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For a reversible reaction $A \rightleftharpoons B$, the $\Delta H _{\text {forward }}$ reaction $=20\,kJ\,mol ^{-1}$. The activation energy of the uncatalysed forward reaction is $300\,kJ\,mol ^{-1}$. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at $27^{\circ}\,C$ is found to be same as that of the uncatalysed reaction at $327^{\circ}\,C$. The activation energy of the cataysed backward reactoion is $....\,kJ\,mol { }^{-1}$.
$130$
$120$
$110$
$100$
Solution
$E _{ a }=300\, kJ\,mol ^{-1}$
$\frac{E_a}{T}=\frac{E_a^{\prime}}{T^{\prime}}$
(Since rate of catalysed and uncatalysed reaction is same)
$\frac{300}{600}=\frac{E_{ a , f }^{\prime}}{300}$
$E _{ a , f }^{\prime}=150$
$20=150- E _{ a , b }^{\prime}$
$E _{ a , b }^{\prime}=130$
