If we plot a graph between log , K and frac{1 | vedclass

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Question

(C) The Arrhenius equation is given by $K = A \, e^{-E_a / RT}$.Taking the natural logarithm on both sides,we get $\ln \, K = \ln \, A - \frac{E_a}{RT

Vedclass · Accepted Answer

$-\frac{E_a}{2.303 \, R}$

Vedclass · Answer

(C) The Arrhenius equation is given by $K = A \, e^{-E_a / RT}$.Taking the natural logarithm on both sides,we get $\ln \, K = \ln \, A - \frac{E_a}{RT}$.To convert this to base $10$,we divide by $2.303$,resulting in $\log \, K = \log \, A - \frac{E_a}{2.303 \, RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \, K$ and $x = \frac{1}{T}$,the slope $m$ is equal to $-\frac{E_a}{2.303 \, R}$.

If we plot a graph between $\log \, K$ and $\frac{1}{T}$ by Arrhenius equation,the slope is

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