At T K,the following equation is obtained fo | vedclass

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Question

(A) According to the Arrhenius equation: $k = A e^{-E_a / RT}$.Taking natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Rearranging th

Vedclass · Accepted Answer

$2.303 x R$

Vedclass · Answer

(A) According to the Arrhenius equation: $k = A e^{-E_a / RT}$.Taking natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Rearranging the terms: $\ln \left(\frac{k}{A}\right) = -\frac{E_a}{RT}$.Converting natural logarithm to base $10$: $2.303 \log \left(\frac{k}{A}\right) = -\frac{E_a}{RT}$.Dividing by $2.303$: $\log \left(\frac{k}{A}\right) = -\frac{E_a}{2.303 RT}$.Given equation: $\log \left(\frac{k}{A}\right) = -\frac{x}{T}$.Comparing the two equations,we get: $\frac{E_a}{2.303 R} = x$.Therefore,the activation energy is: $E_a = 2.303 x R$.

At $T \ K$,the following equation is obtained for a first order reaction: $\log \frac{k}{A} = -\frac{x}{T}$. The activation energy for this reaction is equal to $(R = \text{gas constant})$

Similar Questions

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

Which of the following statements is correct?

Rate constant varies with temperature by the equation $log_{10} K = 5 - 2000 / T$. We can conclude that $(R = 8.314 \ J \ mol^{-1} K^{-1})$

For an endothermic reaction,the energy of activation is $E_a$ and the enthalpy of reaction is $\Delta H$ (both in $kJ/mol$). The minimum value of $E_a$ will be:

Write the Arrhenius equation in the form $\ln \, k = -\frac{E_a}{RT} + \ln \, A$.

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