We are being asked to find the pressure of I_{2} when the system returns to equilibrium. The given reaction is:

**I _{2}(g) **

We're going to calculate the pressure of I_{2} using the following steps:

**Step 1: **Calculate the equilibrium constant, K_{p}.**Step 2:** Calculate the pressure after the system was compressed to half its volume.**Step 3: **Calculate Q and determine the direction of reaction to reestablish the equilibrium.**Step 4: **Set up an ICE chart and calculate the equilibrium pressure of I_{2}.

**Step 1: **Calculate the equilibrium constant, K_{p}.

Given at equilibrium:

P I_{2}(g) = 0.27 atm

P I(g) = 0.26 atm

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{{\mathbf{\left(}{\mathbf{P}}_{\mathbf{I}}\mathbf{\right)}}^{\mathbf{2}}}{{\mathbf{P}}_{{\mathbf{I}}_{\mathbf{2}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{26}\mathbf{)}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{27}}$

A system at equilibrium contains I_{2}(g) at a pressure of 0.27 atm and I(g) at a pressure of 0.26 atm . The system is then compressed to half its volume.

Find the pressure of I when the system returns to equilibrium.

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