CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Activation energy and
the Arrhenius equation
Question Bank - Set 5
Liberty University
Question 1
Question
The rate constant for a particular reaction at 25
°
C is 1.5×10−3s−1, and at 50
°
C
it is 8.0×10−2s−1. Calculate the activation energy for this reaction. (Hint:
Use the Arrhenius equation: k=Ae−Ea
RT , where kis the rate constant, Ais the
pre-exponential factor, Eais the activation energy, R= 8.314 J/mol·K, and T
is the temperature in Kelvin.)
Solution
Step 1: Convert the given temperatures to Kelvin: - For 25
°
C: 25C+ 273.15 =
298.15 K - For 50
°
C: 50C+ 273.15 = 323.15 K
Step 2: Rearrange the Arrhenius equation to solve for the activation energy
Ea:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 3: Substitute the given rate constants and temperatures into the equa-
tion:
ln 8.0×10−2
1.5×10−3=−Ea
8.314 1
323.15 −1
298.15
Step 4: Calculate ln 8.0×10−2
1.5×10−3:
ln 8.0×10−2
1.5×10−3= ln 53.33
3.33 = ln(16) ≈2.77
Step 5: Substitute into the equation and solve for the activation energy Ea:
2.77 = −Ea
8.314 1
323.15 −1
298.15
Step 6: Simplify the equation and solve for Ea:
−2.77 ×8.314
0.02 =Ea
Ea≈ −11472.11 J/mol
Therefore, the activation energy for this reaction is approximately 11472.11
J/mol.
Question 2
Question
For the reaction A →B, the rate constant at 300 K is 1.2×10−3s−1. At 320
K, the rate constant is 2.4×10−3s−1. Calculate the activation energy for this
reaction.
Solution
Step 1: Recall the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant, T= temperature in Kelvin.
Step 2: Let’s first write the Arrhenius equation for the two temperatures
given:
k1=Ae−Ea
RT1
k2=Ae−Ea
RT2
Step 3: Divide the two equations to eliminate A:
k1
k2
=e−Ea
RT1
e−Ea
RT2
Step 4: Simplify the equation:
k1
k2
=e
Ea
R1
T2
−1
T1
Step 5: Plug in the given values:
1.2×10−3
2.4×10−3=eEa
R(1
320 −1
300 )
2
Step 6: Solve for Ea:
eEa
R(1
320 −1
300 )= 0.5
1
Ea
R1
320 −1
300 = ln 0.5
R
320 −300 =1
0.6931 ·Ea
Ea=8.314 ·103
20 ·0.6931
Ea≈60003.5 J/mol
Step 7: Therefore, the activation energy for the reaction A →B is approxi-
mately 60003.5 J/mol.
Question 3
Question
The rate constant for the reaction
2A→B
is found to be 1.5×10−3s−1at 25◦C and 7.8×10−3s−1at 45◦C. Determine
the activation energy for this reaction. (Assume R= 8.314 J ·mol−1·K−1.)
Solution
Step 1: Convert the given rate constants to their corresponding Arrhenius equa-
tion forms using the Arrhenius equation:
k=A×e−Ea
RT
k1=A×e−Ea
RT1
k2=A×e−Ea
RT2
Step 2: Take the ratio of the two rate constants:
k2
k1
=A×e−Ea
RT2
A×e−Ea
RT1
Step 3: Simplify the equation to solve for the activation energy, Ea:
k2
k1
=e
Ea
R1
T1
−1
T2
3
Step 4: Plug in the given values for k1,k2,T1, and T2, and solve for Ea:
7.8×10−3
1.5×10−3=eEa
8.314 (1
298 −1
318 )
Step 5: Calculate the activation energy:
5.2
1=eEa
8.314 (1
298 −1
318 )
5.2 = eEa
8.314 (318−298
298×318 )
ln(5.2) = Ea
8.314 ×20
298 ×318
Ea= 8.314 ×20
298 ×318 ×ln(5.2)
Ea≈48.7 kJ/mol
Therefore, the activation energy for the reaction is approximately 48.7kJ/mol.
Question 4
Question
The rate constant for the decomposition of a certain compound at 298 K is
3.2×10−4s−1. When the temperature is increased to 350 K, the rate constant
becomes 2.1×10−3s−1. Calculate the activation energy for this reaction.
Solution
Let’s first write the Arrhenius equation, which relates the rate constant of a
reaction to the temperature and the activation energy:
k=A·e(−Ea
RT )
Where: - kis the rate constant, - Ais the pre-exponential factor, - Eais
the activation energy, - Ris the universal gas constant (8.314 J/(mol*K)), - T
is the temperature in Kelvin.
We can write two Arrhenius equations for the two temperatures given:
k1=A·e(−Ea
R·298 )and k2=A·e(−Ea
R·350 )
We can take the ratio of these two equations to eliminate A:
k2
k1
=e(−Ea
R·350 )
e(−Ea
R·298 )
Solving for Ea:k2
k1
=e(Ea
R(1
298 −1
350 ))
4
ln k2
k1=Ea
8.314 1
298 −1
350
Ea=
8.314 ×ln k2
k1
1
298 −1
350
Substitute the given rate constants into the equation:
Ea=
8.314 ×ln 2.1×10−3
3.2×10−4
1
298 −1
350
Ea
8.314 ×ln(6.5625)
1
298 −1
350
Ea
8.314 ×1.8808
(0.0034)
Ea
15.6073
0.0034
Ea4597.44 J/mol
Therefore, the activation energy for this reaction is approximately 4597.44
J/mol.
Question 5
Question
The rate constant for a certain reaction is found to be 1.20 ×10−3s−1at 25◦C
and 2.80 ×10−2s−1at 60◦C. Calculate the activation energy (Ea) for this
reaction.
Solution
Step 1: Write 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−1·K−1), T= temperature in Kelvin.
Step 2: Consider the data provided: At 25◦C = 298 K, k1= 1.20 ×10−3s−1
At 60◦C = 333 K, k2= 2.80 ×10−2s−1
Step 3: Take the ratio of the rate constants:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
5
2.80 ×10−2
1.20 ×10−3=e
−Ea
R1
T1
−1
T2
Step 4: Solve for Ea:
2.80 ×10−2
1.20 ×10−3=e−Ea
8.314 (1
298 −1
333 )
23.33 = e−Ea
8.314 (1
298 −1
333 )
Step 5: Simplify the equation and solve for Ea:
23.33 = e−Ea
8.314 (0.00336−0.00299)
23.33 = e−Ea
8.314 ×0.00037
23.33 = e−0.000037Ea
Step 6: Take the natural logarithm of both sides and solve for Ea:
ln(23.33) = lne−0.000037Ea
ln(23.33) = −0.000037Ea
Ea=ln(23.33)
−0.000037
Step 7: Calculate Eausing a calculator:
Ea≈ln(23.33)
−0.000037
Ea≈3.148
−0.000037
Ea≈ −85,134 J/mol
Question 6
Question
Given the rate constant of a chemical reaction at two different temperatures,
determine the activation energy for the reaction. The rate constant at 300 K is
4.32 ×10−3s−1and at 350 K is 1.20 ×10−2s−1. Assume the pre-exponential
factor A= 1.0×108s−1.
6
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 ideal gas constant (8.314 J/(mol
·
K)), - Tis the
temperature in Kelvin.
Step 2: Use the given rate constants to set up two equations based on the
Arrhenius equation for the two temperatures:
4.32 ×10−3= 1.0×108·e−Ea
8.314·300
1.20 ×10−2= 1.0×108·e−Ea
8.314·350
Step 3: Divide the two equations to eliminate Aand solve for Ea:
4.32 ×10−3
1.20 ×10−2=e−Ea
8.314·300
e−Ea
8.314·350
Step 4: Simplify the expression by cross-multiplying and taking the natural
logarithm of both sides to solve for the activation energy Ea.
Step 5: Calculate the activation energy using the formula and the given
temperatures.
Question 7
Question
The rate constant for a reaction at 25
°
C is 4.2×10−3s−1. When the temperature
is increased to 45
°
C, the rate constant becomes 1.2×10−2s−1. Calculate the
activation energy (in kJ/mol) for the reaction.
Solution
Step 1: Convert the given temperatures to kelvin. Given: T1= 25C= 25 +
273 = 298 K T2= 45C= 45 + 273 = 318 K
Step 2: Write the Arrhenius equation. The Arrhenius equation is given by:
k=Ae−Ea
RT
where: k= rate constant A= pre-exponential factor Ea= activation energy R
= gas constant (8.314 J mol−1K−1)T= temperature in kelvin
Step 3: Set up and solve the equations for the two given temperatures. For
T1at 298 K:
4.2×10−3=Ae−Ea
8.314×298
7
For T2at 318 K:
1.2×10−2=Ae−Ea
8.314×318
Step 4: Divide the two equations to eliminate A.
1.2×10−2
4.2×10−3=Ae−Ea
8.314×318
Ae−Ea
8.314×298
Step 5: Solve for the activation energy, Ea.
1.2×10−2
4.2×10−3=e(Ea
8.314 (1
318 −1
298 ))
Step 6: Calculate the activation energy, Ea, in kJ/mol. After solving the
above equation, we find Ea≈72.6 kJ/mol.
Question 8
Question
The rate constant of a reaction at 25
°
C is 1.2×10−3s−1, and its rate constant
at 45
°
C is 5.6×10−2s−1. Calculate the activation energy (Ea) for this reaction.
