CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 1
Liberty University
Question 1
Question
The rate constant for the reaction 2N2O5(g)→4N O2(g) + O2(g) is found to
be 8.00 ×10−2mol−1L s−1at 350 K and 5.00 ×10−1mol−1L s−1at 400 K.
Calculate the activation energy for this reaction.
Solution
To find the activation energy for the reaction, we can use the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the
temperature in Kelvin.
First, we need to find the pre-exponential factor Aby rearranging the Ar-
rhenius equation:
A=k·eEa
RT
Step 1: Calculate Aat 350 K:
A350 = 8.00 ×10−2mol−1L s−1·e
Ea
8.314 J/(mol
·
K)·350 K
A350 = 8.00 ×10−2mol−1L s−1·e
Ea
2920 J/mol
Step 2: Calculate Aat 400 K:
A400 = 5.00 ×10−1mol−1L s−1·e
Ea
8.314 J/(mol
·
K)·400 K
A400 = 5.00 ×10−1mol−1L s−1·e
Ea
3320 J/mol
Step 3: Set up a ratio of the two rate constants and pre-exponential factors
to find the activation energy:
8.00 ×10−2
5.00 ×10−1=A350 ·e−Ea
2920
A400 ·e−Ea
3320
0.16 = A350
A400
·eEa
400 −Ea
350
0.16 = A350
A400
·e50Ea
140000
Step 4: Solve for Eaby isolating it on one side:
e50Ea
140000 =0.16 ·A400
A350
50Ea
140000 = ln 0.16 ·A400
A350
Ea=
140000 ·ln 0.16·A400
A350
50
Therefore, the activation energy for the reaction is Ea=140000·ln0.16·A400
A350
50
J/mol.
Question 2
Question
The rate constant for the reaction 2A →B was found to be 4.5×10−3s−1at
298 K and 2.5×10−2s−1at 318 K. Determine the activation energy for this
reaction.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
where: k1= 4.5×10−3s−1at T1= 298 K, k2= 2.5×10−2s−1at T2= 318 K,
R= 8.314 J ·mol−1·K−1.
2
Step 2: Set up the Arrhenius equation for the two temperatures:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 3: Take the ratio of the two equations to eliminate A:
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 4: Simplify the ratio:
k2
k1
=e
Ea
R1
T1
−1
T2
Step 5: Substitute the given values and solve for Ea:
2.5×10−2
4.5×10−3=eEa
8.314 (1
298 −1
318 )
Step 6: Calculate the activation energy Ea.
Question 3
Question
The rate constant of a reaction at 300 K is 4.5×10−3s−1, and its rate constant
at 340 K is 1.8×10−2s−1. Calculate the activation energy for this reaction.
(Hint: Use the Arrhenius equation: k=Ae−Ea
RT )
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 2: Take the natural logarithm of the equation to linearize it:
ln(k) = ln(A)−Ea
R×1
T
Step 3: Set up two equations using the rate constants and temperatures given:
ln4.5×10−3= ln(A)−Ea
R×1
300
ln1.8×10−2= ln(A)−Ea
R×1
340
3
Step 4: Subtract the two equations to eliminate ln(A):
ln1.8×10−2−ln4.5×10−3=−Ea
R×1
340 −1
300
Step 5: Solve for the activation energy Ea:
∆(ln(k)) = −Ea
R×1
340 −1
300
Ea=−R×∆(ln(k))
1
340 −1
300
Step 6: Plug in the given values of ∆(ln(k)) = ln1.8×10−2−ln4.5×10−3,
R= 8.314 J/(mol·K), and solve for Ea. Remember to convert temperatures to
Kelvin. Step 7: Calculate the activation energy:
Ea=−8.314 ×(ln1.8×10−2−ln4.5×10−3)
1
340 −1
300
Question 4
Question
A certain reaction has an activation energy of 75 kJ/mol and a rate constant of
4.2×10−3s−1at 25◦C. Calculate the rate constant at 50◦C for this reaction.
Solution
Let’s denote the rate constant at 25◦C as k1and the rate constant at 50◦C
as k2. According to the Arrhenius equation, the relationship between the rate
constants and temperature is given by:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy
R= gas constant T= temperature in Kelvin
First, we need to convert the given activation energy to joules:
Ea= 75 kJ/mol ×1000 J/kJ = 75000 J/mol
Step 1: Calculate k1at 25◦C. Using the Arrhenius equation, we have:
k1=A·e−Ea
RT1
Substitute the known values:
4.2×10−3s−1=A·e−75000 J/mol
(8.314 J/mol·K)(25+273.15) K
4
Solve for Ato find k1.
Step 2: Calculate k2at 50◦C. Using the Arrhenius equation again, we have:
k2=A·e−Ea
RT2
Substitute the known values and the previously calculated A:
k2= (Afrom Step 1) ·e−75000 J/mol
(8.314 J/mol·K)(50+273.15) K
Calculate k2to determine the rate constant at 50◦C.
Question 5
Question
The rate constant for a reaction at 25
°
C is 3.67 ×10−3s−1. When the tempera-
ture is increased to 35
°
C, the rate constant becomes 1.45 ×10−2s−1. Calculate
the activation energy for this reaction in kJ/mol.
Solution
Step 1: Write down the Arrhenius equation relating the rate constant to tem-
perature:
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 tem-
perature in Kelvin.
Step 2: Express the ratio of rate constants in terms of the Arrhenius equa-
tion:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
Step 3: Take the natural logarithm of both sides to simplify the equation:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Plug in the values for k1,k2,T1, and T2into the equation. Remember
to convert the temperatures to Kelvin:
ln 1.45 ×10−2
3.67 ×10−3=−Ea
8.314 1
308 −1
298
Step 5: Solve for the activation energy, Ea, in Joules. Then convert this
value to kilojoules:
Ea=−8.314 ×ln 1.45 ×10−2
3.67 ×10−3×1
308 −1
298
5
Ea= [Y OURCALCU LAT EDV ALUE] J/mol
Ea= [Y OURCALCU LAT EDV ALUE] kJ/mol
Thus, the activation energy for this reaction is [YOUR ANSWER] kJ/mol.
Question 6
Question
The rate constant for the reaction A →B is found to be 4.2×10−3s−1at
25◦C. When the temperature is increased to 45◦C, the rate constant becomes
9.6×10−3s−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: Use the data provided to create two equations based on the Arrhenius
equation at 25◦C and 45◦C. At 25◦C:
4.2×10−3=A·e−Ea
8.314·(25+273.15)
At 45◦C:
9.6×10−3=A·e−Ea
8.314·(45+273.15)
Step 3: Divide the two equations to eliminate A:
4.2×10−3
9.6×10−3=e−Ea
8.314·(25+273.15)
e−Ea
8.314·(45+273.15)
Step 4: Simplify the equation to solve for Ea:
4.2
9.6=eEa
8.314 (1
25+273.15 −1
45+273.15 )
Step 5: Solve for Ea:
Ea=−8.314 ·ln 4.17
9.6·1
298 −1
318
Step 6: Calculate the value of Ea:
Ea≈48.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.5
kJ/mol.
6
Question 7
Question
The rate constant for a certain reaction is found to be 2.5×10−3s−1at 25◦C
and 5.0×10−3s−1at 45◦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= temperature in Kelvin.
Step 2: Set up two equations using the rate constants at different tempera-
tures:
2.5×10−3=A·e−Ea
R·(25+273) and 5.0×10−3=A·e−Ea
R·(45+273)
Step 3: Take the ratio of the two equations to eliminate A:
2.5×10−3
5.0×10−3=e−Ea
R·298
e−Ea
R·318
Step 4: Simplify the equation:
1
2=e20Ea
8.314·318 −20Ea
8.314·298
Step 5: Combine exponents and solve for Ea:
1
2=e−20Ea
8.314·9562
Step 6: Solve for Ea:
Ea=−8.314 ·9562
20 ·ln(0.5) ≈39887 J/mol
Therefore, the activation energy for this reaction is approximately 39.9
kJ/mol.
Question 8
Question
The rate constant for a reaction is found to be 6.2×10−3s−1at 335 K and
2.4×10−2s−1at 370 K. Calculate the activation energy for the reaction.
7
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: Using the given data at two different temperatures (335 K and 370
K), set up two separate equations using the Arrhenius equation:
k1=A·e−Ea
R·335
k2=A·e−Ea
R·370
Step 3: Divide the two equations to eliminate the pre-exponential factor A:
k2
k1
=e−Ea
R·370
e−Ea
R·335
Step 4: Simplify the expression by subtracting the exponents:
k2
k1
=eEa
R(1
335 −1
370 )
Step 5: Solve for Ea:k2
k1
=eEa
R(1
335 −1
370 )
ln k2
k1=Ea
R1
335 −1
370
Ea=R·
ln k2
k1
1
335 −1
370
Step 6: Substitute the given values to find Ea:
Ea= 8.314 ·
ln 2.4×10−2
6.2×10−3
1
335 −1
370
Step 7: Calculate Eato find the activation energy for the reaction.
Question 9
Question
The rate constant for a certain reaction at 25
°
C is 3.21 ×10−3s−1. When
the temperature is raised to 65
°
C, the rate constant is found to be 0.0462 s−1.
Calculate the activation energy of the reaction in kJ/mol.
8
Solution
Step 1: Let’s begin by writing the Arrhenius equation, which relates the rate
constant of a reaction to the temperature and 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), - Tis the temperature
in Kelvin.
Step 2: We have rate constants at two temperatures, so we can set up two
equations using the Arrhenius equation:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 3: Taking the ratio of the two rate constants, we can eliminate the
pre-exponential factor A:
k2
k1
=e−Ea
RT2
e−Ea
RT1
k2
k1
=e
Ea
R(1
T1
−1
T2)
Step 4: Now, plug in the values for the rate constants at 25
°
C and 65
°
C:
0.0462
3.21 ×10−3=eEa
8.314 (1
298 −1
338 )
Step 5: Solve for the activation energy Eato find the answer in kJ/mol.
Question 10
Question
The rate constant for the reaction 2A+B→Cat 25
°
C is 3.2×10−3M−1
s−1. When the temperature is increased to 50
°
C, the rate constant becomes
1.2×10−2M−1s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J/(mol K)), - Tis the
temperature in Kelvin.
