CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 10
Liberty University
Question 1
Question
For a certain reaction, the rate constant at 25
°
C is 0.0012 s−1and the rate
constant at 35
°
C is 0.0037 s−1. Determine the activation energy for this reaction
in kJ/mol.
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: Let’s use the data provided at 25
°
C (298 K) and 35
°
C (308 K) to
set up two equations:
0.0012 = Ae−Ea
8.314×298
0.0037 = Ae−Ea
8.314×308
Step 3: Divide the second equation by the first to eliminate A:
0.0037
0.0012 =Ae−Ea
8.314×308
Ae−Ea
8.314×298
Step 4: Solve for Ea:
0.0037
0.0012 =eEa
8.314 (1
298 −1
308 )
Step 5: Simplify and solve for Ea:
ln 0.0037
0.0012=Ea
8.314 1
298 −1
308
Step 6: Calculate Eain J/mol:
Ea= 8.314 × ln 0.0037
0.0012
1
298 −1
308 !
Step 7: Convert Eato kJ/mol:
Ea=
8.314 ×1000 ×ln(0.0037
0.0012 )
1
298 −1
308
1000
Step 8: Calculate the final value for Ea.
Question 2
Question
The activation energy for a reaction is 75 kJ/mol, and the rate constant at 25
°
C
is 1.5×10−3s−1. Calculate the rate constant at 35
°
C for this reaction.
Solution
To calculate the rate constant at 35
°
C, 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.
Step 1: Convert the activation energy to joules: Since 1 kJ = 1000 J, the
activation energy Ea= 75 ×103J/mol.
Step 2: Calculate the rate constant at 35
°
C: Given: - Ea= 75 ×103J/mol,
-R= 8.314 J/(mol
·
K), - T1= 25 + 273 = 298 K, - T2= 35 + 273 = 308 K.
Plugging in these values into the Arrhenius equation:
k2=k1·eEa
R·1
T1
−1
T2
k2= 1.5×10−3·e75×103
8.314 ·(1
298 −1
308 )
k2= 1.5×10−3·e75×103
8.314 ·(0.0034)
2
k2= 1.5×10−3·e32.364
k2≈1.5×10−3·2.6443
k2≈3.9665 ×10−3
Therefore, the rate constant at 35
°
C for this reaction is approximately 3.97×
10−3s−1.
Question 3
Question
The rate constant for a certain reaction is found to be 2.0×10−3s−1at 25◦C.
When the temperature is increased to 50◦C, the rate constant becomes 1.6×
10−2s−1. Determine the activation energy for the 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−1K−1), T= temperature in Kelvin.
Step 2: Use the given data to set up two equations: At 25◦C (298K):
2.0×10−3=A·e−Ea
8.314·298
At 50◦C (323K):
1.6×10−2=A·e−Ea
8.314·323
Step 3: Divide the two equations to eliminate the pre-exponential factor A:
1.6×10−2
2.0×10−3=e−Ea
8.314 (1
323 −1
298 )
Step 4: Solve for the activation energy Ea:
8
2=e1
8.314 (1
298 −1
323 )·Ea
4 = e(323−298
8.314·298·323 )·Ea
Step 5: Calculate the natural log of both sides and solve for Ea:
ln(4) = 323 −298
8.314 ·298 ·323·Ea
Ea=8.314 ·298 ·323
323 −298 ·ln(4)
Ea≈72360 J/mol
3
Question 4
Question
The rate constant of a certain reaction is found to be 3.2×10−4s−1at 25◦Cand
9.8×10−4s−1at 50◦C. Calculate the activation energy (Ea) for the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the equation: T(K) =
T(◦C) + 273.15. - T1= 25◦C+ 273.15 = 298.15K-T2= 50◦C+ 273.15 =
323.15K
Step 2: Now we are going to use the Arrhenius equation:
k=A·e−Ea
RT
where: - k1= 3.2×10−4s−1-k2= 9.8×10−4s−1-T1= 298.15K-T2= 323.15K
We will use the two given rate constants and temperatures to set up two
equations:
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
k2
k1
=e−Ea
R1
T2
−1
T1
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Solve for the activation energy (Ea):
−Ea
R=
ln k2
k1
1
T2
−1
T1
Ea=−R·
ln k2
k1
1
T2
−1
T1
Step 5: Now substitute the known values:
Ea=−(8.314J/mol ·K)·
ln 9.8×10−4
3.2×10−4
1
323.15 −1
298.15
4
Step 6: Calculate the activation energy:
Ea=−(8.314J/mol ·K)·ln(3.0625)
0.0033
Ea≈92.6kJ/mol
Therefore, the activation energy for the reaction is approximately 92.6 kJ/mol.
Question 5
Question
The rate constant for a certain reaction is found to be 5.00 ×10−2s−1at 25◦C
and 1.20 mol L−1. When the temperature is increased to 35◦C, the rate constant
becomes 1.50 ×10−1s−1. Determine the activation energy for this reaction.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
k=A·exp −Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J mol−1K−1), and - Tis the
temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for Ea:
Ea=−
ln k2
k1
1
T2
−1
T1
where: - k1= 5.00 ×10−2s−1at 25◦C, - k2= 1.50 ×10−1s−1at 35◦C, -
T1= 25 + 273 = 298 K, and - T2= 35 + 273 = 308 K.
Step 3: Substitute the given values into the equation:
Ea=−
ln 1.50×10−1
5.00×10−2
1
308 −1
298
Step 4: Calculate Ea:
Ea=−ln(3)
1
308 −1
298
Ea=−ln(3)
298−308
298×308
Ea=−ln(3) ·298 ×308
10
Ea≈47887 J mol−1
Therefore, the activation energy for this reaction is approximately 47887 J
mol−1.
5
Question 6
Question
The rate constant for a certain reaction is found to be 4.55 ×10−3s−1at 25◦C
and 1.80 ×10−2s−1at 35◦C. Calculate the activation energy (Ea) for the
reaction in kJ/mol.
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 activa-
tion energy, Ris the gas constant (8.314 J mol−1K−1), Tis the temperature in
Kelvin.
Step 2: Convert the temperatures to Kelvin: 25◦C = 298 K and 35◦C =
308 K
Step 3: Substitue the given data points into the Arrhenius equation to form
a system of equations:
(4.55 ×10−3=A·e−Ea
8.314×298
1.80 ×10−2=A·e−Ea
8.314×308
Step 4: Divide the two equations to eliminate A:
4.55 ×10−3
1.80 ×10−2=e−Ea
8.314×298
e−Ea
8.314×308
Step 5: Simplify the expression on the right side:
4.55
1.80 =eEa
8.314 (1
298 −1
308 )
Step 6: Solve for the activation energy, Ea, in Joules:
Ea=−8.314 ×ln 4.55
1.80×1
298 −1
308
Step 7: Convert the activation energy to kilojoules per mole:
Ea=−8.314 ×ln 4.55
1.80×1
298 −1
308×1
1000
Calculate the value to find Ea.
6
Question 7
Question
The rate constant for a certain reaction is 2.50×10−3s−1at 25
°
C, and increases
to 5.00 ×10−3s−1at 45
°
C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given: T1= 25C,
T2= 45CAdding 273.15 to each temperature: T1= (25 + 273.15)K= 298.15K
T2= (45 + 273.15)K= 318.15K
Step 2: Use the Arrhenius equation to calculate the activation energy. 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
We know that k1= 2.50 ×10−3s−1,k2= 5.00 ×10−3s−1,T1= 298.15 K,
T2= 318.15 K. We can set up two equations using the Arrhenius equation:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 3: Divide the two equations to eliminate A.
k2
k1
=e−Ea
R·T2
e−Ea
R·T1
k2
k1
=e−Ea
R1
T2
−1
T1
k2
k1
=e−Ea
RT1−T2
T1·T2
Step 4: Solve for the activation energy Ea. Taking the natural logarithm of
both sides:
ln k2
k1=−Ea
RT1−T2
T1·T2
Ea=−R·ln k2
k1T1·T2
T1−T2
Substitute the given values:
Ea=−8.314 J/mol ·ln 5.00 ×10−3
2.50 ×10−3298.15 ·318.15
298.15 −318.15
7
Ea=−8.314 J/mol ·ln(2) ·298.15 ·318.15
−20
Ea≈90160 J/mol
Therefore, the activation energy for this reaction is approximately 90.16
kJ/mol.
Question 8
Question
The rate constant for a certain reaction was found to be 1.20×10−3s−1at 37◦C
and 3.45 ×10−2s−1at 87◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write out the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol·K)), T= temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin: 37◦C = 37+273 = 310 K 87◦C
= 87 + 273 = 360 K
Step 3: Set up two equations using the given rate constants:
(1.20 ×10−3=A·e−Ea
8.314·310
3.45 ×10−2=A·e−Ea
8.314·360
Step 4: Divide the second equation by the first to eliminate A:
3.45 ×10−2
1.20 ×10−3=e−Ea
8.314·360
e−Ea
8.314·310
Step 5: Simplify the equation:
28.75 = eEa
8.314 (1
310 −1
360 )
Step 6: Solve for Ea:
Ea
8.314 1
310 −1
360= ln(28.75)
Ea= 8.314 ×ln(28.75)
1
310 −1
360
Step 7: Calculate Ea:
Ea≈50.3 kJ/mol
Therefore, the activation energy for this reaction is approximately 50.3
kJ/mol.
8
Question 9
Question
The rate constant for the reaction
2A+B→C
is found to be 1.5×10−3M−1s−1at 25
°
C. When the temperature is increased
to 35
°
C, the rate constant becomes 5.0×10−3M−1s−1. Calculate the activation
energy for this reaction. Assume the frequency factor (pre-exponential factor)
is 2.5×1013 s−1.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant at two
different temperatures:
k2=k1×e−Ea
R1
T2
−1
T1
where: k1= 1.5×10−3M−1s−1(at 25
°
C), k2= 5.0×10−3M−1s−1(at 35
°
C),
T1= 25 + 273 K, T2= 35 + 273 K, R= 8.314 J/(mol·K).
