The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$
The activation energy for the reaction is given by $... \, kJ \, mol^{-1}$. (In nearest integer)
(Given: $R = 8.3 \, J \, K^{-1} \, mol^{-1}$)

  • A
    $15$
  • B
    $166$
  • C
    $961$
  • D
    $247$

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Which scientist explained that all the species taking part in a reaction do not possess the same kinetic energy?

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

$A$ reaction takes place in three steps with individual rate constant and activation energy,
Step Rate constant and Activation energy
$Step \ 1$ $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$
$Step \ 2$ $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$
$Step \ 3$ $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.

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For the decomposition reaction of $N_2O_5$,the slope of the graph of $\log K$ versus $1/T$ is $-1.2 \times 10^4 \ K$. Calculate the activation energy $(E_a)$ of the reaction.

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