Question

For a reaction consider the plot of \(\ell n\,k\) versus \(1/T\) given in the figure. If the rate constant of this reaction at 400 K is \(10^{-5}\ \mathrm{s^{-1}}\), then the rate constant at 500 K is –

(The plot of \(\ell n\,k\) against \(1/T\) is a straight line with slope \(= -4606\ \mathrm{K}\).)

• A. \(10^{-4}\ \mathrm{s^{-1}}\)
• B. \(4\times10^{-4}\ \mathrm{s^{-1}}\)
• C. \(10^{-6}\ \mathrm{s^{-1}}\)
• D. \(2\times10^{-4}\ \mathrm{s^{-1}}\)

Hint:

The slope of \(\ln k\) vs \(1/T\) is \(-E_a/R\); use \(\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)\).

Answer 1

The Correct Option is A

Arrhenius: \(\ln k = \ln A - \dfrac{E_a}{RT}\), so slope \(= -E_a/R = -4606\) K.
\(\ln\dfrac{k_2}{k_1} = 4606\left(\dfrac{1}{400}-\dfrac{1}{500}\right) = 4606\times\dfrac{100}{200000} = 4606\times 5\times10^{-4} = 2.303\).
\(\log\dfrac{k_2}{k_1} = 2.303/2.303 = 1\), so \(k_2 = 10\,k_1 = 10\times10^{-5} = 10^{-4}\ \mathrm{s^{-1}}\).
Answer: (1) \(10^{-4}\ \mathrm{s^{-1}}\)