JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Consider \(A \xrightarrow{ k _1} B\) and \(C \xrightarrow{ k _2} D\) are two reactions. If the rate constant \(\left( k _1\right)\) of the \(A \rightarrow B\) reaction can be expressed by the following equation \(\log _{10} k =14.34-\frac{1.5 \times 10^4}{T / K }\) and activation energy of \(C \rightarrow D\) reaction ( \(Ea _2\) ) is \(\frac{1}{5}\) th of the \(A \rightarrow B\) reaction \(\left( Ea _1\right)\), then the value of \(\left( Ea _2\right)\) is _______ \(kJ~mol ^{-1}\). (Nearest Integer)
- A 287
- B 57
- C 114
- D 43
Answer & Solution
Correct Answer
(B) 57
Step-by-step Solution
Detailed explanation
\(\frac{ E _{ a _1}}{2.303 R }=1.5 \times 10^4\) \(Ea_{1} = 1.5 \times 10^{4} \times 2.303 \times 8.314\) \(E _{ a _1}=28.7207 \times 10^4 J\) \(E _{ a _1}=287.207 kJ\) \(E _{ a _2}=\frac{ E _{ a _1}}{5}=\frac{287.207}{5}=57.44 kJ\)
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