The activation energy can be obtained by using the Arrhenius equation: k=
A·e−Ea
RT , where kis the rate constant, Ais the pre-exponential factor, Eais
the activation energy, Ris the gas constant (8.314 J/(mol·K)), and Tis the
temperature in Kelvin.
Solution
Step 1: Convert the temperatures to Kelvin
T1= 25C= 25 + 273 = 298 K
T2= 45C= 45 + 273 = 318 K
Step 2: Calculate the ratio of rate constants and rearrange the Arrhenius
equation to solve for activation energy (Ea)
k1
k2
=A·e−Ea
RT1
A·e−Ea
RT2
1.2×10−3
5.6×10−2=e(Ea
8.314 ×(1
318 −1
298 ))
1.2×10−3
5.6×10−2=e(Ea
8.314 ×(20
63684 ))
21.4286 = e(Ea
1680.5488 )
8
Step 3: Take natural logarithm of both sides to solve for Ea
ln(21.4286) = lne(Ea
1680.5488 )
ln(21.4286) = Ea
1680.5488
Ea= 1680.5488 ×ln(21.4286)
Ea≈65839.3 J/mol
Question 9
Question
The rate constant for a first-order reaction is 6.00 ×10−3s−1at 25
°
C, and
3.00 ×10−2s−1at 50
°
C. Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/mol·K), T= temperature in Kelvin.
Step 2: Write down the equation in terms of k1,k2,T1, and T2.
k2
k1
=A·e−Ea
RT2
A·e−Ea
RT1
Step 3: Solve for Eaby taking the natural logarithm of both sides:
ln k2
k1=−Ea
R1
T2
−1
T1
Step 4: Convert temperatures to Kelvin:
T1= 25C= 273.15K
T2= 50C= 323.15K
Step 5: Subsitute the given values and solve for Ea:
ln 3.00 ×10−2s−1
6.00 ×10−3s−1=−Ea
8.314 1
323.15 −1
273.15
Step 6: Calculate Ea:
ln (5) = −Ea
8.314 1
323.15 −1
273.15
9
ln(5) = −Ea
8.314 1
323.15 −1
273.15
Ea
8.314 =−ln(5)
1
323.15 −1
273.15
Ea= 8.314 ×ln(5)
1
323.15 −1
273.15
Therefore, the activation energy for this reaction is approximately 42,276
J/mol.
Question 10
Question
The rate constant for a reaction at 25
°
C is 3.2×10−3s−1, and the rate constant
at 50
°
C is 2.1×10−2s−1. Calculate the activation energy (in kJ/mol) for this
reaction.
Solution
Step 1: Convert the given rate constants to Arrhenius equation form: Let’s
denote the rate constant at 25
°
C as k1and at 50
°
C as k2. The Arrhenius
equation relates the rate constant to the temperature and the activation energy.
The Arrhenius equation is given by:
k=Ae−Ea
RT
where k= rate constant, A= frequency factor, Ea= activation energy, R=
gas constant (8.314 J/(mol
·
K)), T= temperature in Kelvin.
For k1at 25
°
C: k1=Ae−Ea
RT1
For k2at 50
°
C: k2=Ae−Ea
RT2
Step 2: Take the ratio of the two rate constants to eliminate the frequency
factor: Divide k2by k1to get:
k2
k1
=Ae−Ea
RT2
Ae−Ea
RT1
Step 3: Take natural logarithm of both sides to simplify the equation:
ln k2
k1= ln e−Ea
R1
T2
−1
T1
Step 4: Solve for the activation energy (Ea):
ln k2
k1=−Ea
R1
T2
−1
T1
10
Ea=−R·
ln k2
k1
1
T2
−1
T1
Substitute the given values: T1= 25+273.15 = 298.15 K T2= 50+273.15 =
323.15 K R= 8.314 J/(mol
·
K)
∆Ea=−8.314 ·
ln 2.1×10−2
3.2×10−3
1
323.15 −1
298.15
Calculating this expression will give the activation energy in Joules per mole,
which can then be converted to kilojoules/mol.
Question 11
Question
A reaction has an activation energy of 50 kJ/mol. At 25
°
C, the rate constant
for this reaction is 1.2×10−3s−1. Calculate the rate constant for this reaction
at 75
°
C. (Given: R= 8.314 J/(mol ·K))
Solution
Step 1: Calculate the activation energy in Joules. Using the conversion factor
1 kJ = 1000 J, we have: 50 kJ = 50 ×1000 J = 50000 J.
Step 2: Calculate the activation energy in Reaumur. Given that the acti-
vation energy in Kelvin is the same as in Reaumur, so the activation energy
remains as 50000 J.
Step 3: Calculate the rate constant at 75
°
C using the Arrhenius equation.
The Arrhenius equation is given by:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant, - Tis the temperature in Kelvin.
Step 4: Convert the temperatures to Kelvin. Given temperature at 25
°
C:
25 + 273 = 298 K Given temperature at 75
°
C: 75 + 273 = 348 K
Step 5: Substitute the values and solve for the rate constant at 75
°
C.
k75C=A·e−50000
8.314×348
Step 6: Using the rate constant at 25
°
C to solve for A. Given the rate
constant at 25
°
C is 1.2×10−3s−1and the activation energy (50 kJ) is the same
for both temperatures.
1.2×10−3=A·e−50000
8.314×298
11
Step 7: Solve for A.
A=1.2×10−3
e−50000
8.314×298
Step 8: Substitute the value of Ainto the equation calculated in Step 5 to
find the rate constant at 75
°
C.
Question 12
Question
The rate constant for a certain reaction at 25
°
C is 1.2×10−3s−1. When the
temperature is increased to 35
°
C, the rate constant becomes 3.8×10−3s−1.
Assume the activation energy for the reaction is 55 kJ/mol. Calculate the
frequency factor (pre-exponential factor) for this reaction using the Arrhenius
equation.
Solution
Step 1: Convert the given temperatures to Kelvin:
T1= 25C= 25 + 273 = 298K
T2= 35C= 35 + 273 = 308K
Step 2: Calculate the rate constant ratio:
k2
k1
=3.8×10−3s−1
1.2×10−3s−1= 3.17
Step 3: Use the Arrhenius equation:
k2
k1
= exp −Ea
R1
T1
−1
T2
Step 4: Subsitute the given activation energy and temperatures into the
equation:
3.17 = exp −55000 J/mol
8.314 J/mol K 1
298 K −1
308 K
Step 5: Solve for the frequency factor:
exp −55000 J/mol
8.314 J/mol K 1
298 K −1
308 K= 3.17
1
A= 3.17
A=1
3.17 ≈0.315 s−1
Therefore, the frequency factor for this reaction is approximately 0.315 s−1.
12
Question 13
Question
The rate constant for a reaction at 25
°
C is 1.20 ×10−3s−1, and the activation
energy for the reaction is 75 kJ/mol. Calculate the rate constant at 40
°
C for
the same reaction.
Solution
Step 1: Calculate the activation energy in joules using the given value in kilo-
joules.
Activation energy (J) = 75 ×103J/mol = 75,000 J/mol
Step 2: Calculate the new rate constant at 40
°
C using the Arrhenius equa-
tion:
k2=k1×e
−Ea
R× 1
T2
−
1
T1!
Given: T1= 25C= 298 K, T2= 40C= 313 K, Ea= 75,000 J/mol, R=
8.314 J/mol·K, k1= 1.20 ×10−3s−1
Step 3: Substitute the known values into the Arrhenius equation and solve
for k2.
k2= 1.20 ×10−3×e −75,000
8.314 × 1
313−
1
298!!
Step 4: Calculate the new rate constant k2.
k2≈3.49 ×10−3s−1
Therefore, the rate constant for the reaction at 40
°
C is approximately 3.49×
10−3s−1.
Question 14
Question
The rate constants (k) for a reaction at different temperatures are given in the
table below:
Temperature (K) k(L/mol s)
300 0.005
310 0.008
320 0.013
Calculate the activation energy (Ea) for this reaction. Assume the pre-
exponential factor (A) is 1 ×109L/mol s and the universal gas constant (R) is
8.314 J/(mol K).
13
Solution
Step 1: Calculate the rate constant at 1st temperature using Arrhenius equation.
Given data: - Temperature at 1st point (T1) = 300 K - Rate constant at 1st
point (k1) = 0.005 L/mol s - Pre-exponential factor (A) = 1 ×109L/mol s -
Universal gas constant (R) = 8.314 J/(mol K)
The Arrhenius equation is given as:
k=A·e−Ea
RT
Substitute the known values to find Ea:
k1=A·e−Ea
R·T1
0.005 = 1 ×109·e−Ea
8.314·300
Step 2: Calculate the rate constant at 2nd and 3rd temperatures.
Following the same process as in Step 1, we can calculate the rate constants
at 2nd and 3rd temperatures.
Step 3: Set up equations for Eausing rate constants and temperatures.
We have 3 equations in the form:
ki=A·e−Ea
RTi
where irepresents the temperature point.
Step 4: Solve the system of equations to find the activation energy (Ea).
Solve the system of equations obtained in Step 3 to find the activation energy
(Ea). Substituting the known values for A,R, and the rate constants at different
temperatures, we can find the value of Ea.
Question 15
Question
The rate constant of a reaction at 25◦C is 1.5×10−3s−1, and at 35◦C it is
4.2×10−3s−1. Calculate the activation energy (Ea) of the reaction in kJ mol−1.
(Hint: Assume the activation energy does not change with temperature.)
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures to the activation energy Ea. The Arrhenius equation is
given by:
k=A·exp −Ea
RT
where kis the rate constant, Ais the pre-exponential factor (frequency factor),
Eais the activation energy, Ris the gas constant (8.314 J mol−1K−1), and Tis
the temperature in Kelvin.
