JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The rate of a reaction decreased by \(3.555\) times when the temperature was changed from \(40^{\circ} C\) to \(30^{\circ} C\). The activation energy of the reaction is.........\(kJ\, mol ^{-1}\) [Take; \(R =8.314 \,J\, mol ^{-1}\, K ^{-1}\) In \(3.555=1.268\)]
- A \(85\)
- B \(94\)
- C \(110\)
- D \(100\)
Answer & Solution
Correct Answer
(D) \(100\)
Step-by-step Solution
Detailed explanation
\(\ell n \left(\frac{ K _{ T _{2}}}{ K _{ T _{1}}}\right)=\frac{ E _{ a }}{ R }\left[\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right]\) \(T _{1}=303 K ; T _{2}=313 K\) \(\frac{ K _{ T _{2}}}{ K _{ T _{1}}}=3.555\)…
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