Step 2: Now let’s substitute the values into the Arrhenius equation and solve
for Ea:
5.0×10−3= 1.5×10−3×e−Ea
8.314 (1
308 −1
298 )
Step 3: Simplify the equation:
3.33 = e−Ea
8.314 (1
308 −1
298 )
Step 4: Take the natural logarithm of both sides to solve for Ea:
ln(3.33) = −Ea
8.314 1
308 −1
298
Step 5: Solve for Ea:
Ea=−8.314 ×ln(3.33)∇ · 1
308 −1
298
Step 6: Calculate the activation energy:
Ea≈90.6 kJ/mol
Therefore, the activation energy for the reaction is approximately 90.6 kJ/mol.
Question 10
Question
For a certain reaction, the rate constant is found to be 0.005 s−1at 25◦C and
0.02 s−1at 50◦C. Calculate the activation energy for this reaction.
9
Solution
Step 1: Determine the Arrhenius equation. The Arrhenius equation relates the
rate constant (k) to the temperature (T) and the activation energy (Ea) of a
reaction. It is given by:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Ea is the acti-
vation energy, - Ris the gas constant (8.314 J/(mol·K)), - Tis the temperature
in Kelvin.
Step 2: Determine the pre-exponential factor (A). To solve for A, we can use
the two given rate constants and temperatures: At 25◦C (298 K): k1= 0.005
s−1, At 50◦C (323 K): k2= 0.02 s−1.
Substitute the values into the Arrhenius equation and solve for A:
k1=A·e−Ea
RT1
k2=A·e−Ea
RT2
Divide the second equation by the first:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
e−Ea
R(1
T2
−1
T1)=k2
k1
e−Ea
8.314 (1
323 −1
298 )=0.02
0.005
e−Ea
8.314 (0.0031) = 4
Step 3: Solve for activation energy (Ea). Take the natural logarithm of both
sides of the equation:
lne−Ea
8.314 (0.0031)= ln(4)
−Ea
8.314 ·0.0031 = ln(4)
−0.0031Ea
8.314 = ln(4)
−Ea = 8.314 ·0.0031 ·ln(4)
Ea =−8.314 ·0.0031 ·ln(4)
Ea ≈42.17 kJ/mol
Therefore, the activation energy for this reaction is approximately 42.17
kJ/mol.
10
Question 11
Question
The rate constant for a certain reaction at 25
°
C is 2.0×10−3s−1. When the
temperature is increased to 35
°
C, the rate constant becomes 3.5×10−3s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given: T1= 25C=
25 + 273 = 298 K
T2= 35C= 35 + 273 = 308 K
Step 2: Write the Arrhenius equation relating rate constants and tempera-
tures. The Arrhenius equation is given by:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it.
Taking the natural logarithm of both sides of the Arrhenius equation gives:
ln k= ln A−Ea
R·1
T
Step 4: Write the equation for the first set of data. Using the first set of
data (T1, k1):
ln k1= ln A−Ea
R·1
T1
Step 5: Write the equation for the second set of data. Using the second set
of data (T2, k2):
ln k2= ln A−Ea
R·1
T2
Step 6: Subtract the second equation from the first equation. Subtracting
the second equation from the first equation eliminates the ln A term, giving:
ln k1
k2
=−Ea
R·1
T1
−1
T2
Step 7: Solve for the activation energy Ea. Substitute the given values for
k1,k2,T1, and T2into the equation and solve for Ea:
ln 2.0×10−3
3.5×10−3=−Ea
R·1
298 −1
308
ln 0.57 = −Ea
8.314 ·1
298 −1
308
Ea=−8.314 ·ln 0.57
1
298 −1
308
Thus, the activation energy for this reaction is approximately 50.6 kJ/mol.
11
Question 12
Question
The rate constant for the reaction A →B is 1.2×10−3s−1at 25◦C and 3.6×
10−3s−1at 35◦C. Calculate the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=A·e−Ea
RT , where kis the rate constant, Ais the
pre-exponential factor, Eais the activation energy, R= 8.314 J mol−1K−1is
the gas constant, and Tis the temperature in Kelvin.)
Solution
Step 1: Convert the temperatures to Kelvin At 25◦C, T1= 25 + 273 = 298 K.
At 35◦C, T2= 35 + 273 = 308 K.
Step 2: Write down the given information k1= 1.2×10−3s−1at T1= 298 K.
k2= 3.6×10−3s−1at T2= 308 K. R= 8.314 J mol−1K−1.
Step 3: Calculate the activation energy, EaWe can use the Arrhenius equa-
tion to set up two equations: 1.2×10−3=A·e−Ea
8.314·298 3.6×10−3=A·e−Ea
8.314·308
Step 4: Find the ratio of the two equations Dividing the second equation by
the first gives: 3.6×10−3
1.2×10−3=eEa
8.314 (1
308 −1
298 )3 = e10
2397
Step 5: Solve for the activation energy, EaTaking the natural log of both
sides and rearranging: ln(3) = 10
2397 EaEa=2397
10 ln(3) ≈2397
10 ·1.0986 ≈
262 kJ mol−1
Therefore, the activation energy for the reaction is approximately 262 kJ
mol−1.
Question 13
Question
The rate constant for the reaction A →B is found to be 2.5×10−3s−1at
25◦C. When the temperature is raised to 45◦C, the rate constant becomes
1.6×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures to the activation energy. The Arrhenius equation is given
by:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/(mol
·
K)), and - Tis the
temperature in Kelvin.
Step 2: Let’s first convert the temperatures to Kelvin:
T1= 25 + 273 = 298 K
12
T2= 45 + 273 = 318 K
Step 3: We can write two Arrhenius equations for the two different temper-
atures:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 4: Given that the rate constants are 2.5×10−3s−1at 25◦C and 1.6×
10−2s−1at 45◦C, we have:
2.5×10−3=Ae−Ea
8.314×298
1.6×10−2=Ae−Ea
8.314×318
Step 5: We can divide the second equation by the first equation to eliminate
Aand solve for the activation energy:
1.6×10−2
2.5×10−3=e−Ea
8.314×318
e−Ea
8.314×298
Step 6: Simplifying the above expression, we get:
1.6×10−2
2.5×10−3=eEa
8.314 (1
298 −1
318 )
Step 7: Solve for the activation energy, Ea, by calculating the natural loga-
rithm of both sides and then solving for Ea.
Question 14
Question
The rate constant for a reaction is found to double when the temperature is
increased from 25
°
C to 35
°
C. Calculate the activation energy (Ea) for this re-
action.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant (k) to
temperature and activation energy:
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: Let’s consider the ratio of rate constants at two temperatures:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
13
Given that k2= 2k1and T2= 35 + 273 = 308 K, T1= 25 + 273 = 298 K, we
can substitute these values into the equation.
Step 3:
2k1
k1
=Ae−Ea
8.314×308
Ae−Ea
8.314×298
2 = e−Ea
8.314×308 +Ea
8.314×298
Step 4: Take the natural logarithm on both sides to solve for Ea:
ln(2) = −Ea
8.314 1
308 −1
298
Ea=−8.314 ×(1
308 −1
298)×ln(2)
Step 5: Calculate Ea:
Ea≈ −8.314 ×1
308 −1
298×ln(2)
Ea≈39,223.65 J/mol
Therefore, the activation energy for this reaction is approximately 39.22
kJ/mol.
Question 15
Question
When a certain reaction was run at 25
°
C, the rate constant was found to be
2.0×10−3s−1. When the temperature was increased to 50
°
C, the rate constant
increased to 8.0×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. At 25
°
C: T1= 25C+ 273.15 = 298.15 K
At 50
°
C: T2= 50C+ 273.15 = 323.15 K
Step 2: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 3: Take the natural logarithm of both sides of the Arrhenius equation
to linearize the equation:
ln(k) = ln(A)−Ea
R1
T
14
Step 4: Use the data provided to obtain two equations:
ln2.0×10−3= ln(A)−Ea
R1
298.15
ln8.0×10−2= ln(A)−Ea
R1
323.15
Step 5: Subtract the second equation from the first to eliminate ln(A):
ln 2.0×10−3
8.0×10−2=−Ea
R1
298.15 −1
323.15
Step 6: Solve for the activation energy Ea:
Ea=−R
ln 2.0×10−3
8.0×10−2
1
298.15 −1
323.15
Step 7: Calculate the activation energy using the ideal gas constant R=
8.314 J/(mol
·
K):
Ea=−8.314
ln 2.0×10−3
8.0×10−2
1
298.15 −1
323.15
Question 16
Question
Given the rate constant of a reaction at two temperatures, 25
°
C and 45
°
C,
calculate the activation energy for the reaction. The rate constant at 25
°
C is
5.0×10−3s−1and at 45
°
C is 1.0×10−2s−1. (Assume the activation energy
follows the Arrhenius equation).
Solution
Step 1: Convert the temperatures from Celsius to Kelvin using the formula
T(K) = T(C) + 273. At 25
°
C: T1= 25 + 273 = 298 K At 45
°
C: T2= 45 + 273 =
318 K
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, and T= temperature in Kelvin.
Step 3: Take the natural logarithm of the Arrhenius equation:
ln(k) = ln(A)−Ea
R·1
T
15
This equation can be rewritten as:
ln(k) = −Ea
R·1
T+ ln(A)
Step 4: Set up two equations using the rate constant values at 25
°
C and
45
°
C:
ln5.0×10−3=−Ea
R·1
298 + ln(A)
ln1.0×10−2=−Ea
R·1
318 + ln(A)
Step 5: Subtract the second equation from the first to eliminate ln(A):
ln5.0×10−3−ln1.0×10−2=−Ea
R·1
298 −1
318
Step 6: Solve for the activation energy Eausing the gas constant R= 8.314
J/(mol·K):
Ea=−8.314 ×298 ×318 ×ln5.0×10−3−ln1.0×10−2
Step 7: Calculate the activation energy using the given values and complete
the calculation.
Question 17
Question
The rate constant for a certain reaction is 4.32 ×10−2s−1at 25
°
C and 1.56 ×
10−1s−1at 35
°
C. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin. Given:
T1= 25C= 25 + 273.15 = 298.15 K
T2= 35C= 35 + 273.15 = 308.15 K
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation.
The Arrhenius equation is given by:
k=A×e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol
·
K)), T= temperature in Kelvin.
Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
R×1
T
16
Step 3: Set up the linear equation using the data at two different tempera-
tures. Using the data provided:
ln k1= ln A−Ea
R×1
T1
ln k2= ln A−Ea
R×1
T2
Step 4: Calculate the activation energy. Subtract the two equations to
eliminate ln A:
ln k2−ln k1=Ea
R×1
T1
−1
T2
Step 5: Plug in the values and solve for activation energy. Substitute the
given rate constants and temperatures into the equation:
ln 1.56 ×10−1−ln 4.32 ×10−2=Ea
8.314 ×1
298.15 −1
308.15
Step 6: Calculate Eain joules and convert to kilojoules per mole. After
solving the equation, we get:
Ea= 9.49 kJ/mol
Question 18
Question
The rate constant (k) for a reaction at 25
°
C is 4.2×10−3s−1, and at 35
°
C it is
2.5×10−2s−1. Calculate the activation energy (Ea) for this reaction. (Assume
the frequency factor Ais independent of temperature and is 8.0×1012 s−1)
Solution
Step 1: Recall the Arrhenius equation:
k=A×e−Ea
RT
where: - kis the rate constant - Ais the frequency factor - Eais the activation
energy - Ris the gas constant (8.314 J mol−1K−1) - Tis the temperature in
Kelvin
Step 2: We can rearrange the Arrhenius equation to solve for Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation:
ln 2.5×10−2
4.2×10−3=−Ea
8.314 1
308 −1
298
17
Step 4: Calculate the left-hand side of the equation:
ln 2.5×10−2
4.2×10−3= ln 2.5
0.42 ×10−1= ln (5.952) ≈1.778
Step 5: Plug the values back into the equation and solve for Ea:
1.778 = −Ea
8.314 1
308 −1
298
Step 6: Simplify the equation:
1.778 = −Ea
8.314 ×10
298 ×308
Step 7: Solve for Ea:
Ea=−1.778 ×8.314 ×10
298 ×308
Step 8: Calculate the activation energy Eato find the final answer.
Question 19
Question
For a certain reaction, the rate constant at 25
°
C is 4.0×10−3s−1, and the
activation energy is 60 kJ/mol. Calculate the rate constant at 35
°
C for this
reaction.
Solution
Step 1: Recall the Arrhenius equation, which relates the rate constant of a
reaction to temperature and the activation energy:
k=A·e−Ea
RT
Where: k= rate constant A= pre-exponential factor Ea= activation energy
(J/mol) R= gas constant (8.314 J/mol ·K) T= temperature (K)
Step 2: Let’s first convert the given activation energy from kJ/mol to J/mol,
and both temperatures from
°
C to K:
Ea= 60 ×103J/mol = 60 000 J/mol
T1= 25 + 273 = 298 K
T2= 35 + 273 = 308 K
Step 3: Now, we can rearrange the Arrhenius equation to solve for the pre-
exponential factor A:
A=k
e−Ea
RT
18
Step 4: Substitute the given rate constant kand the values we just calculated
into the equation to solve for Aat 25
°
C:
A=4.0×10−3
e−60 000
8.314×298
Step 5: Calculate the value of Aat 25
°
C:
A≈4.0×10−3
e−23971 ≈4.0×10−3
4.98 ×10−11 ≈8.03 ×107s−1
Step 6: Finally, substitute the newly calculated pre-exponential factor Ainto
the Arrhenius equation for 35
°
C to find the rate constant at this temperature:
k= 8.03 ×107·e−60 000
8.314×308
Step 7: Calculate the rate constant at 35
°
C:
k≈8.03 ×107·e−23971 ≈8.03 ×107·4.98 ×10−11 ≈3.99 ×10−3s−1
Therefore, the rate constant at 35
°
C for this reaction is approximately 3.99×
10−3s−1.
Question 20
Question
For the reaction A →B, the rate constant at 300 K is 2.3×10−3s−1and the
activation energy is 50 kJ/mol. Calculate the rate constant at 350 K for this
reaction.
Solution
Step 1: First, we need to use the Arrhenius equation to relate the rate constants
at two different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1= 2.3×10−3s−1(rate constant at 300 K)
T1= 300 K
Ea= 50 kJ/mol = 50000 J/mol
R= 8.314 J/mol
·
K (gas constant)
T2= 350 K
Step 2: Substitute the values into the equation and solve for k2:
ln k2
2.3×10−3=−50000
8.314 1
350 −1
300
19
ln k2
2.3×10−3=−6049.05
k2
2.3×10−3=e−6049.05
k2≈2.172 ×10−308 s−1
Therefore, the rate constant at 350 K for the reaction is approximately
2.172 ×10−308 s−1.
Question 21
Question
A reaction has an activation energy of 50 kJ/mol and a rate constant of 2.0×
10−3s−1at 25
°
C. Determine the rate constant at 50
°
C for this reaction if the
activation energy does not change.
Solution
Step 1: First, calculate the pre-exponential factor (A) using the Arrhenius equa-
tion:
k=A·e−Ea
RT
Given that k= 2.0×10−3s−1at 25
°
C, we can rearrange the equation to
solve for A:
A=k
e−Ea
RT
Where: - k= 2.0×10−3s−1-Ea= 50 kJ/mol - R= 8.314 J/(mol·K) -
T= 25 + 273.15 K
Step 2: Calculate A
A=2.0×10−3
e−50000
8.314×298.15
A≈4.35 ×1013 s−1
Step 3: Now, we can use the Arrhenius equation to find the rate constant at
50
°
C:
k=A·e−Ea
RT
Where: - A= 4.35 ×1013 s−1-Ea= 50 kJ/mol - R= 8.314 J/(mol·K) -
T= 50 + 273.15 K
Step 4: Calculate the rate constant at 50
°
C
k= 4.35 ×1013 ·e−50000
8.314×323.15
k≈4.15 ×10−3s−1
Therefore, the rate constant at 50
°
C for this reaction is approximately 4.15×
10−3s−1.
20
Question 22
Question
The rate constant for a certain reaction is found to be 4.52 ×10−3s−1at 25◦C
and 1.93 ×10−2s−1at 35◦C. Calculate the activation energy for this reaction.
Assume the frequency factor Ais 1.05 ×1010 s−1.
Solution
Step 1: Write the Arrhenius equation:
k=A·e−Ea
RT
where: kis the rate constant, Ais the frequency factor, Eais the activation
energy, Ris the ideal gas constant (8.314 J/(mol K)), and Tis the temperature
in Kelvin.
Step 2: Set up the Arrhenius equation for both given temperature conditions:
At 25◦C (298 K):
4.52 ×10−3= 1.05 ×1010 ·e−Ea
8.314·298
At 35◦C (308 K):
1.93 ×10−2= 1.05 ×1010 ·e−Ea
8.314·308
Step 3: Take the ratio of the equations to eliminate A:
4.52 ×10−3
1.93 ×10−2=e−Ea
8.314·298
e−Ea
8.314·308
Step 4: Simplify the ratio:
0.234 = eEa
8.314 ·(1
308 −1
298 )
Step 5: Solve for Ea:
ln(0.234) = Ea
8.314 ·1
308 −1
298
Step 6: Calculate Ea:
Ea= 8.314 ·ln(0.234)
1
308 −1
298
Step 7: Plug in the values and calculate Ea:
Ea≈62.1 kJ/mol
Therefore, the activation energy for the reaction is approximately 62.1 kJ/mol.
21
Question 23
Question
The rate constant for a certain reaction is 2.0×10−3s−1at 273 K and 6.5×
10−3s−1at 293 K. Calculate the activation energy of the reaction (given that
the universal gas constant R= 8.314 J mol−1K−1).
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy
(in joules per mole), R= universal gas constant, T= temperature in kelvin.
Step 2: We are given two sets of data points: At 273 K, k1= 2.0×10−3s−1,
At 293 K, k2= 6.5×10−3s−1.
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln(k) = ln(A)−Ea
RT
Step 4: Subtract the second equation from the first to eliminate A:
ln(k2)−ln(k1) = −Ea
R1
T2
−1
T1
Step 5: Plug in the values and solve for Ea:
ln6.5×10−3−ln2.0×10−3=−Ea
8.314 1
293 −1
273
Step 6: Calculate Ea:
−1.0296 = −Ea
8.314 1
293 −1
273
Ea= 54634.6 J/mol
Step 7: Therefore, the activation energy of the reaction is 54.634 kJ/mol.
Question 24
Question
The rate constant for a certain reaction increases from 5.0×10−2s−1at 300 K
to 8.0×10−2s−1at 350 K. Calculate the activation energy for this reaction.
Assume the frequency factor is 1.0×1013 s−1.
22
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to tem-
perature and activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the frequency factor, - Eais the activation
energy, - Ris the gas constant (8.314 J ·mol−1·K−1), and - Tis the absolute
temperature.
Step 2: We can rearrange the Arrhenius equation to solve for the activation
energy Ea:
Ea=−R·1
T1
−1
T2−1
·ln k2
k1
Step 3: Substitute the given values into the equation:
Ea=−(8.314 J ·mol−1·K−1)·1
300 K −1
350 K−1
·ln 8.0×10−2s−1
5.0×10−2s−1
Step 4: Calculate the activation energy:
Ea≈53.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 53.7 kJ/mol.
Question 25
Question
The rate constant for the reaction
2A+B→C
is found to be 3.2×10−3s−1at 25
°
C. When the temperature is increased to
75
°
C, the rate constant becomes 2.4×10−2s−1. Calculate the activation energy
for this reaction.
Solution
Step 1: We start by using the Arrhenius equation, which relates the rate con-
stant of a reaction to the temperature and the activation energy. The Arrhenius
equation is given by:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= gas constant (8.314 J mol−1K−1)T= temperature in Kelvin
23
Step 2: We can rearrange the Arrhenius equation to calculate the activation
energy, Ea. Taking the natural logarithm of both sides gives:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Let’s find the activation energy using the data provided at 25
°
C and
75
°
C. At 25
°
C (298 K), the rate constant k1is 3.2×10−3s−1, and at 75
°
C (348
K), the rate constant k2is 2.4×10−2s−1.
Step 4: Now, we have:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Subtracting the two equations gives:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Plugging in the values:
ln 2.4×10−2
3.2×10−3=−Ea
8.314 ·1
348 −1
298
Step 7: Solving for Ea:
−2.9957 = −Ea
8.314 ·1
348 −1
298
Ea= 51.2 kJ/mol
Therefore, the activation energy for the reaction is 51.2 kJ/mol.
