Question

For the reversible reaction, $A\stackrel{K_1}{\underset{K_2}{\rightleftharpoons }}B$ the initial concentrations of $A$ and $B$ are $a$ $M$ and $b$ $M$, and the equilibrium concentrations are $(a-x)M$ and $(b+x)M$ respectively. The value of $x$ is:

• A. $\frac{K_{1}a+K_{2}b}{K_1+K_2}$
• B. $\frac{K_{1}a-K_{2}b}{K_1+K_2}$
• C. $\frac{K_1+K_2}{K_{1}a+K_{2}b}$
• D. $\frac{K_1+K_2}{K_{1}a-K_{2}b}$

Answer 1

Answer: (B)