Answer
$ P_{ COCl_2 } = 0.703 \space atm$
$ P_{ CO }=0.0570 \space atm$
$ P_{ Cl_2 }= 0.0570 \space atm$
Work Step by Step
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{P_{ CO }P_{ Cl_2 }}{P_{ COCl_2 }}$$
2. At equilibrium, these are the concentrations of each compound:
$ P_{ COCl_2 } = 0.760 \space M - x$
$ P_{ CO } = 0 \space M + x$
$ P_{ Cl_2 } = 0 \space M + x$
$$4.63 \times 10^{-3} = \frac{(x)(x)}{(0.760 - x)}$$
x = 0.0570
$ P_{ COCl_2 } = 0.760 \space M - 0.0570 M = 0.703 atm$
$ P_{ CO } = 0 \space M + 0.0570 M = 0.0570 atm$
$ P_{ Cl_2 } = 0 \space M + 0.0570 M = 0.0570 atm$