14
Step 2: We start by writing the Arrhenius equation for the rate constants
at the two temperatures given:
k1=A·exp −Ea
R·(25 + 273.15)
k2=A·exp −Ea
R·(35 + 273.15)
where k1= 1.5×10−3s−1and k2= 4.2×10−3s−1.
Step 3: Take the ratio of the two equations to eliminate the frequency factor
A:
k2
k1
=
exp −Ea
R·(35+273.15)
exp −Ea
R·(25+273.15)
Step 4: Simplify the equation by taking the natural logarithm of both sides:
ln k2
k1=−Ea
R1
35 + 273.15 −1
25 + 273.15
Step 5: Now, plug in the values for k1,k2, and Rto solve for Ea:
ln 4.2×10−3
1.5×10−3=−Ea
8.314 1
35 + 273.15 −1
25 + 273.15
ln(2.8) = −Ea
8.314 1
308.15 −1
298.15
Step 6: Calculate the activation energy Eain Joules and convert it to
kJ mol−1:
Ea=−8.314 ×10−3×ln(2.8)
1
308.15 −1
298.15
Question 16
Question
For a certain reaction, the rate constant at 25
°
C is 3.21 ×10−2s−1, and at 35
°
C
it is 8.54 ×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Write 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.
15
Step 2: Convert the given temperatures into Kelvin: - For 25
°
C, T= 25 +
273 = 298 K. - For 35
°
C, T= 35 + 273 = 308 K.
Step 3: Substitute the given rate constants and temperatures into the Ar-
rhenius equation to form two equations:
3.21 ×10−2=A e−Ea
8.314×298
8.54 ×10−2=A e−Ea
8.314×308
Step 4: Divide the two equations to eliminate A:
3.21 ×10−2
8.54 ×10−2=e−Ea
8.314×298
e−Ea
8.314×308
Step 5: Simplify the equation:
3.21
8.54 =eEa
8.314 (1
298 −1
308 )
Step 6: Solve for Eausing the natural logarithm:
ln 3.21
8.54=Ea
8.314 1
298 −1
308
Step 7: Calculate the activation energy, Ea, using the calculated logarithm
value:
Ea=−8.314 ×ln 3.21
8.54 1
298 −1
308
Step 8: Perform the calculation to find the activation energy for the reaction.
Question 17
Question
For a certain reaction, the rate constant at 25
°
C is 1.5×10−3s−1and at 45
°
C
it is 7.3×10−3s−1. Calculate the activation energy of the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant at different
temperatures:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/(mol·K)), and - Tis the
temperature in Kelvin.
16
Step 2: Let’s first write the equations for the rate constants at 25
°
C and
45
°
C: At 25
°
C (298 K):
1.5×10−3=A·e−Ea
8.314·298
At 45
°
C (318 K):
7.3×10−3=A·e−Ea
8.314·318
Step 3: Now, let’s divide the equation for 45
°
C by the equation for 25
°
C to
eliminate A:7.3×10−3
1.5×10−3=e−Ea
8.314 (1
318 −1
298 )
Step 4: Solve for the activation energy Ea:
7.3×10−3
1.5×10−3=e−Ea
8.314 (1
318 −1
298 )
4.87 = e−Ea
8.314 (1
318 −1
298 )
Step 5: Taking the natural logarithm of both sides to solve for Ea:
ln(4.87) = −Ea
8.314 1
318 −1
298
Step 6: Calculate Ea:
Ea=−8.314 ×ln(4.87)
1
318 −1
298
Ea≈34.5 kJ/mol
Step 7: Therefore, the activation energy of the reaction is approximately
34.5 kJ/mol.
Question 18
Question
The rate constant for a reaction at 25
°
C is 3.2×10−3s−1, and the activation
energy for the reaction is 85 kJ/mol. Calculate the rate constant at 75
°
C for
the same reaction.
Solution
Step 1: Convert the activation energy to joules: Given: Activation energy,
Ea= 85 kJ/mol
Convert 85 kJ/mol to Joules:
85 kJ/mol ×1000 J
1 kJ = 85000 J/mol
17
Step 2: Use the Arrhenius equation to calculate the rate constant at 75
°
C:
The Arrhenius equation is given by:
k=A·e−Ea
RT
where: k= rate constant at 75
°
CA= pre-exponential factor Ea= activation
energy in J/mol R= gas constant = 8.314 J/(mol
·
K) T= temperature in Kelvin
Given: T1= 25C= 298 K (reference temperature) T2= 75C= 348 K (new
temperature) k1= 3.2×10−3s−1(rate constant at 25
°
C) Ea= 85000 J/mol
First, we need to determine the pre-exponential factor Ausing the rate
constant at 25
°
C:
3.2×10−3=A·e−85000
8.314·298
Step 3: Solve for A:
A= 3.2×10−3·e85000
8.314·298
A≈697149 s−1
Step 4: Now, calculate the rate constant at 75
°
C:
k= 697149 ·e−85000
8.314·348
Step 5: Solve for k:
k≈8.20 s−1
Therefore, the rate constant at 75
°
C for the reaction is approximately 8.20s−1.
Question 19
Question
The rate constant for a reaction was found to be 4.26 ×10−3s−1at 25◦C and
1.29 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 45◦C = 45 + 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures: k2
k1
=e−Ea
R1
T2
−1
T1
Step 3: Plug in the given rate constants and temperatures into the equation
to solve for the activation energy (Ea).
1.29 ×10−2
4.26 ×10−3=e(−Ea
8.314 (1
318.15 −1
298.15 ))
18
Step 4: Simplify and solve for the activation energy:
3.02 = e(−Ea
8.314 (1
318.15 −1
298.15 ))
Step 5: Take the natural logarithm of both sides to isolate the activation
energy:
ln(3.02) = −Ea
8.314 1
318.15 −1
298.15
Step 6: Solve for the activation energy Ea:
Ea=−8.314 ×ln(3.02)
1
318.15 −1
298.15
Step 7: Calculate the activation energy using the values from Step 6.
Question 20
Question
For the reaction A →B, the rate constant at 25
°
C is 5.0×10−3s−1. When
the temperature is increased to 50
°
C, the rate constant becomes 1.0×10−2s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the temperature
in Kelvin.
Step 2: Convert the temperatures to Kelvin: - 25
°
C = 298 K - 50
°
C = 323
K
Step 3: Write down the Arrhenius equation at 25
°
C and 50
°
C:
k1=A·e−Ea
8.314×298
k2=A·e−Ea
8.314×323
Step 4: Divide the two equations to eliminate A:
k2
k1
=e−Ea
8.314×323
e−Ea
8.314×298
Step 5: Simplify the equation:
k2
k1
=e−Ea
8.314 (1
323 −1
298 )
19
Step 6: Solve for Ea:
ln k2
k1=−Ea
8.314 1
323 −1
298
Ea=−8.314 ×1
323 −1
298×ln 1.0×10−2
5.0×10−3
Step 7: Calculate the activation energy:
Ea=−8.314 ×1
323 −1
298×ln 1.0×10−2
5.0×10−3
Ea≈70.6 kJ/mol
Therefore, the activation energy for the reaction is approximately 70.6 kJ/mol.
Question 21
Question
The rate constant of a reaction at 50
°
C is 0.0056 s−1, while at 75
°
C it is 0.046
s−1. Calculate the activation energy for this reaction. (R = 8.314 J/(mol·K))
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea/(RT )
where: k= rate constant, A= pre-exponential factor, Ea= activation energy
(J/mol), R= gas constant (J/(mol·K)), T= temperature (K).
Step 2: Use the given data to set up two equations: For 50
°
C:
0.0056 = A·e−Ea/(8.314·(273+50))
0.0056 = A·e−Ea/3867.3
For 75
°
C:
0.046 = A·e−Ea/(8.314·(273+75))
0.046 = A·e−Ea/4097.05
Step 3: Divide the two equations to eliminate A:
0.046
0.0056 =e−Ea/4097.05
e−Ea/3867.3
8.2143 = e(3867.3−4097.05)/2295458.1
20
8.2143 = e(−229.75)/2295458.1
Step 4: Solve for the activation energy Ea:
8.2143 = e−0.0001
8.2143 = 0.9999
ln(8.2143) = ln(0.9999)
ln(8.2143) = −0.0001
Ea=−0.0001 ×8.314 ×3867.3 = 32.0 kJ/mol
Therefore, the activation energy for this reaction is 32.0 kJ/mol.
Question 22
Question
The rate constant (k) for a reaction is found to be 4.23 ×10−3s−1at 60◦C
and 2.17 ×10−2s−1at 70◦C. Calculate the activation energy (Ea) for this
reaction. (Hint: Use the Arrhenius equation: k=A·e−Ea
RT , where Ais the
pre-exponential factor, Eais the activation energy, Ris the gas constant, and
Tis the temperature in Kelvin.)
Solution
Step 1: Convert the temperatures to Kelvin. Given: - Temperature at 60◦C =
60 + 273 = 333 K - Temperature at 70◦C = 70 + 273 = 343 K
Step 2: Write the Arrhenius equation for both sets of data. At 60◦C (333
K): k1=A·e−Ea
RT1= 4.23 ×10−3s−1
At 70◦C (343 K): k2=A·e−Ea
RT2= 2.17 ×10−2s−1
Step 3: Take the ratio of the two rate constants to eliminate A.k2
k1=A·e−Ea
RT2
A·e−Ea
RT1
⇒k2
k1=e
Ea
R1
T1
−1
T2
Step 4: Solve for the activation energy (Ea). Plugging in the values: 2.17×10−2
4.23×10−3=
eEa
8.314 (1
333 −1
343 )
⇒5.12 = eEa
8.314 (1
333 −1
343 )
Step 5: Solve for Ea. Taking the natural logarithm of both sides: ln(5.12) =
Ea
8.314 1
333 −1
343
⇒Ea= 8.314 ·ln(5.12) 1
333 −1
343
Therefore, the activation energy for this reaction is approximately 38.87 kJ/mol.
