JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For reaction \(A \rightarrow P\), rate constant \(k = 1.5 \times 10^3\) s\(^{-1}\) at \(27°\)C. If activation energy for the above reaction is \(60\) kJ mol\(^{-1}\), then the temperature (in \(°\)C) at which rate constant, \(k = 4.5 \times 10^3\) s\(^{-1}\) is _______. (Nearest integer)
Given : \(\log 2 = 0.30\), \(\log 3 = 0.48\), \(R = 8.3\) J K\(^{-1}\) mol\(^{-1}\), \(\ln 10 = 2.3\)
- A 41
- B 42
- C 43
- D 44
Answer & Solution
Correct Answer
(A) 41
Step-by-step Solution
Detailed explanation
Using the Arrhenius equation: \(\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.3 R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)\) Given: \(k_1 = 1.5 \times 10^3 \text{ s}^{-1}\) \(k_2 = 4.5 \times 10^3 \text{ s}^{-1}\) \(T_1 = 27^\circ\text{C} = 300 \text{ K}\)…
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