For Part a), we're being asked to** determine if the second reaction in equilibrium**. We're given the following reaction:

**H _{2}(g) + I_{2}(g) ⇌ 2 HI(g) **

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

Given at equilibrium:

P_{H}_{2} = 0.958 atm

P_{I}_{2} = 0.877 atm

P_{HI} = 0.020 atm.

$\overline{){{\mathbf{K}}}_{{\mathbf{p}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{{\mathbf{\left(}\mathbf{HI}\mathbf{\right)}}^{\mathbf{2}}}{\left({H}_{2}\right)\left({I}_{2}\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{)}}^{\mathbf{2}}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{958}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{877}\mathbf{)}}$

Consider the following reaction: H_{2}(g) + I_{2}(g) ⇌ 2 HI(g)

A reaction mixture at equilibrium at 175 K contains P_{H}_{2} = 0.958 atm, P_{I}_{2} = 0.877 atm, and P_{HI} = 0.020 atm.

A second reaction mixture, also at 175 K, contains P_{H}_{2} = P_{I}_{2} = 0.614 atm , and P_{HI} = 0.105 atm.

a. Is the second reaction in equilibrium? yes/no

b. If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

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