Question 26
Question
The rate constant for the reaction A →B is found to be 4.75 ×10−4s−1at
25◦C. When the temperature is increased to 50◦C, the rate constant becomes
2.41 ×10−3s−1. Calculate the activation energy (Ea) for this reaction. (Hint:
Use the Arrhenius equation: k=A·e−Ea
RT , where kis the rate constant, Ais
the frequency factor, Eais the activation energy, Ris the gas constant, and T
is the absolute temperature.)
24
Solution
Step 1: Convert the temperatures to Kelvin.
T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 50◦C = 50 + 273.15 = 323.15 K
Step 2: Determine the ratio of rate constants at the two temperatures.
k2
k1
=2.41 ×10−3s−1
4.75 ×10−4s−1= 5.08
Step 3: Substitute into the Arrhenius equation and take the natural loga-
rithm of both sides.
k2
k1
=e−Ea
R1
T2
−1
T1
5.08 = e−Ea
R(1
323.15 −1
298.15 )
ln(5.08) = −Ea
R1
323.15 −1
298.15
Step 4: Solve for the activation energy (Ea).
ln(5.08) = −Ea
8.314 1
323.15 −1
298.15
Ea=−8.314 ×ln(5.08)
1
323.15 −1
298.15
Ea≈83.3 kJ/mol
Therefore, the activation energy for this reaction is approximately 83.3
kJ/mol.
Question 27
Question
The rate constant of a certain reaction is found to be 8.21 ×10−3s−1at 75 ◦C
and 3.54 ×10−2s−1at 100 ◦C. Calculate the activation energy for the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(◦C) + 273.15. At 75 ◦C: T1= 75 + 273.15 = 348.15 K
At 100 ◦C: T2= 100 + 273.15 = 373.15 K
Step 2: Write the Arrhenius equation:
k=A·e−Ea
RT
25
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= gas constant = 8.314 J ·mol−1·K−1T= temperature in Kelvin
Step 3: Set up two equations using the given rate constants and tempera-
tures:
8.21 ×10−3=A·e−Ea
8.314·348.15
3.54 ×10−2=A·e−Ea
8.314·373.15
Step 4: Divide the two equations to eliminate A:
8.21 ×10−3
3.54 ×10−2=e−Ea
8.314·348.15
e−Ea
8.314·373.15
Step 5: Simplify the equation:
8.21 ×10−3
3.54 ×10−2=e373.15−348.15
8.314 ·Ea
8.314·348.15·373.15
Step 6: Solve for Ea:
Ea=−8.314 ·373.15 ·348.15
373.15 −348.15 ·ln 8.21 ×10−3
3.54 ×10−2
Calculating Eagives:
Ea≈35,925 J/mol
Question 28
Question
The rate constant for the reaction
2A→B+C
is 3.21 ×10−3s−1at 25◦C. When the temperature is increased to 55◦C, the
rate constant becomes 6.75 ×10−3s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C + 273 = 298 K
T2= 55◦C + 273 = 328 K
Step 2: Use the Arrhenius equation to relate the rate constants (k1and k2)
at the two temperatures to the activation energy (Ea).
k1=A·e−Ea/(R·T1)
26
k2=A·e−Ea/(R·T2)
Step 3: Divide the two equations to eliminate the pre-exponential factor A.
k2
k1
=e−Ea/(R·T2)
e−Ea/(R·T1)
Step 4: Take the natural logarithm of both sides to simplify the equation.
ln k2
k1=−Ea
R1
T2
−1
T1
Step 5: Now, substitute the given values and constants into the equation
and solve for the activation energy (Ea).
R= 8.314 J ·mol−1·K−1
ln 6.75 ×10−3
3.21 ×10−3=−Ea
8.314 1
328 −1
298
Step 6: Calculate the activation energy using the values obtained.
Ea=−8.314 ×ln 6.75 ×10−3
3.21 ×10−3×1
328 −1
298
Ea≈37.2 kJ/mol
Therefore, the activation energy for the reaction is approximately 37.2 kJ/mol.
Question 29
Question
The rate constant for a certain reaction is observed to be 4.82 ×10−3s−1at
T= 350 K and 1.38 ×10−2s−1at T= 370 K. Calculate the activation energy
for this reaction.
Solution
Step 1: Write the Arrhenius equation: The Arrhenius equation relates the rate
constant (k) of a reaction to the temperature (T) and the activation energy (Ea)
through the equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor or frequency factor, Ea
= activation energy, R= gas constant, and T= temperature in Kelvin.
Step 2: Calculate the ratio of rate constants at the two temperatures: Given
that k1= 4.82×10−3s−1at T1= 350 K and k2= 1.38×10−2s−1at T2= 370 K,
we can find the ratio: k2
k1
=1.38 ×10−2
4.82 ×10−3= 2.86
27
Step 3: Use the ratio to find the activation energy: Taking the natural
logarithm of the Arrhenius equation, we get:
ln(k) = ln(A)−Ea
RT
ln k2
k1=Ea
R1
T1
−1
T2
Plugging in the values and solving for Ea:
ln(2.86) = Ea
R1
350 −1
370
ln(2.86) = Ea
8.314 1
350 −1
370
Ea= 7.58 kJ/mol
Question 30
Question
The rate constant (k) for a certain reaction was found to double when the
temperature was increased from 25
°
C to 35
°
C. Calculate the activation energy
for this reaction. (Assume the pre-exponential factor Ais constant.)
Solution
Step 1: Let’s first recall the Arrhenius equation:
k=Aexp −Ea
RT
where: k= rate constant A= pre-exponential factor (frequency factor) Ea=
activation energy R= gas constant (8.314 J/mol-K) T= temperature in Kelvin
Step 2: Given that the rate constant doubles when the temperature increases
from 25
°
C to 35
°
C, we can set up the following ratio:
k2
k1
= 2
where k2is the rate constant at 35
°
C and k1is the rate constant at 25
°
C.
Step 3: We can express the rate constant in terms of the Arrhenius equation:
k1=Aexp −Ea
RT1
k2=Aexp −Ea
RT2
28
Step 4: Substitute the expressions for k1and k2into the ratio equation:
Aexp −Ea
RT2
Aexp −Ea
RT1= 2
Step 5: Simplify the ratio equation:
exp −Ea
RT2
+Ea
RT1= 2
Step 6: Combine the exponents:
exp Ea
R1
T1
−1
T2= 2
Step 7: Now, substitute the temperatures in Kelvin:
exp Ea
8.314 1
298 −1
308= 2
Step 8: Solve for activation energy Ea:
Ea
8.314 1
298 −1
308= ln(2)
Ea= 8.314 ×(ln(2)) ×298 ×308
10
Step 9: Calculate the activation energy using the given formula.
Question 31
Question
For a certain reaction, the rate constant kat 25
°
C is 1.50 ×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 5.88 ×10−2s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin by adding 273.15.
T1= 25C+ 273.15 = 298.15 K
T2= 50C+ 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at two
different temperatures: k2
k1
=e
Ea
R1
T1
−1
T2
29
Step 3: Substitute the given rate constants and temperatures into the Ar-
rhenius equation and solve for the activation energy Ea:
5.88 ×10−2
1.50 ×10−3=eEa
8.314 (1
298.15 −1
323.15 )
39.2
1.50 ×10−3=eEa
8.314 (1
298.15 −1
323.15 )
26133.33 = eEa
8.314 (1
298.15 −1
323.15 )
Step 4: Simplify the equation and solve for the activation energy Ea:
ln(26133.33) = Ea
8.314 1
298.15 −1
323.15
ln(26133.33)
1
298.15 −1
323.15
=Ea
8.314
Ea= 8.314 ×ln(26133.33)
1
298.15 −1
323.15
Step 5: Calculate the activation energy:
Ea= 8.314 ×ln(26133.33)
1
298.15 −1
323.15
Ea≈62600 J/mol
Question 32
Question
The rate constant for the first-order decomposition of a reactant was found to
be 2.60×10−3s−1at 25◦C and 1.70×10−2s−1at 45◦C. 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 activation energy, temperature, and the pre-exponential factor:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J mol−1K−1), T= temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for activation energy
(Ea):
ln k2
k1=−Ea
R1
T2
−1
T1
30
Step 3: Plug in the given values and solve for the activation energy:
ln 1.70 ×10−2s−1
2.60 ×10−3s−1=−Ea
8.314 1
318 K −1
298 K
ln (6.538) = −Ea
8.314 0.00315 K−1
Step 4: Solve for the activation energy Ea:
Ea=−8.314 J mol−1×0.00315 K−1
ln(6.538)
Ea≈43.22 kJ mol−1
Therefore, the activation energy for this reaction is approximately 43.22 kJ mol−1.
Question 33
Question
The rate constant of a reaction at 25
°
C is 5.0×10−3s−1, and at 35
°
C it is
1.2×10−2s−1. Calculate the activation energy for the reaction in kJ/mol.
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)), and Tis the temperature
in Kelvin.
Step 2: Convert the given temperatures to Kelvin:
T1= 25C= 25 + 273 = 298K
T2= 35C= 35 + 273 = 308K
Step 3: rearrange the Arrhenius equation for two different temperatures to
solve for the activation energy, Ea:
ln k2
k1
T1−T2
=−Ea
R
Step 4: Substitute the given values into the equation:
ln 1.2×10−2
5.0×10−3
298 −308 =−Ea
8.314
31
Step 5: Calculate the activation energy, Ea:
ln(2.4)
−10 =−Ea
8.314
Ea=−8.314 ×ln(2.4)
10
Ea≈41.9 kJ/mol
Therefore, the activation energy for the reaction is approximately 41.9 kJ/mol.
Question 34
Question
The rate constant for a particular reaction is found to be 3.78 ×10−3s−1at
25◦C and 1.32 ×10−2s−1at 50◦C. Calculate the activation energy (Ea) for this
reaction.
Solution
Step 1: Calculate R, the ideal gas constant.
R= 8.314 J mol−1K−1
Step 2: Convert the temperatures to Kelvin.
T1= 25 + 273 = 298 K
T2= 50 + 273 = 323 K
Step 3: Use the Arrhenius equation to find Ea.
k=A·e−Ea
RT
Step 4: Take the natural logarithm of both sides of the equation to linearize
it.
ln k= ln A−Ea
R1
T
Step 5: Substitute the given data and solve for Ea.
ln k1= ln A−Ea
R1
T1
ln k2= ln A−Ea
R1
T2
Step 6: Subtract the two equations to eliminate ln A.
ln k2
k1
=−Ea
R1
T2
−1
T1
32
Step 7: Solve for Ea.
