CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 6
Liberty University
Question 1
Question
The rate constant, k, for a certain reaction is found to be 1.25 ×10−3s−1at
25
°
C and 2.50 ×10−2s−1at 75
°
C. Calculate the activation energy (Ea) for this
reaction.
Solution
Step 1: Convert temperatures from Celsius to Kelvin using the formula T(K) =
T(C) + 273.15.
T1= 25C+ 273.15 = 298.15 K
T2= 75C+ 273.15 = 348.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at two
different temperatures to the activation energy.
k2
k1
=e−Ea
R1
T2
−1
T1
Step 3: Plug in the values for k2,k1,T1, and T2, and the universal gas
constant R= 8.314 J/(mol
·
K).
2.50 ×10−2
1.25 ×10−3=e−Ea
8.314 (1
348.15 −1
298.15 )
Step 4: Solve for the activation energy (Ea).
2.00 ×101
1.25 ×10−3=e−Ea
8.314 (1
348.15 −1
298.15 )
1.60 ×104=e−Ea
8.314 (1
348.15 −1
298.15 )
Step 5: Take the natural logarithm of both sides to solve for Ea.
ln1.60 ×104=−Ea
8.314 1
348.15 −1
298.15
Step 6: Solve for Ea.
Ea=−8.314 ×ln1.60 ×104
1
348.15 −1
298.15
Step 7: Calculate the value of Ea.
Ea≈1.59 ×105J/mol
Question 2
Question
The rate constant for the reaction A →B is 3.5×10−3s−1at 25◦C. If the
activation energy for the reaction is 40 kJ/mol, calculate the rate constant at
55◦C.
Solution
Step 1: Calculate the rate constant at 25◦C 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 ideal gas constant (8.314 J/mol ·K), and - Tis the
temperature in Kelvin.
Given that k= 3.5×10−3s−1and T= 25◦C = 298 K, we can rearrange the
Arrhenius equation to solve for A:
A=k
e−Ea
RT
Step 2: Calculate the pre-exponential factor A.
A=3.5×10−3
e−40×103
8.314×298
Step 3: Calculate the rate constant at 55◦C. Using the same Arrhenius
equation:
k=A·e−Ea
RT
Given that T= 55◦C = 328 K, we can now calculate the rate constant k:
k=A·e−Ea
RT
2
Question 3
Question
The rate constant for a certain reaction is 1.5×10−2s−1at 298 K and 4.5×
10−2s−1at 333 K. Calculate the activation energy for the reaction.
Solution
Step 1: Recall the Arrhenius equation:
k=A·e(−Ea
RT )
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J mol−1K−1), - Tis the tem-
perature in Kelvin.
Step 2: Use the data given to set up two equations:
1.5×10−2=A·e(−Ea
8.314·298 )
4.5×10−2=A·e(−Ea
8.314·333 )
Step 3: Divide the second equation by the first to eliminate A:
4.5×10−2
1.5×10−2=e(−Ea
8.314·333 )
e(−Ea
8.314·298 )
Step 4: Simplify the expression:
3 = e(Ea
8.314 (1
298 −1
333 ))
Step 5: Solve for Ea:
Ea=−8.314 ·1
1
298 −1
333 ·ln(3)
Step 6: Calculate the value of Eato find the activation energy for the reac-
tion.
Question 4
Question
The rate constant of a reaction at 25
°
C is 1.2×10−3s−1and the activation
energy is 80 kJ/mol. Calculate the rate constant of the reaction at 35
°
C. (Given:
R= 8.314 J/mol·K)
3
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 = 1.2×10−3s−1Ea= activation energy = 80
kJ/mol = 80 ×103J/mol R= gas constant = 8.314 J/mol·KT1= temperature
in Kelvin = 25
°
C + 273 = 298 K
Plugging in the values, we get:
1.2×10−3=A·e−80×103
8.314·298
Step 2: Solve for the pre-exponential factor A:
A=k1
e−80×103
8.314·298
Step 3: Calculate the rate constant at 35
°
C using the Arrhenius equation:
k2=A·e−Ea
RT2
where: k2= rate constant at 35
°
CEa= activation energy = 80 kJ/mol =
80 ×103J/mol R= gas constant = 8.314 J/mol·KT2= temperature in Kelvin
= 35
°
C + 273 = 308 K
Plugging in the values and the calculated value of A, we get:
k2=A·e−80×103
8.314·308
Question 5
Question
The rate constant for a certain reaction is 2.50 ×10−2s−1at 350 K, and 1.00 ×
10−1s−1at 400 K. Calculate the activation energy for this reaction in kJ/mol.
Solution
Step 1: Write the Arrhenius equation:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol*K)), - Tis the temperature
in Kelvin.
Step 2: Use the given data to form two equations using the Arrhenius equa-
tion and the two temperatures. At 350 K:
2.50 ×10−2=Ae−Ea
8.314×350
4
At 400 K:
1.00 ×10−1=Ae−Ea
8.314×400
Step 3: Divide the two equations to eliminate A:
2.50 ×10−2
1.00 ×10−1=e−Ea
8.314×350
e−Ea
8.314×400
0.25 = eEa
8.314 (1
350 −1
400 )
Step 4: Solve for the activation energy:
0.25 = eEa
8.314 (1
350 −1
400 )
ln(0.25) = Ea
8.314(1
350 −1
400)
Ea= 8.314 ×(1
350 −1
400)×ln(0.25)
Step 5: Calculate the activation energy in kJ/mol:
Ea= 8.314 ×(1
350 −1
400)×ln(0.25) ≈72.0 kJ/mol
Therefore, the activation energy for this reaction is approximately 72.0
kJ/mol.
Question 6
Question
The rate constant for a certain reaction is found to be 1.25 ×10−2s−1at 25
º
C
and 8.71 ×10−2s−1at 45
º
C. Calculate the activation energy for this reaction.
The activation energy is expressed in units of kJ/mol.
Given: R= 8.314 J/(mol K)
Solution
Step 1: Convert the given temperatures into Kelvin:
T1= 25C+ 273.15 = 298.15 K
T2= 45C+ 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to find the activation energy (Ea):
ln k2
k1=−Ea
R1
T2
−1
T1
5
Plugging in the values:
ln 8.71 ×10−2
1.25 ×10−2=−Ea
8.314 1
318.15 −1
298.15
ln(6.968) = −Ea
8.314 1
318.15 −1
298.15
1.938 = −Ea
8.314 (0.0031 −0.0034)
1.938 = −Ea
8.314(−0.0003)
1.938 = 0.0025Ea
Ea=1.938
0.0025
Ea≈775.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 775.2
kJ/mol.
Question 7
Question
The rate constant for the decomposition of a compound is found to be 9.56 ×
10−4s−1at 30◦C. If the activation energy for this reaction is 78.2 kJ/mol,
calculate the rate constant at 40◦C.
Solution
Step 1: We begin by noting the general form of 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), and - Tis the tem-
perature in Kelvin.
Step 2: First, we must convert the activation energy from kilojoules per mole
to joules per molecule:
Ea= 78.2×103J/mol
Step 3: Next, we calculate the rate constant kat 30◦C (303 K) using the
given values:
k1= 9.56 ×10−4s−1
Step 4: We rearrange the Arrhenius equation to solve for A:
A=k
e−Ea
RT
6
Step 5: Substitute the known values into the equation to find A:
A=9.56 ×10−4
e−78.2×103
8.314×303
Step 6: After calculating A, we use this value along with the activation
energy to find the rate constant k2at 40◦C (313 K) using the Arrhenius equation:
k2=A·e−Ea
RT
Step 7: Substitute the calculated values into the equation to find k2:
k2=A·e−78.2×103
8.314×313
Step 8: Compute the final value of k2to determine the rate constant at
40◦C.
Question 8
Question
The rate constant for a certain reaction is found to be 4.92 ×10−3s−1at 25◦C
and 8.64 ×10−3s−1at 35◦C. Calculate the activation energy (Ea) for this
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−1K−1), - Tis the tem-
perature in Kelvin.
Step 2: Convert temperatures to Kelvin: - 25◦C = 25 + 273 = 298 K -
35◦C = 35 + 273 = 308 K
Step 3: Substitute the given data into the Arrhenius equation for the two
temperatures:
k1=Ae−Ea
R×298
k2=Ae−Ea
R×308
Step 4: Divide the equations to eliminate A:
k2
k1
=e−Ea
R×308
e−Ea
R×298
7
Step 5: Simplify the equation:
k2
k1
=eEa
R(1
298 −1
308 )
Step 6: Substitute the values of k1,k2,R, and solve for Ea:
8.64 ×10−3
4.92 ×10−3=eEa
8.314 (1
298 −1
308 )
Step 7: Calculate Eausing the natural logarithmic function:
ln 8.64 ×10−3
4.92 ×10−3=Ea
8.314 1
298 −1
308
Step 8: Solve for Ea:
Ea= 8.314 ×1
298 −1
308×ln 8.64 ×10−3
4.92 ×10−3
After calculations, Ea≈41.8 kJ/mol.
Question 9
Question
The rate constant of a certain reaction is found to be 4.23 ×10−3s−1at 25◦C
and 1.44 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(◦C) + 273.15. At 25◦C: T1= 25 + 273.15 = 298.15 K
At 45◦C: T2= 45 + 273.15 = 318.15 K
Step 2: Use 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:
(4.23 ×10−3=A×e−Ea
8.314×298.15
1.44 ×10−2=A×e−Ea
8.314×318.15
Step 4: Divide the second equation by the first to eliminate A:
1.44 ×10−2
4.23 ×10−3=A×e−Ea
8.314×318.15
A×e−Ea
8.314×298.15
Step 5: Simplify the equation and solve for Ea. The activation energy Eais
approximately 36.7 kJ/mol.
