ×1000 J
1 kJ = 80000 J/mol
Step 2: Use the Arrhenius equation to relate the rate constants (k1and k2)
at two different temperatures (T1and T2):
k2
k1
=eEa
R1
T1
−1
T2
Step 3: Substitute the given values into the equation above. Remember to
convert the temperatures to Kelvin:
k2
6.0×10−4=e(80000
8.314 (1
298 −1
323 ))
Step 4: Simplify and solve for k2:
k2
6.0×10−4=e(9644.5)
k2= 6.0×10−4×e(9644.5)
k2≈2.765 ×108s−1
Therefore, the value of kwhen the temperature is increased to 50
°
C is ap-
proximately 2.765 ×108s−1.
Question 2
Question
The rate constant for a reaction is found to be 1.25 ×10−3s−1at a temperature
of 25◦C and 1.88 ×10−3s−1at a temperature of 45◦C. Calculate the activation
energy (in kJ/mol) for this reaction.
Solution
Step 1: Convert temperatures to Kelvin using T(K) = T(◦C) + 273.15.
25◦C = 25 + 273.15 = 298.15 K
45◦C = 45 + 273.15 = 318.15 K
Step 2: Write the Arrhenius equation:
k=Ae−Ea
RT
Step 3: Take the natural logarithm of both sides to linearize the Arrhenius
equation:
ln k= ln A−Ea
RT
Step 4: Define two equations from the given data:
ln k1= ln A−Ea
R·T1
ln k2= ln A−Ea
R·T2
Step 5: Subtract the second equation from the first:
ln k1−ln k2=−Ea
R1
T1
−1
T2
ln k1
k2=Ea
R1
T2
−1
T1
2
Step 6: Solve for the activation energy Ea:
Ea=R·
ln k1
k2
1
T2
−1
T1
Ea= 8.314 J/mol ·
ln 1.25×10−3
1.88×10−3
1
318.15 −1
298.15
Ea≈46.59 kJ/mol
Therefore, the activation energy for this reaction is approximately 46.59
kJ/mol.
Question 3
Question
The rate constant for a reaction at 25
°
C is 1.25 ×10−3s−1, and its activation
energy is 75 kJ/mol. Calculate the rate constant at 50
°
C for this reaction.
Solution
Step 1: Calculate the pre-exponential factor Ausing the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant at 25
°
C, - Eais the activation energy, - Ris
the gas constant (8.314 J/mol·K), - Tis the temperature in Kelvin, - Ais the
pre-exponential factor.
Given that k= 1.25 ×10−3s−1at 25
°
C (298 K) and Ea= 75 kJ/mol, we
rearrange the equation to solve for A:
1.25 ×10−3=A·e−75000
8.314·298
A= 1.25 ×10−3∇ · e−28.59
A≈0.00129 s−1
Step 2: Use the Arrhenius equation to find the rate constant at 50
°
C (323
K):
k50 = 0.00129 ·e−75000
8.314·323
k50 ≈3.48 ×10−3s−1
Therefore, the rate constant for the reaction at 50
°
C is approximately 3.48×
10−3s−1.
3
Question 4
Question
The rate constant for a certain reaction is found to be 1.25 ×10−3s−1at 25◦C
and 3.75 ×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation which relates the rate constant to
the temperature and activation energy.
ln k2
k1=−Ea
R1
T2
−1
T1
where: k2and k1are the rate constants at temperatures T2and T1,Eais the
activation energy, R= 8.314 J mol−1K−1is the gas constant.
Step 2: Plug in the given values to the Arrhenius equation.
ln 3.75 ×10−3
1.25 ×10−3=−Ea
8.314 1
318 −1
298
Step 3: Solve for the activation energy Ea. First, simplify the natural loga-
rithm term:
ln (3) = −Ea
8.314 (0.00334)
Step 4: Evaluate the natural logarithm.
1.0986 = −0.00334Ea
Step 5: Solve for the activation energy Ea.
Ea=1.0986
−0.00334 ≈ −328.7 kJ/mol
Therefore, the activation energy for this reaction is approximately -328.7
kJ/mol.
Question 5
Question
The rate constant for the reaction 2A→Bis 2.5×10−3s−1at 25
°
C and
7.5×10−3s−1at 35
°
C. Calculate the activation energy for this reaction. The
gas constant Ris 8.314 J/(mol
·
K).
4
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1= 2.5×10−3s−1at T1= 25C= 298 K, k2= 7.5×10−3s−1at
T2= 35C= 308 K, and R= 8.314 J/(mol
·
K).
Step 2: Substitute the known values into the equation:
ln 7.5×10−3
2.5×10−3=−Ea
8.314 1
308 −1
298
ln(3) = −Ea
8.314 1
308 −1
298
Step 3: Solve for the activation energy (Ea):
Ea=−8.314 ×ln(3)
1
308 −1
298
Ea≈77.09 kJ/mol
Therefore, the activation energy for the reaction is approximately 77.09
kJ/mol.
Question 6
Question
The rate constant for a certain reaction at 35
°
C is 1.8×10−3s−1, and at 45
°
C
it is 5.7×10−2s−1. Determine the activation energy for this reaction.
Given: R= 8.31 J/(mol·K), T1= 35
°
C, T2= 45
°
C
Solution
Step 1: Convert the temperatures from
°
C to K.
T1= 35 + 273 = 308 K
T2= 45 + 273 = 318 K
Step 2: Write the Arrhenius equation:
k=A·e−Ea
RT
5
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
R·1
T
This equation is in the form of y=mx +b, where y= ln k,x= 1/T ,m=−Ea
R,
and b= ln A.
Step 4: Use the data to create two equations based on the Arrhenius equa-
tion:
ln k1= ln A−Ea
R·1
T1
ln k2= ln A−Ea
R·1
T2
Step 5: Subtract the two equations to eliminate ln A:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Solve for the activation energy Ea:
Ea=−R·
ln k2
k1
1
T2
−1
T1
Substitute the given values into the equation to find the activation energy.
Question 7
Question
The rate constant for a certain reaction is found to be 4.32 ×10−2s−1at 25◦C
and 9.28 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
(Hint: Use the Arrhenius equation: k=A·e−Ea
RT )
Solution
Step 1: Convert the temperatures to Kelvin Given:
T1= 25◦C = 25 + 273 = 298 K
T2= 45◦C = 45 + 273 = 318 K
Step 2: Use the Arrhenius equation to set up two equations based on the
rate constants at two temperatures At T1= 298 K:
ln k1= ln A−Ea
RT1
6
ln4.32 ×10−2= ln A−Ea
8.314 J K−1mol−1·298 K
At T2= 318 K:
ln k2= ln A−Ea
RT2
ln9.28 ×10−2= ln A−Ea
8.314 J K−1mol−1·318 K
Step 3: Subtract the two equations to eliminate ln A
ln k2
k1=−Ea
8.314 J K−1mol−11
318 −1
298
Step 4: Solve for the activation energy, Ea
Ea=−8.314 J K−1mol−1·318 ·298
318 −298 ln 9.28 ×10−2
4.32 ×10−2
Calculating this expression gives the activation energy.
Question 8
Question
The rate constant for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 3.75 ×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. At 25
°
C, T1= 25 + 273 = 298 K. At 50
°
C, T2= 50 + 273 = 323 K.
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
Given: k1= 1.25 ×10−3s−1k2= 3.75 ×10−2s−1T1= 298 K T2= 323 K
R= 8.314 J/(mol
·
K)
Substitute the values into the equation:
ln 3.75 ×10−2
1.25 ×10−3=−Ea
8.314 1
323 −1
298
Step 3: Simplify and solve for the activation energy (Ea).
ln (30) = −Ea
8.314 1
323 −1
298
7
ln (30) = −Ea
8.314 1
323 −1
298
Ea
8.314 =−ln (30)
1
323 −1
298
Ea= 8.314 ×ln (30)
1
323 −1
298
Calculating the activation energy:
Ea= 8.314 ×ln (30)
1
323 −1
298
Question 9
Question
For the reaction A →B, the rate constant at 27
°
C is 4.10×10−3s−1and the rate
constant at 57
°
C is 5.73 s−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.
T1= 27C+ 273.15 = 300.15 K
T2= 57C+ 273.15 = 330.15 K
Step 2: Write down the Arrhenius equation:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of both sides of the equation to linearize
it:
ln(k) = ln(A)−Ea
R1
T
Step 4: Set up two equations using the rate constants and temperatures
given:
ln(k1) = ln(A)−Ea
R1
T1
ln(k2) = ln(A)−Ea
R1
T2
Step 5: Subtract the second equation from the first to eliminate ln(A) and
solve for Ea:
ln(k1)−ln(k2) = Ea
R1
T2
−1
T1
8
Step 6: Substitute the given values and solve for Ea:
ln4.10 ×10−3−ln(5.73) = Ea
8.314 1
330.15 −1
300.15
−2.1591 = Ea
8.314 1
330.15 −1
300.15
Step 7: Solve for Ea:
Ea=−2.1591 ×8.314 ×1
330.15 −300.15
Ea≈48.5 kJ/mol
Therefore, the activation energy for the reaction A →B is approximately
48.5 kJ/mol.
