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The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4 s-1, 104.4 kJ mol-1 and 6.0 × 10-4s-1respectively. The value of the rate constant at T → ∞ is
- a)2.0 × 1018 s-1
- b)6.0 × 1014 s-1
- c)infinity
- d)6 × 10-4 s-1
Correct answer is option 'D'. Can you explain this answer?
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The rate constant, the activation energy and the Arrhenius parameter (...
At temperature = ∞
Rate constant = Arrhenious constant.
Rate constant = Arrhenious constant.
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The rate constant, the activation energy and the Arrhenius parameter (...
Given Information:
- Rate constant at 25°C = 3.0 x 10^-4 s^-1
- Activation energy = 104.4 kJ/mol
- Arrhenius parameter (A) = 6.0 x 10^-4 s^-1
Calculating the rate constant at T:
The Arrhenius equation relates the rate constant (k) of a chemical reaction to its activation energy (Ea), the universal gas constant (R), and the temperature (T):
k = A * exp(-Ea/RT)
where:
- A is the Arrhenius parameter
- R is the universal gas constant (8.314 J/mol*K)
To calculate the rate constant at a different temperature (T), we need to plug in the given values and solve for k:
k(T) = A * exp(-Ea/RT)
We are not given a specific temperature (T), so let's assume T = 298 K (25°C). Plugging in the given values, we get:
k(T) = (6.0 x 10^-4 s^-1) * exp(-104400 J/mol / (8.314 J/mol*K * 298 K))
k(T) = 6.0 x 10^-4 s^-1 * exp(-35.14)
k(T) = 6.0 x 10^-4 s^-1 * 3.9 x 10^-16
k(T) = 2.34 x 10^-19 s^-1
Therefore, the rate constant at T (25°C) is 6.0 x 10^-4 s^-1. The correct answer is option D.
- Rate constant at 25°C = 3.0 x 10^-4 s^-1
- Activation energy = 104.4 kJ/mol
- Arrhenius parameter (A) = 6.0 x 10^-4 s^-1
Calculating the rate constant at T:
The Arrhenius equation relates the rate constant (k) of a chemical reaction to its activation energy (Ea), the universal gas constant (R), and the temperature (T):
k = A * exp(-Ea/RT)
where:
- A is the Arrhenius parameter
- R is the universal gas constant (8.314 J/mol*K)
To calculate the rate constant at a different temperature (T), we need to plug in the given values and solve for k:
k(T) = A * exp(-Ea/RT)
We are not given a specific temperature (T), so let's assume T = 298 K (25°C). Plugging in the given values, we get:
k(T) = (6.0 x 10^-4 s^-1) * exp(-104400 J/mol / (8.314 J/mol*K * 298 K))
k(T) = 6.0 x 10^-4 s^-1 * exp(-35.14)
k(T) = 6.0 x 10^-4 s^-1 * 3.9 x 10^-16
k(T) = 2.34 x 10^-19 s^-1
Therefore, the rate constant at T (25°C) is 6.0 x 10^-4 s^-1. The correct answer is option D.
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The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer?
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The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer?.
The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer?.
Solutions for The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10-4s-1, 104.4 kJ mol-1and 6.0 × 10-4s-1respectively. The value of the rate constant at T →∞ isa)2.0 × 1018s-1b)6.0 × 1014s-1c)infinityd)6 × 10-4s-1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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