Question

The rate of the chemical reaction doubles for an increase of \(10 \, \mathrm{K}\) in absolute temperature from \(298 \, \mathrm{K}\). What will be the activation energy?

• A. 52.897 kJ mol\(^{-1}\)
• B. 51.897 kJ mol\(^{-1}\)
• C. 42.897 kJ mol\(^{-1}\)
• D. 41.897 kJ mol\(^{-1}\)

Hint:

If the reaction rate doubles for every \(10 \, \mathrm{K}\) rise, use the Arrhenius equation directly with \(\frac{k_2}{k_1}=2\). Keep temperature in Kelvin and convert the final answer into kJ mol\(^{-1}\) if needed.

Answer 1

The Correct Option is A

Step 1: Use Arrhenius equation in logarithmic form.
When the temperature changes from \(T_1\) to \(T_2\), the Arrhenius equation is written as:
\[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1T_2}\right) \] Since the rate doubles, we take:
\[ \frac{k_2}{k_1} = 2 \] Also,
\[ T_1 = 298 \, \mathrm{K}, \qquad T_2 = 308 \, \mathrm{K} \] Step 2: Substitute the values in the formula.
\[ \log 2 = \frac{E_a}{2.303R}\left(\frac{10}{298 \times 308}\right) \] We know that:
\[ \log 2 = 0.3010 \] and
\[ R = 8.314 \, \mathrm{J \, mol^{-1} \, K^{-1}} \] So,
\[ 0.3010 = \frac{E_a}{2.303 \times 8.314}\left(\frac{10}{298 \times 308}\right) \] Step 3: Calculate the activation energy.
Rearranging,
\[ E_a = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10} \] \[ E_a \approx 52897 \, \mathrm{J \, mol^{-1}} \] \[ E_a \approx 52.897 \, \mathrm{kJ \, mol^{-1}} \] Step 4: Conclusion.
Therefore, the activation energy of the reaction is \(52.897 \, \mathrm{kJ \, mol^{-1}}\). Hence, the correct option is \((A)\).
Final Answer:\(52.897 \, \mathrm{kJ \, mol^{-1}}\).