9
Step 2: Set up the Arrhenius equation at 25
°
C
k1=Ae−Ea
RT1
where k1= 3.2×10−3M−1s−1and T1= 25 + 273 = 298 K.
Step 3: Set up the Arrhenius equation at 50
°
C
k2=Ae−Ea
RT2
where k2= 1.2×10−2M−1s−1and T2= 50 + 273 = 323 K.
Step 4: Take the ratio of the two Arrhenius equations to eliminate A
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R1
T1
−1
T2
Step 5: Solve for the activation energy Ea
k2
k1
=e−Ea
8.314 (1
298 −1
323 )
ln k2
k1=−Ea
8.314 1
298 −1
323
Step 6: Calculate the activation energy Ea
Ea=−8.314 ×ln k2
k1 1
298 −1
323
Ea≈70.1 kJ/mol
Therefore, the activation energy for the reaction is approximately 70.1 kJ/mol.
Question 11
Question
The rate constant of a reaction is 3.25 ×10−3s−1at 25◦C and 1.04 ×10−2s−1
at 35◦C. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Convert the given temperatures from Celsius to Kelvin using the equa-
tion T(K) = T(
°
C) + 273.15. At 25◦C, T= 25 + 273.15 = 298.15 K. At 35◦C,
T= 35 + 273.15 = 308.15 K.
10
Step 2: Calculate the rate constant ratio, k2/k1, using the given rate con-
stants at the two temperatures.
k2
k1
=1.04 ×10−2s−1
3.25 ×10−3s−1= 3.2
Step 3: Use the Arrhenius equation k=Ae−Ea
RT to relate the rate constants
and temperatures at the two temperatures.
3.2 = e−Ea
R(1
308.15 −1
298.15 )
Step 4: Solve for the activation energy, Ea, in J/mol using the ideal gas
constant R= 8.314 J/(mol ·K).
3.2 = e−Ea
8.314 (1
308.15 −1
298.15 )
Step 5: Convert the activation energy from joules to kilojoules by dividing
by 1000.
Ea=−8.314 ×ln(3.2) ×1
308.15 −1
298.15×1000
So, the activation energy for this reaction is approximately 53.5 kJ/mol.
Question 12
Question
The rate constant of a certain reaction is found to be 5.72 ×10−3s−1at 25◦C.
When the temperature is increased to 45◦C, the rate constant becomes 8.91 ×
10−2s−1. Calculate the activation energy (Ea) for this reaction.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15.
T1= 25◦C + 273.15 = 298.15 K
T2= 45◦C + 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures: k2
k1
=e
−Ea
R1
T2
−1
T1
Step 3: Substitute the known values into the equation:
8.91 ×10−2
5.72 ×10−3=e−Ea
8.314 (1
318.15 −1
298.15 )
11
Step 4: Solve for the activation energy (Ea):
8.91 ×10−2
5.72 ×10−3=e−Ea
8.314 (1
318.15 −1
298.15 )
15.5678 = e−Ea
8.314 (1
318.15 −1
298.15 )
ln(15.5678) = −Ea
8.314 1
318.15 −1
298.15
Ea=−8.314 ×ln(15.5678) 1
318.15 −1
298.15
Step 5: Calculate the activation energy (Ea) using the calculated values.
Ea=−8.314 ×ln(15.5678) 1
318.15 −1
298.15
Ea≈36.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 36.2
kJ/mol.
Question 13
Question
The rate constant of a reaction at 25
°
C is found to be 1.75 ×10−2s−1, while
at 50
°
C it is 8.64 ×10−2s−1. Calculate the activation energy for this reaction
using the Arrhenius equation. (Hint: The gas constant, R, is 8.314 J mol−1
K−1)
Solution
Step 1: Convert the temperatures to Kelvin. Given: - Temperature at 25
°
C =
25
°
C + 273.15 = 298.15 K - Temperature at 50
°
C = 50
°
C + 273.15 = 323.15 K
Step 2: Write down the Arrhenius equation. The Arrhenius equation relates
the rate constant of a reaction with the temperature and activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant, - Tis the temperature in Kelvin.
Step 3: Write down the Arrhenius equation for both temperatures.
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
12
Step 4: Take the ratio of the two rate constant equations.
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 5: Simplify the equation and solve for the activation energy.
k2
k1
=e−Ea
R1
T2
−1
T1
ln k2
k1=−Ea
R1
T2
−1
T1
Ea=−R·ln k2
k1·1
T2
−1
T1
Step 6: Calculate the activation energy. Substitute the given values:
Ea=−8.314 J mol−1·ln 8.64 ×10−2
1.75 ×10−2·1
323.15 −1
298.15
Ea≈65641 J mol−1≈65.6 kJ mol−1
Therefore, the activation energy for this reaction is approximately 65.6 kJ
mol−1.
Question 14
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1. When the
temperature is increased to 45
°
C, the rate constant becomes 1.8×10−2s−1.
Determine the activation energy for this reaction.
Solution
Step 1: Calculate the rate constant at 25
°
C using the Arrhenius equation:
k1=A·e−Ea
RT1
Given: k1= 2.5×10−3s−1T1= 25 + 273 = 298 K T2= 45 + 273 = 318 K
Step 2: Calculate the rate constant at 45
°
C using the Arrhenius equation:
k2=A·e−Ea
RT2
Given: k2= 1.8×10−2s−1
Step 3: Take the ratio of the two rate constants to simplify the equation:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
13
1.8×10−2
2.5×10−3=e−Ea
R1
T1
−1
T2
Step 4: Solve for the activation energy Ea:
7.2
mA =e−Ea
R(1
298 −1
318 )
ln 7.2
mA=−Ea
R1
298 −1
318
Step 5: Calculate the activation energy Eausing the gas constant R=
8.314 J mol−1K−1:
Ea=−8.314 ×ln 7.2
mA 1
298 −1
318
Ea≈56.8 kJ mol−1
Therefore, the activation energy for this reaction is approximately 56.8 kJ
mol−1.
Question 15
Question
The rate constant of a first-order reaction is found to be 2.5×10−3s−1at 25◦C
and 0.014 s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: First, we need to calculate the activation energy using the Arrhenius
equation:
The Arrhenius equation is given by:
k=A·e−Ea
RT
Where: - kis the rate constant - Ais the pre-exponential factor - Eais
the activation energy - Ris the gas constant (8.314 J/(mol ·K)) - Tis the
temperature in Kelvin
Step 2: Let’s first convert the temperatures to Kelvin: - 25◦C= 25 + 273 =
298 K- 45◦C= 45 + 273 = 318 K
Step 3: Now, we can write two equations for the rate constant at each
temperature:
k1=A·e−Ea
R·298
k2=A·e−Ea
R·318
Step 4: Substituting the given rate constants into the equations:
2.5×10−3=A·e−Ea
8.314·298
14
0.014 = A·e−Ea
8.314·318
Step 5: Now, we can divide the two equations to eliminate A:
0.014
2.5×10−3=e−Ea
8.314·318
e−Ea
8.314·298
Step 6: Simplifying the equation, we get:
5.6 = eEa
8.314 (1
298 −1
318 )
Step 7: Taking the natural logarithm of both sides to solve for Ea:
ln(5.6) = Ea
8.314 1
298 −1
318
Step 8: Finally, we can solve for Ea:
Ea= 8.314 ×ln(5.6)
1
298 −1
318
Ea≈41.2kJ/mol
Therefore, the activation energy for this reaction is approximately 41.2kJ/mol.
Question 16
Question
The rate constant, k, for a certain reaction is found to be 4.28 ×10−4s−1at
25◦C and 7.31 ×10−3s−1at 35◦C. Calculate the activation energy, Ea, for this
reaction (in kJ/mol). Assume the frequency factor, A, is 1.0×1012 s−1.
Solution
Step 1: Write down the Arrhenius equation which relates the rate constant,
temperature, activation energy, and frequency factor.
k=A·e(−Ea
RT )
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol
·
K)), T= temperature (in Kelvin).
Step 2: Rewrite the Arrhenius equation in terms of the natural logarithm to
simplify calculations.
ln(k) = ln(A)−Ea
R·1
T
Step 3: We have two sets of data: at 25◦C (298K) and at 35◦C (308K). We
will create two equations using these data points.
ln(k1) = ln(A)−Ea
R·1
298
15
ln(k2) = ln(A)−Ea
R·1
308
Step 4: Substitute the given values into the equations and create a system
of equations.
ln4.28 ×10−4= ln1.0×1012−Ea
8.314 ·1
298
ln7.31 ×10−3= ln1.0×1012−Ea
8.314 ·1
308
Step 5: Solve the system of equations simultaneously to find the activation
energy, Ea. Make sure to convert from natural logarithm to base 10 logarithm.
Ea=−8.314 ×
1
ln(4.28×10−4)−ln(7.31×10−3)
1/298−1/308
Step 6: Calculate the activation energy, Ea, to find the answer in kJ/mol.
Question 17
Question
The rate constant of a reaction at 25
°
C is 1.2×10−4s−1, and at 65
°
C it increases
to 3.8×10−2s−1. Calculate the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=Ae−Ea
RT , where kis the rate constant, Ais the
pre-exponential factor, Eais the activation energy, Ris the gas constant, and
Tis the temperature in Kelvin.)
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 2: Take the natural logarithm of both sides to simplify the equation:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Rearrange the equation to solve for activation energy (Ea) in terms
of the rate constants and temperatures given:
1
T2
=1
T1
−Ea
R1
T2
−1
T1
Step 4: Plug in the values given in the question:
1
338 K =1
298 K −Ea
8.314 J/mol ·K1
338 K −1
298 K
16
Step 5: Solve for the activation energy:
Ea=−8.314 J/mol×1
338 K −1
298 K−1
×1
338 K −1
298 K×(ln3.8×10−2−ln1.2×10−4)
Step 6: Calculate the activation energy using the given values and the gas
constant:
Ea= 61.6 kJ/mol
Question 18
Question
The rate constant of a first-order reaction at a certain temperature is 2.5×10−3
s−1. When the temperature is raised by 10
°
C, the rate constant becomes 6.4×
10−3s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperature change to Kelvin. Given that the temperature
is raised by 10
°
C, we need to convert this to Kelvin by adding 273.15. So,
T2=T1+ ∆T=T1+ 10 = T1+ 10 + 273.15 = T1+ 283.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures. The Arrhenius equation is given by:
ln k2
k1=Ea
R1
T1
−1
T2
Step 3: Substitute the given values into the Arrhenius equation. Given:
k1= 2.5×10−3s−1,k2= 6.4×10−3s−1,T1= initial temperature, and
T2=T1+ 283.15 K. Substitute the given values into the Arrhenius equation
and solve for Ea.
