Question

The reaction, $A ( g )+ B ( g ) \rightleftharpoons C ( g )+ D ( g )$ occurs in a single step. The rate constant of forward reaction is $2 \times 10^{-3} mol ^{-1} L sec ^{-1} .$ When the reaction is started with equimolar amounts of $A \& B$, it is found that the concentration of $A$ is twice that of $C$ at equilibrium. The rate constant of backward reaction is:

• A. $5 \times 10^{-4} mol ^{-1} L sec ^{-1}$
• B. $8 \times 10^{-3} mol ^{-1} L \sec ^{-1}$
• C. $1.25 \times 10^{2} mol ^{-1} L sec ^{-1}$
• D. $2 \times 10^{3} mol ^{-1} L sec ^{-1}$

Answer 1

Answer: (B)

At eqm

$(x-y)=2 y$

$x-y=2 y $

$x=3 y$

$K_{C}=\frac{y \times y}{(x-y)(x-y)}=\frac{y^{2}}{2 y \times 2 y}=\frac{y^{2}}{4 y^{2}}$

$K_{c}=\frac{1}{4}$

& $K_{C}=\frac{K_{f}}{K_{b}}$

$\frac{1}{4}=\frac{2 \times 10^{-3}}{K_{b}}$

$K_{b}=8 \times 10^{-3}$