KCET · Chemistry · s Block Elements
A plot of \( \frac{1}{T} \mathrm{Vs} . \mathrm{k} \) for a reaction gives the slope \( -1 \times 10^{4} \mathrm{~K} . \) The energy of activation for the
reaction is
(Given \( R=8.314 J K^{-1} \mathrm{~mol}^{-1} \) )
- A \( 1.202 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
- B \( 83.14 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
- C \( 8314 \mathrm{Jmol}^{-1} \)
- D \( 12.02 \mathrm{~J} \mathrm{~mol}^{-1} \)
Answer & Solution
Correct Answer
(B) \( 83.14 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
Step-by-step Solution
Detailed explanation
Arrhenius equation,
\( k=A e^{-\frac{E_{a}}{R T}} \rightarrow(1) \)
Taking natural log of Eq. (1), we get
\( \ln k=\ln A-\frac{E_{a}}{R} \times \frac{1}{T} \)
By plotting In \( \mathrm{k} \) against \( 1 / \mathrm{T} \), a straight line with slope \( =-\frac{E}{R} \) and intercept \( \ln \mathrm{A} \).
Now,
slope \( =-\frac{E_{a}}{R} \) \( -1 \times 10^{4}=-\frac{E_{a}}{8.314} \) \( E_{a}=8.314 \times 1 \times 10^{4} \) \( E_{a}=83.140 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
\( k=A e^{-\frac{E_{a}}{R T}} \rightarrow(1) \)
Taking natural log of Eq. (1), we get
\( \ln k=\ln A-\frac{E_{a}}{R} \times \frac{1}{T} \)
By plotting In \( \mathrm{k} \) against \( 1 / \mathrm{T} \), a straight line with slope \( =-\frac{E}{R} \) and intercept \( \ln \mathrm{A} \).
Now,
slope \( =-\frac{E_{a}}{R} \) \( -1 \times 10^{4}=-\frac{E_{a}}{8.314} \) \( E_{a}=8.314 \times 1 \times 10^{4} \) \( E_{a}=83.140 \mathrm{~kJ} \mathrm{~mol}^{-1} \)
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