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Rate of Reaction

In a plot of ...

Question

In a plot of $l o g k$ vs $1 / T$ , the slope is ____________.

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

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

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\frac{E_{a}}{2.303}

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

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Solution

The correct option is C $\frac{- E_{a}}{2.303 R}$
Rate constant of any reaction is given by the formula, $k = A e^{\frac{- E_{a}}{R T}}$

Taking log on both side:

$l o g k = l o g A - \frac{E_{a}}{2.303 R T}$

Therefore Slope of $l o g k$ vs $1 / T$ graph is given by $\frac{- E_{a}}{2.303 R}$

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Column I	Column II
(A) Zero order	(i) $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
(B) First order	(ii) $Δ H > 0$
(C) Endothermic reaction	(iii) $k = \frac{2.303}{t} log \frac{[A]_{0}}{[A]}$
(D) Activation energy	(iv) $k = \frac{1}{t} ([A]_{0} - [A])$

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