For two reactions,

$$\begin{align} \ce{A ->[$k_1$] B} \tag{1} \\ \ce{A ->[$k_2$] C} \tag{2} \end{align}$$

you have determined that the activation energies are $\pu{71.9 kJ/mol}$ for $(1)$ and $\pu{142.8 kJ/mol}$ for $(2)$. If the rate constants are equal at a temperature of $\pu{321 K}$, at what temperature will $k_1/k_2$ be equal to $2$?

I know I have to use the Arrhenius equation

$$k=A\exp\left(-\frac{E_\mathrm A}{RT}\right),$$

and I've tried solving for $T$ after filling in my data into the formula as ratios, ie $k_1/k_2$, $E_\mathrm{A}^{(1)}/E_\mathrm{A}^{(2)}$, etc, but I can't seem to get a coherent answer. Any guidance on figuring this out would be much appreciated.


1 Answer 1


To a first approximation, the activation energy of a single given reaction is a constant at all temperatures (see reasons here, although more precisely, there is some temperature variation). However, the activation energies for both reactions need not be the same.

If you consider both reactions at $\pu{321 K}$,

$$\begin{align} \ln k_1 &= \ln k_2 \tag{1}\\ \ln A_1 - \frac{E_{\mathrm A}^{(1)}}{(\pu{321 K})R} &= \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{(\pu{321 K})R} \tag{2}\\[4pt] \ln A_1 - \ln A_2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{(\pu{321 K})R} \tag{3} \\[4pt] &= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{321 K})(\pu{8.31446 J K-1 mol-1})} \tag{3} \\[4pt] &= -26.56 \tag{4} \end{align}$$

Now, denote the desired temperature at which $k_1 = 2k_2$ as $T^*$. At this new temperature, we have

$$\begin{align} \ln k_1 &= \ln 2 + \ln k_2 \tag{5}\\ \ln A_1 - \frac{E_{\mathrm A}^{(1)}}{T^*R} &= \ln 2 + \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{T^*R} \tag{6}\\[4pt] \ln A_1 - \ln A_2 - \ln 2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{T^*R} \tag{7} \end{align}$$

Here we can reuse the previously calculated value for $\ln A_1 - \ln A_2$, and we can also plug in the same activation energies:

$$\begin{align} T^* &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{R(\ln A_1 - \ln A_2 - \ln 2)} \tag{8} \\[4pt] &= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{8.31446 J K-1 mol-1})(-26.56 - 0.693)} \tag{9} \\[4pt] &= \pu{313 K}. \tag{10} \end{align}$$


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