21
Question 23
Question
The rate constant for a certain reaction is measured at two different tempera-
tures, yielding the following data:
Temperature (K) Rate Constant (s−1)
300 1.2×10−3
500 2.5×10−2
Calculate the activation energy (Ea) for this reaction in kJ/mol. Assume
the pre-exponential factor (A) is 1 ×1012 s−1.
Solution
Step 1: 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 (K).
Step 2: We can rearrange the Arrhenius equation to solve for activation
energy (Ea):
Ea=−ln k
A
1/T
Step 3: Let’s calculate the activation energy using the data provided at 300
K:
Ea=−
ln 1.2×10−3
1×1012
1/300
Step 4: Calculating inside the natural logarithm:
Ea=−ln 1.2×10−15
1/300
Step 5: Simplifying the natural logarithm:
Ea=−ln 1.2−ln 10−15
1/300 =−
−0.1823 −(−34.54)
300
Ea=−34.3577
300 = 114.5 kJ/mol (rounded to 3 decimal places)
Therefore, the activation energy (Ea) for this reaction is 114.5 kJ/mol.
22
Question 24
Question
The rate constant for a certain reaction is 4.55 ×10−3s−1at 25◦C and 8.60 ×
10−3s−1at 50◦C. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: First, recall 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: We can write two Arrhenius equations for the given temperatures:
k1=Ae−Ea
R(25+273.15)
k2=Ae−Ea
R(50+273.15)
Step 3: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R(50+273.15)
e−Ea
R(25+273.15)
Step 4: Simplifying the equation gives:
k2
k1
=eEa
R(1
25+273.15 −1
50+273.15 )
Step 5: Plug in the given values for k1and k2, then solve for Ea:
8.60 ×10−3
4.55 ×10−3=eEa
8.314 (1
298.15 −1
323.15 )
Step 6: Solve for Eato get:
Ea=−8.314 ×ln 8.60 ×10−3
4.55 ×10−3 1
298.15 −1
323.15
Step 7: Calculate the activation energy to find the answer in kJ/mol.
Question 25
Question
For the reaction 2A+B→C, the activation energy is 120 kJ/mol and the
frequency factor is 1.5×1011s−1. At a temperature of 400 K, the rate constant
is 3.2×10−5M−1s−1. Calculate the rate constant at a temperature of 450 K.
23
Solution
Step 1: Calculate the rate constant at 400 K using the Arrhenius equation.
k1=A·e−Ea
RT
where: k1= rate constant at 400 K A= frequency factor (1.5×1011s−1)Ea
= activation energy (120 kJ/mol) R= gas constant (8.314 J/(mol*K)) T=
temperature in Kelvin (400 K)
Substitute the given values into the equation to solve for k1:
k1= 1.5×1011 exp −120 ×103
8.314 ×400
k1≈3.23 ×10−5M−1s−1
Step 2: Calculate the rate constant at 450 K using the Arrhenius equation.
k2=A·e−Ea
RT
where: k2= rate constant at 450 K A= frequency factor (1.5×1011s−1)Ea
= activation energy (120 kJ/mol) R= gas constant (8.314 J/(mol*K)) T=
temperature in Kelvin (450 K)
Substitute the values into the equation to find k2:
k2= 1.5×1011 exp −120 ×103
8.314 ×450
k2≈1.34 ×10−4M−1s−1
Therefore, the rate constant at 450K is approximately 1.34 ×10−4M−1s−1.
Question 26
Question
The rate constant for a first-order reaction is 4.15×10−3s−1at 25◦C and 1.24×
10−2s−1at 50◦C. Calculate the activation energy for the reaction.
Solution
Step 1: Convert the temperatures to Kelvin. Given: T1= 25◦C = 298K T2=
50◦C = 323K
Step 2: Write 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−1K−1T= temperature in Kelvin
24
Step 3: Write the Arrhenius equation for the two given temperatures:
k1=Ae−Ea
R·T1
k2=Ae−Ea
R·T2
Step 4: Divide the two equations to eliminate A:
k1
k2
=e−Ea
R·T1
e−Ea
R·T2
Step 5: Simplify the equation:
k1
k2
=e(Ea
R)1
T1
−1
T2
Step 6: Plug in the given values and solve for Ea:
4.15 ×10−3
1.24 ×10−2=e(Ea
8.314 )( 1
298 −1
323 )
Step 7: Solve for Ea:
4.15
1.24 =e(Ea
8.314 )(323−298
298·323 )
3.35 = e(Ea
8.314 )( 25
298·323 )
Step 8: Solve for Eaby taking natural logarithm of both sides:
ln(3.35) = Ea
8.314 25
298 ·323
Step 9: Calculate the activation energy Ea:
Ea= 8.314 ×ln(3.35)
25
298·323
≈60.2 kJ/mol
Therefore, the activation energy for the reaction is approximately 60.2 kJ/mol.
Question 27
Question
The rate constant (k) for a reaction is known to double for every 10
°
C rise
in temperature. If the rate constant at 30
°
C is 5.0×10−3s−1, what is the
activation energy (Ea) for this reaction? (The universal gas constant R= 8.314
J/(mol·K))
25
Solution
Step 1: First, we can represent the relationship between the rate constant and
temperature using the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, and Ea= activation
energy.
Step 2: Given that the rate constant doubles for every 10
°
C rise in temper-
ature, we can express this relationship as:
k2= 2k1
where: k2= rate constant at temperature T2,k1= rate constant at temperature
T1.
Step 3: We are given the rate constant at 30
°
C as 5.0×10−3s−1, which
means:
k1= 5.0×10−3s−1
Step 4: Let’s denote the final temperature as T2= 30 + 10 = 40
°
C. Thus,
the rate constant at 40
°
C is:
k2= 2k1= 2 ×5.0×10−3= 1.0×10−2s−1
Step 5: Now, we can substitute the values of k1,k2,T1, and T2into the
Arrhenius equation to solve for the activation energy (Ea):
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 6: Simplifying the above equation, we get:
1.0×10−2
5.0×10−3=e−Ea
8.314·313.15
e−Ea
8.314·303.15
Step 7: Further simplifying, we find:
2 = e0.0032Ea
Step 8: Taking the natural logarithm of both sides, we get:
ln(2) = 0.0032Ea
Step 9: Finally, solving for Eagives:
Ea=ln(2)
0.0032 ≈0.6931
0.0032 ≈216.59 kJ/mol
Therefore, the activation energy (Ea) for this reaction is approximately
216.59 kJ/mol.
26
Question 28
Question
The rate constant for the decomposition of acetaldehyde at a certain temper-
ature is 3.2×10−3s−1. When the temperature is increased by 10
°
C, the rate
constant becomes 6.8×10−2s−1. Calculate the activation energy for this reac-
tion.
Solution
Step 1: Convert the temperature change to Kelvin.
∆T= 10
°
C = 10 K
Step 2: Apply the Arrhenius equation, which relates the rate constant (k)
to the activation energy (Ea) and temperature (T):
k=A·e−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 3: Set up two equations based on the given information:
3.2×10−3=A·e−Ea
R(T)
6.8×10−2=A·e−Ea
R(T+10)
Step 4: Divide the two equations to eliminate A.
3.2×10−3
6.8×10−2=A·e−Ea
RT
A·e−Ea
R(T+10)
3.2
68 =eEa
R−Ea
R(T+10)
8
170 =eEa
R−Ea
R(T+10)
27
Step 5: Solve for the activation energy (Ea).
ln 8
170=Ea
R−Ea
R(T+ 10)
ln 8
170=Ea1
R−1
R(T+ 10)
ln 8
170=EaT+ 10 −T
R(T+ 10)
ln 8
170=Ea10
R(T+ 10)
Ea=R·10
ln 8
170
Ea= 8.314 J/mol ·10
ln 8
170
Ea≈56.56 kJ/mol
Therefore, the activation energy for the reaction is approximately 56.56
kJ/mol.
Question 29
Question
The rate constant of a certain reaction doubles when the temperature is raised
from 20◦C to 50◦C. Calculate the activation energy for this reaction. Assume
the activation energy is constant over this temperature range.
Solution
Step 1: Write down the Arrhenius equation.
ln k2
k1=Ea
R1
T1
−1
T2
Step 2: Plug in the given values:
ln 2k1
k1=Ea
R1
293 −1
323
]
Step 3: Simplify the equation:
ln(2) = Ea
8.314 1
293 −1
323
]
28
Step 4: Solve for the activation energy Ea:
0.693 = Ea
8.314 323 −293
293 ×323
Step 5: Calculate the activation energy:
Ea= 0.693 ×8.314 ×293 ×323
323 −293 ≈47.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 47.5
kJ/mol.
Question 30
Question
The rate constant of a certain reaction at 25
°
C is 1.20 ×10−3s−1, and at 50
°
C,
the rate constant is 8.50×10−3s−1. Calculate the activation energy (in kJ/mol)
for this reaction.
Solution
Step 1: Recall the Arrhenius equation, which relates the rate constant of a
reaction to its temperature:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/mol-K), and - Tis the tem-
perature in Kelvin.
