On introducing a catalyst at 500 , K, the rat | vedclass

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(D) According to the Arrhenius equation:$k = A \cdot e^{-E_a/RT}$ (Catalyst absent)$k' = A \cdot e^{-E'_a/RT}$ (Catalyst present)Given $E'_a = 4.15 \,

Vedclass · Accepted Answer

$8.3$

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(D) According to the Arrhenius equation:$k = A \cdot e^{-E_a/RT}$ (Catalyst absent)$k' = A \cdot e^{-E'_a/RT}$ (Catalyst present)Given $E'_a = 4.15 \, kJ \, mol^{-1}$ and $\frac{k'}{k} = 2.718 = e^1$.Taking the ratio:$\frac{k'}{k} = e^{(E_a - E'_a)/RT}$$e^1 = e^{(E_a - E'_a)/RT}$Therefore,$1 = \frac{E_a - E'_a}{RT}$$E_a = E'_a + RT$Substituting the values ($R = 8.314 	imes 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$,$T = 500 \, K$):$E_a = 4.15 + (8.314 	imes 10^{-3} 	imes 500)$$E_a = 4.15 + 4.157 = 8.307 \, kJ \, mol^{-1} \approx 8.3 \, kJ \, mol^{-1}$.

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