Question

A container contains three gases, $A,B$ and $C$ in equilibrium, $A\rightleftharpoons 2B+C$
At equilibrium the concentration of $A$ was $3M$ and of $B$ was $4M$. On doubling the volume of container, the new equilibrium concentration of $B$ was $3M$. Calculate $K_C$ of the reaction:

• A. $28.8$
• B. $5.9$
• C. $0.14$
• D. $0.55$

Answer 1

Answer: (A)

$28.8$

Given that $A,B$ and $C$ in equilibrium according to the following reaction:
$A\rightleftharpoons 2B+C$

	$A$	$B$	$C$
Concentration at first equilibrium	$3$	$4$	$x$
Concentration after volume is doubled	$1.5$	$2$	$0.5x$
Concentration at second equilibrium	$(1.5-y)$	$(2+2y)$	$(0.5x+y)$

Given, $2+2y=3;y=0.5$
When volume is doubled, the partial pressure and the concentration are reduced to half.
$K_C=\frac{\left[B{]}^2\right[C]}{\left[A\right]}$ $⟹\frac{16x}{3}=\frac{3^2\times 0.5\times (1+x)}{1}$
$16x=3\times (9\times 0.5\times (1+x\left)\right)$
or $2.5x=13.5$
Therefore, $x=5.4$
$K_C=\frac{16x}{3}=\frac{16\times 5.4}{3}$
Hence, $K_C=28.8$.