Question 10
Question
The rate constant of a reaction at 25
°
C is 1.2×10−3s−1, and the rate constant
at 35
°
C is 4.8×10−3s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Calculate the rate constant ratio using the Arrhenius equation:
k2
k1
=e
Ea
R1
T1
−1
T2
Given: k1= 1.2×10−3s−1at 25◦C, k2= 4.8×10−3s−1at 35◦C, T1= 25+273 =
298 K, T2= 35 + 273 = 308 K, R= 8.314 J mol−1K−1.
Plugging in the values, we get:
4.8×10−3
1.2×10−3=eEa
8.314 (1
298 −1
308 )
Step 2: Solve for the activation energy, Ea.
4.8
1.2=eEa
8.314 (0.00336−0.00325)
Simplify:
4 = eEa
8.314 ×0.00011
Step 3: Take the natural logarithm of both sides to solve for Ea.
ln(4) = Ea
8.314 ×0.00011
9
Step 4: Solve for Ea.
ln(4)
0.00011 ×8.314 = Ea
Step 5: Calculate Eato find the activation energy for the reaction.
Question 11
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1and the
activation energy is 40 kJ/mol. Calculate the rate constant at 45
°
C for this
reaction.
Solution
Step 1: Let’s start by writing down the Arrhenius equation:
k=A·exp −Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - R= 8.314 J/mol ·K is the gas constant, and - Tis the
temperature in Kelvin.
Step 2: We are given that k1= 2.5×10−3s−1at T1= 25C= 298 K, and
Ea= 40 kJ/mol. We need to find k2at T2= 45C= 318 K.
Step 3: First, let’s express the rate constant k1in terms of the Arrhenius
equation using T1and solve for A:
k1=A·exp −Ea
R·T1
2.5×10−3=A·exp −40,000
8.314 ·298
A= 2.5×10−3·exp 40,000
8.314 ·298
A≈9.42 ×109s−1
Step 4: Now, substitute A,Ea,R, and T2into the Arrhenius equation to
find k2:
k2= 9.42 ×109·exp −40,000
8.314 ·318
k2≈6.91 ×10−3s−1
Step 5: Therefore, the rate constant at 45
°
C for this reaction is approxi-
mately 6.91 ×10−3s−1.
10
Question 12
Question
The rate constant for a reaction at 25
°
C is found to be 1.5×10−2s−1, while the
rate constant at 35
°
C is 5.0×10−2s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: We can use the Arrhenius equation to find the activation energy (Ea)
for this reaction. The Arrhenius equation is given by:
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−1K−1),
and - Tis the temperature in Kelvin.
Step 2: First, let’s convert the rate constants given at 25
°
C and 35
°
C to
Kelvin:
T1= 25C+ 273.15 = 298.15 K
T2= 35C+ 273.15 = 308.15 K
Step 3: Now we can set up two Arrhenius equations using the rate constants
and temperatures provided:
k1=A·e−Ea
R·298.15
k2=A·e−Ea
R·308.15
Step 4: Write the ratio of the two rate constant equations:
k1
k2
=A·e−Ea
R·298.15
A·e−Ea
R·308.15
Step 5: Simplify the equation by cancelling out the pre-exponential factors:
k1
k2
=eEa
R(1
298.15 −1
308.15 )
Step 6: Plug in the rate constants given to solve for the activation energy:
1.5×10−2
5.0×10−2=eEa
8.314 (1
298.15 −1
308.15 )
Step 7: Solve for the activation energy (Ea) using natural logarithms:
ln 1.5
5.0=Ea
8.314 1
298.15 −1
308.15
Step 8: Finally, calculate the activation energy from the equation obtained
in the previous step.
11
Question 13
Question
The rate constant for the following reaction is found to be 6.32 ×10−4s−1
at 50◦C and 1.48 ×10−3s−1at 75◦C. Calculate the activation energy of the
reaction.
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. It
is given by:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol K), T= temperature in Kelvin.
Step 2: Plug in the given information. We are given the rate constants kat
two different temperatures: At T1= 50◦C = 323 K, k1= 6.32 ×10−4s−1. At
T2= 75◦C = 348 K, k2= 1.48 ×10−3s−1.
Step 3: Take the natural logarithm of both sides of the Arrhenius equation.
ln(k) = ln(A)−Ea
RT
Step 4: Set up two equations using the data provided. From our given data,
we have two equations:
ln(k1) = ln(A)−Ea
R·T1
ln(k2) = ln(A)−Ea
R·T2
Step 5: Solve for the activation energy. Subtract the second equation from
the first to eliminate ln(A):
ln(k1)−ln(k2) = Ea
R1
T2
−1
T1
Step 6: Substitute the values and calculate the activation energy.
ln6.32 ×10−4−ln1.48 ×10−3=Ea
8.314 1
348 −1
323
1
323 −1
348 =Ea
8.314 1
348 −1
323
1
323 −1
348 =Ea
8.314 1
348 −1
323
Ea≈59993 J/mol
Therefore, the activation energy of the reaction is approximately 59.993 kJ/mol.
12
Question 14
Question
The rate constant for a certain reaction is 2.35×10−3s−1at 25◦C and 6.97×10−3
s−1at 40◦C. Calculate the activation energy for this reaction. The activation en-
ergy for the reaction with respect to the rate constant is given by the Arrhenius
equation:
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), and Tis the temperature
in Kelvin.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15. At 25◦C, T= 25+273.15 = 298.15 K. At 40◦C, T= 40+273.15 = 313.15
K.
Step 2: Substitute the data into the Arrhenius equation to form two equa-
tions: At 25◦C:
2.35 ×10−3=Ae−Ea
8.314·298.15
At 40◦C:
6.97 ×10−3=Ae−Ea
8.314·313.15
Step 3: Divide the two equations to eliminate A:
2.35 ×10−3
6.97 ×10−3=e−Ea
8.314·298.15
e−Ea
8.314·313.15
0.336 = eEa
8.314 (1
313.15 −1
298.15 )
Step 4: Take the natural logarithm of both sides to solve for the activation
energy (Ea):
ln(0.336) = Ea
8.314 1
313.15 −1
298.15
Ea=−8.314 ·ln(0.336) 1
313.15 −1
298.15
Step 5: Calculate Eausing a calculator:
Ea≈55.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 55.2
kJ/mol.
13
Question 15
Question
The rate constant of a certain reaction is found to be 2.5×10−3s−1at 25◦C
and 7.2×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·exp −Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314
Question
The rate constant (k) for a reaction is found to be 6.45 ×10−3s−1at a temper-
ature of 45◦C. When the temperature is increased to 65◦C, the rate constant
becomes 1.24 ×10−2s−1. Calculate the activation energy (in kJ/mol) for this
reaction.
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15. For T= 45◦C: T= 45 + 273.15 = 318.15 K.
For T= 65◦C: T= 65 + 273.15 = 338.15 K.
Step 2: Use the Arrhenius equation k=Ae−Ea
RT to find the activation energy
(Ea) of the reaction. Given: k1= 6.45 ×10−3s−1,T1= 318.15 K, k2= 1.24 ×
10−2s−1,T2= 338.15 K.
We can rewrite the Arrhenius equation as ln k2
k1=Ea
R1
T1
−1
T2where
R= 8.314 J/mol ·K is the gas constant.
Step 3: Substitute the given values and solve for Ea.
ln 1.24 ×10−2
6.45 ×10−3=Ea
8.314 1
318.15 −1
338.15
ln (1.9196) = Ea
8.314 (0.00314 −0.00296)
0.6525 = 0.000191 ∗Ea
Ea=0.6525
0.000191 ≈3418.85 J/mol
Step 4: Convert the activation energy from Joules to kilojoules. Ea=
3418.85 J/mol = 3.41885 kJ/mol.
Therefore, the activation energy for this reaction is approximately 3.42 kJ/mol.
14
Question 17
Question
The rate constant (k) for a certain chemical reaction at 25
°
C is found to be
9.23 ×10−3s−1. When the temperature is increased to 50
°
C, the rate con-
stant increases to 2.76 ×10−2s−1. Calculate the activation energy (Ea) for this
reaction in kJ/mol.
Solution
Step 1: Convert the given rate constants to Arrhenius form.
Step 2: Divide the second equation by the first to eliminate A and solve for
Ea.
Step 3: Solve for Eaby taking the natural logarithm of both sides and then
solve for Eain kJ/mol.
ln(3) = 1
298 −1
323·Ea
Ea=298 ·323
323 −298 ·ln(3)
Ea=96054
25 ·ln(3)
Ea≈58.53 kJ/mol
Therefore, the activation energy for this reaction is approximately 58.53
kJ/mol.
Question 18
Question
For a certain reaction, the rate constant doubles when the temperature is in-
creased from 25
°
C to 35
°
C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol
·
K), T= absolute temperature (in Kelvin).
15
Step 2: Given that the rate constant doubles when the temperature is in-
creased from 25
°
C to 35
°
C, we can use this information to set up a ratio of rate
constants: k2
k1
= 2 = e
Ea
R1
T1
−1
T2
where: k1= rate constant at 25
°
C, k2= rate constant at 35
°
C, T1= 25
°
C =
298 K, T2= 35
°
C = 308 K.
Step 3: Substitute the values into the equation:
2 = eEa
8.314 (1
298 −1
308 )
2 = eEa
8.314 (1
298 −1
308 )
Step 4: Solve for the activation energy Eaby taking the natural logarithm
of both sides:
ln(2) = Ea
8.314 1
298 −1
308
Step 5: Calculate the activation energy by rearranging the equation:
Ea= 8.314 ×1
298 −1
308×ln(2)
Step 6: Perform the calculations to find the activation energy.
