Question

The following equilibrium are given below:
$A_2+3B_2\rightleftharpoons 2A{B}_3...K_1$
$A_2+C_2\rightleftharpoons 2AC...K_2$
$B_2+\frac{1}{2}{C}_2\rightleftharpoons B_{2}C....K_3$
The equilibrium constant of the reaction
$2A{B}_3+\frac{5}{2}{C}_2\rightleftharpoons 2AC+3B_{2}C$. in terms of $K_1,K_2$ and $K_3$ is:

• A. $\frac{K_1{K}_2}{K_3}$
• B. $\frac{K_1{K}_3^2}{K_2}$
• C. $\frac{K_2{K}_3^3}{K_1}$
• D. $K_1{K}_2{K}_3$

Answer 1

Answer: (C)

$\frac{K_2{K}_3^3}{K_1}$

If we multiply the stoichiometric coefficients of a reaction with $n$, then the equilibrium constant becomes $K^n$ .

$A_2+3B_2\rightleftharpoons 2A{B}_3\to K_1\to \left(1\right)$

If we reverse the reaction, then equilibrium constant becomes $1/K$ .

Reverse equation (1) $\Rightarrow 2A{B}_3\rightleftharpoons A_2+3B_2\to 1/K_1\to \left(2\right)$

$A_2+C_2\rightleftharpoons 2AC\to K_2\to \left(3\right)$

$B_2+1/2C_2\rightleftharpoons B_{2}C\to K_3\to \left(4\right)$

Multiply reaction with $3$:

$3B_2+3/2C_2\rightleftharpoons 3B_{2}C\to K_3^3\to \left(5\right)$

Add equation $\left(2\right)$, equation $\left(3\right)$ and equation $\left(5\right)$. If we add reactions, then equilibrium constant get multiplied,

$2A{B}_3\rightleftharpoons A_2+3B_2$ $1/K_1$

$A_2+C_2\rightleftharpoons 2AC$ $K_2$

$3B_2+3/2C_2\rightleftharpoons 3B_{2}C$ $K_3^3$ $\therefore$ New equilibrium constant= $\frac{K_2{K}_3^3}{K_1}$

____________________________

$2A{B}_3+5/2C_2\rightleftharpoons 2AC+3B_{2}C\to \frac{K_2{K}_3^3}{K_1}$