Question

At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms
$I_2\left(g\right)\leftrightarrow 2l\left(g\right)$
Calculate $K_p$ for the equilibrium.

Answer 1

Partial pressure of I atoms,
$P_1=\frac{40}{100}\times p_{total}=\frac{40}{100}\times {10}^5=4\times {10}^{4}Pa$
Partial pressure of $I_2$ molecules,
$P_1=\frac{60}{100}\times p_{total}=\frac{60}{100}\times {10}^5=6\times {10}^{4}Pa$
Now, for the given reaction,
$K_p=\frac{(pI{)}^2}{P_{I_2}}=\frac{(4\times {10}^4{)}^{2}P{a}^2}{6\times {10}^{4}Pa}=2.67\times {10}^{4}Pa$