For a chemical reaction with rise in temperature by 10°, the rate constant becomes nearly ____________________.

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Step 1 — Arrhenius law: The rate constant follows $k=Ae^{-E_a/(RT)}$.

Step 2 — Ratio for a 10° (≈10 K) rise from $T$ to $T+10$:

$\displaystyle \frac{k_{2}}{k_{1}}=\exp\Big(\frac{E_a}{R}\big(\frac{1}{T}-\frac{1}{T+10}\big)\Big)$.

Step 3 — Linearize for typical activation energy. For example, take $E_a\approx50\,	ext{kJ/mol}=5	imes10^{4}\,	ext{J/mol}$, $R=8.314\,	ext{J/mol·K}$ and $T\approx298\,	ext{K}$:

$\displaystyle \ln\frac{k_{2}}{k_{1}}\approx\frac{E_a}{R}\frac{\Delta T}{T^2}=\frac{5	imes10^{4}}{8.314}\frac{10}{298^2}\approx0.677$.

Thus $\displaystyle \frac{k_{2}}{k_{1}}\approx e^{0.677}\approx1.97\approx2$.

Conclusion — For a moderate activation energy and around room temperature, a 10°C rise nearly doubles the rate constant.

FINAL_ANSWER: a

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