# Arrhenius equation using concentration

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.

• 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. – Karsten Theis Jul 1 at 15:55
• Zero order, perhaps, or pseudo zero order? – Karsten Theis Jul 1 at 15:58
• 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? – porphyrin Jul 1 at 18:42