Ea=− R
ln k2
k1!1
T2
−1
T1
Step 8: Plug in the values to calculate Ea.
Ea=− 8.314
ln 1.32×10−2
3.78×10−3!1
323 −1
298
Ea≈65.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 65.2
kJ/mol.
Question 35
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 25◦C and
1.62 ×10−2s−1at 45◦C. Calculate the activation energy for the reaction.
Given: R= 8.314 J/(mol·K)
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
Step 2: Let’s start by taking the natural logarithm of the Arrhenius equation
in order to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 3: We can rewrite the above equation for two different temperatures
T1and T2as follows:
ln(k2) = ln(A)−Ea
R·1
T2
ln(k1) = ln(A)−Ea
R·1
T1
Step 4: Subtracting the two equations gives:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 5: Substitute the given rate constants and temperatures to solve for
the activation energy Ea:
ln 1.62 ×10−2
4.23 ×10−3=−Ea
R·1
318.15 −1
298.15
33
Question 4
Question
The rate constant of a certain reaction is found to be 3.2×10−4s−1at 25◦Cand
9.8×10−4s−1at 50◦C. Calculate the activation energy (Ea) for the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the equation: T(K) =
T(◦C) + 273.15. - T1= 25◦C+ 273.15 = 298.15K-T2= 50◦C+ 273.15 =
323.15K
Step 2: Now we are going to use the Arrhenius equation:
k=A·e−Ea
RT
where: - k1= 3.2×10−4s−1-k2= 9.8×10−4s−1-T1= 298.15K-T2= 323.15K
We will use the two given rate constants and temperatures to set up two
equations:
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
k2
k1
=e−Ea
R1
T2
−1
T1
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Solve for the activation energy (Ea):
−Ea
R=
ln k2
k1
1
T2
−1
T1
Ea=−R·
ln k2
k1
1
T2
−1
T1
Step 5: Now substitute the known values:
Ea=−(8.314J/mol ·K)·
ln 9.8×10−4
3.2×10−4
1
323.15 −1
298.15
4
Step 6: Calculate the activation energy:
Ea=−(8.314J/mol ·K)·ln(3.0625)
0.0033
Ea≈92.6kJ/mol
Therefore, the activation energy for the reaction is approximately 92.6 kJ/mol.
Question 5
Question
The rate constant for a certain reaction is found to be 5.00 ×10−2s−1at 25◦C
and 1.20 mol L−1. When the temperature is increased to 35◦C, the rate constant
becomes 1.50 ×10−1s−1. Determine the activation energy for this reaction.
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
k=A·exp −Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J mol−1K−1), and - Tis the
temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for Ea:
Ea=−
ln k2
k1
1
T2
−1
T1
where: - k1= 5.00 ×10−2s−1at 25◦C, - k2= 1.50 ×10−1s−1at 35◦C, -
T1= 25 + 273 = 298 K, and - T2= 35 + 273 = 308 K.
Step 3: Substitute the given values into the equation:
Ea=−
ln 1.50×10−1
5.00×10−2
1
308 −1
298
Step 4: Calculate Ea:
Ea=−ln(3)
1
308 −1
298
Ea=−ln(3)
298−308
298×308
Ea=−ln(3) ·298 ×308
10
Ea≈47887 J mol−1
Therefore, the activation energy for this reaction is approximately 47887 J
mol−1.
5
Question 6
Question
The rate constant for a certain reaction is found to be 4.55 ×10−3s−1at 25◦C
and 1.80 ×10−2s−1at 35◦C. Calculate the activation energy (Ea) for the
reaction in kJ/mol.
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 activa-
tion energy, Ris the gas constant (8.314 J mol−1K−1), Tis the temperature in
Kelvin.
Step 2: Convert the temperatures to Kelvin: 25◦C = 298 K and 35◦C =
308 K
Step 3: Substitue the given data points into the Arrhenius equation to form
a system of equations:
(4.55 ×10−3=A·e−Ea
8.314×298
1.80 ×10−2=A·e−Ea
8.314×308
Step 4: Divide the two equations to eliminate A:
4.55 ×10−3
1.80 ×10−2=e−Ea
8.314×298
e−Ea
8.314×308
Step 5: Simplify the expression on the right side:
4.55
1.80 =eEa
8.314 (1
298 −1
308 )
Step 6: Solve for the activation energy, Ea, in Joules:
Ea=−8.314 ×ln 4.55
1.80×1
298 −1
308
Step 7: Convert the activation energy to kilojoules per mole:
Ea=−8.314 ×ln 4.55
1.80×1
298 −1
308×1
1000
Calculate the value to find Ea.
6
Question 7
Question
The rate constant for a certain reaction is 2.50×10−3s−1at 25
°
C, and increases
to 5.00 ×10−3s−1at 45
°
C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given: T1= 25C,
T2= 45CAdding 273.15 to each temperature: T1= (25 + 273.15)K= 298.15K
T2= (45 + 273.15)K= 318.15K
Step 2: Use the Arrhenius equation to calculate the activation energy. 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
We know that k1= 2.50 ×10−3s−1,k2= 5.00 ×10−3s−1,T1= 298.15 K,
T2= 318.15 K. We can set up two equations using the Arrhenius equation:
k1=A·e−Ea
R·T1
k2=A·e−Ea
R·T2
Step 3: Divide the two equations to eliminate A.
k2
k1
=e−Ea
R·T2
e−Ea
R·T1
k2
k1
=e−Ea
R1
T2
−1
T1
k2
k1
=e−Ea
RT1−T2
T1·T2
Step 4: Solve for the activation energy Ea. Taking the natural logarithm of
both sides:
ln k2
k1=−Ea
RT1−T2
T1·T2
Ea=−R·ln k2
k1T1·T2
T1−T2
Substitute the given values:
Ea=−8.314 J/mol ·ln 5.00 ×10−3
2.50 ×10−3298.15 ·318.15
298.15 −318.15
7
Ea=−8.314 J/mol ·ln(2) ·298.15 ·318.15
−20
Ea≈90160 J/mol
Therefore, the activation energy for this reaction is approximately 90.16
kJ/mol.
Question 8
Question
The rate constant for a certain reaction was found to be 1.20×10−3s−1at 37◦C
and 3.45 ×10−2s−1at 87◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Write out the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol·K)), T= temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin: 37◦C = 37+273 = 310 K 87◦C
= 87 + 273 = 360 K
Step 3: Set up two equations using the given rate constants:
(1.20 ×10−3=A·e−Ea
8.314·310
3.45 ×10−2=A·e−Ea
8.314·360
Step 4: Divide the second equation by the first to eliminate A:
3.45 ×10−2
1.20 ×10−3=e−Ea
8.314·360
e−Ea
8.314·310
Step 5: Simplify the equation:
28.75 = eEa
8.314 (1
310 −1
360 )
Step 6: Solve for Ea:
Ea
8.314 1
310 −1
360= ln(28.75)
Ea= 8.314 ×ln(28.75)
1
310 −1
360
Step 7: Calculate Ea:
Ea≈50.3 kJ/mol
Therefore, the activation energy for this reaction is approximately 50.3
kJ/mol.
8
Question 9
Question
The rate constant for the reaction
2A+B→C
is found to be 1.5×10−3M−1s−1at 25
°
C. When the temperature is increased
to 35
°
C, the rate constant becomes 5.0×10−3M−1s−1. Calculate the activation
energy for this reaction. Assume the frequency factor (pre-exponential factor)
is 2.5×1013 s−1.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant at two
different temperatures:
k2=k1×e−Ea
R1
T2
−1
T1
where: k1= 1.5×10−3M−1s−1(at 25
°
C), k2= 5.0×10−3M−1s−1(at 35
°
C),
T1= 25 + 273 K, T2= 35 + 273 K, R= 8.314 J/(mol·K).
Step 2: Now let’s substitute the values into the Arrhenius equation and solve
for Ea:
5.0×10−3= 1.5×10−3×e−Ea
8.314 (1
308 −1
298 )
Step 3: Simplify the equation:
3.33 = e−Ea
8.314 (1
308 −1
298 )
Step 4: Take the natural logarithm of both sides to solve for Ea:
ln(3.33) = −Ea
8.314 1
308 −1
298
Step 5: Solve for Ea:
Ea=−8.314 ×ln(3.33)∇ · 1
308 −1
298
Step 6: Calculate the activation energy:
Ea≈90.6 kJ/mol
Therefore, the activation energy for the reaction is approximately 90.6 kJ/mol.
Question 10
Question
For a certain reaction, the rate constant is found to be 0.005 s−1at 25◦C and
0.02 s−1at 50◦C. Calculate the activation energy for this reaction.
9
Solution
Step 1: Determine the Arrhenius equation. The Arrhenius equation relates the
rate constant (k) to the temperature (T) and the activation energy (Ea) of a
reaction. It is given by:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Ea is the acti-
vation energy, - Ris the gas constant (8.314 J/(mol·K)), - Tis the temperature
in Kelvin.
Step 2: Determine the pre-exponential factor (A). To solve for A, we can use
the two given rate constants and temperatures: At 25◦C (298 K): k1= 0.005
s−1, At 50◦C (323 K): k2= 0.02 s−1.
Substitute the values into the Arrhenius equation and solve for A:
k1=A·e−Ea
RT1
k2=A·e−Ea
RT2
Divide the second equation by the first:
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
e−Ea
R(1
T2
−1
T1)=k2
k1
e−Ea
8.314 (1
323 −1
298 )=0.02
0.005
e−Ea
8.314 (0.0031) = 4
Step 3: Solve for activation energy (Ea). Take the natural logarithm of both
sides of the equation:
lne−Ea
8.314 (0.0031)= ln(4)
−Ea
8.314 ·0.0031 = ln(4)
−0.0031Ea
8.314 = ln(4)
−Ea = 8.314 ·0.0031 ·ln(4)
Ea =−8.314 ·0.0031 ·ln(4)
Ea ≈42.17 kJ/mol
Therefore, the activation energy for this reaction is approximately 42.17
kJ/mol.
10
Question 11
Question
The rate constant for a certain reaction at 25
°
C is 2.0×10−3s−1. When the
temperature is increased to 35
°
C, the rate constant becomes 3.5×10−3s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin. Given: T1= 25C=
25 + 273 = 298 K
T2= 35C= 35 + 273 = 308 K
Step 2: Write the Arrhenius equation relating rate constants and tempera-
tures. The Arrhenius equation is given by:
k=A·e−Ea
RT
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it.