Question 19
Question
The rate constant (k) for a certain reaction is described by the Arrhenius equa-
tion:
k=A·e−Ea
RT
where 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.
The rate constant for a reaction is 0.005 s−1at 25
°
C and 0.02 s−1at 45
°
C.
Determine the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25C+ 273.15 = 298.15 K
T2= 45C+ 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to create a system of equations based
on the given data.
k1=A·e−Ea
R·T1
16
k2=A·e−Ea
R·T2
Step 3: Divide the second equation by the first equation to eliminate A.
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
k2
k1
=eEa
R1
T1
−1
T2
Step 4: Substitute the given values for k1,k2,T1, and T2into the equation.
0.02
0.005 =e(Ea
8.314 (1
298.15 −1
318.15 ))
Step 5: Solve for the activation energy Ea.
1
4=e(Ea
8.314 ·(1
298.15 −1
318.15 ))
Step 6: Take the natural logarithm of both sides to solve for Ea.
ln 1
4=Ea
8.314 ·1
298.15 −1
318.15
Step 7: Calculate the activation energy Ea.
Ea= 8.314 ·1
298.15 −1
318.15·ln 1
4
Step 8: After performing the calculations, the activation energy Ea≈42.03
kJ/mol.
Question 20
Question
The rate constant of a certain reaction is found to be 5.0×10−5s−1at 25◦C
and 1.2×10−3s−1at 55◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J mol−1K−1), - Tis the tem-
perature in Kelvin.
Step 2: We have two sets of data points:
17
At 25◦C (298 K):
k1= 5.0×10−5s−1
At 55◦C (328 K):
k2= 1.2×10−3s−1
Step 3: Set up two equations using the Arrhenius equation:
k1=A·e−Ea
R·298
k2=A·e−Ea
R·328
Step 4: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R·328
e−Ea
R·298
Step 5: Simplify the right side of the equation:
k2
k1
=eEa·(1
298 −1
328 )/R
Step 6: Solve for Ea:
Ea=−R·ln k2
k1·1
298 −1
328
Step 7: Plug in the values and calculate the activation energy:
Ea=−8.314 ·ln 1.2×10−3
5.0×10−5·1
298 −1
328
Question 21
Question
The rate constant for a reaction is found to be 1.50 ×10−3s−1at 25
°
C, and
6.32 ×10−3s−1at 45
°
C. Calculate the activation energy of the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. At 25
°
C, T1= 25 + 273.15 = 298.15 K. At 45
°
C, T2= 45 +
273.15 = 318.15 K.
Step 2: Write out the Arrhenius equation and take the ratio of the rate
constants. The Arrhenius equation is 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.
Taking the ratio of the rate constants, we get: k2
k1=Ae−Ea
RT2
Ae−Ea
RT1
=e(Ea
R)1
T1
−1
T2
18
Step 3: Substituting the given values into the equation and solving for Ea.
Given: k1= 1.50 ×10−3s−1,k2= 6.32 ×10−3s−1,T1= 298.15 K, T2= 318.15
K. 6.32×10−3
1.50×10−3=e(Ea
8.314 )( 1
298.15 −1
318.15 )4.2133 = e(Ea
8.314 )( 1
298.15 −1
318.15 )
Step 4: Solve for Ea. From the previous step, Ea
8.314 1
298.15 −1
318.15 =
ln(4.2133). Ea= 8.314 ×ln(4.2133) ×320.88
20×298.15 Ea≈52.1 kJ/mol.
Therefore, the activation energy of the reaction is approximately 52.1 kJ/mol.
Question 22
Question
The rate constant for a certain reaction is found to double when the temperature
is increased from 25
°
C to 45
°
C. Calculate the activation energy for this reaction.
Assume the frequency factor remains constant.
Solution
Step 1: Start by writing out the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol
K)), T= temperature in Kelvin.
Step 2: We are given that the rate constant doubles when the temperature is
increased from 25
°
C (298 K) to 45
°
C (318 K). We can express this relationship
as:
2k298 =k318
Substitute the Arrhenius equation into this relationship:
2Ae−Ea
R(298K)=Ae−Ea
R(318K)
Step 3: Simplify the equation by canceling out the frequency factor, A:
2e−Ea
8.314×298 =e−Ea
8.314×318
Step 4: Take the natural logarithm of both sides to solve for Ea:
ln(2) = −Ea
8.314 1
298 −1
318
Step 5: Calculate the activation energy, Ea, by solving for it in the equation
from the previous step:
Ea=−8.314 ×ln(2) 1
298 −1
318
19
Step 6: Calculate the value of Eausing the formula above:
Ea≈53.34 kJ/mol
Therefore, the activation energy for this reaction is approximately 53.34
kJ/mol.
Question 23
Question
The rate constant for a certain reaction is found to be 5.12 ×10−3s−1at 25◦C
and 0.0180 s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15.
25◦C = 25 + 273.15 = 298.15 K
50◦C = 50 + 273.15 = 323.15 K
Step 2: Take the natural logarithm of the rate constants to simplify the
Arrhenius equation.
ln k1= ln 5.12 ×10−3s−1
ln k2= ln 0.0180 s−1
Step 3: Use the Arrhenius equation k=A·e−Ea
RT and the information
provided to set up a system of equations.
ln k1= ln A−Ea
R·1
T1
ln k2= ln A−Ea
R·1
T2
Step 4: Subtract the second equation from the first to eliminate ln A.
ln k1−ln k2=−Ea
R·1
T1
−1
T2
ln k1
k2=Ea
R·1
T2
−1
T1
20
Step 5: Solve for the activation energy Ea.
Ea=−R·ln k1
k2·1
1
T2
−1
T1
=−8.314 J K−1·ln 5.12 ×10−3
0.0180 ·1
1
323.15 −1
298.15
= 70.2 kJ mol−1
Therefore, the activation energy for this reaction is 70.2 kJ mol−1.
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is 0.0050 mol−1L.s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 0.080 mol−1L.s−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= 25 + 273.15 = 298.15 K
At 50
°
C: T2= 50 + 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures and solve for the activation energy (Ea):
k2
k1
=e−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation:
0.080
0.0050 =e−Ea
8.314 (1
323.15 −1
298.15 )
Step 4: Solve for the activation energy (Ea):
16 = e−Ea
8.314 (1
323.15 −1
298.15 )
Step 5: Simplify the equation further:
16 = e−Ea
8.314 (298.15−323.15
298.15·323.15 )
16 = e−Ea
8.314 (−0.0840)
Step 6: Take the natural logarithm of both sides to solve for Ea:
ln(16) = ln e−0.0840·Ea
8.314
21
ln(16) = −0.0840 ·Ea
8.314
Step 7: Solve for Ea:
Ea=8.314 ·ln(16)
−0.0840
Ea≈8.314 ·2.7726
−0.0840
Ea≈23.075
−0.0840
Ea≈ −274.4 kJ/mol
Therefore, the activation energy for this reaction is approximately 274.4
kJ/mol.
Question 25
Question
The rate constant of a certain reaction was found to be 4.82 ×10−3s−1at
25◦C. When the temperature was increased to 45◦C, the rate constant became
1.59 ×10−2s−1. Calculate the activation energy for this reaction. (R = 8.314
J/(mol
·
K))
Solution
Step 1: Convert temperatures to Kelvin Since temperature must be in Kelvin to
use the Arrhenius equation, we convert 25◦C and 45◦C to Kelvin. 25◦C + 273 =
298 K 45◦C + 273 = 318 K
Step 2: Use the Arrhenius Equation The Arrhenius Equation relates the rate
constant to the activation energy:
k=A·e−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.
We can rearrange the equation to solve for the activation energy:
1
T1
=−Ea
R1
T2
−1
T1
Step 3: Solve for the activation energy Substitute the values into the equa-
tion:
1
298 =−Ea
8.314 1
318 −1
298
22
Solve for Ea:
Ea=−8.314 ×1
298 1
318 −1
298
Ea≈3.45 ×104J/mol
Question 26
Question
The rate constant for a reaction at 25
°
C is 4.82 ×10−3s−1. When the tempera-
ture is raised to 35
°
C, the rate constant becomes 2.56 ×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, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol·K)), - Tis the temperature
in Kelvin.
Step 2: Since the rate constant is dependent on temperature, we can set up
two equations based on the given data:
k1=A·e−Ea
R·298 K
k2=A·e−Ea
R·308 K
where: - k1= 4.82 ×10−3s−1at 25
°
C (298 K), - k2= 2.56 ×10−2s−1at 35
°
C
(308 K).
Step 3: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R·308 K
e−Ea
R·298 K
Step 4: Simplify the equation and solve for the activation energy Ea:
2.56 ×10−2
4.82 ×10−3=e−Ea
R(1
308 K −1
298 K )
Step 5: Calculate the activation energy Eausing the gas constant R=
8.314 J/(mol·K):
Ea=−R·ln 2.56 ×10−2
4.82 ×10−3 1
308 −1
298
Step 6: Plug in the values and calculate the activation energy.
23
Question 27
Question
The rate constant (k) for a certain reaction is found to be 4.5×10−3s−1at 25◦C
and 1.2×10−2s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol ·K)), T= temperature in Kelvin.
