I've been conducting some experiments to determine the temperature dependence of a reaction. To do this, I've been conducting the experiments at different temperatures and generated a graph of $\ln{[\ce{X}]}$ (where $[\ce{X}]$ is the concentration of the product) vs $T^{-1}$. I was told in a passing comment that because the reaction is first order the gradient of this graph is equal to $E_\mathrm a/R$ in the Arrhenius equation

$$ \ln{k} = \ln{A} - \frac{E_\mathrm a}{RT}, $$

but I don't understand why I can substitute $\ln k$ for $\ln [\ce{X}]$. Any help is much appreciated.

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    $\begingroup$ You should get back to the person who told you in passing, and ask for a complete explanation. You say [X] is the concentration of product - but at what time? If [X] is zero at t = 0, and you measure [X] after the same time has passed for all reactions, a larger [X] means a larger k. Whether k is proportional to [X] does depend on the rate law. Could you edit your question to clarify how you did the experiment (initial conditions) and when you measured the product concentration? Otherwise, giving a good answer is impossible. $\endgroup$ – Karsten Theis Jul 1 '19 at 15:55
  • $\begingroup$ Zero order, perhaps, or pseudo zero order? $\endgroup$ – Karsten Theis Jul 1 '19 at 15:58
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    $\begingroup$ Normally one would plot $\ln(X/X_0) $ vs time ($t$) (for a first order reaction and $X_0$ being initial amount ) and obtain $k$ at the prevailing temperature then plot $k$ vs $1/T$, so, are you sure you have'n't got $t$ and $T$ muddled? $\endgroup$ – porphyrin Jul 1 '19 at 18:42

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