I couldn\'t get part B. Anyone can help? Thanks! Consider the following reaction

Question

I couldn't get part B. Anyone can help? Thanks!

Consider the following reaction:
H2(g)+I2(g)2HI(g)
A reaction mixture at equilibrium at 175 K contains PH2=0.958atm, PI2=0.877atm, and PHI=0.020atm. A second reaction mixture, also at 175 K, contains PH2=PI2= 0.615 atm , and PHI= 0.103 atm.

Part A

Is the second reaction at equilibrium?

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Correct

Part B

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

0.0166

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ANSWER 3: Deduction: -3%

0.0134

(wrong again!) =

Explanation / Answer

H2(g)+I2(g)2HI(g)

Kp = pHI^2 / pH2*pI2

    = (0.02^2) / (0.877*0.958)

     = 0.000476

qp = (0.103^2)/(0.615^2) = 0.02805

as qp is different from Kp . second reaction mixture is not at

equilibrium.

kp = 0.000476 = (0.103-x)^2 / (0.615+x)^2

x = 0.088

partial pressure of HI = 0.103-0.088 = 0.015 atm

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