Question

At a certain temperature and total pressure of ${10}^{5}Pa$, iodine vapour contains $40\%$ by volume of I atoms.
$I_2\left(g\right)\rightleftharpoons 2I\left(g\right)$
Calculate $K_p$ for the equilibrium.

Answer 1

From the Dalten law "The mole fraction of a specific gas in a mixture of gases is equal to the ratio of the partial pressure of that gas to the total pressure exerted by the gaseous mixture"

i.e,. Partial pressure = Total pressure × Mole fraction

Given iodine atoms is 40%, hence mole fraction is $\frac{40}{100}$

The partial pressure of iodine atoms is(40% of volume) $P=\frac{{10}^5\times 40}{100}=0.4\times {10}^{5}Pa$.

Given iodine molecule is 40% - 100= 60%, hence mole fraction is $\frac{60}{100}$
The parital ressure of $I_2$ molecules is (40% - 100= 60% of volume) $P^{\prime }=\frac{{10}^5\times 60}{100}=0.6\times {10}^{5}Pa$.

The value of the equilibium constant is $K_P=\frac{P^2}{P^{\prime }}=\frac{(0.4\times {10}^5{)}^2}{0.6\times {10}^5}=2.67\times {10}^4$.