For a first order reaction,a plot of ln k (y- | vedclass

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(A) According to the Arrhenius equation: $\ln k = \ln A - \frac{E_a}{RT}$.Comparing this with the equation of a straight line,$y = mx + c$,where $y =

Vedclass · Accepted Answer

$8.314$

Vedclass · Answer

(A) According to the Arrhenius equation: $\ln k = \ln A - \frac{E_a}{RT}$.Comparing this with the equation of a straight line,$y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$:The slope $(m) = -\frac{E_a}{R}$.Given the slope $(m) = -10^3 \ K$,we have:$-\frac{E_a}{R} = -10^3 \ K$$E_a = 10^3 \ K 	imes R$$E_a = 10^3 \ K 	imes 8.314 \ J \ mol^{-1} \ K^{-1} = 8314 \ J \ mol^{-1}$.To convert to $kJ \ mol^{-1}$,divide by $1000$:$E_a = \frac{8314}{1000} \ kJ \ mol^{-1} = 8.314 \ kJ \ mol^{-1}$.

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