JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
For the equilibrium \(A\left( g \right) \rightleftharpoons B\left( g \right)\), \(\Delta H\) is \(-40\, k\,J/mol\). If the ratio of the activation energies of the forward \((E_f)\) and reverse \((E_b)\) reactions is \(\frac{2}{3}\) then
- A \(E_f\, = 80\, k\,J/mol ; E_b\, = 120\, k\,J/mol\)
- B \(E_f\, = 60\, k\,J/mol ; E_b\, = 100\, k\,J/mol\)
- C \(E_f\, = 30\, k\,J/mol ; E_b\, = 70\, k\,J/mol\)
- D \(E_f\, = 70\, k\,J/mol ; E_b\, = 30\, k\,J/mol\)
Answer & Solution
Correct Answer
(A) \(E_f\, = 80\, k\,J/mol ; E_b\, = 120\, k\,J/mol\)
Step-by-step Solution
Detailed explanation
\(\left( {\frac{{{E_{\mathop {forwad}\limits_{(A \to B)} }}}}{{{E_{\mathop {backward}\limits_{(B \to A)} }}}}} \right) = \frac{2}{3} = \frac{x}{{ - 40 + x}}\) \( \Rightarrow - 80 + 2x = 3x\) \( \Rightarrow x = - 80\,kJ/mol\) for forward For backward…
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