Question

The equilibrium concentration of $[B{]}_{eq}$ for the reversible reaction $A\rightleftharpoons B$ can be evaluated by the expression:

• A. $K_c[A{]}_{eq}$
• B. $\frac{k_f}{k_b}\left[A\right]e_{eq}$
• C. $k_f{k}_b^{-1}[A{]}_{eq}$
• D. $k_f{k}_b[A{]}^{-1}$

Answer 1

The correct option is C $k_f{k}_b^{-1}[A{]}_{eq}$

At equilibrium, rate of forward reaction=rate of backward reaction
$\Rightarrow r_f=r_b$
$A\stackrel{K_f}{\underset{K_b}{\rightleftharpoons }}B$
$\therefore K_f[A{]}_{eq}=K_b[B{]}_{eq}$
$[B{]}_{eq}=K_f{K}_b^{-1}[A{]}_{eq}$
Hence, the correct answer is option (c).