The rate constant of a reaction is increased | vedclass

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(C) According to the Arrhenius equation,$k = Ae^{-E_a / RT}$.Let $E_{a_1}$ be the initial activation energy and $E_{a_2}$ be the activation energy aft

Vedclass · Accepted Answer

$-3.45 \ kJ / mol$

Vedclass · Answer

(C) According to the Arrhenius equation,$k = Ae^{-E_a / RT}$.Let $E_{a_1}$ be the initial activation energy and $E_{a_2}$ be the activation energy after adding the catalyst.$k_1 = Ae^{-E_{a_1} / RT} \dots (1)$$k_2 = 4k_1 = Ae^{-E_{a_2} / RT} \dots (2)$Dividing $(2)$ by $(1)$:$4 = e^{(E_{a_1} - E_{a_2}) / RT}$Taking natural logarithm on both sides:$\ln 4 = \frac{E_{a_1} - E_{a_2}}{RT}$$E_{a_2} - E_{a_1} = -RT \ln 4$Given $T = 27 + 273 = 300 \ K$ and $\ln 4 = 1.386$.$\Delta E_a = E_{a_2} - E_{a_1} = -(8.314 \ J \ K^{-1} \ mol^{-1} 	imes 300 \ K 	imes 1.386)$$\Delta E_a = -3457.3 \ J / mol \approx -3.45 \ kJ / mol$.

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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