Regarding the velocity in a reversible reaction,which is the correct explanation of the effect of a catalyst?

  • A
    It provides a new reaction path of lower activation energy.
  • B
    It increases the kinetic energy of reacting molecules.
  • C
    It displaces the equilibrium state to the right side.
  • D
    It decreases the velocity of the backward reaction.

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The rate constants of a reaction at $500 \, K$ and $700 \, K$ are $0.02 \, s^{-1}$ and $0.07 \, s^{-1}$ respectively. Calculate the values of $E_{a}$ and $A$.

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For reaction $A \to B$,the rate constant $k_1 = A_1 e^{-E_{a_1} / (RT)}$ and for the reaction $X \to Y$ the rate constant $k_2 = A_2 e^{-E_{a_2} / (RT)}$. If $A_1 = 10^8$,$A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal/mol$,$E_{a_2} = 1800 \ cal/mol$,then the temperature at which $k_1 = k_2$ is (Given : $R = 2 \ cal/K \cdot mol$)

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