To determine the equilibrium constant [tex]\( K_c \)[/tex] for the given reaction:
[tex]\[ \text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g) \][/tex]
at equilibrium, you need to use the equilibrium concentrations of the reactants and products. The formula for the equilibrium constant [tex]\( K_c \)[/tex] in terms of concentration (in molarity, M) is given by:
[tex]\[ K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} \][/tex]
where:
- [tex]\([ \text{PCl}_3 ]\)[/tex] is the molar concentration of phosphorus trichloride ([tex]\(\text{PCl}_3\)[/tex]) at equilibrium,
- [tex]\([ \text{Cl}_2 ]\)[/tex] is the molar concentration of chlorine ([tex]\(\text{Cl}_2\)[/tex]) at equilibrium,
- [tex]\([ \text{PCl}_5 ]\)[/tex] is the molar concentration of phosphorus pentachloride ([tex]\(\text{PCl}_5\)[/tex]) at equilibrium.
Given the equilibrium concentrations:
[tex]\[ \left[ \text{PCl}_3 \right] = 0.20 \, \text{M} \][/tex]
[tex]\[ \left[ \text{Cl}_2 \right] = 0.25 \, \text{M} \][/tex]
[tex]\[ \left[ \text{PCl}_5 \right] = 1.20 \, \text{M} \][/tex]
Now substitute these values into the formula for [tex]\( K_c \)[/tex]:
[tex]\[ K_c = \frac{1.20 \, \text{M}}{(0.20 \, \text{M})(0.25 \, \text{M})} \][/tex]
Calculate the denominator first:
[tex]\[ (0.20 \, \text{M}) \times (0.25 \, \text{M}) = 0.05 \, \text{M}^2 \][/tex]
Then, divide the concentration of [tex]\( \text{PCl}_5 \)[/tex] by this result:
[tex]\[ K_c = \frac{1.20 \, \text{M}}{0.05 \, \text{M}^2} \][/tex]
[tex]\[ K_c = 24 \, \text{M}^{-1} \][/tex]
Expressing this answer using two significant figures:
[tex]\[ K_c = 24 \][/tex]
Thus, the equilibrium constant [tex]\( K_c \)[/tex] for this reaction is [tex]\( 24 \)[/tex].
