(A) Given: $\Delta H = 20 \ kJ \ mol^{-1}$,$E_{a,f} = 300 \ kJ \ mol^{-1}$.For the same rate,the activation energy $E_{a}$ is proportional to temperat

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(A) Given: $\Delta H = 20 \ kJ \ mol^{-1}$,$E_{a,f} = 300 \ kJ \ mol^{-1}$.For the same rate,the activation energy $E_{a}$ is proportional to temperature $T$ (assuming the pre-exponential factor $A$ remains constant).$\frac{E_{a,f}}{T_{uncat}} = \frac{E_{a,f}^{\prime}}{T_{cat}}$$T_{uncat} = 327 + 273 = 600 \ K$$T_{cat} = 27 + 273 = 300 \ K$$\frac{300}{600} = \frac{E_{a,f}^{\prime}}{300} \implies E_{a,f}^{\prime} = 150 \ kJ \ mol^{-1}$.We know $\Delta H = E_{a,f}^{\prime} - E_{a,b}^{\prime}$.$20 = 150 - E_{a,b}^{\prime}$.$E_{a,b}^{\prime} = 150 - 20 = 130 \ kJ \ mol^{-1}$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

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For a reversible reaction $A \rightleftharpoons B$,the $\Delta H_{\text{forward}} = 20 \ kJ \ mol^{-1}$. 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 catalysed backward reaction is $.... \ kJ \ mol^{-1}$.

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$130$
B
$120$
C
$110$
D
$100$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The activation energy for the reaction $X \rightarrow Y$ is $150 \text{ kJ mol}^{-1}$. The change in enthalpy for the above reaction is $-135 \text{ kJ mol}^{-1}$. What is the activation energy for the reverse reaction $Y \rightarrow X$?

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Energy of activation of a reactant is reduced by

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The rate of reaction becomes double when the temperature of the reaction is increased from $27\,^{\circ}C$ to $57\,^{\circ}C$. The activation energy of the reaction will be ....... $k\,cal$.

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The activation energy for the reaction $X \rightarrow Y$ is $150 \text{ kJ mol}^{-1}$. The change in enthalpy for the above reaction is $-135 \text{ kJ mol}^{-1}$. What is the activation energy for the reverse reaction $Y \rightarrow X$?

The rate of a chemical reaction doubles for every $10\,^{\circ}C$ rise in temperature. If the temperature is increased to $50\,^{\circ}C$ from $0\,^{\circ}C$,by how many times will the rate of reaction increase?

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