#### Answer

The pressures of the reactants will decrease, and the pressure of the products will increase. That happens because the $Q_p$ calculated is less than the $K_p$ for the reaction.

#### Work Step by Step

1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_P = \frac{[Products]}{[Reactants]} = \frac{P_{ PCl_3 }P_{ Cl_2 }}{P_{ PCl_5 }}$$
2. Calculate the $Q_p$ for these pressures:
$$Q_P = \frac{(0.111)(0.223)}{0.177} = 0.140$$
0.140 is less than 1.05; therefore, more product must be formed in order to reach equilibrium, and some reactant will be consumed in the process.
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We can also see this happen by calculating the change using an ice table:
2. At equilibrium, these are the concentrations of each compound:
$ P_{ PCl_5 } = 0.177 \space M - x$
$ P_{ PCl_3 } = 0.223 \space M + x$
$ P_{ Cl_2 } = 0.111 \space M + x$
$$1.05 = \frac{(0.223 + x)(0.111 + x)}{(0.177 - x)}$$
$x_1$ = -1.49
$x_2$ = 0.108
$x_1$ is invalid, because: $0.223 + (-1.49)$ is a negative number, and a partial pressure cannot be negative.
So $x_2$ = 0.108
$P_{PCl_5} = 0.177 - 0.108 = 0.069$ atm
$P_{PCl_3} = 0.223 + 0.108 = 0.331$ atm
$P_{Cl_2} = 0.111 + 0.108 = 0.219$ atm