# Problem: Consider the following reaction: H2(g) + I2(g) ⇌ 2 HI(g) A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 atm, PI2 = 0.877 atm, and PHI = 0.020 atm. A second reaction mixture, also at 175 K, contains PH2 = PI2 = 0.614 atm , and PHI = 0.105 atm.a. Is the second reaction in equilibrium? yes/nob. If not, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

###### FREE Expert Solution

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

H2(g) + I2(g) ⇌ 2 HI(g)

Step 1: Calculate the equilibrium constant, Kp.

Given at equilibrium:

PH2 = 0.958 atm
PI2 = 0.877 atm
PHI = 0.020 atm.

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

85% (13 ratings) ###### Problem Details

Consider the following reaction: H2(g) + I2(g) ⇌ 2 HI(g)

A reaction mixture at equilibrium at 175 K contains PH2 = 0.958 atm, PI2 = 0.877 atm, and PHI = 0.020 atm.

A second reaction mixture, also at 175 K, contains PH2 = PI2 = 0.614 atm , and PHI = 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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