Question

For a reversible reaction,
$A\rightleftharpoons B$
Pre-exponential factors in forward and backward direction are assumed to be equal.



Select the correct option(s) on the basis of above graph. [Take $R=2Cal/mole-K$]

• A. $(E_a{)}_{forward}=6Kcal/mol$
• B. At $500K$, fraction of molecules of $A$ crossing energy barrier for forward reaction is $e^{-6}$
• C. At $500K$, fraction of molecules of $B$ crossing energy barrier for backward reaction is $e^{-8}$
• D. Equilibrium constant for the reaction is $e^{-2}\text{at}500K$ temperature

Answer 1

Answer: (A), (B), (C)

At $500K$, fraction of molecules of $B$ crossing energy barrier for backward reaction is $e^{-8}$

We know,
$\Delta H=(E_a{)}_f-(E_a{)}_b$
$\Delta H=-2kCal/mol\text{and}(E_a{)}_b=8kCal/mol$
$\therefore (E_a{)}_f=-2kCal/mol+8kCal/mol=6kCal/mol$

Fraction of molecules crossing energy barrier is given as $e^{-\frac{E_a}{RT}}$
Fraction of molecules of $A$ crossing energy barrier for forward reaction at $500K$ $=e^{-\frac{E_a}{RT}}=e^{-\frac{6000}{2\times 500}}=e^{-6}$.
Fraction of molecules of $B$ crossing energy barrier for backward reaction at $500K$ $=e^{-\frac{8000}{2\times 500}}=e^{-8}$.
Equilibrium constant $K_{eq}=\frac{k_f}{k_b}=\frac{A_1{e}^{-\frac{(E_a{)}_f}{RT}}}{A_2{e}^{-\frac{(E_a{)}_b}{RT}}}$
Given that $A_1=A_2$
$\therefore k_{eq}=e^{-(E_a{)}_f-(E_a{)}_b/RT}=e^{-\frac{\Delta H}{RT}}$
$K_{eq}=e^{-(-\frac{2000}{2\times 500})}$
$K_{eq}=e^2$