Step 2: Determine the values needed for the Arrhenius equation. First,
convert the temperatures to Kelvin:
T1= 25 + 273 = 298 K
T2= 50 + 273 = 323 K
Step 3: Substitute the given data into the Arrhenius equation: For T1:
4.5×10−3=Ae−Ea
8.314×298
For T2:
1.2×10−2=Ae−Ea
8.314×323
Step 4: Take the ratio of the two equations to eliminate A:
1.2×10−2
4.5×10−3=e−Ea
8.314×323
e−Ea
8.314×298
Step 5: Simplify the ratio and solve for Ea:
2.67
0.45 =e−Ea
8.314×323 +Ea
8.314×298
5.93 = e25Ea
8.314×298×323
ln(5.93) = 25Ea
8.314 1
298 −1
323
Step 6: Solve for Ea:
Ea=8.314 ×ln(5.93)
25 1
298 −1
323
Ea≈50.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 50.1
kJ/mol.
24
Question 28
Question
The rate constant of a first-order reaction is 6.2×10−4s−1at 25◦C and 1.4×
10−3s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Let’s start by writing the Arrhenius equation, which relates the rate
constant kto the temperature Tand the activation energy Ea:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= gas constant (8.314 J/mol/K) T= temperature in Kelvin
Step 2: We can rearrange the Arrhenius equation to solve for the activation
energy Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given rate constants and temperatures into the equa-
tion:
ln 1.4×10−3s−1
6.2×10−4s−1=−Ea
8.314 1
323 K −1
298 K
Step 4: Calculate the activation energy Ea:
ln 1.4
0.62=−Ea
8.314 298 −323
323 ×298
ln (2.258) = −Ea
8.314 × −0.000202
0.813 = 0.002065 ×Ea
Ea=0.813
0.002065
Ea≈393.74 kJ/mol
Step 5: Therefore, the activation energy for this reaction is approximately
393.74 kJ/mol.
Question 29
Question
The rate constant for the reaction 2A→B+Cis found to be 0.0050 s−1at
300 K and 0.020 s−1at 350 K. Calculate the activation energy (Eain kJ/mol)
for the reaction.
25
Solution
Step 1: Given data: The rate constants at two different temperatures are: -
k1= 0.0050 s−1at 300 K - k2= 0.020 s−1at 350 K
Step 2: The Arrhenius equation relates the rate constant of a reaction to
the temperature and the activation energy. The equation is:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/mol·K), - Tis the temperature
in Kelvin.
Step 3: Taking the natural logarithm of the Arrhenius equation gives:
ln(k) = ln(A)−Ea
R·1
T
Step 4: We have two data points, which can be used to generate two equa-
tions:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Rearranging the equations to solve for activation energy, we get:
Ea
R=1
T2
−1
T1
Ea=R·1
T2
−1
T1
Step 6: Calculate the activation energy using the given temperatures and
gas constant. Note that temperature must be in Kelvin.
Ea= 8.314 J/mol ·K·1
350 K −1
300 K
Step 7: Simplify the expression and convert the activation energy to kJ/mol.
Ea= 8.314 J/mol ·K·1
350 −1
300= 52.43 kJ/mol
Step 8: The activation energy for the reaction is 52.43 kJ/mol.
Question 30
Question
For a certain reaction, the rate constant at 350 K is 4.20 ×10−3s−1and at 400
K is 1.03 ×10−2s−1. Calculate the activation energy of the reaction in kJ/mol.
26
Solution
Step 1: Given that the rate constant k1at temperature T1is 4.20 ×10−3s−1
and at temperature T2is 1.03 ×10−2s−1, we can use the Arrhenius equation
to find the activation energy Ea.
k=A·e−Ea
RT
Step 2: Taking the natural logarithm of both sides, we can rewrite the
Arrhenius equation as:
ln k= ln A−Ea
RT
Step 3: Subtracting the equation at T2from the equation at T1, we get:
ln k2
k1
=Ea
R1
T1
−1
T2
Step 4: Substituting the values given: T1= 350 K, k1= 4.20 ×10−3s−1,
T2= 400 K, k2= 1.03 ×10−2s−1, and R= 8.314 J/(mol*K), we can solve for
Ea.
Step 5: Plugging in the values, we have:
ln 1.03 ×10−2
4.20 ×10−3=Ea
8.314 1
350 −1
400
Step 6: Calculating the left-hand side first:
ln 1.03 ×10−2
4.20 ×10−3= ln 1.03
4.20= ln(0.24524) ≈ −1.405
Step 7: Now, solving for Ea:
−1.405 = Ea
8.314 1
350 −1
400
Step 8: Calculate the difference in the reciprocal temperatures:
1
350 −1
400 = 0.002857 K−1
Step 9: Solve for Ea:
Ea=−1.405 ×8.314∇ · 0.002857 ≈ −4090 J/mol
Step 10: Converting the activation energy to kJ/mol:
Ea=−4090 J/mol = −4.09 kJ/mol
Therefore, the activation energy of the reaction is approximately 4.09 kJ/mol.
27
Question 31
Question
The rate constant for a certain reaction at 25
°
C is 6.2×10−3s−1. When the
temperature is increased to 35
°
C, the rate constant becomes 0.032 s−1. Calculate
the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol K)), T= temperature in Kelvin.
Step 2: Determine the rate constant at each temperature and convert the
temperatures to Kelvin: At 25
°
C: k1= 6.2×10−3s−1,T1= 25 + 273 = 298 K.
At 35
°
C: k2= 0.032 s−1,T2= 35 + 273 = 308 K.
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 4: Set up two equations using the rate constants and temperatures:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Subtract the two equations to eliminate ln(A):
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Solve for the activation energy Ea:
Ea=−R·ln(k2/k1)
1/T2−1/T1
Step 7: Substitute the given values and solve for Ea:
Ea=−8.314 ·ln(0.032/0.0062)
1/308 −1/298
Question 32
Question
For a certain reaction, the rate constant at 37
°
C is found to be 1.20 ×10−2
s−1and at 77
°
C the rate constant is 5.70 ×10−2s−1. Calculate the activation
energy for this reaction.
28
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant k
to 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
Step 2: Rearrange the Arrhenius equation to solve for the activation energy
Ea:
1
T1
=−Ea
R(1
T2
−1
T1
) + lnk
A
Step 3: Convert the temperatures from Celsius to Kelvin: - For 37
°
C, T1=
37C+ 273.15 = 310.15 K - For 77
°
C, T2= 77C+ 273.15 = 350.15 K
Step 4: Substitute known values into the equation:
1
310.15 =−Ea
8.314(1
350.15 −1
310.15) + ln5.70 ×10−2
1.20 ×10−2
Step 5: Solve for the activation energy Ea:
1
310.15 =−Ea
8.314(1
350.15 −1
310.15) + ln5.70 ×10−2
1.20 ×10−2
1
310.15 =−Ea
8.314(1
350.15 −1
310.15) + ln(4.75)
1
310.15 =−Ea
8.314(1
350.15 −1
310.15)+1.5606
1
310.15 =−Ea
8.314(1
350.15 −1
310.15)+1.5606
Ea≈82.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 82.5
kJ/mol.
Question 33
Question
The rate constant for a reaction is found to be 5.0×10−4s−1at 25◦C and
1.0×10−3s−1at 45◦C. Calculate the activation energy for this reaction. (Hint:
You may assume the pre-exponential factor remains constant.)
29
Solution
Step 1: Let’s start by writing the Arrhenius equation, which relates the rate
constant of a reaction to the temperature and 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 set up two equations using the rate constant data at 25◦C
and 45◦C. At 25◦C (298 K):
5.0×10−4=A·e−Ea
8.314×298
At 45◦C (318 K):
1.0×10−3=A·e−Ea
8.314×318
Step 3: Divide the second equation by the first to eliminate the pre-exponential
factor.
1.0×10−3
5.0×10−4=A·e−Ea
8.314×318
A·e−Ea
8.314×298
Step 4: Simplify the equation and solve for the activation energy Ea.
2 = eEa
8.314 (1
298 −1
318 )
Step 5: Take the natural logarithm of both sides to solve for Ea.
ln(2) = Ea
8.314 1
298 −1
318
Step 6: Solve for the activation energy Ea.
Ea= 8.314 ×ln(2)
1
298 −1
318
Step 7: Calculate the activation energy using the given values.
Question 34
Question
The rate constant for a reaction is found to be 4.20 ×10−3s−1at 50◦C and
1.60 ×10−2s−1at 80◦C. Calculate the activation energy of the reaction.
Given: Arrhenius constant, A = 6.50 ×102s−1Gas constant, R = 8.314
J/(mol·K)
30
Solution
Step 1: Convert the temperatures to Kelvin. Given: T1= 50◦C, T2= 80◦C,
R= 8.314 J/(mol·K).
Converting the temperatures to Kelvin: T1= 50 + 273.15 = 323.15 K,
T2= 80 + 273.15 = 353.15 K.
Step 2: Calculate the natural logarithm of the rate constants. Given: k1=
4.20 ×10−3s−1,k2= 1.60 ×10−2s−1.
Taking the natural logarithm of the rate constants: ln k1= ln4.20 ×10−3≈
−5.47, ln k2= ln1.60 ×10−2≈ −4.13.
