Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
MediumAt a certain temperature and total pressure of $10^{5} \ Pa$,iodine vapour contains $40 \%$ by volume of $I$ atoms.
$I_{2(g)} \longleftrightarrow 2I_{(g)}$
Calculate $K_{p}$ for the equilibrium.
$I_{2(g)} \longleftrightarrow 2I_{(g)}$
Calculate $K_{p}$ for the equilibrium.
The partial pressure of a gas is proportional to its volume fraction in the mixture.
Partial pressure of $I$ atoms,$p_{I} = 0.40 \times 10^{5} \ Pa = 4 \times 10^{4} \ Pa$.
Partial pressure of $I_{2}$ molecules,$p_{I_{2}} = 0.60 \times 10^{5} \ Pa = 6 \times 10^{4} \ Pa$.
For the equilibrium $I_{2(g)} \longleftrightarrow 2I_{(g)}$,the equilibrium constant $K_{p}$ is given by:
$K_{p} = \frac{(p_{I})^{2}}{p_{I_{2}}}$
Substituting the values:
$K_{p} = \frac{(4 \times 10^{4})^{2}}{6 \times 10^{4}}$
$K_{p} = \frac{16 \times 10^{8}}{6 \times 10^{4}}$
$K_{p} = 2.67 \times 10^{4} \ Pa$.
Partial pressure of $I$ atoms,$p_{I} = 0.40 \times 10^{5} \ Pa = 4 \times 10^{4} \ Pa$.
Partial pressure of $I_{2}$ molecules,$p_{I_{2}} = 0.60 \times 10^{5} \ Pa = 6 \times 10^{4} \ Pa$.
For the equilibrium $I_{2(g)} \longleftrightarrow 2I_{(g)}$,the equilibrium constant $K_{p}$ is given by:
$K_{p} = \frac{(p_{I})^{2}}{p_{I_{2}}}$
Substituting the values:
$K_{p} = \frac{(4 \times 10^{4})^{2}}{6 \times 10^{4}}$
$K_{p} = \frac{16 \times 10^{8}}{6 \times 10^{4}}$
$K_{p} = 2.67 \times 10^{4} \ Pa$.
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