Question

The reaction of the formation of phosgene gas from $CO$ and ${Cl}_2$ is as follows:
$CO\left(g\right)+{Cl}_2\left(g\right)\rightleftharpoons CO{Cl}_2\left(g\right)$
In an experiment, starting with equimolecular $CO$ and ${Cl}_2$ in a $250mL$ flask, the equilibrium mixture on analysis is found to contain $0.5mole$ $CO$, $0.05$ mole ${Cl}_2$ and $0.15$ mole $CO{Cl}_2$. Calculate the equilibrium constant of the reaction.

Answer 1

The equilibrium concentrations are
$\left[CO\right]=\frac{\text{0.5 mol}}{\text{250 mL}}\times \text{1000 mL/L}=\text{2.0 M}$
$\left[C{l}_2\right]=\frac{\text{0.05 mol}}{\text{250 mL}}\times \text{1000 mL/L}=\text{0.20 M}$
$\left[COC{l}_2\right]=\frac{\text{0.15 mol}}{\text{250 mL}}\times \text{1000 mL/L}=\text{0.60 M}$
The equilibrium constant $K_c=\frac{\left[COC{l}_2\right]}{\left[CO\right]\left[C{l}_2\right]}$
$K_c=\frac{0.60mol/L}{2.0mol/L\times 0.20mol/L}$
$K_c=1.5L/mol$