Find the correct equation among the following.

  • A
    $\ln k - \ln A = \frac{E_a}{RT}$
  • B
    $k = \frac{A E_a}{RT}$
  • C
    $\ln k + \ln A = \frac{E_a}{RT}$
  • D
    $\frac{E_a}{RT} = \ln A - \ln k$

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At $25\,^{\circ}C$,the values of rate constant,activation energy and Arrhenius constant of a reaction are $3 \times 10^{-4}\, s^{-1}$,$129\, kJ/mol$ and $2 \times 10^{15}\, s^{-1}$ respectively. The value of rate constant as $T \to \infty$ is

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Consider the following graph of the kinetic energy distribution among molecules at temperature $T$. If the temperature is increased,how would the resulting graph differ from the one above?

Catalyst $A$ reduces the activation energy for a reaction by $10 \ kJ \ mol^{-1}$ at $300 \ K$. The ratio of rate $\frac{k_{T, \text{Catalysed}}}{k_{T, \text{Uncatalysed}}}$ is $e^{x}$. Find the value of $x$ [nearest integer].
[Assume that the pre-exponential factor is same in both the cases.
Given $R = 8.31 \ J \ K^{-1} \ mol^{-1}$]

For the reaction,$A \rightleftharpoons B$,$E_a = 50 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$. When a catalyst is added,$E_a$ decreases by $10 \ kJ \ mol^{-1}$. What is the $E_a$ for the backward reaction in the presence of the catalyst?

Which graph shows zero activation energy?

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