Taking the natural logarithm of both sides of the Arrhenius equation gives:
ln k= ln A−Ea
R·1
T
Step 4: Write the equation for the first set of data. Using the first set of
data (T1, k1):
ln k1= ln A−Ea
R·1
T1
Step 5: Write the equation for the second set of data. Using the second set
of data (T2, k2):
ln k2= ln A−Ea
R·1
T2
Step 6: Subtract the second equation from the first equation. Subtracting
the second equation from the first equation eliminates the ln A term, giving:
ln k1
k2
=−Ea
R·1
T1
−1
T2
Step 7: Solve for the activation energy Ea. Substitute the given values for
k1,k2,T1, and T2into the equation and solve for Ea:
ln 2.0×10−3
3.5×10−3=−Ea
R·1
298 −1
308
ln 0.57 = −Ea
8.314 ·1
298 −1
308
Ea=−8.314 ·ln 0.57
1
298 −1
308
Thus, the activation energy for this reaction is approximately 50.6 kJ/mol.
11
Question 12
Question
The rate constant for the reaction A →B is 1.2×10−3s−1at 25◦C and 3.6×
10−3s−1at 35◦C. Calculate the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=A·e−Ea
RT , where kis the rate constant, Ais the
pre-exponential factor, Eais the activation energy, R= 8.314 J mol−1K−1is
the gas constant, and Tis the temperature in Kelvin.)
Solution
Step 1: Convert the temperatures to Kelvin At 25◦C, T1= 25 + 273 = 298 K.
At 35◦C, T2= 35 + 273 = 308 K.
Step 2: Write down the given information k1= 1.2×10−3s−1at T1= 298 K.
k2= 3.6×10−3s−1at T2= 308 K. R= 8.314 J mol−1K−1.
Step 3: Calculate the activation energy, EaWe can use the Arrhenius equa-
tion to set up two equations: 1.2×10−3=A·e−Ea
8.314·298 3.6×10−3=A·e−Ea
8.314·308
Step 4: Find the ratio of the two equations Dividing the second equation by
the first gives: 3.6×10−3
1.2×10−3=eEa
8.314 (1
308 −1
298 )3 = e10
2397
Step 5: Solve for the activation energy, EaTaking the natural log of both
sides and rearranging: ln(3) = 10
2397 EaEa=2397
10 ln(3) ≈2397
10 ·1.0986 ≈
262 kJ mol−1
Therefore, the activation energy for the reaction is approximately 262 kJ
mol−1.
Question 13
Question
The rate constant for the reaction A →B is found to be 2.5×10−3s−1at
25◦C. When the temperature is raised to 45◦C, the rate constant becomes
1.6×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures to the activation energy. The Arrhenius equation is given
by:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/(mol
·
K)), and - Tis the
temperature in Kelvin.
Step 2: Let’s first convert the temperatures to Kelvin:
T1= 25 + 273 = 298 K
12
T2= 45 + 273 = 318 K
Step 3: We can write two Arrhenius equations for the two different temper-
atures:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 4: Given that the rate constants are 2.5×10−3s−1at 25◦C and 1.6×
10−2s−1at 45◦C, we have:
2.5×10−3=Ae−Ea
8.314×298
1.6×10−2=Ae−Ea
8.314×318
Step 5: We can divide the second equation by the first equation to eliminate
Aand solve for the activation energy:
1.6×10−2
2.5×10−3=e−Ea
8.314×318
e−Ea
8.314×298
Step 6: Simplifying the above expression, we get:
1.6×10−2
2.5×10−3=eEa
8.314 (1
298 −1
318 )
Step 7: Solve for the activation energy, Ea, by calculating the natural loga-
rithm of both sides and then solving for Ea.
Question 14
Question
The rate constant for a reaction is found to double when the temperature is
increased from 25
°
C to 35
°
C. Calculate the activation energy (Ea) for this re-
action.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant (k) to
temperature and activation energy:
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: Let’s consider the ratio of rate constants at two temperatures:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
13
Given that k2= 2k1and T2= 35 + 273 = 308 K, T1= 25 + 273 = 298 K, we
can substitute these values into the equation.
Step 3:
2k1
k1
=Ae−Ea
8.314×308
Ae−Ea
8.314×298
2 = e−Ea
8.314×308 +Ea
8.314×298
Step 4: Take the natural logarithm on both sides to solve for Ea:
ln(2) = −Ea
8.314 1
308 −1
298
Ea=−8.314 ×(1
308 −1
298)×ln(2)
Step 5: Calculate Ea:
Ea≈ −8.314 ×1
308 −1
298×ln(2)
Ea≈39,223.65 J/mol
Therefore, the activation energy for this reaction is approximately 39.22
kJ/mol.
Question 15
Question
When a certain reaction was run at 25
°
C, the rate constant was found to be
2.0×10−3s−1. When the temperature was increased to 50
°
C, the rate constant
increased to 8.0×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(C) + 273.15. At 25
°
C: T1= 25C+ 273.15 = 298.15 K
At 50
°
C: T2= 50C+ 273.15 = 323.15 K
Step 2: Write down the Arrhenius equation:
k=Ae−Ea
RT
Step 3: Take the natural logarithm of both sides of the Arrhenius equation
to linearize the equation:
ln(k) = ln(A)−Ea
R1
T
14
Step 4: Use the data provided to obtain two equations:
ln2.0×10−3= ln(A)−Ea
R1
298.15
ln8.0×10−2= ln(A)−Ea
R1
323.15
Step 5: Subtract the second equation from the first to eliminate ln(A):
ln 2.0×10−3
8.0×10−2=−Ea
R1
298.15 −1
323.15
Step 6: Solve for the activation energy Ea:
Ea=−R
ln 2.0×10−3
8.0×10−2
1
298.15 −1
323.15
Step 7: Calculate the activation energy using the ideal gas constant R=
8.314 J/(mol
·
K):
Ea=−8.314
ln 2.0×10−3
8.0×10−2
1
298.15 −1
323.15
Question 16
Question
Given the rate constant of a reaction at two temperatures, 25
°
C and 45
°
C,
calculate the activation energy for the reaction. The rate constant at 25
°
C is
5.0×10−3s−1and at 45
°
C is 1.0×10−2s−1. (Assume the activation energy
follows the Arrhenius equation).
Solution
Step 1: Convert the temperatures from Celsius to Kelvin using the formula
T(K) = T(C) + 273. At 25
°
C: T1= 25 + 273 = 298 K At 45
°
C: T2= 45 + 273 =
318 K
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, and T= temperature in Kelvin.
Step 3: Take the natural logarithm of the Arrhenius equation:
ln(k) = ln(A)−Ea
R·1
T
15
This equation can be rewritten as:
ln(k) = −Ea
R·1
T+ ln(A)
Step 4: Set up two equations using the rate constant values at 25
°
C and
45
°
C:
ln5.0×10−3=−Ea
R·1
298 + ln(A)
ln1.0×10−2=−Ea
R·1
318 + ln(A)
Step 5: Subtract the second equation from the first to eliminate ln(A):
ln5.0×10−3−ln1.0×10−2=−Ea
R·1
298 −1
318
Step 6: Solve for the activation energy Eausing the gas constant R= 8.314
J/(mol·K):
Ea=−8.314 ×298 ×318 ×ln5.0×10−3−ln1.0×10−2
Step 7: Calculate the activation energy using the given values and complete
the calculation.
Question 17
Question
The rate constant for a certain reaction is 4.32 ×10−2s−1at 25
°
C and 1.56 ×
10−1s−1at 35
°
C. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin. Given:
T1= 25C= 25 + 273.15 = 298.15 K
T2= 35C= 35 + 273.15 = 308.15 K
Step 2: Calculate the activation energy (Ea) using the Arrhenius equation.
The Arrhenius equation is given by:
k=A×e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol
·
K)), T= temperature in Kelvin.
Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
R×1
T
16
Step 3: Set up the linear equation using the data at two different tempera-
tures. Using the data provided:
ln k1= ln A−Ea
R×1
T1
ln k2= ln A−Ea
R×1
T2
Step 4: Calculate the activation energy. Subtract the two equations to
eliminate ln A:
ln k2−ln k1=Ea
R×1
T1
−1
T2
Step 5: Plug in the values and solve for activation energy. Substitute the
given rate constants and temperatures into the equation:
ln 1.56 ×10−1−ln 4.32 ×10−2=Ea
8.314 ×1
298.15 −1
308.15
Step 6: Calculate Eain joules and convert to kilojoules per mole. After
solving the equation, we get:
Ea= 9.49 kJ/mol
Question 18
Question
The rate constant (k) for a reaction at 25
°
C is 4.2×10−3s−1, and at 35
°
C it is
2.5×10−2s−1. Calculate the activation energy (Ea) for this reaction. (Assume
the frequency factor Ais independent of temperature and is 8.0×1012 s−1)
Solution
Step 1: Recall the Arrhenius equation:
k=A×e−Ea
RT
where: - kis the rate constant - Ais the frequency factor - Eais the activation
energy - Ris the gas constant (8.314 J mol−1K−1) - Tis the temperature in
Kelvin
Step 2: We can rearrange the Arrhenius equation to solve for Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation:
ln 2.5×10−2
4.2×10−3=−Ea
8.314 1
308 −1
298
17
Step 4: Calculate the left-hand side of the equation:
ln 2.5×10−2
4.2×10−3= ln 2.5
0.42 ×10−1= ln (5.952) ≈1.778
Step 5: Plug the values back into the equation and solve for Ea:
1.778 = −Ea
8.314 1
308 −1
298
Step 6: Simplify the equation:
1.778 = −Ea
8.314 ×10
298 ×308
Step 7: Solve for Ea:
Ea=−1.778 ×8.314 ×10
298 ×308
Step 8: Calculate the activation energy Eato find the final answer.
Question 19
Question
For a certain reaction, the rate constant at 25
°
C is 4.0×10−3s−1, and the
activation energy is 60 kJ/mol. Calculate the rate constant at 35
°
C for this
reaction.
