KCET · Chemistry · Chemical Kinetics
For the reaction, \(A \rightleftharpoons B, E_a=50 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta H=-20 \mathrm{~kJ} \mathrm{~mol}^{-1}\). When a catalyst is added \(E_a\) decreases by \(10 \mathrm{~kJ} \mathrm{~mol}^{-1}\). What is the \(E_a\) for the backward reaction in the presence of catalyst?
- A \(60 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(40 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(70 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(20 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(60 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
Given,
\(A \rightleftharpoons B\)
\(\left(E_a\right)_f=50 \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta H=-20 \mathrm{~kJ} / \mathrm{mol}\)
\(\left(E_a\right)_b=?\)
using the formula
\(\Delta H=\left(E_a\right)_f-\left(E_a\right)_b\)
\(-20=40-\left(E_a\right)_h\)
\(\left(E_a\right)_b=40+20=60 \mathrm{~kJ} / \mathrm{mol}\)
\(A \rightleftharpoons B\)
\(\left(E_a\right)_f=50 \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta H=-20 \mathrm{~kJ} / \mathrm{mol}\)
\(\left(E_a\right)_b=?\)
using the formula
\(\Delta H=\left(E_a\right)_f-\left(E_a\right)_b\)
\(-20=40-\left(E_a\right)_h\)
\(\left(E_a\right)_b=40+20=60 \mathrm{~kJ} / \mathrm{mol}\)
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