I'm struggling with this homework question: "For a parallel reaction A goes to B with rate constant $k_1$ and A goes to C with rate constant $k_2$, you determine that the activation energies are 71.9 kJ/mol for $k_1$ and 142.8 kJ/mol for $k_2$. If the rate constants are equal at a temperature of 321 K, at what temperature (in K) will $\frac{k_1}{k_2} = 2$?"

I know I have to use $k=Ae^{\frac{E}{RT}}$, and I've tried solving for $T$ after filling in my data into the formula as ratios, ie $\frac{k_1}{k_2}$, $\frac{E_1}{E_2}$, etc, but I can't seem to get a coherent answer.

I tried $\ln\frac{k_1}{k_2})=-\frac{E}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ as well, but in that case I don't know what $E$ I'm supposed to use.

Any guidance on figuring this out would be much appreciated.

  • 3
    $\begingroup$ This is the 1000th question on Chemistry.SE! $\endgroup$ – ManishEarth Apr 6 '13 at 9:22

Activation energy is constant for a given reaction at any temperature (see reasons here)

So if you consider both reactions at 321K:

$$\ln k_1 = \ln k_2$$ So $$\ln A_1 - \dfrac{E_1}{321R} = \ln A_2 - \dfrac{E_2}{321R}$$ (Since the reactions are different I assume here that $A_1$ and $A_2$ have different values - I think you might have made a mistake here)

Using this you can find the ratio of $A_1$ to $A_2$ and proceed to find the temperature at which $\dfrac{k_1}{k_2} = 2$.


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