Solution
Step 1: Recall the Arrhenius equation, which relates the rate constant of a
reaction to temperature and the activation energy:
k=A·e−Ea
RT
Where: k= rate constant A= pre-exponential factor Ea= activation energy
(J/mol) R= gas constant (8.314 J/mol ·K) T= temperature (K)
Step 2: Let’s first convert the given activation energy from kJ/mol to J/mol,
and both temperatures from
°
C to K:
Ea= 60 ×103J/mol = 60 000 J/mol
T1= 25 + 273 = 298 K
T2= 35 + 273 = 308 K
Step 3: Now, we can rearrange the Arrhenius equation to solve for the pre-
exponential factor A:
A=k
e−Ea
RT
18
Step 4: Substitute the given rate constant kand the values we just calculated
into the equation to solve for Aat 25
°
C:
A=4.0×10−3
e−60 000
8.314×298
Step 5: Calculate the value of Aat 25
°
C:
A≈4.0×10−3
e−23971 ≈4.0×10−3
4.98 ×10−11 ≈8.03 ×107s−1
Step 6: Finally, substitute the newly calculated pre-exponential factor Ainto
the Arrhenius equation for 35
°
C to find the rate constant at this temperature:
k= 8.03 ×107·e−60 000
8.314×308
Step 7: Calculate the rate constant at 35
°
C:
k≈8.03 ×107·e−23971 ≈8.03 ×107·4.98 ×10−11 ≈3.99 ×10−3s−1
Therefore, the rate constant at 35
°
C for this reaction is approximately 3.99×
10−3s−1.
Question 20
Question
For the reaction A →B, the rate constant at 300 K is 2.3×10−3s−1and the
activation energy is 50 kJ/mol. Calculate the rate constant at 350 K for this
reaction.
Solution
Step 1: First, we need to use the Arrhenius equation to relate the rate constants
at two different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1= 2.3×10−3s−1(rate constant at 300 K)
T1= 300 K
Ea= 50 kJ/mol = 50000 J/mol
R= 8.314 J/mol
·
K (gas constant)
T2= 350 K
Step 2: Substitute the values into the equation and solve for k2:
ln k2
2.3×10−3=−50000
8.314 1
350 −1
300
19
ln k2
2.3×10−3=−6049.05
k2
2.3×10−3=e−6049.05
k2≈2.172 ×10−308 s−1
Therefore, the rate constant at 350 K for the reaction is approximately
2.172 ×10−308 s−1.
Question 21
Question
A reaction has an activation energy of 50 kJ/mol and a rate constant of 2.0×
10−3s−1at 25
°
C. Determine the rate constant at 50
°
C for this reaction if the
activation energy does not change.
Solution
Step 1: First, calculate the pre-exponential factor (A) using the Arrhenius equa-
tion:
k=A·e−Ea
RT
Given that k= 2.0×10−3s−1at 25
°
C, we can rearrange the equation to
solve for A:
A=k
e−Ea
RT
Where: - k= 2.0×10−3s−1-Ea= 50 kJ/mol - R= 8.314 J/(mol·K) -
T= 25 + 273.15 K
Step 2: Calculate A
A=2.0×10−3
e−50000
8.314×298.15
A≈4.35 ×1013 s−1
Step 3: Now, we can use the Arrhenius equation to find the rate constant at
50
°
C:
k=A·e−Ea
RT
Where: - A= 4.35 ×1013 s−1-Ea= 50 kJ/mol - R= 8.314 J/(mol·K) -
T= 50 + 273.15 K
Step 4: Calculate the rate constant at 50
°
C
k= 4.35 ×1013 ·e−50000
8.314×323.15
k≈4.15 ×10−3s−1
Therefore, the rate constant at 50
°
C for this reaction is approximately 4.15×
10−3s−1.
20
Question 22
Question
The rate constant for a certain reaction is found to be 4.52 ×10−3s−1at 25◦C
and 1.93 ×10−2s−1at 35◦C. Calculate the activation energy for this reaction.
Assume the frequency factor Ais 1.05 ×1010 s−1.
Solution
Step 1: Write the Arrhenius equation:
k=A·e−Ea
RT
where: kis the rate constant, Ais the frequency factor, Eais the activation
energy, Ris the ideal gas constant (8.314 J/(mol K)), and Tis the temperature
in Kelvin.
Step 2: Set up the Arrhenius equation for both given temperature conditions:
At 25◦C (298 K):
4.52 ×10−3= 1.05 ×1010 ·e−Ea
8.314·298
At 35◦C (308 K):
1.93 ×10−2= 1.05 ×1010 ·e−Ea
8.314·308
Step 3: Take the ratio of the equations to eliminate A:
4.52 ×10−3
1.93 ×10−2=e−Ea
8.314·298
e−Ea
8.314·308
Step 4: Simplify the ratio:
0.234 = eEa
8.314 ·(1
308 −1
298 )
Step 5: Solve for Ea:
ln(0.234) = Ea
8.314 ·1
308 −1
298
Step 6: Calculate Ea:
Ea= 8.314 ·ln(0.234)
1
308 −1
298
Step 7: Plug in the values and calculate Ea:
Ea≈62.1 kJ/mol
Therefore, the activation energy for the reaction is approximately 62.1 kJ/mol.
21
Question 23
Question
The rate constant for a certain reaction is 2.0×10−3s−1at 273 K and 6.5×
10−3s−1at 293 K. Calculate the activation energy of the reaction (given that
the universal gas constant R= 8.314 J mol−1K−1).
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy
(in joules per mole), R= universal gas constant, T= temperature in kelvin.
Step 2: We are given two sets of data points: At 273 K, k1= 2.0×10−3s−1,
At 293 K, k2= 6.5×10−3s−1.
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln(k) = ln(A)−Ea
RT
Step 4: Subtract the second equation from the first to eliminate A:
ln(k2)−ln(k1) = −Ea
R1
T2
−1
T1
Step 5: Plug in the values and solve for Ea:
ln6.5×10−3−ln2.0×10−3=−Ea
8.314 1
293 −1
273
Step 6: Calculate Ea:
−1.0296 = −Ea
8.314 1
293 −1
273
Ea= 54634.6 J/mol
Step 7: Therefore, the activation energy of the reaction is 54.634 kJ/mol.
Question 24
Question
The rate constant for a certain reaction increases from 5.0×10−2s−1at 300 K
to 8.0×10−2s−1at 350 K. Calculate the activation energy for this reaction.
Assume the frequency factor is 1.0×1013 s−1.
22
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant to tem-
perature and activation energy:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the frequency factor, - Eais the activation
energy, - Ris the gas constant (8.314 J ·mol−1·K−1), and - Tis the absolute
temperature.
Step 2: We can rearrange the Arrhenius equation to solve for the activation
energy Ea:
Ea=−R·1
T1
−1
T2−1
·ln k2
k1
Step 3: Substitute the given values into the equation:
Ea=−(8.314 J ·mol−1·K−1)·1
300 K −1
350 K−1
·ln 8.0×10−2s−1
5.0×10−2s−1
Step 4: Calculate the activation energy:
Ea≈53.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 53.7 kJ/mol.
Question 25
Question
The rate constant for the reaction
2A+B→C
is found to be 3.2×10−3s−1at 25
°
C. When the temperature is increased to
75
°
C, the rate constant becomes 2.4×10−2s−1. Calculate the activation energy
for this reaction.
Solution
Step 1: We start by using the Arrhenius equation, which relates the rate con-
stant of a reaction to the temperature and the activation energy. The Arrhenius
equation is given by:
k=A·e−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= gas constant (8.314 J mol−1K−1)T= temperature in Kelvin
23
Step 2: We can rearrange the Arrhenius equation to calculate the activation
energy, Ea. Taking the natural logarithm of both sides gives:
ln(k) = ln(A)−Ea
R·1
T
Step 3: Let’s find the activation energy using the data provided at 25
°
C and
75
°
C. At 25
°
C (298 K), the rate constant k1is 3.2×10−3s−1, and at 75
°
C (348
K), the rate constant k2is 2.4×10−2s−1.
Step 4: Now, we have:
ln(k1) = ln(A)−Ea
R·1
T1
ln(k2) = ln(A)−Ea
R·1
T2
Step 5: Subtracting the two equations gives:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 6: Plugging in the values:
ln 2.4×10−2
3.2×10−3=−Ea
8.314 ·1
348 −1
298
Step 7: Solving for Ea:
−2.9957 = −Ea
8.314 ·1
348 −1
298
Ea= 51.2 kJ/mol
Therefore, the activation energy for the reaction is 51.2 kJ/mol.
Question 26
Question
The rate constant for the reaction A →B is found to be 4.75 ×10−4s−1at
25◦C. When the temperature is increased to 50◦C, the rate constant becomes
2.41 ×10−3s−1. Calculate the activation energy (Ea) for this reaction. (Hint:
Use the Arrhenius equation: k=A·e−Ea
RT , where kis the rate constant, Ais
the frequency factor, Eais the activation energy, Ris the gas constant, and T
is the absolute temperature.)
24
Solution
Step 1: Convert the temperatures to Kelvin.
T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 50◦C = 50 + 273.15 = 323.15 K
Step 2: Determine the ratio of rate constants at the two temperatures.
k2
k1
=2.41 ×10−3s−1
4.75 ×10−4s−1= 5.08
Step 3: Substitute into the Arrhenius equation and take the natural loga-
rithm of both sides.
k2
k1
=e−Ea
R1
T2
−1
T1
5.08 = e−Ea
R(1
323.15 −1
298.15 )
ln(5.08) = −Ea
R1
323.15 −1
298.15
Step 4: Solve for the activation energy (Ea).
ln(5.08) = −Ea
8.314 1
323.15 −1
298.15
Ea=−8.314 ×ln(5.08)
1
323.15 −1
298.15
Ea≈83.3 kJ/mol
Therefore, the activation energy for this reaction is approximately 83.3
kJ/mol.
