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For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:
- a)45.2 JK–1 mol–1
- b)45.2 KJ K–1mol–1
- c)–45.2 JK–1 mol–1
- d)–45.2 KJ K–1mol–1
Correct answer is option 'D'. Can you explain this answer?
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For the first order isomerization of an organic compound at 1300C, the...
Calculation of Standard Entropy of Activation
To calculate the standard entropy of activation, we need to use the Arrhenius equation, which relates the rate constant of a reaction to the activation energy, temperature, and Boltzmann constant. The equation is given as:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
Determination of Boltzmann Constant
The value of the Boltzmann constant is 1.38 x 10^-23 JK^-1. We can use this value to calculate the standard entropy of activation.
Calculation of Temperature in Kelvin
The temperature given in the problem is 1300C. We need to convert it to Kelvin by adding 273.15. Therefore, T = 1573.15 K.
Determination of Pre-exponential Factor
The pre-exponential factor is not given in the problem. However, we can calculate it using the rate constant and activation energy. The equation is given as:
A = k / e^(-Ea/RT)
Substituting the values, we get:
A = 9.12 x 10^4 s^-1 / e^(-108.4 x 10^3 J mol^-1 / (8.314 J K^-1 mol^-1 x 1573.15 K))
A = 2.79 x 10^14 s^-1
Calculation of Standard Entropy of Activation
The standard entropy of activation is given by the equation:
ΔS‡ = (Ea / R) - ln(A/kB)
where ΔS‡ is the standard entropy of activation, R is the gas constant, A is the pre-exponential factor, k is the rate constant, and kB is the Boltzmann constant.
Substituting the values, we get:
ΔS‡ = (108.4 x 10^3 J mol^-1 / 8.314 J K^-1 mol^-1) - ln(2.79 x 10^14 s^-1 / 1.38 x 10^-23 JK^-1 x 1573.15 K)
ΔS‡ = 45.2 kJ K^-1 mol^-1
Therefore, the correct answer is option 'D', which states that the standard entropy of activation for this reaction is 45.2 kJ K^-1 mol^-1.
To calculate the standard entropy of activation, we need to use the Arrhenius equation, which relates the rate constant of a reaction to the activation energy, temperature, and Boltzmann constant. The equation is given as:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
Determination of Boltzmann Constant
The value of the Boltzmann constant is 1.38 x 10^-23 JK^-1. We can use this value to calculate the standard entropy of activation.
Calculation of Temperature in Kelvin
The temperature given in the problem is 1300C. We need to convert it to Kelvin by adding 273.15. Therefore, T = 1573.15 K.
Determination of Pre-exponential Factor
The pre-exponential factor is not given in the problem. However, we can calculate it using the rate constant and activation energy. The equation is given as:
A = k / e^(-Ea/RT)
Substituting the values, we get:
A = 9.12 x 10^4 s^-1 / e^(-108.4 x 10^3 J mol^-1 / (8.314 J K^-1 mol^-1 x 1573.15 K))
A = 2.79 x 10^14 s^-1
Calculation of Standard Entropy of Activation
The standard entropy of activation is given by the equation:
ΔS‡ = (Ea / R) - ln(A/kB)
where ΔS‡ is the standard entropy of activation, R is the gas constant, A is the pre-exponential factor, k is the rate constant, and kB is the Boltzmann constant.
Substituting the values, we get:
ΔS‡ = (108.4 x 10^3 J mol^-1 / 8.314 J K^-1 mol^-1) - ln(2.79 x 10^14 s^-1 / 1.38 x 10^-23 JK^-1 x 1573.15 K)
ΔS‡ = 45.2 kJ K^-1 mol^-1
Therefore, the correct answer is option 'D', which states that the standard entropy of activation for this reaction is 45.2 kJ K^-1 mol^-1.
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For the first order isomerization of an organic compound at 1300C, the...
For the first order isomerization of an organic compoundat 130°c,the activation energy is 108.4KJ mol-1 and the rate constant is 9.12×10^-4s-1. calculate the standard entropy of activation for this reaction. I know the answer is
- 45.2 JK–1 mol–1 .
- 45.2 JK–1 mol–1 .
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For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer?
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For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer?.
For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer?.
Solutions for For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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ample number of questions to practice For the first order isomerization of an organic compound at 1300C, the activation energy is 108.4 KJ mol–1 and the rate constant is 9.12 x 10–4 s–1. The standard entropy of activation for this reaction is:a)45.2 JK–1 mol–1b)45.2 KJ K–1mol–1c)–45.2 JK–1 mol–1d)–45.2 KJ K–1mol–1Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Chemistry tests.
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