Step 3: Use the Arrhenius equation to find the activation energy. The
Arrhenius equation is given by
ln k2
k1=−Ea
R1
T2
−1
T1
Substitute the known values:
−5.47 = −Ea
8.314 1
353.15 −1
323.15
1
8.314 1
353.15 −1
323.15≈6.885 ×10−3
Ea≈6.885 ×10−3×8.314 ≈57.19 kJ/mol
Therefore, the activation energy of the reaction is approximately 57.19 kJ/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 35
°
C, the rate constant becomes 3.75 ×10−3s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures to kelvin using the relationship T(K) =
T(C) + 273.15.
T1= 25C+ 273.15 = 298.15K
T2= 35C+ 273.15 = 308.15K
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation:
k=A×e−Ea
RT
31
where k1= 1.20 ×10−3s−1(rate constant at T1)
k2= 3.75 ×10−3s−1(rate constant at T2)
R= 8.314 J/(mol·K) (gas constant)
T1= 298.15 K
T2= 308.15 K
First, let’s solve for Aby taking the ratio of the rate constants:
k2
k1
=A×e−Ea
RT2
A×e−Ea
RT1
3.75 ×10−3
1.20 ×10−3=e−Ea
8.314×308.15
3.125 = e−Ea
2541.02
ln(3.125) = lne−Ea
2541.02
ln(3.125) = −Ea
2541.02
Ea=−2541.02 ×ln(3.125)
Step 3: Calculate the activation energy, Ea.
Ea=−2541.02 ×ln(3.125)
Ea≈ −2541.02 ×1.139
Ea≈ −2893.03 J/mol
Therefore, the activation energy for this reaction is approximately 2893.03
J/mol.
32
Question 4
Question
The rate constant for a certain reaction is found to be 1.25 ×10−3s−1at 25◦C
and 3.75 ×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation which relates the rate constant to
the temperature and activation energy.
ln k2
k1=−Ea
R1
T2
−1
T1
where: k2and k1are the rate constants at temperatures T2and T1,Eais the
activation energy, R= 8.314 J mol−1K−1is the gas constant.
Step 2: Plug in the given values to the Arrhenius equation.
ln 3.75 ×10−3
1.25 ×10−3=−Ea
8.314 1
318 −1
298
Step 3: Solve for the activation energy Ea. First, simplify the natural loga-
rithm term:
ln (3) = −Ea
8.314 (0.00334)
Step 4: Evaluate the natural logarithm.
1.0986 = −0.00334Ea
Step 5: Solve for the activation energy Ea.
Ea=1.0986
−0.00334 ≈ −328.7 kJ/mol
Therefore, the activation energy for this reaction is approximately -328.7
kJ/mol.
Question 5
Question
The rate constant for the reaction 2A→Bis 2.5×10−3s−1at 25
°
C and
7.5×10−3s−1at 35
°
C. Calculate the activation energy for this reaction. The
gas constant Ris 8.314 J/(mol
·
K).
4
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1= 2.5×10−3s−1at T1= 25C= 298 K, k2= 7.5×10−3s−1at
T2= 35C= 308 K, and R= 8.314 J/(mol
·
K).
Step 2: Substitute the known values into the equation:
ln 7.5×10−3
2.5×10−3=−Ea
8.314 1
308 −1
298
ln(3) = −Ea
8.314 1
308 −1
298
Step 3: Solve for the activation energy (Ea):
Ea=−8.314 ×ln(3)
1
308 −1
298
Ea≈77.09 kJ/mol
Therefore, the activation energy for the reaction is approximately 77.09
kJ/mol.
Question 6
Question
The rate constant for a certain reaction at 35
°
C is 1.8×10−3s−1, and at 45
°
C
it is 5.7×10−2s−1. Determine the activation energy for this reaction.
Given: R= 8.31 J/(mol·K), T1= 35
°
C, T2= 45
°
C
Solution
Step 1: Convert the temperatures from
°
C to K.
T1= 35 + 273 = 308 K
T2= 45 + 273 = 318 K
Step 2: Write the Arrhenius equation:
k=A·e−Ea
RT
5
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
R·1
T
This equation is in the form of y=mx +b, where y= ln k,x= 1/T ,m=−Ea
R,
and b= ln A.
Step 4: Use the data to create two equations based on the Arrhenius equa-
tion:
ln k1= ln A−Ea
R·1
T1
ln k2= ln A−Ea
R·1
T2
Step 5: Subtract the two equations to eliminate ln A:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Solve for the activation energy Ea:
Ea=−R·
ln k2
k1
1
T2
−1
T1
Substitute the given values into the equation to find the activation energy.
Question 7
Question
The rate constant for a certain reaction is found to be 4.32 ×10−2s−1at 25◦C
and 9.28 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
(Hint: Use the Arrhenius equation: k=A·e−Ea
RT )
Solution
Step 1: Convert the temperatures to Kelvin Given:
T1= 25◦C = 25 + 273 = 298 K
T2= 45◦C = 45 + 273 = 318 K
Step 2: Use the Arrhenius equation to set up two equations based on the
rate constants at two temperatures At T1= 298 K:
ln k1= ln A−Ea
RT1
6
ln4.32 ×10−2= ln A−Ea
8.314 J K−1mol−1·298 K
At T2= 318 K:
ln k2= ln A−Ea
RT2
ln9.28 ×10−2= ln A−Ea
8.314 J K−1mol−1·318 K
Step 3: Subtract the two equations to eliminate ln A
ln k2
k1=−Ea
8.314 J K−1mol−11
318 −1
298
Step 4: Solve for the activation energy, Ea
Ea=−8.314 J K−1mol−1·318 ·298
318 −298 ln 9.28 ×10−2
4.32 ×10−2
Calculating this expression gives the activation energy.
Question 8
Question
The rate constant for a certain reaction at 25
°
C is 1.25 ×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 3.75 ×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. At 25
°
C, T1= 25 + 273 = 298 K. At 50
°
C, T2= 50 + 273 = 323 K.
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
Given: k1= 1.25 ×10−3s−1k2= 3.75 ×10−2s−1T1= 298 K T2= 323 K
R= 8.314 J/(mol
·
K)
Substitute the values into the equation:
ln 3.75 ×10−2
1.25 ×10−3=−Ea
8.314 1
323 −1
298
Step 3: Simplify and solve for the activation energy (Ea).
ln (30) = −Ea
8.314 1
323 −1
298
7
ln (30) = −Ea
8.314 1
323 −1
298
Ea
8.314 =−ln (30)
1
323 −1
298
Ea= 8.314 ×ln (30)
1
323 −1
298
Calculating the activation energy:
Ea= 8.314 ×ln (30)
1
323 −1
298
Question 9
Question
For the reaction A →B, the rate constant at 27
°
C is 4.10×10−3s−1and the rate
constant at 57
°
C is 5.73 s−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.
T1= 27C+ 273.15 = 300.15 K
T2= 57C+ 273.15 = 330.15 K
Step 2: Write down the Arrhenius equation:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of both sides of the equation to linearize
it:
ln(k) = ln(A)−Ea
R1
T
Step 4: Set up two equations using the rate constants and temperatures
given:
ln(k1) = ln(A)−Ea
R1
T1
ln(k2) = ln(A)−Ea
R1
T2
Step 5: Subtract the second equation from the first to eliminate ln(A) and
solve for Ea:
ln(k1)−ln(k2) = Ea
R1
T2
−1
T1
8
Step 6: Substitute the given values and solve for Ea:
ln4.10 ×10−3−ln(5.73) = Ea
8.314 1
330.15 −1
300.15
−2.1591 = Ea
8.314 1
330.15 −1
300.15
Step 7: Solve for Ea:
Ea=−2.1591 ×8.314 ×1
330.15 −300.15
Ea≈48.5 kJ/mol
Therefore, the activation energy for the reaction A →B is approximately
48.5 kJ/mol.
Question 10
Question
The rate constant of a reaction at 25
°
C is 1.2×10−3s−1, and the rate constant
at 35
°
C is 4.8×10−3s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Calculate the rate constant ratio using the Arrhenius equation:
k2
k1
=e
Ea
R1
T1
−1
T2
Given: k1= 1.2×10−3s−1at 25◦C, k2= 4.8×10−3s−1at 35◦C, T1= 25+273 =
298 K, T2= 35 + 273 = 308 K, R= 8.314 J mol−1K−1.
Plugging in the values, we get:
4.8×10−3
1.2×10−3=eEa
8.314 (1
298 −1
308 )
Step 2: Solve for the activation energy, Ea.
4.8
1.2=eEa
8.314 (0.00336−0.00325)
Simplify:
4 = eEa
8.314 ×0.00011
Step 3: Take the natural logarithm of both sides to solve for Ea.
ln(4) = Ea
8.314 ×0.00011
9
Step 4: Solve for Ea.
ln(4)
0.00011 ×8.314 = Ea
Step 5: Calculate Eato find the activation energy for the reaction.
Question 11
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1and the
activation energy is 40 kJ/mol. Calculate the rate constant at 45
°
C for this
reaction.
Solution
Step 1: Let’s start by writing down the Arrhenius equation:
k=A·exp −Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - R= 8.314 J/mol ·K is the gas constant, and - Tis the
temperature in Kelvin.
Step 2: We are given that k1= 2.5×10−3s−1at T1= 25C= 298 K, and
Ea= 40 kJ/mol. We need to find k2at T2= 45C= 318 K.