Step 4: Calculate the activation energy. We can rearrange the Arrhenius
equation to solve for Ea:
Ea=
Rln k2
k1
1
T1
−1
T2
Substitute the values of k1,k2,T1, and T2into the equation and calculate Ea.
Remember that the universal gas constant R= 8.314 J/(mol K).
Step 5: Calculate the activation energy. Substitute the values into the equa-
tion:
Ea=
8.314 ×ln 6.4×10−3
2.5×10−3
1
T1
−1
T1+ 283.15
Calculate Eato find the activation energy for this reaction.
17
Question 19
Question
The rate constant, k, for a reaction is known to double when the temperature
is increased from 25
°
C to 35
°
C. Calculate the activation energy, Ea, for this
reaction. Assume a temperature of 25
°
C corresponds to 298 K and the universal
gas constant, R, is 8.314 J/(mol·K).
Solution
Step 1: We start by using the Arrhenius equation, which relates the rate con-
stant, k, to the temperature and activation energy:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= universal gas constant T= temperature in Kelvin
Step 2: Given that the rate constant doubles when the temperature increases
from 25
°
C to 35
°
C, we can write:
k2
k1
=2=A·e−Ea
R·308K
A·e−Ea
R·298K
2 = e−Ea
8.314·308 ∇ · e−Ea
8.314·298
Step 3: Simplifying the equation:
2 = e−Ea
2547.392 ∇ · e−Ea
2473.572
2 = e+0.02733·Ea
Step 4: Taking the natural logarithm of both sides:
ln(2) = 0.02733 ·Ea
Step 5: Solve for the activation energy, Ea:
Ea=ln(2)
0.02733 ≈0.6931
0.02733 ≈25.38 kJ/mol
Step 6: Therefore, the activation energy for this reaction is approximately
25.38 kJ/mol.
Question 20
Question
The rate constant of a certain reaction at 25
°
C is 1.5×10−3s−1. When the tem-
perature is raised to 35
°
C, the rate constant becomes 3.8×10−3s−1. Calculate
the activation energy (in kJ/mol) for this reaction.
18
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor (frequency
factor), - Eais the activation energy, - Ris the gas constant (8.314 J/(mol·K)),
-Tis the temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin: - T1= 25 + 273 = 298 K -
T2= 35 + 273 = 308 K
Step 3: Write down the Arrhenius equation at each temperature:
k1=A·e−Ea
R·298
k2=A·e−Ea
R·308
Step 4: Divide the two equations to eliminate Aand solve for Ea:
k2
k1
=e−Ea
R·308
e−Ea
R·298
3.8×10−3
1.5×10−3=e−Ea
8.314·308 ·eEa
8.314·298
2.5333 = eEa
8.314 (1
298 −1
308 )
Step 5: Solve for Ea:
ln(2.5333) = Ea
8.314 1
298 −1
308
Ea= 8.314 ·ln(2.5333) ·1
298 −1
308
Calculating Ea:
Ea≈60.81 kJ/mol
Therefore, the activation energy for this reaction is approximately 60.81
kJ/mol.
Question 21
Question
The rate constant for a certain reaction is found to be 1.25 ×10−3s−1at 25◦C
and 3.92 ×10−2s−1at 35◦C. Calculate the activation energy for this reaction.
19
Solution
Step 1: Let’s first write down the Arrhenius equation, which relates the rate
constant of a reaction to temperature:
k=A·e−Ea
RT
where: k= rate constant of the reaction, A= pre-exponential factor, Ea=
activation energy, R= gas constant (8.314 J/(mol
·
K)), and T= temperature
in Kelvin.
Step 2: Next, we will take the natural logarithm of the Arrhenius equation
to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 3: We are given two sets of data at different temperatures, so we have
two equations:
(ln1.25 ×10−3= ln(A)−Ea
8.314 ·1
25+273.15
ln3.92 ×10−2= ln(A)−Ea
8.314 ·1
35+273.15
Step 4: Now, we will solve the system of equations to find the values of ln(A)
and Ea.
Step 5: Using the data provided, we find that ln(A)≈ −8.90 and Ea≈52400
J/mol.
Question 22
Question
The rate constant for the decomposition of compound X at 25
°
C is 1.2×10−3
s−1. When the temperature is increased to 55
°
C, the rate constant becomes
5.8×10−2s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
Where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/(mol*K)), - Tis the
temperature in Kelvin.
Step 2: Take the natural logarithm of both sides to simplify the equation:
ln k= ln A−Ea
RT
20
Step 3: Let’s denote the two different sets of data as Set 1 (25
°
C) and Set 2
(55
°
C): For Set 1:
ln k1= ln A−Ea
R(25 + 273)
ln k1= ln A−Ea
298R
For Set 2:
ln k2= ln A−Ea
R(55 + 273)
ln k2= ln A−Ea
328R
Step 4: Subtract the Set 1 equation from the Set 2 equation to eliminate
ln A:
ln k2−ln k1=−Ea
328R+Ea
298R
ln k2
k1=Ea
298R−Ea
328R
Step 5: Determine the ratio of rate constants:
ln 5.8×10−2
1.2×10−3=Ea
298 ×8.314 −Ea
328 ×8.314
Step 6: Solve for Ea:
ln 5.8×10−2
1.2×10−3=Ea1
298 ×8.314 −1
328 ×8.314
Step 7: Calculate the activation energy Ea.
Question 23
Question
The rate constant for a certain reaction is 6.5×10−3s−1at 25◦C and 1.8×
10−2s−1at 35◦C. Calculate the activation energy for this reaction. Assume the
frequency factor, A, is 8.0×1010 s−1.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C + 273.15 = 298.15 K
T2= 35◦C + 273.15 = 308.15 K
21
Step 2: Calculate the activation energy, Ea, using the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
where: - k1= 6.5×10−3s−1-k2= 1.8×10−2s−1-R= 8.314 J ·mol−1·K−1
Substitute the values and solve for Ea:
ln 1.8×10−2
6.5×10−3=−Ea
8.314 1
308.15 −1
298.15
Step 3: Simplify the equation and solve for Ea.
ln 2.769
0.65 =−Ea
8.314 1
308.15 −1
298.15
Step 4: Calculate Eato find the activation energy for this reaction.
Question 24
Question
The rate constant for a reaction at 25◦C is 8.25 ×10−3s−1. When the temper-
ature is increased to 50◦C, the rate constant becomes 0.045 s−1. Calculate the
activation energy for this reaction.
Solution
Step 1: Convert the temperatures to Kelvin using T(K) = T(◦C) + 273.
Initial temperature: T1= 25◦C + 273 = 298K
Final temperature: T2= 50◦C + 273 = 323K
Step 2: Write 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 3: Set up two equations using the given data: at T1= 298K, k1=
8.25 ×10−3s−1
8.25 ×10−3=A·e−Ea
8.314·298
at T2= 323K, k2= 0.045 s−1
0.045 = A·e−Ea
8.314·323
22
Step 4: Divide the two equations to eliminate A:
8.25 ×10−3
0.045 =e−Ea
8.314·298
e−Ea
8.314·323
Step 5: Simplify the equation and solve for Ea:
8.25 ×10−3
0.045 =e(1
8.314 ·(1
323 −1
298 ))·Ea
Step 6: Calculate the activation energy:
Ea=−8.314 ×1
323 −1
298×ln 8.25 ×10−3
0.045
After solving the equation, you should find the activation energy for this
reaction.
Question 25
Question
The rate constant for the decomposition of a certain compound at 25
°
C is 3.2×
10−4s−1. When the temperature is raised to 50
°
C, the rate constant is 7.1×10−3
s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Calculate the rate constant at 25
°
C using the Arrhenius equation:
k1=A·e−Ea
RT1
where: k1= rate constant at 25
°
C = 3.2×10−4s−1A= pre-exponential factor
(frequency factor, with correct units) Ea= activation energy R= gas constant
= 8.314 J/(mol·K) T1= temperature in Kelvin = 25
°
C + 273.15
Step 2: Calculate the rate constant at 50
°
C using the Arrhenius equation:
k2=A·e−Ea
RT2
where: k2= rate constant at 50
°
C = 7.1×10−3s−1T2= temperature in Kelvin
= 50
°
C + 273.15
Step 3: Set up the ratio of the rate constants:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
Step 4: Simplify the ratio by dividing out the pre-exponential factors:
k2
k1
=e
Ea
R1
T1
−1
T2
23
Step 5: Solve for the activation energy (Ea) by rearranging the equation:
Ea=−R·ln k2
k1·1
T1
−1
T2
Step 6: Substitute the given values and solve for Ea. Remember to convert
temperatures to Kelvin:
Ea=−8.314 J/mol ·ln 7.1×10−3
3.2×10−4·1
25 + 273.15 −1
50 + 273.15
Question 26
Question
The rate constant for the reaction A →B at 25◦C is 2.17 ×10−3s−1. When the
temperature is increased to 55◦C, the rate constant becomes 9.95 ×10−2s−1.
Calculate the activation energy of the reaction.
Solution
Step 1: Let’s start by writing 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−1·K−1), and - Tis the
temperature in Kelvin.
Step 2: We have two sets of data points: At 25◦C (298 K):
k1= 2.17 ×10−3s−1
At 55◦C (328 K):
k2= 9.95 ×10−2s−1
Step 3: Let’s rewrite the Arrhenius equation for the two temperature data
points:
k1=A·e−Ea/(8.314·298)
k2=A·e−Ea/(8.314·328)
Step 4: Divide the second equation by the first to eliminate A:
k2
k1
=e−Ea/(8.314·328)
e−Ea/(8.314·298)
k2
k1
=eEa(1/298−1/328)/8.314
24
Step 5: Solve for Ea:
Ea=8.314 ·298 ·328
328 −298 ·ln 9.95 ×10−2
2.17 ×10−3
Ea≈42.9 kJ ·mol−1
Therefore, the activation energy of the reaction is approximately 42.9 kJ/mol.
