Question

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

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

Answer 1

Answer: (C)

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

$K_C=\frac{K_f}{K_b}=\frac{\left[A\right]}{\left[B\right]}$

$\Rightarrow \left[B\right]=K_1{K}_b^{-1}[A{]}_e$