Doubling the absolute temperature from $T_1 = 300\,\text{K}$ to $T_2 = 600\,\text{K}$ for a reaction with activation energy $E_a = 100\,\text{kJ/mol}$ approximately INCREASES the rate constant by a factor of (Use $R = 8.314\,\text{J/(K mol)}$.)

1. Use the ratio form of the Arrhenius equation: $\ln\dfrac{k_2}{k_1} = -\dfrac{E_a}{R}\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)$. 2. Compute $\dfrac{1}{T_2} - \dfrac{1}{T_1} = \dfrac{1}{600} - \dfrac{1}{300} = \dfrac{1 - 2}{600} = -\dfrac{1}{600}\,\text{K}^{-1}$. 3. So $\ln(k_2/k_1) = -\dfrac{E_a}{R}\cdot(-1/600) = \dfrac{E_a}{600\,R}$. 4. Plug in $E_a = 100\,000\,\text{J/mol}$ and $R = 8.314\,\text{J/(K mol)}$: $\ln(k_2/k_1) = \dfrac{100\,000}{600 \times 8.314} = \dfrac{100\,000}{4988.4} = 20.05$. 5. Exponentiate: $k_2/k_1 = e^{20.05} \approx 5\times 10^{8}$, closest to $7\times 10^{8}$ (the slight difference comes from rounding in step 4). 6. This astonishingly large factor illustrates why reaction rates can change dramatically over a moderate temperature swing for high-activation-energy reactions. 7. Option A is the popular textbook rule "$k$ doubles for every $10\,\text{K}$" applied for $300\,\text{K}$ rise — not justified for high $E_a$. Option B is way too small. Option D is the wrong-units result. _Source: NCERT Class 12 Chemistry Part 1, Ch 3, §3.5 (Arrhenius equation, applied to ratio of $k$ at two temperatures), p. 19–20._