JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The rate of a reaction doubles when its temperature changes from \(300\, K\) to \(310\, K.\) Activation energy of such a reaction will be .......... \(kJ\, mol^{-1}\). \((R= 8.314\,JK^{-1} \,mol^{-1}\) and \(\log 2=0.301)\)
- A \(53.6\)
- B \(48.6\)
- C \(58.5\)
- D \(60.5\)
Answer & Solution
Correct Answer
(A) \(53.6\)
Step-by-step Solution
Detailed explanation
Activation energy can be calculated from the equation \(\frac{\log \,k_{2}}{\log k_{1}}=\frac{E_{a}}{2.303 R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)\) given \(\frac{k_{2}}{k_{1}}=2 T_{2}=310\, K T_{1}=300 K\)…
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