8
Question 10
Question
The rate constant for a reaction at 25
°
C is 1.67 ×10−4s−1. When the tempera-
ture is increased to 35
°
C, the rate constant becomes 5.29 ×10−3s−1. Calculate
the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin. Given temperatures: T1=
25Cand T2= 35CConverting to Kelvin: T1= 273 + 25 = 298Kand T2=
273 + 35 = 308K
Step 2: Use the Arrhenius equation to relate the rate constants with tem-
perature. The Arrhenius equation is given by:
k=Ae−Ea
RT
Step 3: Take the natural logarithm of the Arrhenius equation for both sets
of data. For the first set of data:
ln(k1) = ln(A)−Ea
RT1
ln(k1) = ln(A)−Ea
R×298
For the second set of data:
ln(k2) = ln(A)−Ea
RT2
ln(k2) = ln(A)−Ea
R×308
Step 4: Subtract the two equations to eliminate ln(A). Subtracting the two
equations:
ln(k2)−ln(k1) = −Ea
R(1
308 −1
298)
Step 5: Solve for the activation energy, Ea.
ln(k1)
k1
−ln(k2)
k2
=Ea
R(1
298 −1
308)
Ea=−R×ln(k1)×k2−ln(k2)×k1
k1−k2
Substitute the given values to solve for activation energy.
9
Question 11
Question
The rate constant (k) of a certain reaction is found to be 2.5×10−2s−1at 300
K and 1.0×10−1s−1at 350 K. Calculate the activation energy (Ea) for this
reaction. The activation energy is in units of 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: Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
RT
Step 3: Rewrite the linearized equation as a linear equation (y = mx + b):
y=mx +b
where: - y= ln k, - m=−Ea
R, - x=1
T, - b= ln A.
Step 4: Use the data given at 300 K and 350 K to set up two equations:
ln2.5×10−2= ln A−Ea
R×300
ln1.0×10−1= ln A−Ea
R×350
Step 5: Solve the system of equations to find Ea.
Let’s calculate the activation energy:
From the given data: - At 300 K: k1= 2.5×10−2s−1- At 350 K: k2=
1.0×10−1s−1
Step 6: Calculate the activation energy We have the two equations:
ln k1= ln A−Ea
R×300
ln k2= ln A−Ea
R×350
Substitute the given values and simplify to find Ea.
10
Question 12
Question
The reaction rate constant for the decomposition of a particular compound at
25
°
C is 4.5×10−3s−1. When the temperature is raised to 50
°
C, the rate constant
increases to 2.8×10−2s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Calculate the ratio of rate constants using the Arrhenius equation:
k2
k1
= exp Ea
R1
T1
−1
T2
Where: k1= 4.5×10−3s−1(rate constant at 25
°
C), k2= 2.8×10−2s−1(rate
constant at 50
°
C), Eais the activation energy (unknown), R= 8.314 J/(mol
·
K)
(gas constant), T1= 25 + 273.15 K (initial temperature), T2= 50 + 273.15 K
(final temperature).
Step 2: Substitute the given values into the equation and solve for Ea:
2.8×10−2
4.5×10−3= exp Ea
8.314 1
298.15 −1
323.15
6.22 = exp Ea
8.314 (0.003352 −0.003097)
6.22 = exp Ea
8.314 ×0.000255
Step 3: Take the natural logarithm of both sides to solve for Ea:
ln(6.22) = ln exp Ea
8.314 ×0.000255
ln(6.22) = Ea
8.314 ×0.000255
Step 4: Rearrange the equation to solve for Ea:
Ea= 8.314 ×ln(6.22)
0.000255
Step 5: Calculate Eato find the activation energy for the reaction.
Question 13
Question
The rate constant for a reaction at 40
°
C is 5.0×10−3s−1, and at 100
°
C the rate
constant is 1.2 s−1. Calculate the activation energy for the reaction. (Note: The
11
activation energy can be calculated using 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)), and Tis the temperature in
Kelvin.)
Solution
Step 1: Convert temperatures to Kelvin. At 40
°
C: T1= 40 + 273 = 313 K
At 100
°
C: T2= 100 + 273 = 373 K
Step 2: Write the Arrhenius equation for both temperatures. At 40
°
C:
5.0×10−3=Ae−Ea
8.314×313
At 100
°
C: 1.2 = Ae−Ea
8.314×373
Step 3: Divide the two Arrhenius equations to eliminate A.1.2
5.0×10−3=
e−Ea
8.314×373
e−Ea
8.314×313
240 = eEa
8.314 (1
313 −1
373 )
Step 4: Solve for Ea. Taking the natural logarithm of both sides to remove
the exponential term:
ln(240) = Ea
8.314 (1
313 −1
373 )
Ea= 8.314 ×ln(240)
(1
313 −1
373 )
Ea≈8.314 ×ln(240)
0.0038
Ea≈104,000 J/mol
Therefore, the activation energy for the reaction is approximately 104,000
J/mol.
Question 14
Question
The rate constant for a certain reaction is 4.20 ×10−3s−1at 25◦C and 1.00 ×
10−2s−1at 35◦C. Calculate the activation energy for this reaction.
Solution
Step 1: First, we need to calculate the activation energy (Ea) using the Arrhe-
nius 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: Let’s start by converting the given temperatures to Kelvin:
T1= 25◦C = 25 + 273 = 298 K
T2= 35◦C = 35 + 273 = 308 K
12
Step 3: Next, we can set up two equations using the Arrhenius equation and
the given rate constants:
k1=A·e−Ea
R·T1= 4.20 ×10−3s−1
k2=A·e−Ea
R·T2= 1.00 ×10−2s−1
Step 4: Divide the two equations to eliminate the pre-exponential factor A:
k1
k2
=e−Ea
R·T1
e−Ea
R·T2
Step 5: Simplify the equation further by using the property of exponents:
k1
k2
=e
Ea
R1
T2
−1
T1
Step 6: Plug in the known values and solve for the activation energy Ea:
4.20 ×10−3
1.00 ×10−2=eEa
8.314 (1
308 −1
298 )
Step 7: Solve for Ea:
Ea=−8.314 ×ln 4.20 ×10−3
1.00 ×10−2 1
308 −1
298
Step 8: Calculate Eato find the activation energy for this reaction.
Question 15
Question
The rate constant of a certain reaction is found to be 6.32×10−3s−1at 25◦C and
1.25 ×10−2s−1at 35◦C. Calculate the activation energy (Ea) for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
Given: T1= 25◦C= 298 K and T2= 35◦C= 308 K.
Step 2: Use the Arrhenius equation to relate the rate constants and temper-
atures to the activation energy:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Plug in the given values:
ln 1.25 ×10−2s−1
6.32 ×10−3s−1=−Ea
8.314 1
308 −1
298
13
Step 4: Solve for the activation energy (Ea).
ln 1.25
0.632=−Ea
8.314 1
308 −1
298
Step 5: Simplify and solve for Ea.
ln (1.979) = −Ea
8.314 1
308 −1
298
ln(1.979) = −Ea
8.314 ×0.00336
Ea=−8.314 ×ln(1.979)
0.00336
Ea≈52.3 kJ/mol
Step 6: Therefore, the activation energy for this reaction is approximately
52.3 kJ/mol.
Question 16
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 25◦C and
9.31 ×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to solve for the activation energy
(Ea):
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol ·K), and T= temperature (in Kelvin).
Step 2: First, we need to convert the temperatures to Kelvin:
T1= 25◦C + 273.15 = 298.15 K
T2= 45◦C + 273.15 = 318.15 K
Step 3: We can set up two equations using the given rate constants and
temperatures:
4.23 ×10−3=Ae−Ea
8.314×298.15
9.31 ×10−3=Ae−Ea
8.314×318.15
Step 4: Divide the second equation by the first to eliminate A:
9.31 ×10−3
4.23 ×10−3=Ae−Ea
8.314×318.15
Ae−Ea
8.314×298.15
14
2.20 = eEa
8.314 (1
298.15 −1
318.15 )
Step 5: Solve for Ea:
ln(2.20) = Ea
8.314 1
298.15 −1
318.15
Ea= 8.314 ×ln(2.20)
1
298.15 −1
318.15
Ea≈45.67 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.67 kJ/mol.
Question 17
Question
The rate constant (k) for a reaction was found to be 1.5×10−3s−1at 25
°
C.
When the temperature was increased to 55
°
C, the rate constant increased to
8.5×10−2s−1. Calculate the activation energy of the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol
·
K)), T= temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin:
T1= 25 + 273 = 298K
T2= 55 + 273 = 328K
Step 3: Plug in the given values to find the pre-exponential factor A: When
T= 298K,k= 1.5×10−3s−1
1.5×10−3=A·e−Ea
8.314×298
Solve for A.
Step 4: Substitute the new values when T= 328Kand k= 8.5×10−2s−1
into the Arrhenius equation:
8.5×10−2=A·e−Ea
8.314×328
Step 5: Take the ratio of the two rate constant equations:
8.5×10−2
1.5×10−3=eEa
8.314 (1
328 −1
298 )
Solve for Eato find the activation energy of the reaction.
15
Question 18
Question
The rate constant, k, for the reaction
A→B
is found to be 1.20 ×10−3s−1at 25◦C and 4.70 ×10−3s−1at 45◦C. Calculate
the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at
different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1= 1.20 ×10−3s−1at 25◦C = 298 K k2= 4.70 ×10−3s−1at 45◦C =
318 K R= 8.314 J K−1mol−1
Step 2: Substitute the values into the equation:
ln 4.70 ×10−3
1.20 ×10−3=−Ea
8.314 1
318 −1
298
Step 3: Simplify the equation:
ln (3.92) = −Ea
8.314 1
318 −1
298
Step 4: Calculate the activation energy, Ea:
Ea=−8.314 ×ln (3.92) ×1
318 −1
298
Step 5: Therefore, the activation energy for this reaction is approximately
40.6 kJ/mol.
Question 19
Question
The rate constant for a certain reaction is found to be 3.12 ×10−4s−1at 25◦C
and 4.79 ×10−2s−1at 75◦C. Calculate the activation energy for this reaction.
16
Solution
Step 1: Write the Arrhenius equation:
The Arrhenius equation relates the rate constant of a reaction to the tem-
perature and the activation energy. It 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)), - Tis the
temperature in Kelvin.
