Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.
The correct answer is

  • A
    Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
  • B
    Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$
  • C
    $(A)$ is true, but $(R)$ is not true
  • D
    $(A)$ is not true, but $(R)$ is true

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Similar Questions

How are the values of $E_a$ and $A$ obtained from the Arrhenius equation?

$A$ reaction takes place in three steps with individual rate constants and activation energies. The overall rate constant is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$. The overall activation energy of the reaction in $kJ/mol$ is:
$Step$ $Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$ $k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$ $k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$ $k_3, E_{a_3} = 50\ kJ/mol$

Difficult
View Solution

The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.
Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$

For the reaction $CH_{3}CH_{2}CH_{2}I + OH^{-} \rightarrow CH_{3}CH_{2}CH_{2}OH + I^{-}$,the rate constant is $1.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$ at $27^{\circ}C$ $(300 \ K)$ and $38.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$ at $327 \ K$. Calculate the activation energy $(E_{a})$ in $cal \ mol^{-1}$.

Activation energy of a reaction is

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