Question 27
Question
The rate constant of a certain reaction is found to be 8.21 ×10−3s−1at 75 ◦C
and 3.54 ×10−2s−1at 100 ◦C. Calculate the activation energy for the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(◦C) + 273.15. At 75 ◦C: T1= 75 + 273.15 = 348.15 K
At 100 ◦C: T2= 100 + 273.15 = 373.15 K
Step 2: Write the Arrhenius equation:
k=A·e−Ea
RT
25
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= gas constant = 8.314 J ·mol−1·K−1T= temperature in Kelvin
Step 3: Set up two equations using the given rate constants and tempera-
tures:
8.21 ×10−3=A·e−Ea
8.314·348.15
3.54 ×10−2=A·e−Ea
8.314·373.15
Step 4: Divide the two equations to eliminate A:
8.21 ×10−3
3.54 ×10−2=e−Ea
8.314·348.15
e−Ea
8.314·373.15
Step 5: Simplify the equation:
8.21 ×10−3
3.54 ×10−2=e373.15−348.15
8.314 ·Ea
8.314·348.15·373.15
Step 6: Solve for Ea:
Ea=−8.314 ·373.15 ·348.15
373.15 −348.15 ·ln 8.21 ×10−3
3.54 ×10−2
Calculating Eagives:
Ea≈35,925 J/mol
Question 28
Question
The rate constant for the reaction
2A→B+C
is 3.21 ×10−3s−1at 25◦C. When the temperature is increased to 55◦C, the
rate constant becomes 6.75 ×10−3s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C + 273 = 298 K
T2= 55◦C + 273 = 328 K
Step 2: Use the Arrhenius equation to relate the rate constants (k1and k2)
at the two temperatures to the activation energy (Ea).
k1=A·e−Ea/(R·T1)
26
k2=A·e−Ea/(R·T2)
Step 3: Divide the two equations to eliminate the pre-exponential factor A.
k2
k1
=e−Ea/(R·T2)
e−Ea/(R·T1)
Step 4: Take the natural logarithm of both sides to simplify the equation.
ln k2
k1=−Ea
R1
T2
−1
T1
Step 5: Now, substitute the given values and constants into the equation
and solve for the activation energy (Ea).
R= 8.314 J ·mol−1·K−1
ln 6.75 ×10−3
3.21 ×10−3=−Ea
8.314 1
328 −1
298
Step 6: Calculate the activation energy using the values obtained.
Ea=−8.314 ×ln 6.75 ×10−3
3.21 ×10−3×1
328 −1
298
Ea≈37.2 kJ/mol
Therefore, the activation energy for the reaction is approximately 37.2 kJ/mol.
Question 29
Question
The rate constant for a certain reaction is observed to be 4.82 ×10−3s−1at
T= 350 K and 1.38 ×10−2s−1at T= 370 K. Calculate the activation energy
for this reaction.
Solution
Step 1: Write the Arrhenius equation: The Arrhenius equation relates the rate
constant (k) of a reaction to the temperature (T) and the activation energy (Ea)
through the equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor or frequency factor, Ea
= activation energy, R= gas constant, and T= temperature in Kelvin.
Step 2: Calculate the ratio of rate constants at the two temperatures: Given
that k1= 4.82×10−3s−1at T1= 350 K and k2= 1.38×10−2s−1at T2= 370 K,
we can find the ratio: k2
k1
=1.38 ×10−2
4.82 ×10−3= 2.86
27
Step 3: Use the ratio to find the activation energy: Taking the natural
logarithm of the Arrhenius equation, we get:
ln(k) = ln(A)−Ea
RT
ln k2
k1=Ea
R1
T1
−1
T2
Plugging in the values and solving for Ea:
ln(2.86) = Ea
R1
350 −1
370
ln(2.86) = Ea
8.314 1
350 −1
370
Ea= 7.58 kJ/mol
Question 30
Question
The rate constant (k) for a certain reaction was found to double when the
temperature was increased from 25
°
C to 35
°
C. Calculate the activation energy
for this reaction. (Assume the pre-exponential factor Ais constant.)
Solution
Step 1: Let’s first recall the Arrhenius equation:
k=Aexp −Ea
RT
where: k= rate constant A= pre-exponential factor (frequency factor) Ea=
activation energy R= gas constant (8.314 J/mol-K) T= temperature in Kelvin
Step 2: Given that the rate constant doubles when the temperature increases
from 25
°
C to 35
°
C, we can set up the following ratio:
k2
k1
= 2
where k2is the rate constant at 35
°
C and k1is the rate constant at 25
°
C.
Step 3: We can express the rate constant in terms of the Arrhenius equation:
k1=Aexp −Ea
RT1
k2=Aexp −Ea
RT2
28
Step 4: Substitute the expressions for k1and k2into the ratio equation:
Aexp −Ea
RT2
Aexp −Ea
RT1= 2
Step 5: Simplify the ratio equation:
exp −Ea
RT2
+Ea
RT1= 2
Step 6: Combine the exponents:
exp Ea
R1
T1
−1
T2= 2
Step 7: Now, substitute the temperatures in Kelvin:
exp Ea
8.314 1
298 −1
308= 2
Step 8: Solve for activation energy Ea:
Ea
8.314 1
298 −1
308= ln(2)
Ea= 8.314 ×(ln(2)) ×298 ×308
10
Step 9: Calculate the activation energy using the given formula.
Question 31
Question
For a certain reaction, the rate constant kat 25
°
C is 1.50 ×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 5.88 ×10−2s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Convert the temperatures from Celsius to Kelvin by adding 273.15.
T1= 25C+ 273.15 = 298.15 K
T2= 50C+ 273.15 = 323.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at two
different temperatures: k2
k1
=e
Ea
R1
T1
−1
T2
29
Step 3: Substitute the given rate constants and temperatures into the Ar-
rhenius equation and solve for the activation energy Ea:
5.88 ×10−2
1.50 ×10−3=eEa
8.314 (1
298.15 −1
323.15 )
39.2
1.50 ×10−3=eEa
8.314 (1
298.15 −1
323.15 )
26133.33 = eEa
8.314 (1
298.15 −1
323.15 )
Step 4: Simplify the equation and solve for the activation energy Ea:
ln(26133.33) = Ea
8.314 1
298.15 −1
323.15
ln(26133.33)
1
298.15 −1
323.15
=Ea
8.314
Ea= 8.314 ×ln(26133.33)
1
298.15 −1
323.15
Step 5: Calculate the activation energy:
Ea= 8.314 ×ln(26133.33)
1
298.15 −1
323.15
Ea≈62600 J/mol
Question 32
Question
The rate constant for the first-order decomposition of a reactant was found to
be 2.60×10−3s−1at 25◦C and 1.70×10−2s−1at 45◦C. 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 activation energy, temperature, and the pre-exponential factor:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J mol−1K−1), T= temperature in Kelvin.
Step 2: Rearrange the Arrhenius equation to solve for activation energy
(Ea):
ln k2
k1=−Ea
R1
T2
−1
T1
30
Step 3: Plug in the given values and solve for the activation energy:
ln 1.70 ×10−2s−1
2.60 ×10−3s−1=−Ea
8.314 1
318 K −1
298 K
ln (6.538) = −Ea
8.314 0.00315 K−1
Step 4: Solve for the activation energy Ea:
Ea=−8.314 J mol−1×0.00315 K−1
ln(6.538)
Ea≈43.22 kJ mol−1
Therefore, the activation energy for this reaction is approximately 43.22 kJ mol−1.
Question 33
Question
The rate constant of a reaction at 25
°
C is 5.0×10−3s−1, and at 35
°
C it is
1.2×10−2s−1. Calculate the activation energy for the reaction in kJ/mol.
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)), and Tis the temperature
in Kelvin.
Step 2: Convert the given temperatures to Kelvin:
T1= 25C= 25 + 273 = 298K
T2= 35C= 35 + 273 = 308K
Step 3: rearrange the Arrhenius equation for two different temperatures to
solve for the activation energy, Ea:
ln k2
k1
T1−T2
=−Ea
R
Step 4: Substitute the given values into the equation:
ln 1.2×10−2
5.0×10−3
298 −308 =−Ea
8.314
31
Step 5: Calculate the activation energy, Ea:
ln(2.4)
−10 =−Ea
8.314
Ea=−8.314 ×ln(2.4)
10
Ea≈41.9 kJ/mol
Therefore, the activation energy for the reaction is approximately 41.9 kJ/mol.
Question 34
Question
The rate constant for a particular reaction is found to be 3.78 ×10−3s−1at
25◦C and 1.32 ×10−2s−1at 50◦C. Calculate the activation energy (Ea) for this
reaction.
Solution
Step 1: Calculate R, the ideal gas constant.
R= 8.314 J mol−1K−1
Step 2: Convert the temperatures to Kelvin.
T1= 25 + 273 = 298 K
T2= 50 + 273 = 323 K
Step 3: Use the Arrhenius equation to find Ea.
k=A·e−Ea
RT
Step 4: Take the natural logarithm of both sides of the equation to linearize
it.
ln k= ln A−Ea
R1
T
Step 5: Substitute the given data and solve for Ea.
ln k1= ln A−Ea
R1
T1
ln k2= ln A−Ea
R1
T2
Step 6: Subtract the two equations to eliminate ln A.
ln k2
k1
=−Ea
R1
T2
−1
T1
32
Step 7: Solve for Ea.
Ea=− R
ln k2
k1!1
T2
−1
T1
Step 8: Plug in the values to calculate Ea.
Ea=− 8.314
ln 1.32×10−2
3.78×10−3!1
323 −1
298
Ea≈65.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 65.2
kJ/mol.
Question 35
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 25◦C and
1.62 ×10−2s−1at 45◦C. Calculate the activation energy for the reaction.
Given: R= 8.314 J/(mol·K)
Solution
Step 1: Calculate the activation energy using the Arrhenius equation:
k=A·e−Ea
RT
Step 2: Let’s start by taking the natural logarithm of the Arrhenius equation
in order to linearize it:
ln(k) = ln(A)−Ea
R·1
T
Step 3: We can rewrite the above equation for two different temperatures
T1and T2as follows:
ln(k2) = ln(A)−Ea
R·1
T2
ln(k1) = ln(A)−Ea
R·1
T1
Step 4: Subtracting the two equations gives:
ln k2
k1=−Ea
R·1
T2
−1
T1
Step 5: Substitute the given rate constants and temperatures to solve for
the activation energy Ea:
ln 1.62 ×10−2
4.23 ×10−3=−Ea
R·1
318.15 −1
298.15
33
Step 6: Solving the above equation for Eagives:
Ea=−R·
ln 1.62×10−2
4.23×10−3
1
318.15 −1
298.15
Step 7: Now, substitute the given value of Rand calculate Ea:
Ea=−8.314 ·
ln 1.62×10−2
4.23×10−3
1
318.15 −1
298.15
Step 8: Calculate the value of Eato find the activation energy for the reac-
tion.
34
