Decomposition of a hydrocarbon follows the eq | vedclass

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Question

(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.Comparing the given equation $k = (5.5 	imes 10^{11} 	ext{ s}^{-1}) e^{-28000 / T}$ wit

Vedclass · Accepted Answer

$232$

Vedclass · Answer

(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.Comparing the given equation $k = (5.5 	imes 10^{11} 	ext{ s}^{-1}) e^{-28000 / T}$ with the Arrhenius equation, we identify the exponent term:$\frac{E_a}{R} = 28000 	ext{ K}$.Given $R = 8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1}$, we calculate the activation energy $E_a$:$E_a = 28000 	imes 8.3 = 232400 	ext{ J mol}^{-1}$.To convert this into $	ext{kJ mol}^{-1}$, we divide by $1000$:$E_a = \frac{232400}{1000} = 232.4 	ext{ kJ mol}^{-1}$.The nearest integer value is $232 	ext{ kJ mol}^{-1}$.

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

Similar Questions

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