Question 27
Question
The rate constant for a reaction at 298 K is 1.80×10−3s−1. When the tempera-
ture is raised to 373 K, the rate constant increases to 2.75×10−2s−1. Determine
the activation energy for the reaction.
Solution
Step 1: Write the Arrhenius equation for the temperature dependence of the
rate constant:
k=Ae−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol K)), T= temperature (K).
Step 2: Set up the equation for two different temperatures:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 3: Take the ratio of the two rate constants:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R(1
T2
−1
T1)
Step 4: Substitute the given rate constants and temperatures into the equa-
tion: 2.75 ×10−2
1.80 ×10−3=e−Ea
8.314 (1
373 −1
298 )
Step 5: Solve for the activation energy (Ea):
2.75 ×10−2
1.80 ×10−3=e−Ea
8.314 (0.00268)
25
2.75 ×10−2
1.80 ×10−3=e−0.02188Ea
Step 6: Take the natural logarithm of both sides to solve for Ea:
ln 2.75 ×10−2
1.80 ×10−3=−0.02188Ea
Ea=
ln 2.75×10−2
1.80×10−3
−0.02188
Step 7: Calculate Eausing a calculator:
Ea=
ln 2.75×10−2
1.80×10−3
−0.02188
Question 28
Question
The rate constant of a reaction at 298 K is 5.0×10−3s−1, and the rate constant
at 350 K is 0.10 s−1. Calculate the activation energy (in kJ/mol) for the reaction.
(Hint: Use the Arrhenius equation k=A·e−Ea
RT , where Ais the pre-exponential
factor, Eais the activation energy, Ris the gas constant in J/mol
·
K, and Tis
the temperature in Kelvin.)
Solution
Step 1: Convert the given temperatures to Kelvin. At 298 K, T1= 298 K
At 350 K, T2= 350 K
Step 2: Calculate the activation energy. We can rearrange the Arrhenius
equation to solve for Ea:
ln k2
k1=Ea
R1
T1
−1
T2
Plugging in the given values:
ln 0.10
5.0×10−3=Ea
8.314 1
298 −1
350
ln(20) = Ea
8.314 ×0.0016
Ea= 8.314 ×ln(20) ×0.0016
1000
Ea≈48.74 kJ/mol
Therefore, the activation energy for the reaction is approximately 48.74
kJ/mol.
26
Question 29
Question
The rate constant for a certain reaction is found to be 4.2×10−3s−1at 25◦C
and 1.36 ×10−2s−1at 45◦C. Calculate the activation energy (in kJ/mol) 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 45◦C, T2= 45 + 273.15 = 318.15 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: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J mol−1K−1), T= temperature (in Kelvin).
Taking the ratio of the two rate constants:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
Step 3: Simplify the equation by taking the natural logarithm of both sides
and rearrange to solve for the activation energy Ea.
ln k2
k1=Ea
R1
T1
−1
T2
Step 4: Substitute the given values into the equation and solve for the acti-
vation energy Ea.
ln 1.36 ×10−2
4.2×10−3=Ea
8.314 1
298.15 −1
318.15
Step 5: Finally, calculate the activation energy Eain kilojoules per mole
using the result from step 4.
Ea= ln 1.36 ×10−2
4.2×10−3×8.314 ×1
298.15 −1
318.15×10−3
27
Question 30
Question
The rate constant for a certain reaction is found to be 1.25 ×10−3s−1at 25◦C
and 3.45 ×10−3s−1at 50◦C. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant
(k), the pre-exponential factor (A), the activation energy (Ea), the gas constant
(R), and the temperature (T):
k=A·e−Ea
RT
Step 2: Take the natural logarithm of the Arrhenius equation to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Identify two sets of data needed to solve for the activation energy.
We are given k1= 1.25 ×10−3s−1and T1= 25◦C (298 K), as well as k2=
3.45 ×10−3s−1and T2= 50◦C (323 K).
Step 4: Rewrite the linearized Arrhenius equation using the two sets of data:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Subtract the two equations obtained in Step 4 to eliminate ln(A):
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Plug in the values and solve for the activation energy Ea:
ln 3.45 ×10−3
1.25 ×10−3=−Ea
8.314 ·1
323 −1
298
ln(2.76) = −Ea
8.314 ·(0.00309 −0.00336)
ln(2.76) = −Ea
8.314 ·(−0.00027)
Ea=8.314 ×ln(2.76)
0.00027 ≈62.4 kJ/mol
Therefore, the activation energy for the reaction is approximately 62.4 kJ/mol.
28
Question 31
Question
The rate constant for the decomposition of a compound at 25
°
C is 6.0×10−3
s−1. When the temperature is increased to 45
°
C, the rate constant becomes
2.4×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant of
a reaction to the temperature and the activation energy.
k=A·e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J/mol*K), and Tis the temperature in
Kelvin.
Step 2: Set up a system of equations using the Arrhenius equation for the
two given temperature points. At 25
°
C (298 K):
6.0×10−3=A·e−Ea
8.314·298
At 45
°
C (318 K):
2.4×10−2=A·e−Ea
8.314·318
Step 3: Take the ratio of the two equations to eliminate A.
2.4×10−2
6.0×10−3=eEa
8.314 (1
298 −1
318 )
Step 4: Solve for Eaby simplifying and taking the natural logarithm of both
sides.
ln 2.4×10−2
6.0×10−3=Ea
8.314 1
298 −1
318
Step 5: Calculate the activation energy Eausing the obtained value from
the natural logarithm.
Ea= 8.314 ×ln 2.4×10−2
6.0×10−3/1
298 −1
318
Step 6: Substitute the values and calculate the activation energy.
Ea= 8.314 ×ln 2.4×10−2
6.0×10−3/1
298 −1
318
Ea≈43.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 43.1
kJ/mol.
29
Question 32
Question
The rate constant for a certain reaction is found to be 4.2×10−3s−1at 25◦C and
1.3×10−2s−1at 35◦C. Calculate the activation energy (Ea) for this reaction.
Given: R= 8.314 J mol−1K−1.
Solution
Step 1: Calculate the activation energy 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, and Tis the temperature in Kelvin.
Step 2: First, we need to calculate the pre-exponential factor Afor the
reaction. We can use the rate constants at two different temperatures (k1and
k2) to find A:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
Step 3: Divide the equation by Aand take the natural logarithm of both
sides to simplify:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Plug in the given values of rate constants and temperatures to solve
for Ea:
Ea=−R·ln k2
k1 1
T2
−1
T1
Step 5: Substituting k1= 4.2×10−3s−1,T1= 298 K, k2= 1.3×10−2s−1,
and T2= 308 K into the equation gives:
Ea=−8.314 ·ln 1.3×10−2
4.2×10−3 1
308 −1
298
Step 6: Calculate the activation energy Eausing the given values:
Ea≈45.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.7 kJ/mol.
Question 33
Question
The rate constant for the reaction A →B + C is 3.2×10−4s−1at 25◦C and
9.2×10−3s−1at 55◦C. Calculate the activation energy (Ea) for this reaction.
30
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant with
temperature:
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: Convert the temperatures to Kelvin: 25◦C = 25 + 273 = 298 K,
55◦C = 55 + 273 = 328 K.
Step 3: Set up two Arrhenius equations using the given data:
(3.2×10−4=Ae−Ea
8.314×298
9.2×10−3=Ae−Ea
8.314×328
Step 4: Divide the two equations to eliminate A:
3.2×10−4
9.2×10−3=e−Ea
8.314×298
e−Ea
8.314×328
Step 5: Simplify the equation:
3.2×10−4
9.2×10−3=e−Ea
8.314 (1
298 −1
328 )
Step 6: Solve for the activation energy Ea:
3.2×10−4
9.2×10−3=e−Ea
8.314 (1
298 −1
328 )
Step 7: Calculate Eausing the natural logarithm:
Ea=−8.314 ×ln 3.2×10−4
9.2×10−3×1
1
298 −1
328
Step 8: Plug in the values and calculate Eato get the final answer.
Question 34
Question
For a certain reaction, the rate constant at 300 K is 0.025 s−1and the rate
constant at 350 K is 0.075 s−1. Calculate the activation energy for this reaction.
31
Solution
Step 1: Write down 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)), - Tis the temperature
in Kelvin.
Step 2: We are given two sets of data points: At 300 K, k1= 0.025 s−1, At
350 K, k2= 0.075 s−1.
Step 3: Rewrite the Arrhenius equation for the two data points:
k1=Ae−Ea
R×300
k2=Ae−Ea
R×350
Step 4: Divide the second equation by the first to eliminate A:
k2
k1
=e−Ea
R×350
e−Ea
R×300
Step 5: Simplify the expression:
k2
k1
=e−Ea
R(1
350 −1
300 )
Step 6: Calculate the activation energy Ea:
k2
k1
=e−Ea
8.314 (1
350 −1
300 )
ln k2
k1=−Ea
8.314 1
350 −1
300
Ea=−8.314 ×ln 0.075
0.025
1
350 −1
300
Step 7: Calculate the activation energy:
Ea=−8.314 ×ln(3)
1
350 −1
300
≈81.6 kJ/mol
Therefore, the activation energy for the reaction is approximately 81.6 kJ/mol.
Question 35
Question
For a certain reaction, the rate constant at 25
°
C is found to be 4.28 ×10−3
s−1. When the temperature is increased to 55
°
C, the rate constant becomes
2.17 ×10−2s−1. Calculate the activation energy for this reaction.
32
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 55
°
C: T2= 55 + 273.15 = 328.15 K
Step 2: Use the Arrhenius equation to relate the rate constants and temper-
atures to the activation energy. The Arrhenius equation is given by:
k=A·e−Ea
RT
where: k= rate constant
A= pre-exponential factor (constant)
Ea= activation energy
R= gas constant (8.314 J/(mol ·K))
T= temperature in Kelvin
Step 3: Write out the Arrhenius equation for both temperatures:
(4.28 ×10−3=A·e−Ea
8.314·298.15
2.17 ×10−2=A·e−Ea
8.314·328.15
Step 4: Take the ratio of the rate constants to simplify the equation:
4.28 ×10−3
2.17 ×10−2=A·e−Ea
8.314·298.15
A·e−Ea
8.314·328.15
Step 5: Simplify the equation and solve for the activation energy (Ea):
4.28 ×10−3
2.17 ×10−2=e(1
8.314 ·(1
328.15 −1
298.15 ))·Ea
Step 6: Calculate the activation energy using the simplified equation above.