Step 3: First, let’s express the rate constant k1in terms of the Arrhenius
equation using T1and solve for A:
k1=A·exp −Ea
R·T1
2.5×10−3=A·exp −40,000
8.314 ·298
A= 2.5×10−3·exp 40,000
8.314 ·298
A≈9.42 ×109s−1
Step 4: Now, substitute A,Ea,R, and T2into the Arrhenius equation to
find k2:
k2= 9.42 ×109·exp −40,000
8.314 ·318
k2≈6.91 ×10−3s−1
Step 5: Therefore, the rate constant at 45
°
C for this reaction is approxi-
mately 6.91 ×10−3s−1.
10
Question 12
Question
The rate constant for a reaction at 25
°
C is found to be 1.5×10−2s−1, while the
rate constant at 35
°
C is 5.0×10−2s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: We can use the Arrhenius equation to find the activation energy (Ea)
for this reaction. The Arrhenius equation is given by:
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−1K−1),
and - Tis the temperature in Kelvin.
Step 2: First, let’s convert the rate constants given at 25
°
C and 35
°
C to
Kelvin:
T1= 25C+ 273.15 = 298.15 K
T2= 35C+ 273.15 = 308.15 K
Step 3: Now we can set up two Arrhenius equations using the rate constants
and temperatures provided:
k1=A·e−Ea
R·298.15
k2=A·e−Ea
R·308.15
Step 4: Write the ratio of the two rate constant equations:
k1
k2
=A·e−Ea
R·298.15
A·e−Ea
R·308.15
Step 5: Simplify the equation by cancelling out the pre-exponential factors:
k1
k2
=eEa
R(1
298.15 −1
308.15 )
Step 6: Plug in the rate constants given to solve for the activation energy:
1.5×10−2
5.0×10−2=eEa
8.314 (1
298.15 −1
308.15 )
Step 7: Solve for the activation energy (Ea) using natural logarithms:
ln 1.5
5.0=Ea
8.314 1
298.15 −1
308.15
Step 8: Finally, calculate the activation energy from the equation obtained
in the previous step.
11
Question 13
Question
The rate constant for the following reaction is found to be 6.32 ×10−4s−1
at 50◦C and 1.48 ×10−3s−1at 75◦C. Calculate the activation energy of the
reaction.
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. It
is given by:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol K), T= temperature in Kelvin.
Step 2: Plug in the given information. We are given the rate constants kat
two different temperatures: At T1= 50◦C = 323 K, k1= 6.32 ×10−4s−1. At
T2= 75◦C = 348 K, k2= 1.48 ×10−3s−1.
Step 3: Take the natural logarithm of both sides of the Arrhenius equation.
ln(k) = ln(A)−Ea
RT
Step 4: Set up two equations using the data provided. From our given data,
we have two equations:
ln(k1) = ln(A)−Ea
R·T1
ln(k2) = ln(A)−Ea
R·T2
Step 5: Solve for the activation energy. Subtract the second equation from
the first to eliminate ln(A):
ln(k1)−ln(k2) = Ea
R1
T2
−1
T1
Step 6: Substitute the values and calculate the activation energy.
ln6.32 ×10−4−ln1.48 ×10−3=Ea
8.314 1
348 −1
323
1
323 −1
348 =Ea
8.314 1
348 −1
323
1
323 −1
348 =Ea
8.314 1
348 −1
323
Ea≈59993 J/mol
Therefore, the activation energy of the reaction is approximately 59.993 kJ/mol.
12
Question 14
Question
The rate constant for a certain reaction is 2.35×10−3s−1at 25◦C and 6.97×10−3
s−1at 40◦C. Calculate the activation energy for this reaction. The activation en-
ergy for the reaction with respect to the rate constant is given by the Arrhenius
equation:
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), and Tis the temperature
in Kelvin.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15. At 25◦C, T= 25+273.15 = 298.15 K. At 40◦C, T= 40+273.15 = 313.15
K.
Step 2: Substitute the data into the Arrhenius equation to form two equa-
tions: At 25◦C:
2.35 ×10−3=Ae−Ea
8.314·298.15
At 40◦C:
6.97 ×10−3=Ae−Ea
8.314·313.15
Step 3: Divide the two equations to eliminate A:
2.35 ×10−3
6.97 ×10−3=e−Ea
8.314·298.15
e−Ea
8.314·313.15
0.336 = eEa
8.314 (1
313.15 −1
298.15 )
Step 4: Take the natural logarithm of both sides to solve for the activation
energy (Ea):
ln(0.336) = Ea
8.314 1
313.15 −1
298.15
Ea=−8.314 ·ln(0.336) 1
313.15 −1
298.15
Step 5: Calculate Eausing a calculator:
Ea≈55.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 55.2
kJ/mol.
13
Question 15
Question
The rate constant of a certain reaction is found to be 2.5×10−3s−1at 25◦C
and 7.2×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·exp −Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314
Question
The rate constant (k) for a reaction is found to be 6.45 ×10−3s−1at a temper-
ature of 45◦C. When the temperature is increased to 65◦C, the rate constant
becomes 1.24 ×10−2s−1. Calculate the activation energy (in kJ/mol) for this
reaction.
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15. For T= 45◦C: T= 45 + 273.15 = 318.15 K.
For T= 65◦C: T= 65 + 273.15 = 338.15 K.
Step 2: Use the Arrhenius equation k=Ae−Ea
RT to find the activation energy
(Ea) of the reaction. Given: k1= 6.45 ×10−3s−1,T1= 318.15 K, k2= 1.24 ×
10−2s−1,T2= 338.15 K.
We can rewrite the Arrhenius equation as ln k2
k1=Ea
R1
T1
−1
T2where
R= 8.314 J/mol ·K is the gas constant.
Step 3: Substitute the given values and solve for Ea.
ln 1.24 ×10−2
6.45 ×10−3=Ea
8.314 1
318.15 −1
338.15
ln (1.9196) = Ea
8.314 (0.00314 −0.00296)
0.6525 = 0.000191 ∗Ea
Ea=0.6525
0.000191 ≈3418.85 J/mol
Step 4: Convert the activation energy from Joules to kilojoules. Ea=
3418.85 J/mol = 3.41885 kJ/mol.
Therefore, the activation energy for this reaction is approximately 3.42 kJ/mol.
14
Question 17
Question
The rate constant (k) for a certain chemical reaction at 25
°
C is found to be
9.23 ×10−3s−1. When the temperature is increased to 50
°
C, the rate con-
stant increases to 2.76 ×10−2s−1. Calculate the activation energy (Ea) for this
reaction in kJ/mol.
Solution
Step 1: Convert the given rate constants to Arrhenius form.
Step 2: Divide the second equation by the first to eliminate A and solve for
Ea.
Step 3: Solve for Eaby taking the natural logarithm of both sides and then
solve for Eain kJ/mol.
ln(3) = 1
298 −1
323·Ea
Ea=298 ·323
323 −298 ·ln(3)
Ea=96054
25 ·ln(3)
Ea≈58.53 kJ/mol
Therefore, the activation energy for this reaction is approximately 58.53
kJ/mol.
Question 18
Question
For a certain reaction, the rate constant doubles when the temperature is in-
creased from 25
°
C to 35
°
C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol
·
K), T= absolute temperature (in Kelvin).
15
Step 2: Given that the rate constant doubles when the temperature is in-
creased from 25
°
C to 35
°
C, we can use this information to set up a ratio of rate
constants: k2
k1
= 2 = e
Ea
R1
T1
−1
T2
where: k1= rate constant at 25
°
C, k2= rate constant at 35
°
C, T1= 25
°
C =
298 K, T2= 35
°
C = 308 K.
Step 3: Substitute the values into the equation:
2 = eEa
8.314 (1
298 −1
308 )
2 = eEa
8.314 (1
298 −1
308 )
Step 4: Solve for the activation energy Eaby taking the natural logarithm
of both sides:
ln(2) = Ea
8.314 1
298 −1
308
Step 5: Calculate the activation energy by rearranging the equation:
Ea= 8.314 ×1
298 −1
308×ln(2)
Step 6: Perform the calculations to find the activation energy.
Question 19
Question
The rate constant (k) for a certain reaction is described by the Arrhenius equa-
tion:
k=A·e−Ea
RT
where 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.
The rate constant for a reaction is 0.005 s−1at 25
°
C and 0.02 s−1at 45
°
C.
Determine the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25C+ 273.15 = 298.15 K
T2= 45C+ 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to create a system of equations based
on the given data.
k1=A·e−Ea
R·T1
16
k2=A·e−Ea
R·T2
Step 3: Divide the second equation by the first equation to eliminate A.
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
k2
k1
=eEa
R1
T1
−1
T2
Step 4: Substitute the given values for k1,k2,T1, and T2into the equation.
0.02
0.005 =e(Ea
8.314 (1
298.15 −1
318.15 ))
Step 5: Solve for the activation energy Ea.
1
4=e(Ea
8.314 ·(1
298.15 −1
318.15 ))
Step 6: Take the natural logarithm of both sides to solve for Ea.
ln 1
4=Ea
8.314 ·1
298.15 −1
318.15
Step 7: Calculate the activation energy Ea.
Ea= 8.314 ·1
298.15 −1
318.15·ln 1
4
Step 8: After performing the calculations, the activation energy Ea≈42.03
kJ/mol.
Question 20
Question
The rate constant of a certain reaction is found to be 5.0×10−5s−1at 25◦C
and 1.2×10−3s−1at 55◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J mol−1K−1), - Tis the tem-
perature in Kelvin.
