The following figure shows a graph of log_{10 | vedclass

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(C) According to the Arrhenius equation: $\log_{10}K = \log_{10}A - \frac{E_a}{2.303RT}$.Comparing this with the equation of a straight line $y = mx +

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$2$

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(C) According to the Arrhenius equation: $\log_{10}K = \log_{10}A - \frac{E_a}{2.303RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log_{10}K$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_a}{2.303R}$.Given the slope $	an 	heta = -\frac{1}{2.303}$,we have:$-\frac{E_a}{2.303R} = -\frac{1}{2.303}$.Canceling $-1/2.303$ from both sides,we get $\frac{E_a}{R} = 1$,which implies $E_a = R \ cal/mol$.Since the value of $R$ is approximately $2 \ cal \ K^{-1} \ mol^{-1}$,the activation energy $E_a = 2 \ cal/mol$.

The following figure shows a graph of $\log_{10}K$ vs $\frac{1}{T}$,where $K$ is the rate constant and $T$ is the temperature. The straight line $BC$ has a slope,$\tan \theta = -\frac{1}{2.303}$,and an intercept of $5$ on the $Y$-axis. Thus,$E_a$,the energy of activation,is ....... $cal$.

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