Question:
At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms
$\mathrm{I}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{I}(\mathrm{g})$
Calculate Kpfor the equilibrium.
Solution:
Partial pressure of I atoms,
$p_{1}=\frac{40}{100} \times p_{\text {total }}$
$=\frac{40}{100} \times 10^{5}$
$=4 \times 10^{4} \mathrm{~Pa}$
Partial pressure of I2 molecules,
$p_{\mathrm{l}_{2}}=\frac{60}{100} \times p_{\text {total }}$
$=\frac{60}{100} \times 10^{5}$
$=6 \times 10^{4} \mathrm{~Pa}$
Now, for the given reaction,
$K_{p}=\frac{(p \mathrm{I})^{2}}{p_{1_{z}}}$
$=\frac{\left(4 \times 10^{4}\right)^{2} \mathrm{~Pa}^{2}}{6 \times 10^{4} \mathrm{~Pa}}$
$=2.67 \times 10^{4} \mathrm{~Pa}$