Question

The reaction, $A(g)+B(g)\rightleftharpoons C(g)+D(g)$ occurs in a single step. The rate constant of forward reaction is $2\times 10^{-3}mo{l}^{-1}Lse{c}^{-1}.$ 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. $5\times 10^{-4}mo{l}^{-1}Lse{c}^{-1}$
• B. $8\times 10^{-3}mo{l}^{-1}L{\sec }^{-1}$
• C. $1.25\times 10^{2}mo{l}^{-1}Lse{c}^{-1}$
• D. $2\times 10^{3}mo{l}^{-1}Lse{c}^{-1}$

Answer 1

Answer: (B)

At eqm

$(x-y)=2y$

$x-y=2y$

$x=3y$

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

$K_c=\frac{1}{4}$

& $K_C=\frac{K_f}{K_b}$

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

$K_b=8\times 10^{-3}$