A graph plotted between log ,K vs frac{1}{T} | vedclass

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(B) The Arrhenius equation is given by: $k = A \,e^{-\frac{E_A}{RT}}$Taking the logarithm on both sides:$\ln \,k = \ln \,A - \frac{E_A}{RT}$Converting

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(B) The Arrhenius equation is given by: $k = A \,e^{-\frac{E_A}{RT}}$Taking the logarithm on both sides:$\ln \,k = \ln \,A - \frac{E_A}{RT}$Converting to base $10$ logarithm:$\log \,k = \log \,A - \frac{E_A}{2.303 \,RT}$Comparing this with the equation of a straight line,$y = mx + c$:Here,$y = \log \,k$,$x = \frac{1}{T}$,$m = -\frac{E_A}{2.303 \,R}$ (slope),and $c = \log \,A$ (intercept).Since the slope is negative $(-\frac{E_A}{2.303 \,R})$,the graph of $\log \,k$ versus $\frac{1}{T}$ is a straight line with a negative slope,which corresponds to option $B$.

$A$ graph plotted between $\log \,K$ vs $\frac{1}{T}$ for calculating activation energy is shown by:

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