Step 2: We have two sets of data points:
17
At 25◦C (298 K):
k1= 5.0×10−5s−1
At 55◦C (328 K):
k2= 1.2×10−3s−1
Step 3: Set up two equations using the Arrhenius equation:
k1=A·e−Ea
R·298
k2=A·e−Ea
R·328
Step 4: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R·328
e−Ea
R·298
Step 5: Simplify the right side of the equation:
k2
k1
=eEa·(1
298 −1
328 )/R
Step 6: Solve for Ea:
Ea=−R·ln k2
k1·1
298 −1
328
Step 7: Plug in the values and calculate the activation energy:
Ea=−8.314 ·ln 1.2×10−3
5.0×10−5·1
298 −1
328
Question 21
Question
The rate constant for a reaction is found to be 1.50 ×10−3s−1at 25
°
C, and
6.32 ×10−3s−1at 45
°
C. Calculate the activation energy of the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. At 25
°
C, T1= 25 + 273.15 = 298.15 K. At 45
°
C, T2= 45 +
273.15 = 318.15 K.
Step 2: Write out the Arrhenius equation and take the ratio of the rate
constants. The Arrhenius equation is 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.
Taking the ratio of the rate constants, we get: k2
k1=Ae−Ea
RT2
Ae−Ea
RT1
=e(Ea
R)1
T1
−1
T2
18
Step 3: Substituting the given values into the equation and solving for Ea.
Given: k1= 1.50 ×10−3s−1,k2= 6.32 ×10−3s−1,T1= 298.15 K, T2= 318.15
K. 6.32×10−3
1.50×10−3=e(Ea
8.314 )( 1
298.15 −1
318.15 )4.2133 = e(Ea
8.314 )( 1
298.15 −1
318.15 )
Step 4: Solve for Ea. From the previous step, Ea
8.314 1
298.15 −1
318.15 =
ln(4.2133). Ea= 8.314 ×ln(4.2133) ×320.88
20×298.15 Ea≈52.1 kJ/mol.
Therefore, the activation energy of the reaction is approximately 52.1 kJ/mol.
Question 22
Question
The rate constant for a certain reaction is found to double when the temperature
is increased from 25
°
C to 45
°
C. Calculate the activation energy for this reaction.
Assume the frequency factor remains constant.
Solution
Step 1: Start by writing out the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol
K)), T= temperature in Kelvin.
Step 2: We are given that the rate constant doubles when the temperature is
increased from 25
°
C (298 K) to 45
°
C (318 K). We can express this relationship
as:
2k298 =k318
Substitute the Arrhenius equation into this relationship:
2Ae−Ea
R(298K)=Ae−Ea
R(318K)
Step 3: Simplify the equation by canceling out the frequency factor, A:
2e−Ea
8.314×298 =e−Ea
8.314×318
Step 4: Take the natural logarithm of both sides to solve for Ea:
ln(2) = −Ea
8.314 1
298 −1
318
Step 5: Calculate the activation energy, Ea, by solving for it in the equation
from the previous step:
Ea=−8.314 ×ln(2) 1
298 −1
318
19
Step 6: Calculate the value of Eausing the formula above:
Ea≈53.34 kJ/mol
Therefore, the activation energy for this reaction is approximately 53.34
kJ/mol.
Question 23
Question
The rate constant for a certain reaction is found to be 5.12 ×10−3s−1at 25◦C
and 0.0180 s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15.
25◦C = 25 + 273.15 = 298.15 K
50◦C = 50 + 273.15 = 323.15 K
Step 2: Take the natural logarithm of the rate constants to simplify the
Arrhenius equation.
ln k1= ln 5.12 ×10−3s−1
ln k2= ln 0.0180 s−1
Step 3: Use the Arrhenius equation k=A·e−Ea
RT and the information
provided to set up a system of equations.
ln k1= ln A−Ea
R·1
T1
ln k2= ln A−Ea
R·1
T2
Step 4: Subtract the second equation from the first to eliminate ln A.
ln k1−ln k2=−Ea
R·1
T1
−1
T2
ln k1
k2=Ea
R·1
T2
−1
T1
20
Step 5: Solve for the activation energy Ea.
Ea=−R·ln k1
k2·1
1
T2
−1
T1
=−8.314 J K−1·ln 5.12 ×10−3
0.0180 ·1
1
323.15 −1
298.15
= 70.2 kJ mol−1
Therefore, the activation energy for this reaction is 70.2 kJ mol−1.
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is 0.0050 mol−1L.s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 0.080 mol−1L.s−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= 25 + 273.15 = 298.15 K
At 50
°
C: T2= 50 + 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures and solve for the activation energy (Ea):
k2
k1
=e−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation:
0.080
0.0050 =e−Ea
8.314 (1
323.15 −1
298.15 )
Step 4: Solve for the activation energy (Ea):
16 = e−Ea
8.314 (1
323.15 −1
298.15 )
Step 5: Simplify the equation further:
16 = e−Ea
8.314 (298.15−323.15
298.15·323.15 )
16 = e−Ea
8.314 (−0.0840)
Step 6: Take the natural logarithm of both sides to solve for Ea:
ln(16) = ln e−0.0840·Ea
8.314
21
ln(16) = −0.0840 ·Ea
8.314
Step 7: Solve for Ea:
Ea=8.314 ·ln(16)
−0.0840
Ea≈8.314 ·2.7726
−0.0840
Ea≈23.075
−0.0840
Ea≈ −274.4 kJ/mol
Therefore, the activation energy for this reaction is approximately 274.4
kJ/mol.
Question 25
Question
The rate constant of a certain reaction was found to be 4.82 ×10−3s−1at
25◦C. When the temperature was increased to 45◦C, the rate constant became
1.59 ×10−2s−1. Calculate the activation energy for this reaction. (R = 8.314
J/(mol
·
K))
Solution
Step 1: Convert temperatures to Kelvin Since temperature must be in Kelvin to
use the Arrhenius equation, we convert 25◦C and 45◦C to Kelvin. 25◦C + 273 =
298 K 45◦C + 273 = 318 K
Step 2: Use the Arrhenius Equation The Arrhenius Equation relates the rate
constant to the activation energy:
k=A·e−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.
We can rearrange the equation to solve for the activation energy:
1
T1
=−Ea
R1
T2
−1
T1
Step 3: Solve for the activation energy Substitute the values into the equa-
tion:
1
298 =−Ea
8.314 1
318 −1
298
22
Solve for Ea:
Ea=−8.314 ×1
298 1
318 −1
298
Ea≈3.45 ×104J/mol
Question 26
Question
The rate constant for a reaction at 25
°
C is 4.82 ×10−3s−1. When the tempera-
ture is raised to 35
°
C, the rate constant becomes 2.56 ×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, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol·K)), - Tis the temperature
in Kelvin.
Step 2: Since the rate constant is dependent on temperature, we can set up
two equations based on the given data:
k1=A·e−Ea
R·298 K
k2=A·e−Ea
R·308 K
where: - k1= 4.82 ×10−3s−1at 25
°
C (298 K), - k2= 2.56 ×10−2s−1at 35
°
C
(308 K).
Step 3: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R·308 K
e−Ea
R·298 K
Step 4: Simplify the equation and solve for the activation energy Ea:
2.56 ×10−2
4.82 ×10−3=e−Ea
R(1
308 K −1
298 K )
Step 5: Calculate the activation energy Eausing the gas constant R=
8.314 J/(mol·K):
Ea=−R·ln 2.56 ×10−2
4.82 ×10−3 1
308 −1
298
Step 6: Plug in the values and calculate the activation energy.
23
Question 27
Question
The rate constant (k) for a certain reaction is found to be 4.5×10−3s−1at 25◦C
and 1.2×10−2s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol ·K)), T= temperature in Kelvin.
Step 2: Determine the values needed for the Arrhenius equation. First,
convert the temperatures to Kelvin:
T1= 25 + 273 = 298 K
T2= 50 + 273 = 323 K
Step 3: Substitute the given data into the Arrhenius equation: For T1:
4.5×10−3=Ae−Ea
8.314×298
For T2:
1.2×10−2=Ae−Ea
8.314×323
Step 4: Take the ratio of the two equations to eliminate A:
1.2×10−2
4.5×10−3=e−Ea
8.314×323
e−Ea
8.314×298
Step 5: Simplify the ratio and solve for Ea:
2.67
0.45 =e−Ea
8.314×323 +Ea
8.314×298
5.93 = e25Ea
8.314×298×323
ln(5.93) = 25Ea
8.314 1
298 −1
323
Step 6: Solve for Ea:
Ea=8.314 ×ln(5.93)
25 1
298 −1
323
Ea≈50.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 50.1
kJ/mol.
24
Question 28
Question
The rate constant of a first-order reaction is 6.2×10−4s−1at 25◦C and 1.4×
10−3s−1at 50◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Let’s start by writing the Arrhenius equation, which relates the rate
constant kto the temperature Tand the activation energy Ea:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= gas constant (8.314 J/mol/K) T= temperature in Kelvin
Step 2: We can rearrange the Arrhenius equation to solve for the activation
energy Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given rate constants and temperatures into the equa-
tion:
ln 1.4×10−3s−1
6.2×10−4s−1=−Ea
8.314 1
323 K −1
298 K
Step 4: Calculate the activation energy Ea:
ln 1.4
0.62=−Ea
8.314 298 −323
323 ×298
ln (2.258) = −Ea
8.314 × −0.000202
0.813 = 0.002065 ×Ea
Ea=0.813
0.002065
Ea≈393.74 kJ/mol
Step 5: Therefore, the activation energy for this reaction is approximately
393.74 kJ/mol.
