NEET · Chemistry · STD 12 - 3. Chemical kinetics
The rate of a reaction quadruples when temperature changes from \(27^{\circ} \mathrm{C}\) to \(57^{\circ} \mathrm{C}\). Calculate the energy of activation. Given \(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021\)
- A \(380.4 \mathrm{~kJ} / \mathrm{mol}\)
- B \(3.80 \mathrm{~kJ} / \mathrm{mol}\)
- C \(3804 \mathrm{~kJ} / \mathrm{mol}\)
- D \(38.04 \mathrm{~kJ} / \mathrm{mol}\)
Answer & Solution
Correct Answer
(D) \(38.04 \mathrm{~kJ} / \mathrm{mol}\)
Step-by-step Solution
Detailed explanation
\( \text { Sol. } \log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \) \( \log \left(\frac{4}{1}\right)=\frac{E_a}{2.303 R}\left(\frac{1}{300}-\frac{1}{330}\right) \)
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