JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The rate constant of a reaction increases by five times on increase in temperature from \(27^{\circ} C\) to \(52^{\circ} C\). The value of activation energy in \(kJ mol ^{-1}\) is \(.... .\) (Rounded-off to the nearest integer) \(\left[ R =8.314\, J \,K ^{-1} \,mol ^{-1}\right]\)
- A \(26\)
- B \(10\)
- C \(52\)
- D \(48\)
Answer & Solution
Correct Answer
(C) \(52\)
Step-by-step Solution
Detailed explanation
\(T _{1}=300 \,K , T _{2}=325\, K , K _{2}=5\, K\) \(\ln \frac{ K _{2}}{ K _{1}}=\frac{ Ea }{ R }\left[\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right]\) or, \(\ln 5=\frac{ Ea }{8.314}\left[\frac{1}{300}-\frac{1}{325}\right]\) or,…
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