Question 29
Question
The rate constant for the reaction 2A→B+Cis found to be 0.0050 s−1at
300 K and 0.020 s−1at 350 K. Calculate the activation energy (Eain kJ/mol)
for the reaction.
25
Solution
Step 1: Given data: The rate constants at two different temperatures are: -
k1= 0.0050 s−1at 300 K - k2= 0.020 s−1at 350 K
Step 2: The Arrhenius equation relates the rate constant of a reaction to
the temperature and the activation energy. The equation is:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/mol·K), - Tis the temperature
in Kelvin.
Step 3: Taking the natural logarithm of the Arrhenius equation gives:
ln(k) = ln(A)−Ea
R·1
T
Step 4: We have two data points, which can be used to generate two equa-
tions:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Rearranging the equations to solve for activation energy, we get:
Ea
R=1
T2
−1
T1
Ea=R·1
T2
−1
T1
Step 6: Calculate the activation energy using the given temperatures and
gas constant. Note that temperature must be in Kelvin.
Ea= 8.314 J/mol ·K·1
350 K −1
300 K
Step 7: Simplify the expression and convert the activation energy to kJ/mol.
Ea= 8.314 J/mol ·K·1
350 −1
300= 52.43 kJ/mol
Step 8: The activation energy for the reaction is 52.43 kJ/mol.
Question 30
Question
For a certain reaction, the rate constant at 350 K is 4.20 ×10−3s−1and at 400
K is 1.03 ×10−2s−1. Calculate the activation energy of the reaction in kJ/mol.
26
Solution
Step 1: Given that the rate constant k1at temperature T1is 4.20 ×10−3s−1
and at temperature T2is 1.03 ×10−2s−1, we can use the Arrhenius equation
to find the activation energy Ea.
k=A·e−Ea
RT
Step 2: Taking the natural logarithm of both sides, we can rewrite the
Arrhenius equation as:
ln k= ln A−Ea
RT
Step 3: Subtracting the equation at T2from the equation at T1, we get:
ln k2
k1
=Ea
R1
T1
−1
T2
Step 4: Substituting the values given: T1= 350 K, k1= 4.20 ×10−3s−1,
T2= 400 K, k2= 1.03 ×10−2s−1, and R= 8.314 J/(mol*K), we can solve for
Ea.
Step 5: Plugging in the values, we have:
ln 1.03 ×10−2
4.20 ×10−3=Ea
8.314 1
350 −1
400
Step 6: Calculating the left-hand side first:
ln 1.03 ×10−2
4.20 ×10−3= ln 1.03
4.20= ln(0.24524) ≈ −1.405
Step 7: Now, solving for Ea:
−1.405 = Ea
8.314 1
350 −1
400
Step 8: Calculate the difference in the reciprocal temperatures:
1
350 −1
400 = 0.002857 K−1
Step 9: Solve for Ea:
Ea=−1.405 ×8.314∇ · 0.002857 ≈ −4090 J/mol
Step 10: Converting the activation energy to kJ/mol:
Ea=−4090 J/mol = −4.09 kJ/mol
Therefore, the activation energy of the reaction is approximately 4.09 kJ/mol.
27
Question 31
Question
The rate constant for a certain reaction at 25
°
C is 6.2×10−3s−1. When the
temperature is increased to 35
°
C, the rate constant becomes 0.032 s−1. Calculate
the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol K)), T= temperature in Kelvin.
Step 2: Determine the rate constant at each temperature and convert the
temperatures to Kelvin: At 25
°
C: k1= 6.2×10−3s−1,T1= 25 + 273 = 298 K.
At 35
°
C: k2= 0.032 s−1,T2= 35 + 273 = 308 K.
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 4: Set up two equations using the rate constants and temperatures:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Subtract the two equations to eliminate ln(A):
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Solve for the activation energy Ea:
Ea=−R·ln(k2/k1)
1/T2−1/T1
Step 7: Substitute the given values and solve for Ea:
Ea=−8.314 ·ln(0.032/0.0062)
1/308 −1/298
Question 32
Question
For a certain reaction, the rate constant at 37
°
C is found to be 1.20 ×10−2
s−1and at 77
°
C the rate constant is 5.70 ×10−2s−1. Calculate the activation
energy for this reaction.
28
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant k
to 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
Step 2: Rearrange the Arrhenius equation to solve for the activation energy
Ea:
1
T1
=−Ea
R(1
T2
−1
T1
) + lnk
A
Step 3: Convert the temperatures from Celsius to Kelvin: - For 37
°
C, T1=
37C+ 273.15 = 310.15 K - For 77
°
C, T2= 77C+ 273.15 = 350.15 K
Step 4: Substitute known values into the equation:
1
310.15 =−Ea
8.314(1
350.15 −1
310.15) + ln5.70 ×10−2
1.20 ×10−2
Step 5: Solve for the activation energy Ea:
1
310.15 =−Ea
8.314(1
350.15 −1
310.15) + ln5.70 ×10−2
1.20 ×10−2
1
310.15 =−Ea
8.314(1
350.15 −1
310.15) + ln(4.75)
1
310.15 =−Ea
8.314(1
350.15 −1
310.15)+1.5606
1
310.15 =−Ea
8.314(1
350.15 −1
310.15)+1.5606
Ea≈82.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 82.5
kJ/mol.
Question 33
Question
The rate constant for a reaction is found to be 5.0×10−4s−1at 25◦C and
1.0×10−3s−1at 45◦C. Calculate the activation energy for this reaction. (Hint:
You may assume the pre-exponential factor remains constant.)
29
Solution
Step 1: Let’s start by writing the Arrhenius equation, which relates the rate
constant of a reaction to the temperature and 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 set up two equations using the rate constant data at 25◦C
and 45◦C. At 25◦C (298 K):
5.0×10−4=A·e−Ea
8.314×298
At 45◦C (318 K):
1.0×10−3=A·e−Ea
8.314×318
Step 3: Divide the second equation by the first to eliminate the pre-exponential
factor.
1.0×10−3
5.0×10−4=A·e−Ea
8.314×318
A·e−Ea
8.314×298
Step 4: Simplify the equation and solve for the activation energy Ea.
2 = eEa
8.314 (1
298 −1
318 )
Step 5: Take the natural logarithm of both sides to solve for Ea.
ln(2) = Ea
8.314 1
298 −1
318
Step 6: Solve for the activation energy Ea.
Ea= 8.314 ×ln(2)
1
298 −1
318
Step 7: Calculate the activation energy using the given values.
Question 34
Question
The rate constant for a reaction is found to be 4.20 ×10−3s−1at 50◦C and
1.60 ×10−2s−1at 80◦C. Calculate the activation energy of the reaction.
Given: Arrhenius constant, A = 6.50 ×102s−1Gas constant, R = 8.314
J/(mol·K)
30
Solution
Step 1: Convert the temperatures to Kelvin. Given: T1= 50◦C, T2= 80◦C,
R= 8.314 J/(mol·K).
Converting the temperatures to Kelvin: T1= 50 + 273.15 = 323.15 K,
T2= 80 + 273.15 = 353.15 K.
Step 2: Calculate the natural logarithm of the rate constants. Given: k1=
4.20 ×10−3s−1,k2= 1.60 ×10−2s−1.
Taking the natural logarithm of the rate constants: ln k1= ln4.20 ×10−3≈
−5.47, ln k2= ln1.60 ×10−2≈ −4.13.
Step 3: Use the Arrhenius equation to find the activation energy. The
Arrhenius equation is given by
ln k2
k1=−Ea
R1
T2
−1
T1
Substitute the known values:
−5.47 = −Ea
8.314 1
353.15 −1
323.15
1
8.314 1
353.15 −1
323.15≈6.885 ×10−3
Ea≈6.885 ×10−3×8.314 ≈57.19 kJ/mol
Therefore, the activation energy of the reaction is approximately 57.19 kJ/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 35
°
C, the rate constant becomes 3.75 ×10−3s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures to kelvin using the relationship T(K) =
T(C) + 273.15.
T1= 25C+ 273.15 = 298.15K
T2= 35C+ 273.15 = 308.15K
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation:
k=A×e−Ea
RT
31
where k1= 1.20 ×10−3s−1(rate constant at T1)
k2= 3.75 ×10−3s−1(rate constant at T2)
R= 8.314 J/(mol·K) (gas constant)
T1= 298.15 K
T2= 308.15 K
First, let’s solve for Aby taking the ratio of the rate constants:
k2
k1
=A×e−Ea
RT2
A×e−Ea
RT1
3.75 ×10−3
1.20 ×10−3=e−Ea
8.314×308.15
3.125 = e−Ea
2541.02
ln(3.125) = lne−Ea
2541.02
ln(3.125) = −Ea
2541.02
Ea=−2541.02 ×ln(3.125)
Step 3: Calculate the activation energy, Ea.
Ea=−2541.02 ×ln(3.125)
Ea≈ −2541.02 ×1.139
Ea≈ −2893.03 J/mol
Therefore, the activation energy for this reaction is approximately 2893.03
J/mol.
32
