Arrhenius equation

For a reaction, $k = A e^{-E_a/(RT)}$

Taking natural logarithm: $\ln k = \ln A - \frac{E_a}{RT}$

Form of the graph

The graph is plotted between $\ln k$ and $\dfrac{10^3}{T}$.

Let $x = \frac{10^3}{T}$ Then, $\frac{1}{T} = \frac{x}{10^3}$

So the Arrhenius equation becomes $\ln k = \ln A - \frac{E_a}{R}\cdot \frac{x}{10^3}$

Hence slope of the graph is $m = -\frac{E_a}{1000R}$

Using the slope

From the given analyzed plot of $\ln k$ vs $\dfrac{10^3}{T}$, the slope is approximately $m \approx -18.5$

Therefore, $E_a = -m(1000R)$

Substituting $R = 8.31\,\text{J K}^{-1}\text{mol}^{-1}$, $E_a = 18.5 \times 1000 \times 8.31$ $E_a = 153735\,\text{J mol}^{-1}$

Convert to kJ mol$^{-1}$

$E_a = 153.735\,\text{kJ mol}^{-1}$

Nearest integer: $E_a \approx 154\,\text{kJ mol}^{-1}$

Comparison with stored answer

Derived answer = $154$

Stored correct answer = $154$

Both match.