For a first order reaction, the plot of log K against 1T is a straight line. The slope of the line is equal to

k=Ae−Ea/RT log k=log A−Ea2.303R.1T Equation of straight line slope =−Ea2.303R

For a first order reaction, the plot of log K against 1T is a straight line. The slope of the line is equal to

k=Ae−Ea/RT log k=log A−Ea2.303R.1T Equation of straight line slope =−Ea2.303R

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For a first order reaction, the plot of $\log K$ against $\frac{1}{T}$ is a straight line. The slope of the line is equal to

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$- \frac{E_{a}}{2.303}$

$- \frac{E_{a}}{R}$

$- \frac{2.303}{E_{a} R}$

$\frac{- E_{a}}{2.303 R}$

answer is D.

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$k = A e^{- E_{a} / R T}$ $\log k = \log A - \frac{E_{a}}{2.303 R} . \frac{1}{T}$

Equation of straight line slope $= \frac{- E_{a}}{2.303 R}$

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For a first order reaction, the plot of log K against 1T is a straight line. The slope of the line is equal to

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For a first order reaction, the plot of $\log K$ against $\frac{1}{T}$ is a straight line. The slope of the line is equal to