Question

Temperature of a reaction with an activation energy value of 15 kcal mol\(^{-1}\) is increased from 300 K to 310 K. If the value of the ideal gas constant (R) is 1.9872 cal mol\(^{-1}\) K\(^{-1}\), the ratio of the reaction rate constants \(\left(\frac{k_{310}}{k_{300}}\right)\) (rounded off to two decimal places) is ____________.

Hint:

Arrhenius equation is temperature-sensitive: even a small increase in temperature can cause exponential increases in rate constant.
Always convert activation energy units to match R before calculation.

Answer 1

Correct Answer: 2.24

Step 1: Using Arrhenius equation ratio form:
\[ \ln \left( \frac{k_{2}}{k_{1}} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Step 2: Substituting values:
\(E_a = 15 \, \text{kcal mol}^{-1} = 15000 \, \text{cal mol}^{-1}\),
\(R = 1.9872 \, \text{cal mol}^{-1} K^{-1}\),
\(T_1 = 300 K, \, T_2 = 310 K\).
Step 3: Compute:
\(\frac{1}{300} - \frac{1}{310} = 0.003333 - 0.003226 = 0.000107\).
\(\frac{E_a}{R} = \frac{15000}{1.9872} = 7550.83\).
\(\ln \left(\frac{k_{310}}{k_{300}}\right) = 7550.83 \times 0.000107 = 0.807\).
Step 4: Therefore,
\(\frac{k_{310}}{k_{300}} = e^{0.807} \approx 2.24 \approx 2.31\).
\(\therefore\) The ratio is approximately \(\mathbf{2.31}\).