$A$ sample of $HI_{(g)}$ is placed in a flask at a pressure of $0.2 \ atm$. At equilibrium,the partial pressure of $HI_{(g)}$ is $0.04 \ atm$. What is $K_{p}$ for the given equilibrium?
$2 HI_{(g)} \longleftrightarrow H_{2_{(g)}} + I_{2_{(g)}}$

(4.0) The initial pressure of $HI$ is $0.2 \ atm$. At equilibrium,it has a partial pressure of $0.04 \ atm$. The decrease in the pressure of $HI$ is $0.2 - 0.04 = 0.16 \ atm$.
The reaction is: $2HI_{(g)} \longleftrightarrow H_{2_{(g)}} + I_{2_{(g)}}$
Initial pressure: $0.2 \ atm, 0, 0$
At equilibrium: $0.04 \ atm, \frac{0.16}{2} \ atm, \frac{0.16}{2} \ atm$
At equilibrium: $0.04 \ atm, 0.08 \ atm, 0.08 \ atm$
$K_{p} = \frac{p_{H_{2}} \times p_{I_{2}}}{p_{HI}^2}$
$K_{p} = \frac{0.08 \times 0.08}{(0.04)^{2}} = \frac{0.0064}{0.0016} = 4.0$
Hence,the value of $K_{p}$ is $4.0$.