JEE Advanced · Chemistry · 18. Chemical Kinetics
For a first order reaction, \(A \rightarrow P\), the temperature \((T)\) dependent rate constant \((k)\) was found to follow the equation, \(\log k=-(2000) / T+6.0\)
The pre-exponential factor \(A\) and the activation energy \(\left(E_a\right)\), respectively, are
- A \(1.0 \times 10^6 \mathrm{~s}^{-1}\) and \(9.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(6.0 \mathrm{~s}^{-1}\) and \(16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(1.0 \times 10^6 \mathrm{~s}^{-1}\) and \(16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(1.0 \times 10^6 \mathrm{~s}^{-1}\) and \(38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(1.0 \times 10^6 \mathrm{~s}^{-1}\) and \(38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
Comparing the slope and intercept of the given equation with the following Arrhenius equation :
\(
\log k=-\frac{E_a}{2303 R T}+\log A
\)
Hence, \(\log A=6\) i.e. \(A=10^6 \mathrm{~s}^{-1}\).
Comparing slope gives \(E_a=38.3 \mathrm{~kJ} / \mathrm{mol}\).
\(
\log k=-\frac{E_a}{2303 R T}+\log A
\)
Hence, \(\log A=6\) i.e. \(A=10^6 \mathrm{~s}^{-1}\).
Comparing slope gives \(E_a=38.3 \mathrm{~kJ} / \mathrm{mol}\).
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