Step 2: Determine the two rate constants:
At 25◦C, T= 298 K and k1= 3.12 ×10−4s−1.
At 75◦C, T= 348 K and k2= 4.79 ×10−2s−1.
Step 3: Take the natural logarithm of the Arrhenius equation:
ln(k) = ln(A)−Ea
RT
Step 4: Set up the equation for the two temperatures:
ln(k1) = ln(A)−Ea
R·298
ln(k2) = ln(A)−Ea
R·348
Step 5: Calculate the difference between the two equations:
ln(k2)−ln(k1) = −Ea
R1
348 −1
298
Step 6: Solve for the activation energy:
Ea=−R·ln(k2)−ln(k1)
1
348 −1
298
Ea=−8.314 ×103·ln4.79 ×10−2−ln3.12 ×10−4
1
348 −1
298
Calculating the value gives Ea≈77.8 kJ/mol.
Question 20
Question
The rate constant for a certain reaction is 2.5×10−3s−1at 25
°
C and 7.8×
10−3s−1at 45
°
C. Calculate the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=A e−Ea
RT )
17
Solution
Step 1: Convert temperatures to Kelvin Given: - T1= 25C-T2= 45CWe
convert these temperatures to Kelvin using the formula T(K) = T(C) + 273.15:
-T1= 25 + 273.15 = 298.15 K-T2= 45 + 273.15 = 318.15 K
Step 2: Determine the ratio of rate constants We use the ratio of rate con-
stants for the two temperatures:
k2
k1
=A e−Ea
RT2
A e−Ea
RT1
Step 3: Solve for activation energy, EaSubstitute the rate constants and
temperatures into the equation above:
7.8×10−3
2.5×10−3=e−Ea
8.314×318.15
e−Ea
8.314×298.15
Step 4: Simplify the equation Divide the left side of the equation to get:
3.12 = eEa
2981.23 −Ea
2481.85
Step 5: Take the natural logarithm of both sides
ln(3.12) = lneEa
2981.23 −Ea
2481.85
Step 6: Simplify the natural logarithm Using the properties of logarithms:
ln(3.12) = Ea
2981.23 −Ea
2481.85
Step 7: Solve for activation energy, EaMultiply both sides by 2981.23 ×
2481.85 to solve for Ea:
Ea=2981.23 ×2481.85
ln(3.12)
Calculate Eato find the activation energy for this reaction.
Question 21
Question
Given the following data for a reaction at two different temperatures, determine
the activation energy (Ea) for the reaction.
Temperature (K) Rate Constant (s−1)
300 1.5×10−3
350 4.0×10−2
18
Solution
Step 1: Write the Arrhenius Equation: The Arrhenius equation relates the rate
constant of a reaction (k) to the temperature (T) and the activation energy (Ea)
of the reaction. The equation is given as:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the temperature
in Kelvin.
Step 2: Calculate k/T for both temperature points: For the data provided:
- At 300 K:
k1/T1=1.5×10−3
300 = 5 ×10−6s−1·K−1
- At 350 K:
k2/T2=4.0×10−2
350 ≈1.14 ×10−4s−1·K−1
Step 3: Set up the ratio to determine the activation energy (Ea): Taking
the natural logarithm (ln) of the Arrhenius equation, we get:
ln(k) = ln(A)−Ea
RT
This equation can be rearranged to get the slope-intercept form:
ln(k) = −Ea
R·1
T+ln(A)
Comparing this equation with y=mx +c, where y=ln(k), m=−Ea
R,x=1
T,
and c=ln(A), we see that the slope of the line (m) is −Ea
R.
Thus, we can set up the following equation using the two data points:
slope = −Ea
R=ln(k2)−ln(k1)
(1/T2)−(1/T1)
Step 4: Calculate the activation energy (Ea): Substitute the calculated
values: ln(1.14 ×10−4)−ln(5 ×10−6)
(1/350) −(1/300) =−Ea
8.314
⇒Ea=−8.314 ×ln(1.14 ×10−4)−ln(5 ×10−6)
(1/350) −(1/300)
⇒Ea≈74,200 J/mol
Therefore, the activation energy for the reaction is approximately 74,200
J/mol.
19
Question 22
Question
The rate constant of a certain reaction is 4.32×10−3s−1at 350 K and 1.23×10−2
s−1at 400 K. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A×e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J/mol·K), and Tis the temperature in
Kelvin.
Step 2: Rearrange the equation to solve for the activation energy Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation.
Given: k1= 4.32 ×10−3s−1(at 350 K)
k2= 1.23 ×10−2s−1(at 400 K)
R= 8.314 J/mol·K
T1= 350 K
T2= 400 K
Step 4: Calculate the activation energy Ea:
ln 1.23 ×10−2
4.32 ×10−3=−Ea
8.314 1
400 −1
350
ln(2.85) = −Ea
8.314 1
400 −1
350
Step 5: Solve for the activation energy Ea:
1
8.314 1
400 −1
350ln(2.85) = Ea
Ea≈65.68 kJ/mol
Therefore, the activation energy for this reaction is approximately 65.68
kJ/mol.
20
Question 23
Question
The rate constant for a certain reaction is observed to double when the temper-
ature is increased from 25
°
C to 45
°
C. If the activation energy for the reaction
is 50 kJ/mol, calculate the rate constant at 25
°
C and the rate constant at 80
°
C
for this reaction.
Solution
Step 1: Calculate the rate constant at 25
°
C using the Arrhenius equation. The
Arrhenius equation is given by:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the temperature
in Kelvin.
Given that the rate constant doubles when the temperature increases from
25
°
C to 45
°
C, we can use this information to find the pre-exponential factor, A.
Step 2: Calculate the rate constant at 25
°
C. Let’s denote the rate constant
at 25
°
C as k25, and the rate constant at 45
°
C as k45. Since the rate constant
doubles when the temperature increases from 25
°
C to 45
°
C, we have:
k45
k25
= 2
Next, we will use the relationship between rate constant and temperature to
find the pre-exponential factor, A.
Step 3: Solve for the rate constant at 80
°
C using the Arrhenius equation.
Now that we have the pre-exponential factor, A, and the activation energy, Ea,
we can calculate the rate constant at 80
°
C using the Arrhenius equation.
Let’s denote the rate constant at 80
°
C as k80. We need to convert 80
°
C to
Kelvin by adding 273.15:
T= 80C+ 273.15 = 353.15K
Now, we can use the Arrhenius equation to find the rate constant at 80
°
C:
k80 =Ae−Ea
RT
Calculating k80 will give us the rate constant at 80
°
C for this reaction.
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is found to be 1.5×10−3M/s.
When the temperature is increased to 50
°
C, the rate constant is measured to
be 3.0×10−2M/s. Calculate the activation energy (Ea) for this reaction.
21
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at
different temperatures: k2
k1
=e−Ea
R1
T2
−1
T1
where: - k1and k2are the rate constants at temperatures T1and T2, respectively,
-Eais the activation energy, - R= 8.314 J/(mol·K) is the gas constant, - T1
and T2are the temperatures in Kelvin.
Step 2: Substituting the known values into the equation, we have:
3.0×10−2
1.5×10−3=e(−Ea
8.314 (1
323 −1
298 ))
Step 3: Simplifying the equation gives:
20 = e(−Ea
8.314 (1
323 −1
298 ))
Step 4: Taking the natural logarithm of both sides to solve for Ea, we get:
ln(20) = −Ea
8.314 1
323 −1
298
Step 5: Multiplying through by 8.314 and rearranging the equation yields:
Ea=−8.314 ×1
323 −298 ln(20)
Step 6: Calculating the value of Ea, we find:
Ea≈61.31 kJ/mol
Thus, the activation energy for this reaction is approximately 61.31 kJ/mol.
Question 25
Question
The rate constant for a reaction at 25
°
C is 2.0×10−2s−1. When the temperature
is increased to 35
°
C, the rate constant becomes 7.0×10−2s−1. Calculate the
activation energy for this reaction.
Solution
Step 1: Recall the Arrhenius equation:
k=Ae−Ea
RT
22
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol K)), T= temperature (in Kelvin).
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
ln k2
k1
1
T2
−1
T1
=−Ea
R
Step 3: Plug in the given values: k1= 2.0×10−2s−1,k2= 7.0×10−2s−1,
T1= 25 + 273 = 298 K, T2= 35 + 273 = 308 K.
Step 4: Calculate the activation energy:
Ea=−R×
ln 7.0×10−2
2.0×10−2
1
308 −1
298
Step 5: Simplify the equation and solve for Ea:
Ea=−8.314 ×ln 3.5
1
308 −1
298
Ea=−8.314 ×ln 3.5
1
308 −1
298
Step 6: Calculate the activation energy Eato find its value.
Question 26
Question
The rate constant for a certain reaction quadruples when the temperature is
increased from 25
°
C to 50
°
C. Calculate the activation energy for this reaction.
Solution
Step 1: Start by writing down the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol K), T= temperature in Kelvin.
Step 2: Given that the rate constant quadruples when the temperature is
increased from 25
°
C to 50
°
C, we can write this relationship as:
4k1=k2
where: k1is the rate constant at 25
°
C (298K), and k2is the rate constant at
50
°
C (323K).
23
Step 3: We can rewrite the Arrhenius equation for both temperatures as:
k1=Ae−Ea
R·298
k2=Ae−Ea
R·323
Step 4: Substitute the expressions for k1and k2from Step 3 into the rela-
tionship from Step 2:
4Ae−Ea
R·298 =Ae−Ea
R·323
Step 5: Simplify the equation by dividing both sides by Aand then taking
the natural logarithm:
4eEa
R·323 −Ea
R·298 =e0
Step 6: Simplify the exponents and solve for Eato find the activation energy.