33
A400 = 5.00 ×10−1mol−1L s−1·e
Ea
8.314 J/(mol
·
K)·400 K
A400 = 5.00 ×10−1mol−1L s−1·e
Ea
3320 J/mol
Step 3: Set up a ratio of the two rate constants and pre-exponential factors
to find the activation energy:
8.00 ×10−2
5.00 ×10−1=A350 ·e−Ea
2920
A400 ·e−Ea
3320
0.16 = A350
A400
·eEa
400 −Ea
350
0.16 = A350
A400
·e50Ea
140000
Step 4: Solve for Eaby isolating it on one side:
e50Ea
140000 =0.16 ·A400
A350
50Ea
140000 = ln 0.16 ·A400
A350
Ea=
140000 ·ln 0.16·A400
A350
50
Therefore, the activation energy for the reaction is Ea=140000·ln0.16·A400
A350
50
J/mol.
Question 2
Question
The rate constant for the reaction 2A →B was found to be 4.5×10−3s−1at
298 K and 2.5×10−2s−1at 318 K. Determine the activation energy for this
reaction.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
where: k1= 4.5×10−3s−1at T1= 298 K, k2= 2.5×10−2s−1at T2= 318 K,
R= 8.314 J ·mol−1·K−1.
2
Step 2: Set up the Arrhenius equation for the two temperatures:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 3: Take the ratio of the two equations to eliminate A:
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 4: Simplify the ratio:
k2
k1
=e
Ea
R1
T1
−1
T2
Step 5: Substitute the given values and solve for Ea:
2.5×10−2
4.5×10−3=eEa
8.314 (1
298 −1
318 )
Step 6: Calculate the activation energy Ea.
Question 3
Question
The rate constant of a reaction at 300 K is 4.5×10−3s−1, and its rate constant
at 340 K is 1.8×10−2s−1. Calculate the activation energy for this reaction.
(Hint: Use the Arrhenius equation: k=Ae−Ea
RT )
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 2: Take the natural logarithm of the equation to linearize it:
ln(k) = ln(A)−Ea
R×1
T
Step 3: Set up two equations using the rate constants and temperatures given:
ln4.5×10−3= ln(A)−Ea
R×1
300
ln1.8×10−2= ln(A)−Ea
R×1
340
3
Step 4: Subtract the two equations to eliminate ln(A):
ln1.8×10−2−ln4.5×10−3=−Ea
R×1
340 −1
300
Step 5: Solve for the activation energy Ea:
∆(ln(k)) = −Ea
R×1
340 −1
300
Ea=−R×∆(ln(k))
1
340 −1
300
Step 6: Plug in the given values of ∆(ln(k)) = ln1.8×10−2−ln4.5×10−3,
R= 8.314 J/(mol·K), and solve for Ea. Remember to convert temperatures to
Kelvin. Step 7: Calculate the activation energy:
Ea=−8.314 ×(ln1.8×10−2−ln4.5×10−3)
1
340 −1
300
Question 4
Question
A certain reaction has an activation energy of 75 kJ/mol and a rate constant of
4.2×10−3s−1at 25◦C. Calculate the rate constant at 50◦C for this reaction.
Solution
Let’s denote the rate constant at 25◦C as k1and the rate constant at 50◦C
as k2. According to the Arrhenius equation, the relationship between the rate
constants and temperature is given by:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy
R= gas constant T= temperature in Kelvin
First, we need to convert the given activation energy to joules:
Ea= 75 kJ/mol ×1000 J/kJ = 75000 J/mol
Step 1: Calculate k1at 25◦C. Using the Arrhenius equation, we have:
k1=A·e−Ea
RT1
Substitute the known values:
4.2×10−3s−1=A·e−75000 J/mol
(8.314 J/mol·K)(25+273.15) K
4
Solve for Ato find k1.
Step 2: Calculate k2at 50◦C. Using the Arrhenius equation again, we have:
k2=A·e−Ea
RT2
Substitute the known values and the previously calculated A:
k2= (Afrom Step 1) ·e−75000 J/mol
(8.314 J/mol·K)(50+273.15) K
Calculate k2to determine the rate constant at 50◦C.
Question 5
Question
The rate constant for a reaction at 25
°
C is 3.67 ×10−3s−1. When the tempera-
ture is increased to 35
°
C, the rate constant becomes 1.45 ×10−2s−1. Calculate
the activation energy for this reaction in kJ/mol.
Solution
Step 1: Write down the Arrhenius equation relating the rate constant to tem-
perature:
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 tem-
perature in Kelvin.
Step 2: Express the ratio of rate constants in terms of the Arrhenius equa-
tion:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
Step 3: Take the natural logarithm of both sides to simplify the equation:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Plug in the values for k1,k2,T1, and T2into the equation. Remember
to convert the temperatures to Kelvin:
ln 1.45 ×10−2
3.67 ×10−3=−Ea
8.314 1
308 −1
298
Step 5: Solve for the activation energy, Ea, in Joules. Then convert this
value to kilojoules:
Ea=−8.314 ×ln 1.45 ×10−2
3.67 ×10−3×1
308 −1
298
5
Ea= [Y OURCALCULAT EDV ALU E] J/mol
Ea= [Y OURCALCULAT EDV ALU E] kJ/mol
Thus, the activation energy for this reaction is [YOUR ANSWER] kJ/mol.
Question 6
Question
The rate constant for the reaction A →B is found to be 4.2×10−3s−1at
25◦C. When the temperature is increased to 45◦C, the rate constant becomes
9.6×10−3s−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: Use the data provided to create two equations based on the Arrhenius
equation at 25◦C and 45◦C. At 25◦C:
4.2×10−3=A·e−Ea
8.314·(25+273.15)
At 45◦C:
9.6×10−3=A·e−Ea
8.314·(45+273.15)
Step 3: Divide the two equations to eliminate A:
4.2×10−3
9.6×10−3=e−Ea
8.314·(25+273.15)
e−Ea
8.314·(45+273.15)
Step 4: Simplify the equation to solve for Ea:
4.2
9.6=eEa
8.314 (1
25+273.15 −1
45+273.15 )
Step 5: Solve for Ea:
Ea=−8.314 ·ln 4.17
9.6·1
298 −1
318
Step 6: Calculate the value of Ea:
Ea≈48.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 48.5
kJ/mol.
6
Question 7
Question
The rate constant for a certain reaction is found to be 2.5×10−3s−1at 25◦C
and 5.0×10−3s−1at 45◦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= temperature in Kelvin.
Step 2: Set up two equations using the rate constants at different tempera-
tures:
2.5×10−3=A·e−Ea
R·(25+273) and 5.0×10−3=A·e−Ea
R·(45+273)
Step 3: Take the ratio of the two equations to eliminate A:
2.5×10−3
5.0×10−3=e−Ea
R·298
e−Ea
R·318
Step 4: Simplify the equation:
1
2=e20Ea
8.314·318 −20Ea
8.314·298
Step 5: Combine exponents and solve for Ea:
1
2=e−20Ea
8.314·9562
Step 6: Solve for Ea:
Ea=−8.314 ·9562
20 ·ln(0.5) ≈39887 J/mol
Therefore, the activation energy for this reaction is approximately 39.9
kJ/mol.
Question 8
Question
The rate constant for a reaction is found to be 6.2×10−3s−1at 335 K and
2.4×10−2s−1at 370 K. Calculate the activation energy for the reaction.
7
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: Using the given data at two different temperatures (335 K and 370
K), set up two separate equations using the Arrhenius equation:
k1=A·e−Ea
R·335
k2=A·e−Ea
R·370
Step 3: Divide the two equations to eliminate the pre-exponential factor A:
k2
k1
=e−Ea
R·370
e−Ea
R·335
Step 4: Simplify the expression by subtracting the exponents:
k2
k1
=eEa
R(1
335 −1
370 )
Step 5: Solve for Ea:k2
k1
=eEa
R(1
335 −1
370 )
ln k2
k1=Ea
R1
335 −1
370
Ea=R·
ln k2
k1
1
335 −1
370
Step 6: Substitute the given values to find Ea:
Ea= 8.314 ·
ln 2.4×10−2
6.2×10−3
1
335 −1
370
Step 7: Calculate Eato find the activation energy for the reaction.
Question 9
Question
The rate constant for a certain reaction at 25
°
C is 3.21 ×10−3s−1. When
the temperature is raised to 65
°
C, the rate constant is found to be 0.0462 s−1.
Calculate the activation energy of the reaction in kJ/mol.
8
Solution
Step 1: Let’s begin by writing the Arrhenius equation, which relates the rate
constant of a reaction to the temperature and 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), - Tis the temperature
in Kelvin.
Step 2: We have rate constants at two temperatures, so we can set up two
equations using the Arrhenius equation:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 3: Taking the ratio of the two rate constants, we can eliminate the
pre-exponential factor A:
k2
k1
=e−Ea
RT2
e−Ea
RT1
k2
k1
=e
Ea
R(1
T1
−1
T2)
Step 4: Now, plug in the values for the rate constants at 25
°
C and 65
°
C:
0.0462
3.21 ×10−3=eEa
8.314 (1
298 −1
338 )
Step 5: Solve for the activation energy Eato find the answer in kJ/mol.
Question 10
Question
The rate constant for the reaction 2A+B→Cat 25
°
C is 3.2×10−3M−1
s−1. When the temperature is increased to 50
°
C, the rate constant becomes
1.2×10−2M−1s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the ideal gas constant (8.314 J/(mol K)), - Tis the
temperature in Kelvin.
9
Step 2: Set up the Arrhenius equation at 25
°
C
k1=Ae−Ea
RT1
where k1= 3.2×10−3M−1s−1and T1= 25 + 273 = 298 K.
Step 3: Set up the Arrhenius equation at 50
°
C
k2=Ae−Ea
RT2
where k2= 1.2×10−2M−1s−1and T2= 50 + 273 = 323 K.
