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The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:
(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)
Correct answer is '54'. Can you explain this answer?
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The rate of a reaction doubles when its temperature changes from 300 ...
≈ 54 KJ mol-1
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The rate of a reaction doubles when its temperature changes from 300 ...
Given information:
- The rate of a reaction doubles when the temperature changes from 300 K to 310 K.
- Activation energy (Ea) = ?
- R = 8.314 J K-1 mol-1
- log 2 = 0.301
Formula:
The rate constant (k) of a reaction can be calculated using the Arrhenius equation:
k = A * e(-Ea/RT)
where:
- k is the rate constant
- A is the pre-exponential factor or the frequency factor
- Ea is the activation energy
- R is the gas constant
- T is the temperature in Kelvin
Approach:
1. Use the given information to find the ratio of rate constants (k2/k1) at two different temperatures.
2. Use the Arrhenius equation to express the ratio of rate constants in terms of temperature and activation energy.
3. Solve the equation to find the value of activation energy (Ea).
Solution:
Let's assume the rate constant at 300 K is k1 and at 310 K is k2.
Step 1: Finding the ratio of rate constants
The rate of a reaction is directly proportional to the rate constant. Therefore, the ratio of rate constants can be calculated as follows:
k2/k1 = 2
Step 2: Expressing the ratio of rate constants in terms of temperature and activation energy
Using the Arrhenius equation, we have:
k2/k1 = e(-Ea/R) * (1/T2 - 1/T1)
Substituting the given values:
2 = e(-Ea/8.314) * (1/310 - 1/300)
Step 3: Solving the equation to find the activation energy (Ea)
Taking the natural logarithm of both sides of the equation:
ln(2) = -Ea/8.314 * (1/310 - 1/300)
Rearranging the equation to solve for Ea:
Ea = -8.314 * ln(2) / (1/310 - 1/300)
Ea ≈ 54 kJ mol-1 (rounded to the nearest integer)
Conclusion:
The activation energy (Ea) of the reaction is approximately 54 kJ mol-1.
- The rate of a reaction doubles when the temperature changes from 300 K to 310 K.
- Activation energy (Ea) = ?
- R = 8.314 J K-1 mol-1
- log 2 = 0.301
Formula:
The rate constant (k) of a reaction can be calculated using the Arrhenius equation:
k = A * e(-Ea/RT)
where:
- k is the rate constant
- A is the pre-exponential factor or the frequency factor
- Ea is the activation energy
- R is the gas constant
- T is the temperature in Kelvin
Approach:
1. Use the given information to find the ratio of rate constants (k2/k1) at two different temperatures.
2. Use the Arrhenius equation to express the ratio of rate constants in terms of temperature and activation energy.
3. Solve the equation to find the value of activation energy (Ea).
Solution:
Let's assume the rate constant at 300 K is k1 and at 310 K is k2.
Step 1: Finding the ratio of rate constants
The rate of a reaction is directly proportional to the rate constant. Therefore, the ratio of rate constants can be calculated as follows:
k2/k1 = 2
Step 2: Expressing the ratio of rate constants in terms of temperature and activation energy
Using the Arrhenius equation, we have:
k2/k1 = e(-Ea/R) * (1/T2 - 1/T1)
Substituting the given values:
2 = e(-Ea/8.314) * (1/310 - 1/300)
Step 3: Solving the equation to find the activation energy (Ea)
Taking the natural logarithm of both sides of the equation:
ln(2) = -Ea/8.314 * (1/310 - 1/300)
Rearranging the equation to solve for Ea:
Ea = -8.314 * ln(2) / (1/310 - 1/300)
Ea ≈ 54 kJ mol-1 (rounded to the nearest integer)
Conclusion:
The activation energy (Ea) of the reaction is approximately 54 kJ mol-1.
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The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)Correct answer is '54'. Can you explain this answer?
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The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)Correct answer is '54'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)Correct answer is '54'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)Correct answer is '54'. Can you explain this answer?.
The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)Correct answer is '54'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)Correct answer is '54'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy ( in kJ mol-1) of such a reaction will be:(R = 8.314 JK-1 mol-1 and log 2 = 0.301) (Nearest integer)Correct answer is '54'. Can you explain this answer?.
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