Remember to convert temperatures to Kelvin:
4eEa
8.314·323 −Ea
8.314·298 = 1
e25Ea
8.314·323·298 = 4
25Ea
8.314 ·323 ·298 = ln(4)
Ea=8.314 ·323 ·298 ·ln(4)
25
Ea≈59.77 kJ/mol
Therefore, the activation energy for this reaction is approximately 59.77
kJ/mol.
Question 27
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1and the
activation energy is 70 kJ/mol. Calculate the rate constant at 40
°
C for this
reaction.
24
Solution
Step 1: Convert the activation energy to the correct units. Given activation
energy = 70 kJ/mol, convert it to Joules per mole. 1 kJ = 1000 J, so 70 kJ =
70 ×1000 J = 70000 J. Now, 1 J/mol = 1 mol/J, so the activation energy in
J/mol is 70000 J/mol.
Step 2: Use the Arrhenius equation to find the rate constant at 40
°
C. The
Arrhenius equation is given by:
k=A·exp −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)
At 25
°
C = 298 K, we have:
k1= 2.5×10−3s−1
Ea= 70000 J/mol
Now, at 40
°
C = 313 K, we need to find k2:
k2=A·exp −70000
8.314 ×313
Step 3: Calculate the rate constant at 40
°
C.
k2=A·exp −70000
2608.282
Since we don’t know the pre-exponential factor A, we cannot directly calcu-
late k2without additional information.
Question 28
Question
The rate constant of a reaction is found to be 2.70 ×10−3s−1at 50
°
C and
8.60 ×10−3s−1at 100
°
C. Calculate the activation energy (in kJ/mol) 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.
25
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Plug in the given values: k1= 2.70 ×10−3s−1(at 50
°
C = 323 K),
k2= 8.60 ×10−3s−1(at 100
°
C = 373 K), R= 8.314 J/(mol·K).
Step 4: Calculate the activation energy Ea:
ln 8.60 ×10−3
2.70 ×10−3=−Ea
8.314 1
373 −1
323
ln (3.1852) = −Ea
8.314 1
373 −1
323
1.1563 = −Ea
8.314 1
373 −1
323
Step 5: Solve for activation energy Ea:
Ea=−8.314 ×1.1563 1
373 −1
323
Ea≈64.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 64.7
kJ/mol.
Question 29
Question
The rate constant of a certain reaction is found to be 3.2×10−3s−1at 300 K
and 1.2×10−2s−1at 350 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):
k=A·e−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: Use the given data to set up two equations:
From the given data, we have:
k1= 3.2×10−3s−1at 300 K
26
k2= 1.2×10−2s−1at 350 K
Substitute these values into the Arrhenius equation to get two equations:
3.2×10−3=A·e−Ea
8.314·300
1.2×10−2=A·e−Ea
8.314·350
Step 3: Take the ratio of the two equations:
Divide the second equation by the first to eliminate A:
1.2×10−2
3.2×10−3=eEa
8.314 (1
300 −1
350 )
Step 4: Solve for Ea:
Calculate the left side of the equation:
1.2×10−2
3.2×10−3= 3.75
Next, solve for Ea:
3.75 = eEa
8.314 (1
300 −1
350 )
ln(3.75) = Ea
8.314 1
300 −1
350
Ea= 8.314 ×ln(3.75)
1
300 −1
350
Ea≈47.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 47.2
kJ/mol.
Question 30
Question
The rate constant for the reaction 2NOBr(g)→2NO(g)+Br2(g) is found to be
3.28×10−2s−1at 1000 K and 1.08×10−3s−1at 900 K. Calculate the activation
energy for this reaction. (Hint: Use the Arrhenius equation, k=Aexp −Ea
RT )
27
Solution
Step 1: Write the Arrhenius equation in logarithmic form:
ln(k) = ln(A)−Ea
RT
Step 2: Write two equations using the given data:
ln(k1) = ln(A)−Ea
R×1000
ln(k2) = ln(A)−Ea
R×900
Step 3: Subtract the second equation from the first equation:
ln(k1)−ln(k2) = Ea
R1
900 −1
1000
Step 4: Simplify the equation:
ln k1
k2=Ea
R×1
9000
Step 5: Solve for the activation energy Ea:
Ea=R×9000 ×ln k1
k2
Step 6: Plug in the given values for k1and k2, and the Universal Gas
Constant R= 8.314 J/mol ·K:
Ea= 8.314 ×9000 ×ln 3.28 ×10−2
1.08 ×10−3
Step 7: Calculate the activation energy:
Ea= 8.314 ×9000 ×ln 30.37
1.08
Ea= 8.314 ×9000 ×ln(28.10)
Ea≈8.314 ×9000 ×3.337
Ea≈249,173 J/mol
Question 31
Question
The rate constant for a certain reaction is found to be 3.81 ×10−2s−1at 280◦C
and 2.46 s−1at 300◦C. Determine the activation energy for this reaction. The
activation energy constant Ais 1.00 ×109s−1.
28
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: Rearrange the Arrhenius equation for two sets of conditions to solve
for the activation energy Ea:
k1
k2
=Ae−Ea
R·T1
Ae−Ea
R·T2
Step 3: Plug in the given values:
3.81 ×10−2
2.46 =1.00 ×109e−Ea
8.314·280
1.00 ×109e−Ea
8.314·300
Step 4: Simplify the equation:
3.81 ×10−2
2.46 =e−Ea
8.314·280 +Ea
8.314·300
Step 5: Further simplify and solve for Ea:
ln 3.81 ×10−2
2.46 =−Ea
8.314 1
280 −1
300
Ea=−8.314 ×300 ×280
300 −280 ×ln 3.81 ×10−2
2.46
Step 6: Calculate the activation energy:
Ea≈1.62 ×105J/mol
Question 32
Question
The rate constant of a reaction at 25
°
C is 2.5×10−3s−1and the activation
energy is 50 kJ mol−1. Calculate the rate constant at 35
°
C using the Arrhenius
equation.
29
Solution
Step 1: Convert the activation energy from kJ to J: Given activation energy
Ea= 50 kJ mol−1, we convert this to joules:
Ea= 50 ×103J mol−1= 5.0×104J mol−1
Step 2: Determine the new rate constant at 35
°
C using the Arrhenius equa-
tion: 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.
Given that the rate constant at 25
°
C is 2.5×10−3s−1and the activation
energy is 5.0×104J mol−1, we can calculate the pre-exponential factor Ausing
the rate constant at 25
°
C:
2.5×10−3s−1=A×e−5.0×104
8.314×(25+273.15)
2.5×10−3s−1=A×e−5.0×104
8.314×298.15
2.5×10−3s−1=A×e−5.0×104
2478.8091
A= 2.5×10−3×e5.0×104
2478.8091
A≈2.723 ×10−2s−1
Now, we use the Arrhenius equation to find the rate constant at 35
°
C:
k= 2.723 ×10−2×e−5.0×104
8.314×(35+273.15)
k= 2.723 ×10−2×e−5.0×104
8.314×308.15
k= 2.723 ×10−2×e−5.0×104
2550.7291
k≈7.368 ×10−3s−1
Therefore, the rate constant at 35
°
C is approximately 7.368 ×10−3s−1.
Question 33
Question
The rate constant of a first-order reaction at 25
°
C is 4.0×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.6×10−2s−1.
Calculate the activation energy of the reaction. (Hint: Use the Arrhenius equa-
tion 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.)
30
Solution
Step 1: Convert the temperatures to Kelvin: 25C+273 = 298 K 50C+273 = 323
K
Step 2: Write the Arrhenius equation for the two sets of conditions: For the
first condition (25
°
C): k1=A·e−Ea
RT1
For the second condition (50
°
C): k2=A·e−Ea
RT2
Step 3: Divide the two Arrhenius equations to eliminate the pre-exponential
factor A:k2
k1=e−Ea
RT2
e−Ea
RT1
Step 4: Simplify the equation: k2
k1=e
Ea
R1
T1
−1
T2
Step 5: Plug in the values: 1.6×10−2
4.0×10−3=eEa
8.314 (1
298 −1
323 )
Step 6: Solve for the activation energy Ea:1.6×10−2
4.0×10−3=eEa
8.314 (1
298 −1
323 )
4 = eEa
8.314 (1
298 −1
323 )
Step 7: Take the natural logarithm (ln) of both sides and solve for Ea:
ln(4) = Ea
8.314 1
298 −1
323
Ea= 8.314 ×ln(4)∇ · 1
323 −1
298
Step 8: Calculate the value of activation energy Eawith the known values.
Question 34
Question
For a certain reaction, the rate constant at 25
°
C is 4.0×10−3s−1, and at 35
°
C it
is 2.0×10−2s−1. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/(mol
·
K)), and - Tis the
temperature in Kelvin.
Step 2: Set up the Arrhenius equation for the two given temperatures: At
25
°
C (298 K):
4.0×10−3=A·e−Ea
8.314·298
At 35
°
C (308 K):
2.0×10−2=A·e−Ea
8.314·308
Step 3: Divide the equations to eliminate the pre-exponential factor:
4.0×10−3
2.0×10−2=e−Ea
8.314·298
e−Ea
8.314·308
31
Step 4: Simplify the equation:
0.2 = eEa
8.314 (1
308 −1
298 )
Step 5: Solve for the activation energy Ea:
Ea=−8.314 ·ln(0.2) ·1
308 −1
298
Step 6: Calculate the activation energy in kJ/mol:
Ea=−8.314 ·ln(0.2) ·1
308 −1
298= 76.15 kJ/mol
Question 35
Question
The rate constant for a certain reaction is found to be 4.63 ×10−3s−1at 25◦C.
When the temperature is increased to 55◦C, the rate constant becomes 6.82 ×
10−2s−1. Calculate the activation energy of this reaction.
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(
°
C) +
273.15.