Step 2: We are given two sets of data points to plug into the Arrhenius
equation. We can set up two equations:
k1=Ae−Ea
R·298
k2=Ae−Ea
R·323
Step 3: Substitute the given rate constants and convert the temperatures to
Kelvin:
k1= 1.20 ×10−3=Ae−Ea
8.314·298
k2= 8.50 ×10−3=Ae−Ea
8.314·323
Step 4: Divide the two equations to eliminate A:
k2
k1
=8.50 ×10−3
1.20 ×10−3=e−Ea
8.314·323
e−Ea
8.314·298
29
Step 5: Take the natural logarithm of both sides to eliminate the exponential
terms:
ln 8.50 ×10−3
1.20 ×10−3=−Ea
8.314 1
323 −1
298
Step 6: Solve for Eato find the activation energy of the reaction in Joules.
Convert to kilojoules by dividing by 1000.
Ea=−8.314 ×ln 8.50 ×10−3
1.20 ×10−3×1
323 −1
298
Step 7: Calculate the activation energy:
Ea=
−8.314 ×ln 8.50×10−3
1.20×10−3
323 −298 ×1000
Therefore, the activation energy for this reaction is approximately 76.3
kJ/mol.
Question 31
Question
For a certain reaction, the rate constant at 25
°
C is 8.0×10−4s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.6×10−2s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol·K)), - Tis the temperature
in Kelvin.
Step 2: Determine the rate constant at both temperatures in terms of the
Arrhenius equation: At 25
°
C (298 K):
8.0×10−4=A·e−Ea
8.314×298
At 50
°
C (323 K):
1.6×10−2=A·e−Ea
8.314×323
Step 3: Divide the two rate constant expressions to eliminate A:
1.6×10−2
8.0×10−4=e−Ea
8.314×323
e−Ea
8.314×298
30
Step 4: Simplify the equation:
20 = eEa
8.314 (1
298 −1
323 )
Step 5: Solve for Ea:
ln(20) = Ea
8.314 1
298 −1
323
Ea= 8.314 ×ln(20)
1
298 −1
323
Step 6: Calculate the activation energy:
Ea= 8.314 ×ln(20)
1
298 −1
323
≈44.6 kJ/mol
Question 32
Question
The rate constant for a certain reaction is found to be 2.54 ×10−3s−1at 25◦C
and 1.63 ×10−2s−1at 50◦C. Calculate the activation energy for this reaction
in kJ/mol. (Hint: Use the Arrhenius equation: k=Ae−Ea/RT ).
Solution
Step 1: Convert the given temperatures to Kelvin. Given: T1= 25◦C=
25 + 273.15 = 298.15K T2= 50◦C= 50 + 273.15 = 323.15K
Step 2: Write the Arrhenius equation for the rate constant. The Arrhenius
equation is: k=Ae−Ea/RT
Step 3: Determine the ratio of rate constants at the two temperatures.
Given: k1= 2.54 ×10−3s−1k2= 1.63 ×10−2s−1
k2
k1=1.63×10−2
2.54×10−3= 6.417
Step 4: Substitute the known values into the ratio and simplify. k2
k1=
Ae−Ea/RT2
Ae−Ea/RT1=e−Ea/R(1
T2
−1
T1)
We can rewrite this as: ln k2
k1=−Ea
R1
T2
−1
T1
Step 5: Calculate the activation energy Eain kJ/mol. Using the ideal gas
constant R= 8.314J/(mol ·K):
Ea=−ln 6.417
8.314 ×8.314 ×1
323.15 −1
298.15
Ea=−0.792 ×8.314 ×(0.0031) = −0.0207kJ/mol
Therefore, the activation energy for this reaction is approximately 20.7kJ/mol .
31
Question 33
Question
The rate constant of a first-order reaction at 25
°
C is 2.8×10−3s−1. When
the temperature is raised to 45
°
C, the rate constant becomes 6.4×10−2s−1.
Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol*K)), T= temperature (in Kelvin).
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
Ea=−ln(k)−ln(A)
1/T
Step 3: Calculate the activation energy for 25
°
C:
Ea=−ln2.8×10−3−ln(A)
1/(25 + 273.15)
Step 4: Calculate the activation energy for 45
°
C:
Ea=−ln6.4×10−2−ln(A)
1/(45 + 273.15)
Step 5: Set up a system of equations using the two calculated activation
energies:
−ln2.8×10−3−ln(A)
1/(25 + 273.15) =−ln6.4×10−2−ln(A)
1/(45 + 273.15)
Step 6: Solve the system of equations to find the value of A.
Step 7: Calculate the activation energy Eausing the value of Aand any of
the previously calculated activation energies.
Question 34
Question
The rate constant (k) for a first-order reaction is found to be 1.25 ×10−2s−1
at 300 K and 2.50 ×10−2s−1at 325 K. Calculate the activation energy (Ea)
for this reaction. Assume the pre-exponential factor (A) is 8.314 ×108s−1.
32
Solution
Step 1: Start with 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: For the first temperature (300 K):
1.25 ×10−2= 8.314 ×108·e−Ea
8.314·300
Step 3: Solve for Ea:
1.25 ×10−2
8.314 ×108=e−Ea
8.314·300
ln 1.25 ×10−2
8.314 ×108=−Ea
8.314 ·300
Step 4: Calculate the value inside the natural logarithm:
ln 1.25 ×10−2
8.314 ×108≈ln1.504 ×10−11≈ −24.300
Step 5: Solve for Ea:
Ea=−24.300 ×8.314 ·300 ≈6.06 ×104J/mol
Step 6: Repeat Steps 2-5 for the second temperature (325 K), using 2.50 ×
10−2in the Arrhenius equation.
Step 7: Finally, calculate Eausing the two sets of data to get a more accurate
value. The activation energy is the same regardless of temperature.
Question 35
Question
The rate constant for a certain reaction is found to be 4.23 ×10−3s−1at 27
°
C
and 2.64 ×10−2s−1at 37
°
C. Calculate the activation energy of the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(
°
C) + 273.15. At 27
°
C: T1= 27 + 273.15 = 300.15 K At 37
°
C: T2= 37 +
273.15 = 310.15 K
Step 2: Write the Arrhenius equation, which relates the rate constant kto
the temperature T:
k=A·e−Ea
RT
33
where: - kis the rate constant - Ais the pre-exponential factor - Eais the
activation energy - Ris the gas constant (8.314 J/(mol K)) - Tis the temperature
in Kelvin
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
R·1
T
Step 4: Calculate the value of ln kfor both temperatures: At 27
°
C:
ln4.23 ×10−3= ln A−Ea
8.314 ·1
300.15
−5.465 ≈ln A−Ea
2498.511
At 37
°
C:
ln2.64 ×10−2= ln A−Ea
8.314 ·1
310.15
−3.629 ≈ln A−Ea
2581.379
Step 5: Subtract the two equations to eliminate ln Aand solve for Ea:
−3.629 −(−5.465) ≈Ea
2581.379 −Ea
2498.511
1.836 ≈82.869 ·Ea
Ea≈1.836
82.869
Ea≈0.0221 kJ/mol
34
Question 13
Question
The rate constant for a reaction at 25
°
C is 1.20 ×10−3s−1, and the activation
energy for the reaction is 75 kJ/mol. Calculate the rate constant at 40
°
C for
the same reaction.
Solution
Step 1: Calculate the activation energy in joules using the given value in kilo-
joules.
Activation energy (J) = 75 ×103J/mol = 75,000 J/mol
Step 2: Calculate the new rate constant at 40
°
C using the Arrhenius equa-
tion:
k2=k1×e
−Ea
R× 1
T2
−
1
T1!
Given: T1= 25C= 298 K, T2= 40C= 313 K, Ea= 75,000 J/mol, R=
8.314 J/mol·K, k1= 1.20 ×10−3s−1
Step 3: Substitute the known values into the Arrhenius equation and solve
for k2.
k2= 1.20 ×10−3×e −75,000
8.314 × 1
313−
1
298!!
Step 4: Calculate the new rate constant k2.
k2≈3.49 ×10−3s−1
Therefore, the rate constant for the reaction at 40
°
C is approximately 3.49×
10−3s−1.
Question 14
Question
The rate constants (k) for a reaction at different temperatures are given in the
table below:
Temperature (K) k(L/mol s)
300 0.005
310 0.008
320 0.013
Calculate the activation energy (Ea) for this reaction. Assume the pre-
exponential factor (A) is 1 ×109L/mol s and the universal gas constant (R) is
8.314 J/(mol K).
13
Solution
Step 1: Calculate the rate constant at 1st temperature using Arrhenius equation.
Given data: - Temperature at 1st point (T1) = 300 K - Rate constant at 1st
point (k1) = 0.005 L/mol s - Pre-exponential factor (A) = 1 ×109L/mol s -
Universal gas constant (R) = 8.314 J/(mol K)
The Arrhenius equation is given as:
k=A·e−Ea
RT
Substitute the known values to find Ea:
k1=A·e−Ea
R·T1
0.005 = 1 ×109·e−Ea
8.314·300
Step 2: Calculate the rate constant at 2nd and 3rd temperatures.
Following the same process as in Step 1, we can calculate the rate constants
at 2nd and 3rd temperatures.
Step 3: Set up equations for Eausing rate constants and temperatures.
We have 3 equations in the form:
ki=A·e−Ea
RTi
where irepresents the temperature point.
Step 4: Solve the system of equations to find the activation energy (Ea).