Step 4: Take the ratio of the two Arrhenius equations to eliminate A
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R1
T1
−1
T2
Step 5: Solve for the activation energy Ea
k2
k1
=e−Ea
8.314 (1
298 −1
323 )
ln k2
k1=−Ea
8.314 1
298 −1
323
Step 6: Calculate the activation energy Ea
Ea=−8.314 ×ln k2
k1 1
298 −1
323
Ea≈70.1 kJ/mol
Therefore, the activation energy for the reaction is approximately 70.1 kJ/mol.
Question 11
Question
The rate constant of a reaction is 3.25 ×10−3s−1at 25◦C and 1.04 ×10−2s−1
at 35◦C. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Convert the given temperatures from Celsius to Kelvin using the equa-
tion T(K) = T(
°
C) + 273.15. At 25◦C, T= 25 + 273.15 = 298.15 K. At 35◦C,
T= 35 + 273.15 = 308.15 K.
10
Step 2: Calculate the rate constant ratio, k2/k1, using the given rate con-
stants at the two temperatures.
k2
k1
=1.04 ×10−2s−1
3.25 ×10−3s−1= 3.2
Step 3: Use the Arrhenius equation k=Ae−Ea
RT to relate the rate constants
and temperatures at the two temperatures.
3.2 = e−Ea
R(1
308.15 −1
298.15 )
Step 4: Solve for the activation energy, Ea, in J/mol using the ideal gas
constant R= 8.314 J/(mol ·K).
3.2 = e−Ea
8.314 (1
308.15 −1
298.15 )
Step 5: Convert the activation energy from joules to kilojoules by dividing
by 1000.
Ea=−8.314 ×ln(3.2) ×1
308.15 −1
298.15×1000
So, the activation energy for this reaction is approximately 53.5 kJ/mol.
Question 12
Question
The rate constant of a certain reaction is found to be 5.72 ×10−3s−1at 25◦C.
When the temperature is increased to 45◦C, the rate constant becomes 8.91 ×
10−2s−1. Calculate the activation energy (Ea) for this reaction.
Solution
Step 1: Convert the temperatures to Kelvin using the formula T(K) = T(◦C) +
273.15.
T1= 25◦C + 273.15 = 298.15 K
T2= 45◦C + 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures: k2
k1
=e
−Ea
R1
T2
−1
T1
Step 3: Substitute the known values into the equation:
8.91 ×10−2
5.72 ×10−3=e−Ea
8.314 (1
318.15 −1
298.15 )
11
Step 4: Solve for the activation energy (Ea):
8.91 ×10−2
5.72 ×10−3=e−Ea
8.314 (1
318.15 −1
298.15 )
15.5678 = e−Ea
8.314 (1
318.15 −1
298.15 )
ln(15.5678) = −Ea
8.314 1
318.15 −1
298.15
Ea=−8.314 ×ln(15.5678) 1
318.15 −1
298.15
Step 5: Calculate the activation energy (Ea) using the calculated values.
Ea=−8.314 ×ln(15.5678) 1
318.15 −1
298.15
Ea≈36.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 36.2
kJ/mol.
Question 13
Question
The rate constant of a reaction at 25
°
C is found to be 1.75 ×10−2s−1, while
at 50
°
C it is 8.64 ×10−2s−1. Calculate the activation energy for this reaction
using the Arrhenius equation. (Hint: The gas constant, R, is 8.314 J mol−1
K−1)
Solution
Step 1: Convert the temperatures to Kelvin. Given: - Temperature at 25
°
C =
25
°
C + 273.15 = 298.15 K - Temperature at 50
°
C = 50
°
C + 273.15 = 323.15 K
Step 2: Write down the Arrhenius equation. The Arrhenius equation relates
the rate constant of a reaction with the temperature and activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant, - Tis the temperature in Kelvin.
Step 3: Write down the Arrhenius equation for both temperatures.
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
12
Step 4: Take the ratio of the two rate constant equations.
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 5: Simplify the equation and solve for the activation energy.
k2
k1
=e−Ea
R1
T2
−1
T1
ln k2
k1=−Ea
R1
T2
−1
T1
Ea=−R·ln k2
k1·1
T2
−1
T1
Step 6: Calculate the activation energy. Substitute the given values:
Ea=−8.314 J mol−1·ln 8.64 ×10−2
1.75 ×10−2·1
323.15 −1
298.15
Ea≈65641 J mol−1≈65.6 kJ mol−1
Therefore, the activation energy for this reaction is approximately 65.6 kJ
mol−1.
Question 14
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1. When the
temperature is increased to 45
°
C, the rate constant becomes 1.8×10−2s−1.
Determine the activation energy for this reaction.
Solution
Step 1: Calculate the rate constant at 25
°
C using the Arrhenius equation:
k1=A·e−Ea
RT1
Given: k1= 2.5×10−3s−1T1= 25 + 273 = 298 K T2= 45 + 273 = 318 K
Step 2: Calculate the rate constant at 45
°
C using the Arrhenius equation:
k2=A·e−Ea
RT2
Given: k2= 1.8×10−2s−1
Step 3: Take the ratio of the two rate constants to simplify the equation:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
13
1.8×10−2
2.5×10−3=e−Ea
R1
T1
−1
T2
Step 4: Solve for the activation energy Ea:
7.2
mA =e−Ea
R(1
298 −1
318 )
ln 7.2
mA=−Ea
R1
298 −1
318
Step 5: Calculate the activation energy Eausing the gas constant R=
8.314 J mol−1K−1:
Ea=−8.314 ×ln 7.2
mA 1
298 −1
318
Ea≈56.8 kJ mol−1
Therefore, the activation energy for this reaction is approximately 56.8 kJ
mol−1.
Question 15
Question
The rate constant of a first-order reaction is found to be 2.5×10−3s−1at 25◦C
and 0.014 s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: First, we need to calculate the activation energy using the Arrhenius
equation:
The Arrhenius equation is given by:
k=A·e−Ea
RT
Where: - kis the rate constant - Ais the pre-exponential factor - Eais
the activation energy - Ris the gas constant (8.314 J/(mol ·K)) - Tis the
temperature in Kelvin
Step 2: Let’s first convert the temperatures to Kelvin: - 25◦C= 25 + 273 =
298 K- 45◦C= 45 + 273 = 318 K
Step 3: Now, we can write two equations for the rate constant at each
temperature:
k1=A·e−Ea
R·298
k2=A·e−Ea
R·318
Step 4: Substituting the given rate constants into the equations:
2.5×10−3=A·e−Ea
8.314·298
14
0.014 = A·e−Ea
8.314·318
Step 5: Now, we can divide the two equations to eliminate A:
0.014
2.5×10−3=e−Ea
8.314·318
e−Ea
8.314·298
Step 6: Simplifying the equation, we get:
5.6 = eEa
8.314 (1
298 −1
318 )
Step 7: Taking the natural logarithm of both sides to solve for Ea:
ln(5.6) = Ea
8.314 1
298 −1
318
Step 8: Finally, we can solve for Ea:
Ea= 8.314 ×ln(5.6)
1
298 −1
318
Ea≈41.2kJ/mol
Therefore, the activation energy for this reaction is approximately 41.2kJ/mol.
Question 16
Question
The rate constant, k, for a certain reaction is found to be 4.28 ×10−4s−1at
25◦C and 7.31 ×10−3s−1at 35◦C. Calculate the activation energy, Ea, for this
reaction (in kJ/mol). Assume the frequency factor, A, is 1.0×1012 s−1.
Solution
Step 1: Write down the Arrhenius equation which relates the rate constant,
temperature, activation energy, and frequency factor.
k=A·e(−Ea
RT )
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol
·
K)), T= temperature (in Kelvin).
Step 2: Rewrite the Arrhenius equation in terms of the natural logarithm to
simplify calculations.
ln(k) = ln(A)−Ea
R·1
T
Step 3: We have two sets of data: at 25◦C (298K) and at 35◦C (308K). We
will create two equations using these data points.
ln(k1) = ln(A)−Ea
R·1
298
15
ln(k2) = ln(A)−Ea
R·1
308
Step 4: Substitute the given values into the equations and create a system
of equations.
ln4.28 ×10−4= ln1.0×1012−Ea
8.314 ·1
298
ln7.31 ×10−3= ln1.0×1012−Ea
8.314 ·1
308
Step 5: Solve the system of equations simultaneously to find the activation
energy, Ea. Make sure to convert from natural logarithm to base 10 logarithm.
Ea=−8.314 ×
1
ln(4.28×10−4)−ln(7.31×10−3)
1/298−1/308
Step 6: Calculate the activation energy, Ea, to find the answer in kJ/mol.
Question 17
Question
The rate constant of a reaction at 25
°
C is 1.2×10−4s−1, and at 65
°
C it increases
to 3.8×10−2s−1. Calculate the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=Ae−Ea
RT , where kis the rate constant, Ais the
pre-exponential factor, Eais the activation energy, Ris the gas constant, and
Tis the temperature in Kelvin.)
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 2: Take the natural logarithm of both sides to simplify the equation:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Rearrange the equation to solve for activation energy (Ea) in terms
of the rate constants and temperatures given:
1
T2
=1
T1
−Ea
R1
T2
−1
T1
Step 4: Plug in the values given in the question:
1
338 K =1
298 K −Ea
8.314 J/mol ·K1
338 K −1
298 K
16
Step 5: Solve for the activation energy:
Ea=−8.314 J/mol×1
338 K −1
298 K−1
×1
338 K −1
298 K×(ln3.8×10−2−ln1.2×10−4)
Step 6: Calculate the activation energy using the given values and the gas
constant:
Ea= 61.6 kJ/mol
Question 18
Question
The rate constant of a first-order reaction at a certain temperature is 2.5×10−3
s−1. When the temperature is raised by 10
°
C, the rate constant becomes 6.4×
10−3s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperature change to Kelvin. Given that the temperature
is raised by 10
°
C, we need to convert this to Kelvin by adding 273.15. So,
T2=T1+ ∆T=T1+ 10 = T1+ 10 + 273.15 = T1+ 283.15 K.
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures. The Arrhenius equation is given by:
ln k2
k1=Ea
R1
T1
−1
T2
Step 3: Substitute the given values into the Arrhenius equation. Given:
k1= 2.5×10−3s−1,k2= 6.4×10−3s−1,T1= initial temperature, and
T2=T1+ 283.15 K. Substitute the given values into the Arrhenius equation
and solve for Ea.