T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 55◦C = 55 + 273.15 = 328.15 K
Step 2: Apply the Arrhenius equation to relate the rate constants to the
temperatures.
ln k2
k1=−Ea
R1
T2
−1
T1
where: - k1= 4.63 ×10−3s−1-k2= 6.82 ×10−2s−1-R= 8.314 J ·mol−1·K−1
Step 3: Substitute the given values into the equation above.
ln 6.82 ×10−2
4.63 ×10−3=−Ea
8.314 1
328.15 −1
298.15
Step 4: Solve for the activation energy Ea.
ln 6.82 ×10−2
4.63 ×10−3=−Ea
8.314 1
328.15 −1
298.15
ln 6.82 ×10−2
4.63 ×10−3=−Ea
8.314 0.00305 −0.00335
(328.15)(298.15)
32
Question 3
Question
The rate constant for a certain reaction is 1.5×10−2s−1at 298 K and 4.5×
10−2s−1at 333 K. Calculate the activation energy for the reaction.
Solution
Step 1: Recall the Arrhenius equation:
k=A·e(−Ea
RT )
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J mol−1K−1), - Tis the tem-
perature in Kelvin.
Step 2: Use the data given to set up two equations:
1.5×10−2=A·e(−Ea
8.314·298 )
4.5×10−2=A·e(−Ea
8.314·333 )
Step 3: Divide the second equation by the first to eliminate A:
4.5×10−2
1.5×10−2=e(−Ea
8.314·333 )
e(−Ea
8.314·298 )
Step 4: Simplify the expression:
3 = e(Ea
8.314 (1
298 −1
333 ))
Step 5: Solve for Ea:
Ea=−8.314 ·1
1
298 −1
333 ·ln(3)
Step 6: Calculate the value of Eato find the activation energy for the reac-
tion.
Question 4
Question
The rate constant of a reaction at 25
°
C is 1.2×10−3s−1and the activation
energy is 80 kJ/mol. Calculate the rate constant of the reaction at 35
°
C. (Given:
R= 8.314 J/mol·K)
3
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 = 1.2×10−3s−1Ea= activation energy = 80
kJ/mol = 80 ×103J/mol R= gas constant = 8.314 J/mol·KT1= temperature
in Kelvin = 25
°
C + 273 = 298 K
Plugging in the values, we get:
1.2×10−3=A·e−80×103
8.314·298
Step 2: Solve for the pre-exponential factor A:
A=k1
e−80×103
8.314·298
Step 3: Calculate the rate constant at 35
°
C using the Arrhenius equation:
k2=A·e−Ea
RT2
where: k2= rate constant at 35
°
CEa= activation energy = 80 kJ/mol =
80 ×103J/mol R= gas constant = 8.314 J/mol·KT2= temperature in Kelvin
= 35
°
C + 273 = 308 K
Plugging in the values and the calculated value of A, we get:
k2=A·e−80×103
8.314·308
Question 5
Question
The rate constant for a certain reaction is 2.50 ×10−2s−1at 350 K, and 1.00 ×
10−1s−1at 400 K. Calculate the activation energy for this reaction in kJ/mol.
Solution
Step 1: Write the Arrhenius equation:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol*K)), - Tis the temperature
in Kelvin.
Step 2: Use the given data to form two equations using the Arrhenius equa-
tion and the two temperatures. At 350 K:
2.50 ×10−2=Ae−Ea
8.314×350
4
At 400 K:
1.00 ×10−1=Ae−Ea
8.314×400
Step 3: Divide the two equations to eliminate A:
2.50 ×10−2
1.00 ×10−1=e−Ea
8.314×350
e−Ea
8.314×400
0.25 = eEa
8.314 (1
350 −1
400 )
Step 4: Solve for the activation energy:
0.25 = eEa
8.314 (1
350 −1
400 )
ln(0.25) = Ea
8.314(1
350 −1
400)
Ea= 8.314 ×(1
350 −1
400)×ln(0.25)
Step 5: Calculate the activation energy in kJ/mol:
Ea= 8.314 ×(1
350 −1
400)×ln(0.25) ≈72.0 kJ/mol
Therefore, the activation energy for this reaction is approximately 72.0
kJ/mol.
Question 6
Question
The rate constant for a certain reaction is found to be 1.25 ×10−2s−1at 25
º
C
and 8.71 ×10−2s−1at 45
º
C. Calculate the activation energy for this reaction.
The activation energy is expressed in units of kJ/mol.
Given: R= 8.314 J/(mol K)
Solution
Step 1: Convert the given temperatures into Kelvin:
T1= 25C+ 273.15 = 298.15 K
T2= 45C+ 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to find the activation energy (Ea):
ln k2
k1=−Ea
R1
T2
−1
T1
5
Plugging in the values:
ln 8.71 ×10−2
1.25 ×10−2=−Ea
8.314 1
318.15 −1
298.15
ln(6.968) = −Ea
8.314 1
318.15 −1
298.15
1.938 = −Ea
8.314 (0.0031 −0.0034)
1.938 = −Ea
8.314(−0.0003)
1.938 = 0.0025Ea
Ea=1.938
0.0025
Ea≈775.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 775.2
kJ/mol.
Question 7
Question
The rate constant for the decomposition of a compound is found to be 9.56 ×
10−4s−1at 30◦C. If the activation energy for this reaction is 78.2 kJ/mol,
calculate the rate constant at 40◦C.
Solution
Step 1: We begin by noting the general form of 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), and - Tis the tem-
perature in Kelvin.
Step 2: First, we must convert the activation energy from kilojoules per mole
to joules per molecule:
Ea= 78.2×103J/mol
Step 3: Next, we calculate the rate constant kat 30◦C (303 K) using the
given values:
k1= 9.56 ×10−4s−1
Step 4: We rearrange the Arrhenius equation to solve for A:
A=k
e−Ea
RT
6
Step 5: Substitute the known values into the equation to find A:
A=9.56 ×10−4
e−78.2×103
8.314×303
Step 6: After calculating A, we use this value along with the activation
energy to find the rate constant k2at 40◦C (313 K) using the Arrhenius equation:
k2=A·e−Ea
RT
Step 7: Substitute the calculated values into the equation to find k2:
k2=A·e−78.2×103
8.314×313
Step 8: Compute the final value of k2to determine the rate constant at
40◦C.
Question 8
Question
The rate constant for a certain reaction is found to be 4.92 ×10−3s−1at 25◦C
and 8.64 ×10−3s−1at 35◦C. Calculate the activation energy (Ea) for this
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−1K−1), - Tis the tem-
perature in Kelvin.
Step 2: Convert temperatures to Kelvin: - 25◦C = 25 + 273 = 298 K -
35◦C = 35 + 273 = 308 K
Step 3: Substitute the given data into the Arrhenius equation for the two
temperatures:
k1=Ae−Ea
R×298
k2=Ae−Ea
R×308
Step 4: Divide the equations to eliminate A:
k2
k1
=e−Ea
R×308
e−Ea
R×298
7
Step 5: Simplify the equation:
k2
k1
=eEa
R(1
298 −1
308 )
Step 6: Substitute the values of k1,k2,R, and solve for Ea:
8.64 ×10−3
4.92 ×10−3=eEa
8.314 (1
298 −1
308 )
Step 7: Calculate Eausing the natural logarithmic function:
ln 8.64 ×10−3
4.92 ×10−3=Ea
8.314 1
298 −1
308
Step 8: Solve for Ea:
Ea= 8.314 ×1
298 −1
308×ln 8.64 ×10−3
4.92 ×10−3
After calculations, Ea≈41.8 kJ/mol.
Question 9
Question
The rate constant of a certain reaction is found to be 4.23 ×10−3s−1at 25◦C
and 1.44 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(◦C) + 273.15. At 25◦C: T1= 25 + 273.15 = 298.15 K
At 45◦C: T2= 45 + 273.15 = 318.15 K
Step 2: Use 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:
(4.23 ×10−3=A×e−Ea
8.314×298.15
1.44 ×10−2=A×e−Ea
8.314×318.15
Step 4: Divide the second equation by the first to eliminate A:
1.44 ×10−2
4.23 ×10−3=A×e−Ea
8.314×318.15
A×e−Ea
8.314×298.15
Step 5: Simplify the equation and solve for Ea. The activation energy Eais
approximately 36.7 kJ/mol.
8
Question 10
Question
The rate constant for a reaction at 25
°
C is 1.67 ×10−4s−1. When the tempera-
ture is increased to 35
°
C, the rate constant becomes 5.29 ×10−3s−1. Calculate
the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin. Given temperatures: T1=
25Cand T2= 35CConverting to Kelvin: T1= 273 + 25 = 298Kand T2=
273 + 35 = 308K
Step 2: Use the Arrhenius equation to relate the rate constants with tem-
perature. The Arrhenius equation is given by:
k=Ae−Ea
RT
Step 3: Take the natural logarithm of the Arrhenius equation for both sets
of data. For the first set of data:
ln(k1) = ln(A)−Ea
RT1
ln(k1) = ln(A)−Ea
R×298
For the second set of data:
ln(k2) = ln(A)−Ea
RT2
ln(k2) = ln(A)−Ea
R×308
Step 4: Subtract the two equations to eliminate ln(A). Subtracting the two
equations:
ln(k2)−ln(k1) = −Ea
R(1
308 −1
298)
Step 5: Solve for the activation energy, Ea.
ln(k1)
k1
−ln(k2)
k2
=Ea
R(1
298 −1
308)
Ea=−R×ln(k1)×k2−ln(k2)×k1
k1−k2
Substitute the given values to solve for activation energy.
9
Question 11
Question
The rate constant (k) of a certain reaction is found to be 2.5×10−2s−1at 300
K and 1.0×10−1s−1at 350 K. Calculate the activation energy (Ea) for this
reaction. The activation energy is in units of 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: Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
RT
Step 3: Rewrite the linearized equation as a linear equation (y = mx + b):
y=mx +b
where: - y= ln k, - m=−Ea
R, - x=1
T, - b= ln A.
Step 4: Use the data given at 300 K and 350 K to set up two equations:
ln2.5×10−2= ln A−Ea
R×300
ln1.0×10−1= ln A−Ea
R×350
Step 5: Solve the system of equations to find Ea.
