CUET · CHEMISTRY · PYQ PAPER 2023
The rate constant for the first order decomposition of \(H _2 O _2\) is given by:\[ \log k=15-\frac{1.2 \times 10^2}{T}\] The activation energy ( \(E_a\) ) would be:
- A 6 kJ
- B 2.3 kJ
- C 4.5 kJ
- D 3.2 kJ
Answer & Solution
Correct Answer
(B) 2.3 kJ
Step-by-step Solution
Detailed explanation
\( \log k = \log A - \frac{E_a}{2.303 R T} \) \( \frac{E_a}{2.303 R} = 1.2 \times 10^2 \) \( E_a = 1.2 \times 10^2 \times 2.303 \times 8.314 \, J/mol \) \( E_a = 2307.696 \, J/mol \) \( E_a = 2.307696 \, kJ/mol \) \( E_a \approx 2.3 \, kJ \)
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