The rate constants for the decomposition of a | vedclass

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(B) According to the Arrhenius equation,$\ln \ k = \ln \ A - \frac{E_a}{RT}$.We are plotting $\ln \ k$ versus $\frac{10^{3}}{T}$.Rewriting the equatio

Vedclass · Accepted Answer

$154$

Vedclass · Answer

(B) According to the Arrhenius equation,$\ln \ k = \ln \ A - \frac{E_a}{RT}$.We are plotting $\ln \ k$ versus $\frac{10^{3}}{T}$.Rewriting the equation: $\ln \ k = \ln \ A - \left( \frac{E_a}{R 	imes 10^{3}} ight) 	imes \left( \frac{10^{3}}{T} ight)$.Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R 	imes 10^{3}}$.Given the slope is $-18.5$,we have: $-18.5 = -\frac{E_a}{8.31 	imes 10^{3}}$.$E_a = 18.5 	imes 8.31 	imes 10^{3} \ J \ mol^{-1} = 153735 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $E_a = 153.735 \ kJ \ mol^{-1}$.Rounding to the nearest integer,we get $154 \ kJ \ mol^{-1}$.

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