Let’s calculate the activation energy:
From the given data: - At 300 K: k1= 2.5×10−2s−1- At 350 K: k2=
1.0×10−1s−1
Step 6: Calculate the activation energy We have the two equations:
ln k1= ln A−Ea
R×300
ln k2= ln A−Ea
R×350
Substitute the given values and simplify to find Ea.
10
Question 12
Question
The reaction rate constant for the decomposition of a particular compound at
25
°
C is 4.5×10−3s−1. When the temperature is raised to 50
°
C, the rate constant
increases to 2.8×10−2s−1. Calculate the activation energy for the reaction.
Solution
Step 1: Calculate the ratio of rate constants using the Arrhenius equation:
k2
k1
= exp Ea
R1
T1
−1
T2
Where: k1= 4.5×10−3s−1(rate constant at 25
°
C), k2= 2.8×10−2s−1(rate
constant at 50
°
C), Eais the activation energy (unknown), R= 8.314 J/(mol
·
K)
(gas constant), T1= 25 + 273.15 K (initial temperature), T2= 50 + 273.15 K
(final temperature).
Step 2: Substitute the given values into the equation and solve for Ea:
2.8×10−2
4.5×10−3= exp Ea
8.314 1
298.15 −1
323.15
6.22 = exp Ea
8.314 (0.003352 −0.003097)
6.22 = exp Ea
8.314 ×0.000255
Step 3: Take the natural logarithm of both sides to solve for Ea:
ln(6.22) = ln exp Ea
8.314 ×0.000255
ln(6.22) = Ea
8.314 ×0.000255
Step 4: Rearrange the equation to solve for Ea:
Ea= 8.314 ×ln(6.22)
0.000255
Step 5: Calculate Eato find the activation energy for the reaction.
Question 13
Question
The rate constant for a reaction at 40
°
C is 5.0×10−3s−1, and at 100
°
C the rate
constant is 1.2 s−1. Calculate the activation energy for the reaction. (Note: The
11
activation energy can be calculated using 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)), and Tis the temperature in
Kelvin.)
Solution
Step 1: Convert temperatures to Kelvin. At 40
°
C: T1= 40 + 273 = 313 K
At 100
°
C: T2= 100 + 273 = 373 K
Step 2: Write the Arrhenius equation for both temperatures. At 40
°
C:
5.0×10−3=Ae−Ea
8.314×313
At 100
°
C: 1.2 = Ae−Ea
8.314×373
Step 3: Divide the two Arrhenius equations to eliminate A.1.2
5.0×10−3=
e−Ea
8.314×373
e−Ea
8.314×313
240 = eEa
8.314 (1
313 −1
373 )
Step 4: Solve for Ea. Taking the natural logarithm of both sides to remove
the exponential term:
ln(240) = Ea
8.314 (1
313 −1
373 )
Ea= 8.314 ×ln(240)
(1
313 −1
373 )
Ea≈8.314 ×ln(240)
0.0038
Ea≈104,000 J/mol
Therefore, the activation energy for the reaction is approximately 104,000
J/mol.
Question 14
Question
The rate constant for a certain reaction is 4.20 ×10−3s−1at 25◦C and 1.00 ×
10−2s−1at 35◦C. Calculate the activation energy for this reaction.
Solution
Step 1: First, we need to calculate the activation energy (Ea) using the Arrhe-
nius 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: Let’s start by converting the given temperatures to Kelvin:
T1= 25◦C = 25 + 273 = 298 K
T2= 35◦C = 35 + 273 = 308 K
12
Step 3: Next, we can set up two equations using the Arrhenius equation and
the given rate constants:
k1=A·e−Ea
R·T1= 4.20 ×10−3s−1
k2=A·e−Ea
R·T2= 1.00 ×10−2s−1
Step 4: Divide the two equations to eliminate the pre-exponential factor A:
k1
k2
=e−Ea
R·T1
e−Ea
R·T2
Step 5: Simplify the equation further by using the property of exponents:
k1
k2
=e
Ea
R1
T2
−1
T1
Step 6: Plug in the known values and solve for the activation energy Ea:
4.20 ×10−3
1.00 ×10−2=eEa
8.314 (1
308 −1
298 )
Step 7: Solve for Ea:
Ea=−8.314 ×ln 4.20 ×10−3
1.00 ×10−2 1
308 −1
298
Step 8: Calculate Eato find the activation energy for this reaction.
Question 15
Question
The rate constant of a certain reaction is found to be 6.32×10−3s−1at 25◦C and
1.25 ×10−2s−1at 35◦C. Calculate the activation energy (Ea) for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
Given: T1= 25◦C= 298 K and T2= 35◦C= 308 K.
Step 2: Use the Arrhenius equation to relate the rate constants and temper-
atures to the activation energy:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Plug in the given values:
ln 1.25 ×10−2s−1
6.32 ×10−3s−1=−Ea
8.314 1
308 −1
298
13
Step 4: Solve for the activation energy (Ea).
ln 1.25
0.632=−Ea
8.314 1
308 −1
298
Step 5: Simplify and solve for Ea.
ln (1.979) = −Ea
8.314 1
308 −1
298
ln(1.979) = −Ea
8.314 ×0.00336
Ea=−8.314 ×ln(1.979)
0.00336
Ea≈52.3 kJ/mol
Step 6: Therefore, the activation energy for this reaction is approximately
52.3 kJ/mol.
Question 16
Question
The rate constant for a reaction is found to be 4.23 ×10−3s−1at 25◦C and
9.31 ×10−3s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to solve for the activation energy
(Ea):
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol ·K), and T= temperature (in Kelvin).
Step 2: First, we need to convert the temperatures to Kelvin:
T1= 25◦C + 273.15 = 298.15 K
T2= 45◦C + 273.15 = 318.15 K
Step 3: We can set up two equations using the given rate constants and
temperatures:
4.23 ×10−3=Ae−Ea
8.314×298.15
9.31 ×10−3=Ae−Ea
8.314×318.15
Step 4: Divide the second equation by the first to eliminate A:
9.31 ×10−3
4.23 ×10−3=Ae−Ea
8.314×318.15
Ae−Ea
8.314×298.15
14
2.20 = eEa
8.314 (1
298.15 −1
318.15 )
Step 5: Solve for Ea:
ln(2.20) = Ea
8.314 1
298.15 −1
318.15
Ea= 8.314 ×ln(2.20)
1
298.15 −1
318.15
Ea≈45.67 kJ/mol
Therefore, the activation energy for this reaction is approximately 45.67 kJ/mol.
Question 17
Question
The rate constant (k) for a reaction was found to be 1.5×10−3s−1at 25
°
C.
When the temperature was increased to 55
°
C, the rate constant increased to
8.5×10−2s−1. Calculate the activation energy of the reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol
·
K)), T= temperature in Kelvin.
Step 2: Convert the temperatures to Kelvin:
T1= 25 + 273 = 298K
T2= 55 + 273 = 328K
Step 3: Plug in the given values to find the pre-exponential factor A: When
T= 298K,k= 1.5×10−3s−1
1.5×10−3=A·e−Ea
8.314×298
Solve for A.
Step 4: Substitute the new values when T= 328Kand k= 8.5×10−2s−1
into the Arrhenius equation:
8.5×10−2=A·e−Ea
8.314×328
Step 5: Take the ratio of the two rate constant equations:
8.5×10−2
1.5×10−3=eEa
8.314 (1
328 −1
298 )
Solve for Eato find the activation energy of the reaction.
15
Question 18
Question
The rate constant, k, for the reaction
A→B
is found to be 1.20 ×10−3s−1at 25◦C and 4.70 ×10−3s−1at 45◦C. Calculate
the activation energy for this reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at
different temperatures:
ln k2
k1=−Ea
R1
T2
−1
T1
where: k1= 1.20 ×10−3s−1at 25◦C = 298 K k2= 4.70 ×10−3s−1at 45◦C =
318 K R= 8.314 J K−1mol−1
Step 2: Substitute the values into the equation:
ln 4.70 ×10−3
1.20 ×10−3=−Ea
8.314 1
318 −1
298
Step 3: Simplify the equation:
ln (3.92) = −Ea
8.314 1
318 −1
298
Step 4: Calculate the activation energy, Ea:
Ea=−8.314 ×ln (3.92) ×1
318 −1
298
Step 5: Therefore, the activation energy for this reaction is approximately
40.6 kJ/mol.
Question 19
Question
The rate constant for a certain reaction is found to be 3.12 ×10−4s−1at 25◦C
and 4.79 ×10−2s−1at 75◦C. Calculate the activation energy for this reaction.
16
Solution
Step 1: Write the Arrhenius equation:
The Arrhenius equation relates the rate constant of a reaction to the tem-
perature and the activation energy. It 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)), - Tis the
temperature in Kelvin.
Step 2: Determine the two rate constants:
At 25◦C, T= 298 K and k1= 3.12 ×10−4s−1.
At 75◦C, T= 348 K and k2= 4.79 ×10−2s−1.
Step 3: Take the natural logarithm of the Arrhenius equation:
ln(k) = ln(A)−Ea
RT
Step 4: Set up the equation for the two temperatures:
ln(k1) = ln(A)−Ea
R·298
ln(k2) = ln(A)−Ea
R·348
Step 5: Calculate the difference between the two equations:
ln(k2)−ln(k1) = −Ea
R1
348 −1
298
Step 6: Solve for the activation energy:
Ea=−R·ln(k2)−ln(k1)
1
348 −1
298
Ea=−8.314 ×103·ln4.79 ×10−2−ln3.12 ×10−4
1
348 −1
298
Calculating the value gives Ea≈77.8 kJ/mol.
