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Q.

$K = A e^{- \frac{E_{a}}{RT}}$ is Arrhenius equation. It gives variation of rate constant of reaction with temperature. If a graph is plotted between $lnK$ vs $\frac{1}{T}$ , the slope of the graph will be (K= rate constant, T= absolute temperature)

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a

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

b

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

c

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

d

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

answer is D.

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Detailed Solution

$\begin{array}{l} lnk = lnA - \frac{E_{a}}{R} (\frac{1}{T}) \end{array}; y c m x$

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