Question

The equilibrium concentration of $B\left[{\left(B\right)}_e\right]$ for the reversible reaction $A\rightleftharpoons B$ can be evaluated by the expression:

• A. $K_c{\left[A\right]}_e^{-1}$
• B. $\frac{K_f}{K_b}{\left[A\right]}_e^{-1}$
• C. $K_f{K_b}^{-1}{\left[A\right]}_e$
• D. $K_f{K}_b{\left[A\right]}^{-1}$

Answer 1

The correct option is C $K_f{K_b}^{-1}{\left[A\right]}_e$

For given reaction, $K_C=\frac{K_f}{K_b}=\frac{[B{]}_e}{[A{]}_e}$

$\therefore [B{]}_e=\frac{K_f}{K_b}[A{]}_e$

$=K_f{K}_b^{-1}[A{]}_e$