Solve the system of equations obtained in Step 3 to find the activation energy
(Ea). Substituting the known values for A,R, and the rate constants at different
temperatures, we can find the value of Ea.
Question 15
Question
The rate constant of a reaction at 25◦C is 1.5×10−3s−1, and at 35◦C it is
4.2×10−3s−1. Calculate the activation energy (Ea) of the reaction in kJ mol−1.
(Hint: Assume the activation energy does not change with temperature.)
Solution
Step 1: We can use the Arrhenius equation to relate the rate constants at two
different temperatures to the activation energy Ea. The Arrhenius equation is
given by:
k=A·exp −Ea
RT
where kis the rate constant, Ais the pre-exponential factor (frequency factor),
Eais the activation energy, Ris the gas constant (8.314 J mol−1K−1), and Tis
the temperature in Kelvin.
14
Step 2: We start by writing the Arrhenius equation for the rate constants
at the two temperatures given:
k1=A·exp −Ea
R·(25 + 273.15)
k2=A·exp −Ea
R·(35 + 273.15)
where k1= 1.5×10−3s−1and k2= 4.2×10−3s−1.
Step 3: Take the ratio of the two equations to eliminate the frequency factor
A:
k2
k1
=
exp −Ea
R·(35+273.15)
exp −Ea
R·(25+273.15)
Step 4: Simplify the equation by taking the natural logarithm of both sides:
ln k2
k1=−Ea
R1
35 + 273.15 −1
25 + 273.15
Step 5: Now, plug in the values for k1,k2, and Rto solve for Ea:
ln 4.2×10−3
1.5×10−3=−Ea
8.314 1
35 + 273.15 −1
25 + 273.15
ln(2.8) = −Ea
8.314 1
308.15 −1
298.15
Step 6: Calculate the activation energy Eain Joules and convert it to
kJ mol−1:
Ea=−8.314 ×10−3×ln(2.8)
1
308.15 −1
298.15
Question 16
Question
For a certain reaction, the rate constant at 25
°
C is 3.21 ×10−2s−1, and at 35
°
C
it is 8.54 ×10−2s−1. Calculate the activation energy for this reaction.
Solution
Step 1: Write 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.
15
Step 2: Convert the given temperatures into Kelvin: - For 25
°
C, T= 25 +
273 = 298 K. - For 35
°
C, T= 35 + 273 = 308 K.
Step 3: Substitute the given rate constants and temperatures into the Ar-
rhenius equation to form two equations:
3.21 ×10−2=A e−Ea
8.314×298
8.54 ×10−2=A e−Ea
8.314×308
Step 4: Divide the two equations to eliminate A:
3.21 ×10−2
8.54 ×10−2=e−Ea
8.314×298
e−Ea
8.314×308
Step 5: Simplify the equation:
3.21
8.54 =eEa
8.314 (1
298 −1
308 )
Step 6: Solve for Eausing the natural logarithm:
ln 3.21
8.54=Ea
8.314 1
298 −1
308
Step 7: Calculate the activation energy, Ea, using the calculated logarithm
value:
Ea=−8.314 ×ln 3.21
8.54 1
298 −1
308
Step 8: Perform the calculation to find the activation energy for the reaction.
Question 17
Question
For a certain reaction, the rate constant at 25
°
C is 1.5×10−3s−1and at 45
°
C
it is 7.3×10−3s−1. Calculate the activation energy of the reaction.
Solution
Step 1: We can use the Arrhenius equation to relate the rate constant at different
temperatures:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/(mol·K)), and - Tis the
temperature in Kelvin.
16
Step 2: Let’s first write the equations for the rate constants at 25
°
C and
45
°
C: At 25
°
C (298 K):
1.5×10−3=A·e−Ea
8.314·298
At 45
°
C (318 K):
7.3×10−3=A·e−Ea
8.314·318
Step 3: Now, let’s divide the equation for 45
°
C by the equation for 25
°
C to
eliminate A:7.3×10−3
1.5×10−3=e−Ea
8.314 (1
318 −1
298 )
Step 4: Solve for the activation energy Ea:
7.3×10−3
1.5×10−3=e−Ea
8.314 (1
318 −1
298 )
4.87 = e−Ea
8.314 (1
318 −1
298 )
Step 5: Taking the natural logarithm of both sides to solve for Ea:
ln(4.87) = −Ea
8.314 1
318 −1
298
Step 6: Calculate Ea:
Ea=−8.314 ×ln(4.87)
1
318 −1
298
Ea≈34.5 kJ/mol
Step 7: Therefore, the activation energy of the reaction is approximately
34.5 kJ/mol.
Question 18
Question
The rate constant for a reaction at 25
°
C is 3.2×10−3s−1, and the activation
energy for the reaction is 85 kJ/mol. Calculate the rate constant at 75
°
C for
the same reaction.
Solution
Step 1: Convert the activation energy to joules: Given: Activation energy,
Ea= 85 kJ/mol
Convert 85 kJ/mol to Joules:
85 kJ/mol ×1000 J
1 kJ = 85000 J/mol
17
Step 2: Use the Arrhenius equation to calculate the rate constant at 75
°
C:
The Arrhenius equation is given by:
k=A·e−Ea
RT
where: k= rate constant at 75
°
CA= pre-exponential factor Ea= activation
energy in J/mol R= gas constant = 8.314 J/(mol
·
K) T= temperature in Kelvin
Given: T1= 25C= 298 K (reference temperature) T2= 75C= 348 K (new
temperature) k1= 3.2×10−3s−1(rate constant at 25
°
C) Ea= 85000 J/mol
First, we need to determine the pre-exponential factor Ausing the rate
constant at 25
°
C:
3.2×10−3=A·e−85000
8.314·298
Step 3: Solve for A:
A= 3.2×10−3·e85000
8.314·298
A≈697149 s−1
Step 4: Now, calculate the rate constant at 75
°
C:
k= 697149 ·e−85000
8.314·348
Step 5: Solve for k:
k≈8.20 s−1
Therefore, the rate constant at 75
°
C for the reaction is approximately 8.20s−1.
Question 19
Question
The rate constant for a reaction was found to be 4.26 ×10−3s−1at 25◦C and
1.29 ×10−2s−1at 45◦C. Calculate the activation energy for this reaction.
Solution
Step 1: Convert the given temperatures to Kelvin.
T1= 25◦C = 25 + 273.15 = 298.15 K
T2= 45◦C = 45 + 273.15 = 318.15 K
Step 2: Use the Arrhenius equation to relate the rate constants at the two
temperatures: k2
k1
=e−Ea
R1
T2
−1
T1
Step 3: Plug in the given rate constants and temperatures into the equation
to solve for the activation energy (Ea).
1.29 ×10−2
4.26 ×10−3=e(−Ea
8.314 (1
318.15 −1
298.15 ))
18
Step 4: Simplify and solve for the activation energy:
3.02 = e(−Ea
8.314 (1
318.15 −1
298.15 ))
Step 5: Take the natural logarithm of both sides to isolate the activation
energy:
ln(3.02) = −Ea
8.314 1
318.15 −1
298.15
Step 6: Solve for the activation energy Ea:
Ea=−8.314 ×ln(3.02)
1
318.15 −1
298.15
Step 7: Calculate the activation energy using the values from Step 6.
Question 20
Question
For the reaction A →B, the rate constant at 25
°
C is 5.0×10−3s−1. When
the temperature is increased to 50
°
C, the rate constant becomes 1.0×10−2s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol
·
K)), - Tis the temperature
in Kelvin.
Step 2: Convert the temperatures to Kelvin: - 25
°
C = 298 K - 50
°
C = 323
K
Step 3: Write down the Arrhenius equation at 25
°
C and 50
°
C:
k1=A·e−Ea
8.314×298
k2=A·e−Ea
8.314×323
Step 4: Divide the two equations to eliminate A:
k2
k1
=e−Ea
8.314×323
e−Ea
8.314×298
Step 5: Simplify the equation:
k2
k1
=e−Ea
8.314 (1
323 −1
298 )
19
Step 6: Solve for Ea:
ln k2
k1=−Ea
8.314 1
323 −1
298
Ea=−8.314 ×1
323 −1
298×ln 1.0×10−2
5.0×10−3
Step 7: Calculate the activation energy:
Ea=−8.314 ×1
323 −1
298×ln 1.0×10−2
5.0×10−3
Ea≈70.6 kJ/mol
Therefore, the activation energy for the reaction is approximately 70.6 kJ/mol.
Question 21
Question
The rate constant of a reaction at 50
°
C is 0.0056 s−1, while at 75
°
C it is 0.046
s−1. Calculate the activation energy for this reaction. (R = 8.314 J/(mol·K))
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea/(RT )
where: k= rate constant, A= pre-exponential factor, Ea= activation energy
(J/mol), R= gas constant (J/(mol·K)), T= temperature (K).
Step 2: Use the given data to set up two equations: For 50
°
C:
0.0056 = A·e−Ea/(8.314·(273+50))
0.0056 = A·e−Ea/3867.3
For 75
°
C:
0.046 = A·e−Ea/(8.314·(273+75))
0.046 = A·e−Ea/4097.05
Step 3: Divide the two equations to eliminate A:
0.046
0.0056 =e−Ea/4097.05
e−Ea/3867.3
8.2143 = e(3867.3−4097.05)/2295458.1
20
8.2143 = e(−229.75)/2295458.1
Step 4: Solve for the activation energy Ea:
8.2143 = e−0.0001
8.2143 = 0.9999
ln(8.2143) = ln(0.9999)
ln(8.2143) = −0.0001
Ea=−0.0001 ×8.314 ×3867.3 = 32.0 kJ/mol
Therefore, the activation energy for this reaction is 32.0 kJ/mol.
