NEET · Chemistry · STD 12 - 3. Chemical kinetics
Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, \(T(K)\).
\(\ln k=14.34-\frac{1.25 \times 10^4}{T}\)
The energy of activation in \(kcal\ mol ^{-1}\) for the reaction is :
(Given : \(k\) in s \(^{-1}, R=1.987 cal\ mol ^{-1} K^{-1}\))
- A 24.84
- B 14.34
- C 18.63
- D 12.42
Answer & Solution
Correct Answer
(A) 24.84
Step-by-step Solution
Detailed explanation
(A) The Arrhenius equation is given by:
\(\ln k=\ln A-\frac{E_a}{R T}\)
The given equation is:
\(\ln k=1(D)34-\frac{(A)25 \times 10^4}{T}\)
Comparing the two equations:
\(\frac{E_a}{R}=(A)25 \times 10^4\)
\(E_a=(A)25 \times 10^4 \times R\)
Substituting \(R=(A)987 cal mol ^{-1} K^{-1}\) :
\(E_a=(A)25 \times 10^4 \times (A)987\)
\(E_a=24837.5 cal mol^{-1}\)
\(E_a=2(D)8375 kcal mol^{-1} \approx 2(D)84 kcal mol^{-1}\)
\(\ln k=\ln A-\frac{E_a}{R T}\)
The given equation is:
\(\ln k=1(D)34-\frac{(A)25 \times 10^4}{T}\)
Comparing the two equations:
\(\frac{E_a}{R}=(A)25 \times 10^4\)
\(E_a=(A)25 \times 10^4 \times R\)
Substituting \(R=(A)987 cal mol ^{-1} K^{-1}\) :
\(E_a=(A)25 \times 10^4 \times (A)987\)
\(E_a=24837.5 cal mol^{-1}\)
\(E_a=2(D)8375 kcal mol^{-1} \approx 2(D)84 kcal mol^{-1}\)
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