Question

In the reaction $P{Cl}_5\rightleftharpoons P{Cl}_3+{Cl}_2$ the partial pressure of $P{Cl}_3,{Cl}_2$ and $P{Cl}_5$ are 0.3,0.2 and 0.6 $atm$ respectively.If partial pressure of $P{Cl}_3$ and ${Cl}_2$ was increased twice,what will be the partial pressure of $P{Cl}_5$ in $atm$ :

• A. 0.2
• B. 1.2
• C. 2.4
• D. 0.15

Answer 1

Answer: (C)

2.4

	$PC{l}_5$	$PC{l}_3$	$C{l}_2$
First equilibrium	0.6	0.3	0.2
Second equilibrium	x	0.6	0.4

From the first equilibrium data, calculate $K$.
$K=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$
$K=\frac{0.3\times 0.2}{0.6}=0.1$
From the second equilibrium data, calculate $x$.
$K=\frac{0.6\times 0.4}{x}=0.1$
$x=2.4$