Question

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What of $K_P$ for the given equilibrium?
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$ :

• A. 6
• B. 16
• C. 4
• D. 2

Answer 1

Answer: (C)

The equilibrium reaction is shown below.
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$

	$HI$	$H_2$	$I_2$
Initial	0.2	0	0
Change	-2x	x	x
Equilibrium	0.2-2x	x	x

But the equilibrium pressure of $HI$ is $0.04$ atm.
$0.2-2x=0.04$
Hence, $2x=0.2-0.04=0.16$ or $x=0.08$.
The expression for the equilibrium constant is $K_P=\frac{P_{H_2}{P}_{I_2}}{P_{HI}^2}=\frac{0.08\times 0.08}{{0.04}^2}=4$.