The graph of log K versus 1/T is given below. Here ‘x’ is equal to
[Where K = rate constant and T = absolute temperature]

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

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

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

$\frac{- 2 .303 R}{E_{a}}$

answer is B.

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Detailed Solution

$\ln K = \frac{- Ea}{RT} + \ln A$ (from Arrhenius equation)
$\begin{array}{l} \log K = (\frac{- Ea}{2 .303 R}) \times \frac{1}{T} + \log A, y = mx + c \\ ∴ x = \frac{- Ea}{2 .303 R} \end{array}$

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