Discussion
A reaction has an activation energy of about 50 kilojoules per mole. Roughly what happens to its rate for a ten degree rise near room temperature?
Answer: B
Correct answer: it roughly doubles
From the Arrhenius equation, \[ \frac{k_2}{k_1} = \exp\left[\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right] = \exp\left[\frac{50{,}000}{8.314}\left(\frac{1}{298}-\frac{1}{308}\right)\right] = e^{0.655} = 1.9. \] The familiar rule that rate doubles for a ten degree rise is therefore not a law but a coincidence of the activation energies most ordinary reactions happen to have - a reaction of 25 kJ/mol would rise by only about forty per cent.
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