Question 22
Question
The rate constant (k) for a reaction is found to be 4.23 ×10−3s−1at 60◦C
and 2.17 ×10−2s−1at 70◦C. Calculate the activation energy (Ea) for this
reaction. (Hint: Use the Arrhenius equation: k=A·e−Ea
RT , where Ais the
pre-exponential factor, Eais the activation energy, Ris the gas constant, and
Tis the temperature in Kelvin.)
Solution
Step 1: Convert the temperatures to Kelvin. Given: - Temperature at 60◦C =
60 + 273 = 333 K - Temperature at 70◦C = 70 + 273 = 343 K
Step 2: Write the Arrhenius equation for both sets of data. At 60◦C (333
K): k1=A·e−Ea
RT1= 4.23 ×10−3s−1
At 70◦C (343 K): k2=A·e−Ea
RT2= 2.17 ×10−2s−1
Step 3: Take the ratio of the two rate constants to eliminate A.k2
k1=A·e−Ea
RT2
A·e−Ea
RT1
⇒k2
k1=e
Ea
R1
T1
−1
T2
Step 4: Solve for the activation energy (Ea). Plugging in the values: 2.17×10−2
4.23×10−3=
eEa
8.314 (1
333 −1
343 )
⇒5.12 = eEa
8.314 (1
333 −1
343 )
Step 5: Solve for Ea. Taking the natural logarithm of both sides: ln(5.12) =
Ea
8.314 1
333 −1
343
⇒Ea= 8.314 ·ln(5.12) 1
333 −1
343
Therefore, the activation energy for this reaction is approximately 38.87 kJ/mol.
21
Question 23
Question
The rate constant for a certain reaction is measured at two different tempera-
tures, yielding the following data:
Temperature (K) Rate Constant (s−1)
300 1.2×10−3
500 2.5×10−2
Calculate the activation energy (Ea) for this reaction in kJ/mol. Assume
the pre-exponential factor (A) is 1 ×1012 s−1.
Solution
Step 1: 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 (K).
Step 2: We can rearrange the Arrhenius equation to solve for activation
energy (Ea):
Ea=−ln k
A
1/T
Step 3: Let’s calculate the activation energy using the data provided at 300
K:
Ea=−
ln 1.2×10−3
1×1012
1/300
Step 4: Calculating inside the natural logarithm:
Ea=−ln 1.2×10−15
1/300
Step 5: Simplifying the natural logarithm:
Ea=−ln 1.2−ln 10−15
1/300 =−
−0.1823 −(−34.54)
300
Ea=−34.3577
300 = 114.5 kJ/mol (rounded to 3 decimal places)
Therefore, the activation energy (Ea) for this reaction is 114.5 kJ/mol.
22
Question 24
Question
The rate constant for a certain reaction is 4.55 ×10−3s−1at 25◦C and 8.60 ×
10−3s−1at 50◦C. Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: First, recall 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: We can write two Arrhenius equations for the given temperatures:
k1=Ae−Ea
R(25+273.15)
k2=Ae−Ea
R(50+273.15)
Step 3: Divide the two equations to eliminate A:
k2
k1
=e−Ea
R(50+273.15)
e−Ea
R(25+273.15)
Step 4: Simplifying the equation gives:
k2
k1
=eEa
R(1
25+273.15 −1
50+273.15 )
Step 5: Plug in the given values for k1and k2, then solve for Ea:
8.60 ×10−3
4.55 ×10−3=eEa
8.314 (1
298.15 −1
323.15 )
Step 6: Solve for Eato get:
Ea=−8.314 ×ln 8.60 ×10−3
4.55 ×10−3 1
298.15 −1
323.15
Step 7: Calculate the activation energy to find the answer in kJ/mol.
Question 25
Question
For the reaction 2A+B→C, the activation energy is 120 kJ/mol and the
frequency factor is 1.5×1011s−1. At a temperature of 400 K, the rate constant
is 3.2×10−5M−1s−1. Calculate the rate constant at a temperature of 450 K.
23
Solution
Step 1: Calculate the rate constant at 400 K using the Arrhenius equation.
k1=A·e−Ea
RT
where: k1= rate constant at 400 K A= frequency factor (1.5×1011s−1)Ea
= activation energy (120 kJ/mol) R= gas constant (8.314 J/(mol*K)) T=
temperature in Kelvin (400 K)
Substitute the given values into the equation to solve for k1:
k1= 1.5×1011 exp −120 ×103
8.314 ×400
k1≈3.23 ×10−5M−1s−1
Step 2: Calculate the rate constant at 450 K using the Arrhenius equation.
k2=A·e−Ea
RT
where: k2= rate constant at 450 K A= frequency factor (1.5×1011s−1)Ea
= activation energy (120 kJ/mol) R= gas constant (8.314 J/(mol*K)) T=
temperature in Kelvin (450 K)
Substitute the values into the equation to find k2:
k2= 1.5×1011 exp −120 ×103
8.314 ×450
k2≈1.34 ×10−4M−1s−1
Therefore, the rate constant at 450K is approximately 1.34 ×10−4M−1s−1.
Question 26
Question
The rate constant for a first-order reaction is 4.15×10−3s−1at 25◦C and 1.24×
10−2s−1at 50◦C. Calculate the activation energy for the reaction.
Solution
Step 1: Convert the temperatures to Kelvin. Given: T1= 25◦C = 298K T2=
50◦C = 323K
Step 2: Write 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−1K−1T= temperature in Kelvin
24
Step 3: Write the Arrhenius equation for the two given temperatures:
k1=Ae−Ea
R·T1
k2=Ae−Ea
R·T2
Step 4: Divide the two equations to eliminate A:
k1
k2
=e−Ea
R·T1
e−Ea
R·T2
Step 5: Simplify the equation:
k1
k2
=e(Ea
R)1
T1
−1
T2
Step 6: Plug in the given values and solve for Ea:
4.15 ×10−3
1.24 ×10−2=e(Ea
8.314 )( 1
298 −1
323 )
Step 7: Solve for Ea:
4.15
1.24 =e(Ea
8.314 )(323−298
298·323 )
3.35 = e(Ea
8.314 )( 25
298·323 )
Step 8: Solve for Eaby taking natural logarithm of both sides:
ln(3.35) = Ea
8.314 25
298 ·323
Step 9: Calculate the activation energy Ea:
Ea= 8.314 ×ln(3.35)
25
298·323
≈60.2 kJ/mol
Therefore, the activation energy for the reaction is approximately 60.2 kJ/mol.
Question 27
Question
The rate constant (k) for a reaction is known to double for every 10
°
C rise
in temperature. If the rate constant at 30
°
C is 5.0×10−3s−1, what is the
activation energy (Ea) for this reaction? (The universal gas constant R= 8.314
J/(mol·K))
25
Solution
Step 1: First, we can represent the relationship between the rate constant and
temperature using the Arrhenius equation:
k=A·e−Ea
RT
where: k= rate constant, A= pre-exponential factor, and Ea= activation
energy.
Step 2: Given that the rate constant doubles for every 10
°
C rise in temper-
ature, we can express this relationship as:
k2= 2k1
where: k2= rate constant at temperature T2,k1= rate constant at temperature
T1.
Step 3: We are given the rate constant at 30
°
C as 5.0×10−3s−1, which
means:
k1= 5.0×10−3s−1
Step 4: Let’s denote the final temperature as T2= 30 + 10 = 40
°
C. Thus,
the rate constant at 40
°
C is:
k2= 2k1= 2 ×5.0×10−3= 1.0×10−2s−1
Step 5: Now, we can substitute the values of k1,k2,T1, and T2into the
Arrhenius equation to solve for the activation energy (Ea):
k2
k1
=A·e−Ea
R·T2
A·e−Ea
R·T1
Step 6: Simplifying the above equation, we get:
1.0×10−2
5.0×10−3=e−Ea
8.314·313.15
e−Ea
8.314·303.15
Step 7: Further simplifying, we find:
2 = e0.0032Ea
Step 8: Taking the natural logarithm of both sides, we get:
ln(2) = 0.0032Ea
Step 9: Finally, solving for Eagives:
Ea=ln(2)
0.0032 ≈0.6931
0.0032 ≈216.59 kJ/mol
Therefore, the activation energy (Ea) for this reaction is approximately
216.59 kJ/mol.
26
Question 28
Question
The rate constant for the decomposition of acetaldehyde at a certain temper-
ature is 3.2×10−3s−1. When the temperature is increased by 10
°
C, the rate
constant becomes 6.8×10−2s−1. Calculate the activation energy for this reac-
tion.
Solution
Step 1: Convert the temperature change to Kelvin.
∆T= 10
°
C = 10 K
Step 2: Apply the Arrhenius equation, which relates the rate constant (k)
to the activation energy (Ea) and temperature (T):
k=A·e−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 3: Set up two equations based on the given information:
3.2×10−3=A·e−Ea
R(T)
6.8×10−2=A·e−Ea
R(T+10)
Step 4: Divide the two equations to eliminate A.
3.2×10−3
6.8×10−2=A·e−Ea
RT
A·e−Ea
R(T+10)
3.2
68 =eEa
R−Ea
R(T+10)
8
170 =eEa
R−Ea
R(T+10)
27
Step 5: Solve for the activation energy (Ea).
ln 8
170=Ea
R−Ea
R(T+ 10)
ln 8
170=Ea1
R−1
R(T+ 10)
ln 8
170=EaT+ 10 −T
R(T+ 10)
ln 8
170=Ea10
R(T+ 10)
Ea=R·10
ln 8
170
Ea= 8.314 J/mol ·10
ln 8
170
Ea≈56.56 kJ/mol
Therefore, the activation energy for the reaction is approximately 56.56
kJ/mol.
