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The rate constant (k) of a reaction is measured at different temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ $m o l^{- 1}$ is : (R is gas constant)

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2/R

1/R

answer is C.

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Detailed Solution

$\ln K = \ln A - \frac{E a}{R T}$

$\to \frac{10^{3}}{T} = 0 i . e T \to \infty$

$i . e \ln K = \ln A$

at origin $\ln K = 10 = \ln A$

$\to a t \frac{10^{3}}{T} = 5 \Rightarrow T = 200 K \ln K = 0$

$\ln A = \frac{E a}{R T}$

$10 = \frac{E a}{R 200}$

Ea = 2000 R J

= 2R kJ/mole

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