Question

Consider the following reaction:
H2(g)+I2(g)⇌2HI(g)
A reaction mixture at equilibrium at 175 Kcontains PH2=0.958atm, PI2=0.877atm, and PHI=0.020atm. A second reaction mixture, also at 175 K, contains PH2=PI2=0.622 atm , and PHI= 0.109 atm

Answer 1

1) H2(g) + I2(g) ⇌ 2HI(g)

partial pressure   0.958   0.877   0.020

Kp = [pHI]² / [ pH2] [ pI2] = [0.020]² / [ 0.958] [ 0.877] = 0.0004 / 0.84 = 0.00047

2)

H2(g) + I2(g) ⇌ 2HI(g)

partial pressure 0.622 0.622 p HI

Kp = [pHI]² / [ pH2] [ pI2] =

0.00047 =[pHI]² / [0.622] [0.622]

[pHI]² =  0.00047 *  [0.622]* [0.622] = 0.0002

pHI = 0.0002 = 0.014

pHI = 0.014 is the partial pressure of HI when the reaction reaches equilibrium at 175K