Question 20
Question
The rate constant for a certain reaction is 2.5×10−3s−1at 25
°
C and 7.8×
10−3s−1at 45
°
C. Calculate the activation energy for this reaction. (Hint: Use
the Arrhenius equation: k=A e−Ea
RT )
17
Solution
Step 1: Convert temperatures to Kelvin Given: - T1= 25C-T2= 45CWe
convert these temperatures to Kelvin using the formula T(K) = T(C) + 273.15:
-T1= 25 + 273.15 = 298.15 K-T2= 45 + 273.15 = 318.15 K
Step 2: Determine the ratio of rate constants We use the ratio of rate con-
stants for the two temperatures:
k2
k1
=A e−Ea
RT2
A e−Ea
RT1
Step 3: Solve for activation energy, EaSubstitute the rate constants and
temperatures into the equation above:
7.8×10−3
2.5×10−3=e−Ea
8.314×318.15
e−Ea
8.314×298.15
Step 4: Simplify the equation Divide the left side of the equation to get:
3.12 = eEa
2981.23 −Ea
2481.85
Step 5: Take the natural logarithm of both sides
ln(3.12) = lneEa
2981.23 −Ea
2481.85
Step 6: Simplify the natural logarithm Using the properties of logarithms:
ln(3.12) = Ea
2981.23 −Ea
2481.85
Step 7: Solve for activation energy, EaMultiply both sides by 2981.23 ×
2481.85 to solve for Ea:
Ea=2981.23 ×2481.85
ln(3.12)
Calculate Eato find the activation energy for this reaction.
Question 21
Question
Given the following data for a reaction at two different temperatures, determine
the activation energy (Ea) for the reaction.
Temperature (K) Rate Constant (s−1)
300 1.5×10−3
350 4.0×10−2
18
Solution
Step 1: Write the Arrhenius Equation: The Arrhenius equation relates the rate
constant of a reaction (k) to the temperature (T) and the activation energy (Ea)
of the reaction. The equation is given as:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the temperature
in Kelvin.
Step 2: Calculate k/T for both temperature points: For the data provided:
- At 300 K:
k1/T1=1.5×10−3
300 = 5 ×10−6s−1·K−1
- At 350 K:
k2/T2=4.0×10−2
350 ≈1.14 ×10−4s−1·K−1
Step 3: Set up the ratio to determine the activation energy (Ea): Taking
the natural logarithm (ln) of the Arrhenius equation, we get:
ln(k) = ln(A)−Ea
RT
This equation can be rearranged to get the slope-intercept form:
ln(k) = −Ea
R·1
T+ln(A)
Comparing this equation with y=mx +c, where y=ln(k), m=−Ea
R,x=1
T,
and c=ln(A), we see that the slope of the line (m) is −Ea
R.
Thus, we can set up the following equation using the two data points:
slope = −Ea
R=ln(k2)−ln(k1)
(1/T2)−(1/T1)
Step 4: Calculate the activation energy (Ea): Substitute the calculated
values: ln(1.14 ×10−4)−ln(5 ×10−6)
(1/350) −(1/300) =−Ea
8.314
⇒Ea=−8.314 ×ln(1.14 ×10−4)−ln(5 ×10−6)
(1/350) −(1/300)
⇒Ea≈74,200 J/mol
Therefore, the activation energy for the reaction is approximately 74,200
J/mol.
19
Question 22
Question
The rate constant of a certain reaction is 4.32×10−3s−1at 350 K and 1.23×10−2
s−1at 400 K. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A×e−Ea
RT
where kis the rate constant, Ais the pre-exponential factor, Eais the activation
energy, Ris the gas constant (8.314 J/mol·K), and Tis the temperature in
Kelvin.
Step 2: Rearrange the equation to solve for the activation energy Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given values into the equation.
Given: k1= 4.32 ×10−3s−1(at 350 K)
k2= 1.23 ×10−2s−1(at 400 K)
R= 8.314 J/mol·K
T1= 350 K
T2= 400 K
Step 4: Calculate the activation energy Ea:
ln 1.23 ×10−2
4.32 ×10−3=−Ea
8.314 1
400 −1
350
ln(2.85) = −Ea
8.314 1
400 −1
350
Step 5: Solve for the activation energy Ea:
1
8.314 1
400 −1
350ln(2.85) = Ea
Ea≈65.68 kJ/mol
Therefore, the activation energy for this reaction is approximately 65.68
kJ/mol.
20
Question 23
Question
The rate constant for a certain reaction is observed to double when the temper-
ature is increased from 25
°
C to 45
°
C. If the activation energy for the reaction
is 50 kJ/mol, calculate the rate constant at 25
°
C and the rate constant at 80
°
C
for this reaction.
Solution
Step 1: Calculate the rate constant at 25
°
C using the Arrhenius equation. The
Arrhenius equation is given by:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the temperature
in Kelvin.
Given that the rate constant doubles when the temperature increases from
25
°
C to 45
°
C, we can use this information to find the pre-exponential factor, A.
Step 2: Calculate the rate constant at 25
°
C. Let’s denote the rate constant
at 25
°
C as k25, and the rate constant at 45
°
C as k45. Since the rate constant
doubles when the temperature increases from 25
°
C to 45
°
C, we have:
k45
k25
= 2
Next, we will use the relationship between rate constant and temperature to
find the pre-exponential factor, A.
Step 3: Solve for the rate constant at 80
°
C using the Arrhenius equation.
Now that we have the pre-exponential factor, A, and the activation energy, Ea,
we can calculate the rate constant at 80
°
C using the Arrhenius equation.
Let’s denote the rate constant at 80
°
C as k80. We need to convert 80
°
C to
Kelvin by adding 273.15:
T= 80C+ 273.15 = 353.15K
Now, we can use the Arrhenius equation to find the rate constant at 80
°
C:
k80 =Ae−Ea
RT
Calculating k80 will give us the rate constant at 80
°
C for this reaction.
Question 24
Question
For a certain reaction, the rate constant at 25
°
C is found to be 1.5×10−3M/s.
When the temperature is increased to 50
°
C, the rate constant is measured to
be 3.0×10−2M/s. Calculate the activation energy (Ea) for this reaction.
21
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at
different temperatures: k2
k1
=e−Ea
R1
T2
−1
T1
where: - k1and k2are the rate constants at temperatures T1and T2, respectively,
-Eais the activation energy, - R= 8.314 J/(mol·K) is the gas constant, - T1
and T2are the temperatures in Kelvin.
Step 2: Substituting the known values into the equation, we have:
3.0×10−2
1.5×10−3=e(−Ea
8.314 (1
323 −1
298 ))
Step 3: Simplifying the equation gives:
20 = e(−Ea
8.314 (1
323 −1
298 ))
Step 4: Taking the natural logarithm of both sides to solve for Ea, we get:
ln(20) = −Ea
8.314 1
323 −1
298
Step 5: Multiplying through by 8.314 and rearranging the equation yields:
Ea=−8.314 ×1
323 −298 ln(20)
Step 6: Calculating the value of Ea, we find:
Ea≈61.31 kJ/mol
Thus, the activation energy for this reaction is approximately 61.31 kJ/mol.
Question 25
Question
The rate constant for a reaction at 25
°
C is 2.0×10−2s−1. When the temperature
is increased to 35
°
C, the rate constant becomes 7.0×10−2s−1. Calculate the
activation energy for this reaction.
Solution
Step 1: Recall the Arrhenius equation:
k=Ae−Ea
RT
22
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol K)), T= temperature (in Kelvin).
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
ln k2
k1
1
T2
−1
T1
=−Ea
R
Step 3: Plug in the given values: k1= 2.0×10−2s−1,k2= 7.0×10−2s−1,
T1= 25 + 273 = 298 K, T2= 35 + 273 = 308 K.
Step 4: Calculate the activation energy:
Ea=−R×
ln 7.0×10−2
2.0×10−2
1
308 −1
298
Step 5: Simplify the equation and solve for Ea:
Ea=−8.314 ×ln 3.5
1
308 −1
298
Ea=−8.314 ×ln 3.5
1
308 −1
298
Step 6: Calculate the activation energy Eato find its value.
Question 26
Question
The rate constant for a certain reaction quadruples when the temperature is
increased from 25
°
C to 50
°
C. Calculate the activation energy for this reaction.
Solution
Step 1: Start by writing down the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol K), T= temperature in Kelvin.
Step 2: Given that the rate constant quadruples when the temperature is
increased from 25
°
C to 50
°
C, we can write this relationship as:
4k1=k2
where: k1is the rate constant at 25
°
C (298K), and k2is the rate constant at
50
°
C (323K).
23
Step 3: We can rewrite the Arrhenius equation for both temperatures as:
k1=Ae−Ea
R·298
k2=Ae−Ea
R·323
Step 4: Substitute the expressions for k1and k2from Step 3 into the rela-
tionship from Step 2:
4Ae−Ea
R·298 =Ae−Ea
R·323
Step 5: Simplify the equation by dividing both sides by Aand then taking
the natural logarithm:
4eEa
R·323 −Ea
R·298 =e0
Step 6: Simplify the exponents and solve for Eato find the activation energy.
Remember to convert temperatures to Kelvin:
4eEa
8.314·323 −Ea
8.314·298 = 1
e25Ea
8.314·323·298 = 4
25Ea
8.314 ·323 ·298 = ln(4)
Ea=8.314 ·323 ·298 ·ln(4)
25
Ea≈59.77 kJ/mol
Therefore, the activation energy for this reaction is approximately 59.77
kJ/mol.
Question 27
Question
For a certain reaction, the rate constant at 25
°
C is 2.5×10−3s−1and the
activation energy is 70 kJ/mol. Calculate the rate constant at 40
°
C for this
reaction.
24
Solution
Step 1: Convert the activation energy to the correct units. Given activation
energy = 70 kJ/mol, convert it to Joules per mole. 1 kJ = 1000 J, so 70 kJ =
70 ×1000 J = 70000 J. Now, 1 J/mol = 1 mol/J, so the activation energy in
J/mol is 70000 J/mol.
Step 2: Use the Arrhenius equation to find the rate constant at 40
°
C. The
Arrhenius equation is given by:
k=A·exp −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)
At 25
°
C = 298 K, we have:
k1= 2.5×10−3s−1
Ea= 70000 J/mol
Now, at 40
°
C = 313 K, we need to find k2:
k2=A·exp −70000
8.314 ×313
Step 3: Calculate the rate constant at 40
°
C.
k2=A·exp −70000
2608.282
Since we don’t know the pre-exponential factor A, we cannot directly calcu-
late k2without additional information.
