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 is K_p for the given equilibrium?
$2HI\left(g\right)\leftrightarrow H_2\left(g\right)+I_2\left(g\right)$

Answer 1

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 - 0.04 = 0.16. The given reaction is:
$\begin{array}{cccccc}& 2H{I}_{\left(g\right)}& \leftrightarrow & H_{2\left(g\right)}& +& I_{2\left(g\right)}\\ Initialconc.& 0.2atm& & 0& & 0\\ Atequilibrium& 0.04atm& & \frac{0.16}{2}& & \frac{2.15}{2}\\ & & & =0.08atm& & =0.08atm\end{array}$
Therefore,
$K_p=\frac{P_{H_2}\times P_{I_2}}{P_{HI}}=\frac{0.08\times 0.08}{(0.04{)}^2}=\frac{0.0064}{0.0016}=4.0$
Hence, the value of $K_p$ for the given equilibrium is 4.0.