Step 4: Calculate the activation energy. We can rearrange the Arrhenius
equation to solve for Ea:
Ea=
Rln k2
k1
1
T1
−1
T2
Substitute the values of k1,k2,T1, and T2into the equation and calculate Ea.
Remember that the universal gas constant R= 8.314 J/(mol K).
Step 5: Calculate the activation energy. Substitute the values into the equa-
tion:
Ea=
8.314 ×ln 6.4×10−3
2.5×10−3
1
T1
−1
T1+ 283.15
Calculate Eato find the activation energy for this reaction.
17
Question 19
Question
The rate constant, k, for a reaction is known to double when the temperature
is increased from 25
°
C to 35
°
C. Calculate the activation energy, Ea, for this
reaction. Assume a temperature of 25
°
C corresponds to 298 K and the universal
gas constant, R, is 8.314 J/(mol·K).
Solution
Step 1: We start by using the Arrhenius equation, which relates the rate con-
stant, k, to the temperature and activation energy:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= universal gas constant T= temperature in Kelvin
Step 2: Given that the rate constant doubles when the temperature increases
from 25
°
C to 35
°
C, we can write:
k2
k1
=2=A·e−Ea
R·308K
A·e−Ea
R·298K
2 = e−Ea
8.314·308 ∇ · e−Ea
8.314·298
Step 3: Simplifying the equation:
2 = e−Ea
2547.392 ∇ · e−Ea
2473.572
2 = e+0.02733·Ea
Step 4: Taking the natural logarithm of both sides:
ln(2) = 0.02733 ·Ea
Step 5: Solve for the activation energy, Ea:
Ea=ln(2)
0.02733 ≈0.6931
0.02733 ≈25.38 kJ/mol
Step 6: Therefore, the activation energy for this reaction is approximately
25.38 kJ/mol.
Question 20
Question
The rate constant of a certain reaction at 25
°
C is 1.5×10−3s−1. When the tem-
perature is raised to 35
°
C, the rate constant becomes 3.8×10−3s−1. Calculate
the activation energy (in kJ/mol) for this reaction.
18
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor (frequency
factor), - Eais the activation energy, - Ris the gas constant (8.314 J/(mol·K)),
-Tis the temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin: - T1= 25 + 273 = 298 K -
T2= 35 + 273 = 308 K
Step 3: Write down the Arrhenius equation at each temperature:
k1=A·e−Ea
R·298
k2=A·e−Ea
R·308
Step 4: Divide the two equations to eliminate Aand solve for Ea:
k2
k1
=e−Ea
R·308
e−Ea
R·298
3.8×10−3
1.5×10−3=e−Ea
8.314·308 ·eEa
8.314·298
2.5333 = eEa
8.314 (1
298 −1
308 )
Step 5: Solve for Ea:
ln(2.5333) = Ea
8.314 1
298 −1
308
Ea= 8.314 ·ln(2.5333) ·1
298 −1
308
Calculating Ea:
Ea≈60.81 kJ/mol
Therefore, the activation energy for this reaction is approximately 60.81
kJ/mol.
Question 21
Question
The rate constant for a certain reaction is found to be 1.25 ×10−3s−1at 25◦C
and 3.92 ×10−2s−1at 35◦C. Calculate the activation energy for this reaction.
19
Solution
Step 1: Let’s first write down the Arrhenius equation, which relates the rate
constant of a reaction to temperature:
k=A·e−Ea
RT
where: k= rate constant of the reaction, A= pre-exponential factor, Ea=
activation energy, R= gas constant (8.314 J/(mol
·
K)), and T= temperature
in Kelvin.
Step 2: Next, we will take the natural logarithm of the Arrhenius equation
to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 3: We are given two sets of data at different temperatures, so we have
two equations:
(ln1.25 ×10−3= ln(A)−Ea
8.314 ·1
25+273.15
ln3.92 ×10−2= ln(A)−Ea
8.314 ·1
35+273.15
Step 4: Now, we will solve the system of equations to find the values of ln(A)
and Ea.
Step 5: Using the data provided, we find that ln(A)≈ −8.90 and Ea≈52400
J/mol.
Question 22
Question
The rate constant for the decomposition of compound X at 25
°
C is 1.2×10−3
s−1. When the temperature is increased to 55
°
C, the rate constant becomes
5.8×10−2s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
Where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the gas constant (8.314 J/(mol*K)), - Tis the
temperature in Kelvin.
Step 2: Take the natural logarithm of both sides to simplify the equation:
ln k= ln A−Ea
RT
20
Step 3: Let’s denote the two different sets of data as Set 1 (25
°
C) and Set 2
(55
°
C): For Set 1:
ln k1= ln A−Ea
R(25 + 273)
ln k1= ln A−Ea
298R
For Set 2:
ln k2= ln A−Ea
R(55 + 273)
ln k2= ln A−Ea
328R
Step 4: Subtract the Set 1 equation from the Set 2 equation to eliminate
ln A:
ln k2−ln k1=−Ea
328R+Ea
298R
ln k2
k1=Ea
298R−Ea
328R
Step 5: Determine the ratio of rate constants:
ln 5.8×10−2
1.2×10−3=Ea
298 ×8.314 −Ea
328 ×8.314
Step 6: Solve for Ea:
ln 5.8×10−2
1.2×10−3=Ea1
298 ×8.314 −1
328 ×8.314
Step 7: Calculate the activation energy Ea.
Question 23
Question
The rate constant for a certain reaction is 6.5×10−3s−1at 25◦C and 1.8×
10−2s−1at 35◦C. Calculate the activation energy for this reaction. Assume the
frequency factor, A, is 8.0×1010 s−1.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C + 273.15 = 298.15 K
T2= 35◦C + 273.15 = 308.15 K
21
Step 2: Calculate the activation energy, Ea, using the Arrhenius equation:
ln k2
k1=−Ea
R1
T2
−1
T1
where: - k1= 6.5×10−3s−1-k2= 1.8×10−2s−1-R= 8.314 J ·mol−1·K−1
Substitute the values and solve for Ea:
ln 1.8×10−2
6.5×10−3=−Ea
8.314 1
308.15 −1
298.15
Step 3: Simplify the equation and solve for Ea.
ln 2.769
0.65 =−Ea
8.314 1
308.15 −1
298.15
Step 4: Calculate Eato find the activation energy for this reaction.
Question 24
Question
The rate constant for a reaction at 25◦C is 8.25 ×10−3s−1. When the temper-
ature is increased to 50◦C, the rate constant becomes 0.045 s−1. Calculate the
activation energy for this reaction.
Solution
Step 1: Convert the temperatures to Kelvin using T(K) = T(◦C) + 273.
Initial temperature: T1= 25◦C + 273 = 298K
Final temperature: T2= 50◦C + 273 = 323K
Step 2: Write 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 3: Set up two equations using the given data: at T1= 298K, k1=
8.25 ×10−3s−1
8.25 ×10−3=A·e−Ea
8.314·298
at T2= 323K, k2= 0.045 s−1
0.045 = A·e−Ea
8.314·323
22
Step 4: Divide the two equations to eliminate A:
8.25 ×10−3
0.045 =e−Ea
8.314·298
e−Ea
8.314·323
Step 5: Simplify the equation and solve for Ea:
8.25 ×10−3
0.045 =e(1
8.314 ·(1
323 −1
298 ))·Ea
Step 6: Calculate the activation energy:
Ea=−8.314 ×1
323 −1
298×ln 8.25 ×10−3
0.045
After solving the equation, you should find the activation energy for this
reaction.
Question 25
Question
The rate constant for the decomposition of a certain compound at 25
°
C is 3.2×
10−4s−1. When the temperature is raised to 50
°
C, the rate constant is 7.1×10−3
s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Calculate the rate constant at 25
°
C using the Arrhenius equation:
k1=A·e−Ea
RT1
where: k1= rate constant at 25
°
C = 3.2×10−4s−1A= pre-exponential factor
(frequency factor, with correct units) Ea= activation energy R= gas constant
= 8.314 J/(mol·K) T1= temperature in Kelvin = 25
°
C + 273.15
Step 2: Calculate the rate constant at 50
°
C using the Arrhenius equation:
k2=A·e−Ea
RT2
where: k2= rate constant at 50
°
C = 7.1×10−3s−1T2= temperature in Kelvin
= 50
°
C + 273.15
Step 3: Set up the ratio of the rate constants:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
Step 4: Simplify the ratio by dividing out the pre-exponential factors:
k2
k1
=e
Ea
R1
T1
−1
T2
23
Step 5: Solve for the activation energy (Ea) by rearranging the equation:
Ea=−R·ln k2
k1·1
T1
−1
T2
Step 6: Substitute the given values and solve for Ea. Remember to convert
temperatures to Kelvin:
Ea=−8.314 J/mol ·ln 7.1×10−3
3.2×10−4·1
25 + 273.15 −1
50 + 273.15
Question 26
Question
The rate constant for the reaction A →B at 25◦C is 2.17 ×10−3s−1. When the
temperature is increased to 55◦C, the rate constant becomes 9.95 ×10−2s−1.
Calculate the activation energy of the reaction.
Solution
Step 1: Let’s start by writing 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−1·K−1), and - Tis the
temperature in Kelvin.
Step 2: We have two sets of data points: At 25◦C (298 K):
k1= 2.17 ×10−3s−1
At 55◦C (328 K):
k2= 9.95 ×10−2s−1
Step 3: Let’s rewrite the Arrhenius equation for the two temperature data
points:
k1=A·e−Ea/(8.314·298)
k2=A·e−Ea/(8.314·328)
Step 4: Divide the second equation by the first to eliminate A:
k2
k1
=e−Ea/(8.314·328)
e−Ea/(8.314·298)
k2
k1
=eEa(1/298−1/328)/8.314
24
Step 5: Solve for Ea:
Ea=8.314 ·298 ·328
328 −298 ·ln 9.95 ×10−2
2.17 ×10−3
Ea≈42.9 kJ ·mol−1
Therefore, the activation energy of the reaction is approximately 42.9 kJ/mol.
Question 27
Question
The rate constant for a reaction at 298 K is 1.80×10−3s−1. When the tempera-
ture is raised to 373 K, the rate constant increases to 2.75×10−2s−1. Determine
the activation energy for the reaction.
Solution
Step 1: Write the Arrhenius equation for the temperature dependence of the
rate constant:
k=Ae−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol K)), T= temperature (K).