Question 28
Question
The rate constant of a reaction is found to be 2.70 ×10−3s−1at 50
°
C and
8.60 ×10−3s−1at 100
°
C. Calculate the activation energy (in kJ/mol) 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.
25
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Plug in the given values: k1= 2.70 ×10−3s−1(at 50
°
C = 323 K),
k2= 8.60 ×10−3s−1(at 100
°
C = 373 K), R= 8.314 J/(mol·K).
Step 4: Calculate the activation energy Ea:
ln 8.60 ×10−3
2.70 ×10−3=−Ea
8.314 1
373 −1
323
ln (3.1852) = −Ea
8.314 1
373 −1
323
1.1563 = −Ea
8.314 1
373 −1
323
Step 5: Solve for activation energy Ea:
Ea=−8.314 ×1.1563 1
373 −1
323
Ea≈64.7 kJ/mol
Therefore, the activation energy for this reaction is approximately 64.7
kJ/mol.
Question 29
Question
The rate constant of a certain reaction is found to be 3.2×10−3s−1at 300 K
and 1.2×10−2s−1at 350 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):
k=A·e−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: Use the given data to set up two equations:
From the given data, we have:
k1= 3.2×10−3s−1at 300 K
26
k2= 1.2×10−2s−1at 350 K
Substitute these values into the Arrhenius equation to get two equations:
3.2×10−3=A·e−Ea
8.314·300
1.2×10−2=A·e−Ea
8.314·350
Step 3: Take the ratio of the two equations:
Divide the second equation by the first to eliminate A:
1.2×10−2
3.2×10−3=eEa
8.314 (1
300 −1
350 )
Step 4: Solve for Ea:
Calculate the left side of the equation:
1.2×10−2
3.2×10−3= 3.75
Next, solve for Ea:
3.75 = eEa
8.314 (1
300 −1
350 )
ln(3.75) = Ea
8.314 1
300 −1
350
Ea= 8.314 ×ln(3.75)
1
300 −1
350
Ea≈47.2 kJ/mol
Therefore, the activation energy for this reaction is approximately 47.2
kJ/mol.
Question 30
Question
The rate constant for the reaction 2NOBr(g)→2NO(g)+Br2(g) is found to be
3.28×10−2s−1at 1000 K and 1.08×10−3s−1at 900 K. Calculate the activation
energy for this reaction. (Hint: Use the Arrhenius equation, k=Aexp −Ea
RT )
27
Solution
Step 1: Write the Arrhenius equation in logarithmic form:
ln(k) = ln(A)−Ea
RT
Step 2: Write two equations using the given data:
ln(k1) = ln(A)−Ea
R×1000
ln(k2) = ln(A)−Ea
R×900
Step 3: Subtract the second equation from the first equation:
ln(k1)−ln(k2) = Ea
R1
900 −1
1000
Step 4: Simplify the equation:
ln k1
k2=Ea
R×1
9000
Step 5: Solve for the activation energy Ea:
Ea=R×9000 ×ln k1
k2
Step 6: Plug in the given values for k1and k2, and the Universal Gas
Constant R= 8.314 J/mol ·K:
Ea= 8.314 ×9000 ×ln 3.28 ×10−2
1.08 ×10−3
Step 7: Calculate the activation energy:
Ea= 8.314 ×9000 ×ln 30.37
1.08
Ea= 8.314 ×9000 ×ln(28.10)
Ea≈8.314 ×9000 ×3.337
Ea≈249,173 J/mol
Question 31
Question
The rate constant for a certain reaction is found to be 3.81 ×10−2s−1at 280◦C
and 2.46 s−1at 300◦C. Determine the activation energy for this reaction. The
activation energy constant Ais 1.00 ×109s−1.
28
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: Rearrange the Arrhenius equation for two sets of conditions to solve
for the activation energy Ea:
k1
k2
=Ae−Ea
R·T1
Ae−Ea
R·T2
Step 3: Plug in the given values:
3.81 ×10−2
2.46 =1.00 ×109e−Ea
8.314·280
1.00 ×109e−Ea
8.314·300
Step 4: Simplify the equation:
3.81 ×10−2
2.46 =e−Ea
8.314·280 +Ea
8.314·300
Step 5: Further simplify and solve for Ea:
ln 3.81 ×10−2
2.46 =−Ea
8.314 1
280 −1
300
Ea=−8.314 ×300 ×280
300 −280 ×ln 3.81 ×10−2
2.46
Step 6: Calculate the activation energy:
Ea≈1.62 ×105J/mol
Question 32
Question
The rate constant of a reaction at 25
°
C is 2.5×10−3s−1and the activation
energy is 50 kJ mol−1. Calculate the rate constant at 35
°
C using the Arrhenius
equation.
29
Solution
Step 1: Convert the activation energy from kJ to J: Given activation energy
Ea= 50 kJ mol−1, we convert this to joules:
Ea= 50 ×103J mol−1= 5.0×104J mol−1
Step 2: Determine the new rate constant at 35
°
C using the Arrhenius equa-
tion: 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.
Given that the rate constant at 25
°
C is 2.5×10−3s−1and the activation
energy is 5.0×104J mol−1, we can calculate the pre-exponential factor Ausing
the rate constant at 25
°
C:
2.5×10−3s−1=A×e−5.0×104
8.314×(25+273.15)
2.5×10−3s−1=A×e−5.0×104
8.314×298.15
2.5×10−3s−1=A×e−5.0×104
2478.8091
A= 2.5×10−3×e5.0×104
2478.8091
A≈2.723 ×10−2s−1
Now, we use the Arrhenius equation to find the rate constant at 35
°
C:
k= 2.723 ×10−2×e−5.0×104
8.314×(35+273.15)
k= 2.723 ×10−2×e−5.0×104
8.314×308.15
k= 2.723 ×10−2×e−5.0×104
2550.7291
k≈7.368 ×10−3s−1
Therefore, the rate constant at 35
°
C is approximately 7.368 ×10−3s−1.
Question 33
Question
The rate constant of a first-order reaction at 25
°
C is 4.0×10−3s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.6×10−2s−1.
Calculate the activation energy of the reaction. (Hint: Use the Arrhenius equa-
tion 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.)
30
Solution
Step 1: Convert the temperatures to Kelvin: 25C+273 = 298 K 50C+273 = 323
K
Step 2: Write the Arrhenius equation for the two sets of conditions: For the
first condition (25
°
C): k1=A·e−Ea
RT1
For the second condition (50
°
C): k2=A·e−Ea
RT2
Step 3: Divide the two Arrhenius equations to eliminate the pre-exponential
factor A:k2
k1=e−Ea
RT2
e−Ea
RT1
Step 4: Simplify the equation: k2
k1=e
Ea
R1
T1
−1
T2
Step 5: Plug in the values: 1.6×10−2
4.0×10−3=eEa
8.314 (1
298 −1
323 )
Step 6: Solve for the activation energy Ea:1.6×10−2
4.0×10−3=eEa
8.314 (1
298 −1
323 )
4 = eEa
8.314 (1
298 −1
323 )
Step 7: Take the natural logarithm (ln) of both sides and solve for Ea:
ln(4) = Ea
8.314 1
298 −1
323
Ea= 8.314 ×ln(4)∇ · 1
323 −1
298
Step 8: Calculate the value of activation energy Eawith the known values.
Question 34
Question
For a certain reaction, the rate constant at 25
°
C is 4.0×10−3s−1, and at 35
°
C it
is 2.0×10−2s−1. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/(mol
·
K)), and - Tis the
temperature in Kelvin.
Step 2: Set up the Arrhenius equation for the two given temperatures: At
25
°
C (298 K):
4.0×10−3=A·e−Ea
8.314·298
At 35
°
C (308 K):
2.0×10−2=A·e−Ea
8.314·308
Step 3: Divide the equations to eliminate the pre-exponential factor:
4.0×10−3
2.0×10−2=e−Ea
8.314·298
e−Ea
8.314·308
31
Step 4: Simplify the equation:
0.2 = eEa
8.314 (1
308 −1
298 )
Step 5: Solve for the activation energy Ea:
Ea=−8.314 ·ln(0.2) ·1
308 −1
298
Step 6: Calculate the activation energy in kJ/mol:
Ea=−8.314 ·ln(0.2) ·1
308 −1
298= 76.15 kJ/mol
Question 35
Question
The rate constant for a certain reaction is found to be 4.63 ×10−3s−1at 25◦C.
When the temperature is increased to 55◦C, the rate constant becomes 6.82 ×
10−2s−1. Calculate the activation energy of this reaction.
Solution
Step 1: Convert temperatures to Kelvin using the formula T(K) = T(
°
C) +
273.15.
T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 55◦C = 55 + 273.15 = 328.15 K
Step 2: Apply the Arrhenius equation to relate the rate constants to the
temperatures.
ln k2
k1=−Ea
R1
T2
−1
T1
where: - k1= 4.63 ×10−3s−1-k2= 6.82 ×10−2s−1-R= 8.314 J ·mol−1·K−1
Step 3: Substitute the given values into the equation above.
ln 6.82 ×10−2
4.63 ×10−3=−Ea
8.314 1
328.15 −1
298.15
Step 4: Solve for the activation energy Ea.
ln 6.82 ×10−2
4.63 ×10−3=−Ea
8.314 1
328.15 −1
298.15
ln 6.82 ×10−2
4.63 ×10−3=−Ea
8.314 0.00305 −0.00335
(328.15)(298.15)
32
ln 6.82 ×10−2
4.63 ×10−3=−Ea
8.314 ×(−3.63 ×10−7)
−Ea
8.314 = ln 6.82 ×10−2
4.63 ×10−3∇ · −3.63 ×10−7
Ea=−8.314 ×ln 6.82 ×10−2
4.63 ×10−3∇ · −3.63 ×10−7
Step 5: Calculate the activation energy Eausing a calculator.
33