Question 29
Question
The rate constant of a certain reaction doubles when the temperature is raised
from 20◦C to 50◦C. Calculate the activation energy for this reaction. Assume
the activation energy is constant over this temperature range.
Solution
Step 1: Write down the Arrhenius equation.
ln k2
k1=Ea
R1
T1
−1
T2
Step 2: Plug in the given values:
ln 2k1
k1=Ea
R1
293 −1
323
]
Step 3: Simplify the equation:
ln(2) = Ea
8.314 1
293 −1
323
]
28
Step 4: Solve for the activation energy Ea:
0.693 = Ea
8.314 323 −293
293 ×323
Step 5: Calculate the activation energy:
Ea= 0.693 ×8.314 ×293 ×323
323 −293 ≈47.5 kJ/mol
Therefore, the activation energy for this reaction is approximately 47.5
kJ/mol.
Question 30
Question
The rate constant of a certain reaction at 25
°
C is 1.20 ×10−3s−1, and at 50
°
C,
the rate constant is 8.50×10−3s−1. Calculate the activation energy (in kJ/mol)
for this reaction.
Solution
Step 1: Recall the Arrhenius equation, which relates the rate constant of a
reaction to its temperature:
k=Ae−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the
activation energy, - Ris the gas constant (8.314 J/mol-K), and - Tis the tem-
perature in Kelvin.
Step 2: We are given two sets of data points to plug into the Arrhenius
equation. We can set up two equations:
k1=Ae−Ea
R·298
k2=Ae−Ea
R·323
Step 3: Substitute the given rate constants and convert the temperatures to
Kelvin:
k1= 1.20 ×10−3=Ae−Ea
8.314·298
k2= 8.50 ×10−3=Ae−Ea
8.314·323
Step 4: Divide the two equations to eliminate A:
k2
k1
=8.50 ×10−3
1.20 ×10−3=e−Ea
8.314·323
e−Ea
8.314·298
29
Step 5: Take the natural logarithm of both sides to eliminate the exponential
terms:
ln 8.50 ×10−3
1.20 ×10−3=−Ea
8.314 1
323 −1
298
Step 6: Solve for Eato find the activation energy of the reaction in Joules.
Convert to kilojoules by dividing by 1000.
Ea=−8.314 ×ln 8.50 ×10−3
1.20 ×10−3×1
323 −1
298
Step 7: Calculate the activation energy:
Ea=
−8.314 ×ln 8.50×10−3
1.20×10−3
323 −298 ×1000
Therefore, the activation energy for this reaction is approximately 76.3
kJ/mol.
Question 31
Question
For a certain reaction, the rate constant at 25
°
C is 8.0×10−4s−1. When the
temperature is increased to 50
°
C, the rate constant becomes 1.6×10−2s−1.
Calculate the activation energy for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=A·e−Ea
RT
where: - kis the rate constant, - Ais the pre-exponential factor, - Eais the acti-
vation energy, - Ris the gas constant (8.314 J/(mol·K)), - Tis the temperature
in Kelvin.
Step 2: Determine the rate constant at both temperatures in terms of the
Arrhenius equation: At 25
°
C (298 K):
8.0×10−4=A·e−Ea
8.314×298
At 50
°
C (323 K):
1.6×10−2=A·e−Ea
8.314×323
Step 3: Divide the two rate constant expressions to eliminate A:
1.6×10−2
8.0×10−4=e−Ea
8.314×323
e−Ea
8.314×298
30
Step 4: Simplify the equation:
20 = eEa
8.314 (1
298 −1
323 )
Step 5: Solve for Ea:
ln(20) = Ea
8.314 1
298 −1
323
Ea= 8.314 ×ln(20)
1
298 −1
323
Step 6: Calculate the activation energy:
Ea= 8.314 ×ln(20)
1
298 −1
323
≈44.6 kJ/mol
Question 32
Question
The rate constant for a certain reaction is found to be 2.54 ×10−3s−1at 25◦C
and 1.63 ×10−2s−1at 50◦C. Calculate the activation energy for this reaction
in kJ/mol. (Hint: Use the Arrhenius equation: k=Ae−Ea/RT ).
Solution
Step 1: Convert the given temperatures to Kelvin. Given: T1= 25◦C=
25 + 273.15 = 298.15K T2= 50◦C= 50 + 273.15 = 323.15K
Step 2: Write the Arrhenius equation for the rate constant. The Arrhenius
equation is: k=Ae−Ea/RT
Step 3: Determine the ratio of rate constants at the two temperatures.
Given: k1= 2.54 ×10−3s−1k2= 1.63 ×10−2s−1
k2
k1=1.63×10−2
2.54×10−3= 6.417
Step 4: Substitute the known values into the ratio and simplify. k2
k1=
Ae−Ea/RT2
Ae−Ea/RT1=e−Ea/R(1
T2
−1
T1)
We can rewrite this as: ln k2
k1=−Ea
R1
T2
−1
T1
Step 5: Calculate the activation energy Eain kJ/mol. Using the ideal gas
constant R= 8.314J/(mol ·K):
Ea=−ln 6.417
8.314 ×8.314 ×1
323.15 −1
298.15
Ea=−0.792 ×8.314 ×(0.0031) = −0.0207kJ/mol
Therefore, the activation energy for this reaction is approximately 20.7kJ/mol .
31
Question 33
Question
The rate constant of a first-order reaction at 25
°
C is 2.8×10−3s−1. When
the temperature is raised to 45
°
C, the rate constant becomes 6.4×10−2s−1.
Calculate the activation energy (in kJ/mol) for this reaction.
Solution
Step 1: Write down the Arrhenius equation:
k=Ae−Ea
RT
where: k= rate constant, A= pre-exponential factor, Ea= activation energy,
R= gas constant (8.314 J/(mol*K)), T= temperature (in Kelvin).
Step 2: Rearrange the Arrhenius equation to solve for activation energy Ea:
Ea=−ln(k)−ln(A)
1/T
Step 3: Calculate the activation energy for 25
°
C:
Ea=−ln2.8×10−3−ln(A)
1/(25 + 273.15)
Step 4: Calculate the activation energy for 45
°
C:
Ea=−ln6.4×10−2−ln(A)
1/(45 + 273.15)
Step 5: Set up a system of equations using the two calculated activation
energies:
−ln2.8×10−3−ln(A)
1/(25 + 273.15) =−ln6.4×10−2−ln(A)
1/(45 + 273.15)
Step 6: Solve the system of equations to find the value of A.
Step 7: Calculate the activation energy Eausing the value of Aand any of
the previously calculated activation energies.
Question 34
Question
The rate constant (k) for a first-order reaction is found to be 1.25 ×10−2s−1
at 300 K and 2.50 ×10−2s−1at 325 K. Calculate the activation energy (Ea)
for this reaction. Assume the pre-exponential factor (A) is 8.314 ×108s−1.
32
Solution
Step 1: Start with 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: For the first temperature (300 K):
1.25 ×10−2= 8.314 ×108·e−Ea
8.314·300
Step 3: Solve for Ea:
1.25 ×10−2
8.314 ×108=e−Ea
8.314·300
ln 1.25 ×10−2
8.314 ×108=−Ea
8.314 ·300
Step 4: Calculate the value inside the natural logarithm:
ln 1.25 ×10−2
8.314 ×108≈ln1.504 ×10−11≈ −24.300
Step 5: Solve for Ea:
Ea=−24.300 ×8.314 ·300 ≈6.06 ×104J/mol
Step 6: Repeat Steps 2-5 for the second temperature (325 K), using 2.50 ×
10−2in the Arrhenius equation.
Step 7: Finally, calculate Eausing the two sets of data to get a more accurate
value. The activation energy is the same regardless of temperature.
Question 35
Question
The rate constant for a certain reaction is found to be 4.23 ×10−3s−1at 27
°
C
and 2.64 ×10−2s−1at 37
°
C. Calculate the activation energy of the reaction.
Solution
Step 1: Convert the given temperatures to Kelvin using the formula T(K) =
T(
°
C) + 273.15. At 27
°
C: T1= 27 + 273.15 = 300.15 K At 37
°
C: T2= 37 +
273.15 = 310.15 K
Step 2: Write the Arrhenius equation, which relates the rate constant kto
the temperature T:
k=A·e−Ea
RT
33
where: - kis the rate constant - Ais the pre-exponential factor - Eais the
activation energy - Ris the gas constant (8.314 J/(mol K)) - Tis the temperature
in Kelvin
Step 3: Take the natural logarithm of the Arrhenius equation to linearize it:
ln k= ln A−Ea
R·1
T
Step 4: Calculate the value of ln kfor both temperatures: At 27
°
C:
ln4.23 ×10−3= ln A−Ea
8.314 ·1
300.15
−5.465 ≈ln A−Ea
2498.511
At 37
°
C:
ln2.64 ×10−2= ln A−Ea
8.314 ·1
310.15
−3.629 ≈ln A−Ea
2581.379
Step 5: Subtract the two equations to eliminate ln Aand solve for Ea:
−3.629 −(−5.465) ≈Ea
2581.379 −Ea
2498.511
1.836 ≈82.869 ·Ea
Ea≈1.836
82.869
Ea≈0.0221 kJ/mol
34