Step 2: Set up the equation for two different temperatures:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 3: Take the ratio of the two rate constants:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
k2
k1
=e−Ea
R(1
T2
−1
T1)
Step 4: Substitute the given rate constants and temperatures into the equa-
tion: 2.75 ×10−2
1.80 ×10−3=e−Ea
8.314 (1
373 −1
298 )
Step 5: Solve for the activation energy (Ea):
2.75 ×10−2
1.80 ×10−3=e−Ea
8.314 (0.00268)
25
2.75 ×10−2
1.80 ×10−3=e−0.02188Ea
Step 6: Take the natural logarithm of both sides to solve for Ea:
ln 2.75 ×10−2
1.80 ×10−3=−0.02188Ea
Ea=
ln 2.75×10−2
1.80×10−3
−0.02188
Step 7: Calculate Eausing a calculator:
Ea=
ln 2.75×10−2
1.80×10−3
−0.02188
Question 28
Question
The rate constant of a reaction at 298 K is 5.0×10−3s−1, and the rate constant
at 350 K is 0.10 s−1. Calculate the activation energy (in kJ/mol) for the reaction.
(Hint: Use the Arrhenius equation k=A·e−Ea
RT , where Ais the pre-exponential
factor, Eais the activation energy, Ris the gas constant in J/mol
·
K, and Tis
the temperature in Kelvin.)
Solution
Step 1: Convert the given temperatures to Kelvin. At 298 K, T1= 298 K
At 350 K, T2= 350 K
Step 2: Calculate the activation energy. We can rearrange the Arrhenius
equation to solve for Ea:
ln k2
k1=Ea
R1
T1
−1
T2
Plugging in the given values:
ln 0.10
5.0×10−3=Ea
8.314 1
298 −1
350
ln(20) = Ea
8.314 ×0.0016
Ea= 8.314 ×ln(20) ×0.0016
1000
Ea≈48.74 kJ/mol
Therefore, the activation energy for the reaction is approximately 48.74
kJ/mol.
26
Question 29
Question
The rate constant for a certain reaction is found to be 4.2×10−3s−1at 25◦C
and 1.36 ×10−2s−1at 45◦C. Calculate the activation energy (in kJ/mol) 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 45◦C, T2= 45 + 273.15 = 318.15 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: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J mol−1K−1), T= temperature (in Kelvin).
Taking the ratio of the two rate constants:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
Step 3: Simplify the equation by taking the natural logarithm of both sides
and rearrange to solve for the activation energy Ea.
ln k2
k1=Ea
R1
T1
−1
T2
Step 4: Substitute the given values into the equation and solve for the acti-
vation energy Ea.
ln 1.36 ×10−2
4.2×10−3=Ea
8.314 1
298.15 −1
318.15
Step 5: Finally, calculate the activation energy Eain kilojoules per mole
using the result from step 4.
Ea= ln 1.36 ×10−2
4.2×10−3×8.314 ×1
298.15 −1
318.15×10−3
27
Question 30
Question
The rate constant for a certain reaction is found to be 1.25 ×10−3s−1at 25◦C
and 3.45 ×10−3s−1at 50◦C. Calculate the activation energy for the reaction.
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant
(k), the pre-exponential factor (A), the activation energy (Ea), the gas constant
(R), and the temperature (T):
k=A·e−Ea
RT
Step 2: Take the natural logarithm of the Arrhenius equation to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Identify two sets of data needed to solve for the activation energy.
We are given k1= 1.25 ×10−3s−1and T1= 25◦C (298 K), as well as k2=
3.45 ×10−3s−1and T2= 50◦C (323 K).
Step 4: Rewrite the linearized Arrhenius equation using the two sets of data:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Subtract the two equations obtained in Step 4 to eliminate ln(A):
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Plug in the values and solve for the activation energy Ea:
ln 3.45 ×10−3
1.25 ×10−3=−Ea
8.314 ·1
323 −1
298
ln(2.76) = −Ea
8.314 ·(0.00309 −0.00336)
ln(2.76) = −Ea
8.314 ·(−0.00027)
Ea=8.314 ×ln(2.76)
0.00027 ≈62.4 kJ/mol
Therefore, the activation energy for the reaction is approximately 62.4 kJ/mol.
28
Question 31
Question
The rate constant for the decomposition of a compound at 25
°
C is 6.0×10−3
s−1. When the temperature is increased to 45
°
C, the rate constant becomes
2.4×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation, which relates the rate constant of
a reaction to the temperature and the activation energy.
k=A·e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J/mol*K), and Tis the temperature in
Kelvin.
Step 2: Set up a system of equations using the Arrhenius equation for the
two given temperature points. At 25
°
C (298 K):
6.0×10−3=A·e−Ea
8.314·298
At 45
°
C (318 K):
2.4×10−2=A·e−Ea
8.314·318
Step 3: Take the ratio of the two equations to eliminate A.
2.4×10−2
6.0×10−3=eEa
8.314 (1
298 −1
318 )
Step 4: Solve for Eaby simplifying and taking the natural logarithm of both
sides.
ln 2.4×10−2
6.0×10−3=Ea
8.314 1
298 −1
318
Step 5: Calculate the activation energy Eausing the obtained value from
the natural logarithm.
Ea= 8.314 ×ln 2.4×10−2
6.0×10−3/1
298 −1
318
Step 6: Substitute the values and calculate the activation energy.
Ea= 8.314 ×ln 2.4×10−2
6.0×10−3/1
298 −1
318
Ea≈43.1 kJ/mol
Therefore, the activation energy for this reaction is approximately 43.1
kJ/mol.
29
Question 32
Question
The rate constant for a certain reaction is found to be 4.2×10−3s−1at 25◦C and
1.3×10−2s−1at 35◦C. Calculate the activation energy (Ea) for this reaction.
Given: R= 8.314 J mol−1K−1.
Solution
Step 1: Calculate the activation energy 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, and Tis the temperature in Kelvin.
Step 2: First, we need to calculate the pre-exponential factor Afor the
reaction. We can use the rate constants at two different temperatures (k1and
k2) to find A:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
Step 3: Divide the equation by Aand take the natural logarithm of both
sides to simplify:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Plug in the given values of rate constants and temperatures to solve
for Ea:
Ea=−R·ln k2
k1 1
T2
−1
T1
Step 5: Substituting k1= 4.2×10−3s−1,T1= 298 K, k2= 1.3×10−2s−1,
and T2= 308 K into the equation gives:
Ea=−8.314 ·ln 1.3×10−2
4.2×10−3 1
308 −1
298
Step 6: Calculate the activation energy Eausing the given values:
Ea≈45.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.7 kJ/mol.
Question 33
Question
The rate constant for the reaction A →B + C is 3.2×10−4s−1at 25◦C and
9.2×10−3s−1at 55◦C. Calculate the activation energy (Ea) for this reaction.
30
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant with
temperature:
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: Convert the temperatures to Kelvin: 25◦C = 25 + 273 = 298 K,
55◦C = 55 + 273 = 328 K.
Step 3: Set up two Arrhenius equations using the given data:
(3.2×10−4=Ae−Ea
8.314×298
9.2×10−3=Ae−Ea
8.314×328
Step 4: Divide the two equations to eliminate A:
3.2×10−4
9.2×10−3=e−Ea
8.314×298
e−Ea
8.314×328
Step 5: Simplify the equation:
3.2×10−4
9.2×10−3=e−Ea
8.314 (1
298 −1
328 )
Step 6: Solve for the activation energy Ea:
3.2×10−4
9.2×10−3=e−Ea
8.314 (1
298 −1
328 )
Step 7: Calculate Eausing the natural logarithm:
Ea=−8.314 ×ln 3.2×10−4
9.2×10−3×1
1
298 −1
328
Step 8: Plug in the values and calculate Eato get the final answer.
Question 34
Question
For a certain reaction, the rate constant at 300 K is 0.025 s−1and the rate
constant at 350 K is 0.075 s−1. Calculate the activation energy for this reaction.
31
Solution
Step 1: Write down 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)), - Tis the temperature
in Kelvin.
Step 2: We are given two sets of data points: At 300 K, k1= 0.025 s−1, At
350 K, k2= 0.075 s−1.
Step 3: Rewrite the Arrhenius equation for the two data points:
k1=Ae−Ea
R×300
k2=Ae−Ea
R×350
Step 4: Divide the second equation by the first to eliminate A:
k2
k1
=e−Ea
R×350
e−Ea
R×300
Step 5: Simplify the expression:
k2
k1
=e−Ea
R(1
350 −1
300 )
Step 6: Calculate the activation energy Ea:
k2
k1
=e−Ea
8.314 (1
350 −1
300 )
ln k2
k1=−Ea
8.314 1
350 −1
300
Ea=−8.314 ×ln 0.075
0.025
1
350 −1
300
Step 7: Calculate the activation energy:
Ea=−8.314 ×ln(3)
1
350 −1
300
≈81.6 kJ/mol
Therefore, the activation energy for the reaction is approximately 81.6 kJ/mol.
Question 35
Question
For a certain reaction, the rate constant at 25
°
C is found to be 4.28 ×10−3
s−1. When the temperature is increased to 55
°
C, the rate constant becomes
2.17 ×10−2s−1. Calculate the activation energy for this reaction.
32
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 55
°
C: T2= 55 + 273.15 = 328.15 K
Step 2: Use the Arrhenius equation to relate the rate constants and temper-
atures to the activation energy. The Arrhenius equation is given by:
k=A·e−Ea
RT
where: k= rate constant
A= pre-exponential factor (constant)
Ea= activation energy
R= gas constant (8.314 J/(mol ·K))
T= temperature in Kelvin
Step 3: Write out the Arrhenius equation for both temperatures:
(4.28 ×10−3=A·e−Ea
8.314·298.15
2.17 ×10−2=A·e−Ea
8.314·328.15
Step 4: Take the ratio of the rate constants to simplify the equation:
4.28 ×10−3
2.17 ×10−2=A·e−Ea
8.314·298.15
A·e−Ea
8.314·328.15
Step 5: Simplify the equation and solve for the activation energy (Ea):
4.28 ×10−3
2.17 ×10−2=e(1
8.314 ·(1
328.15 −1
298.15 ))·Ea
Step 6: Calculate the activation energy using the simplified equation above.
33
