Consider the equilibrium below. H2S(g) + I2 (s) 2HI(g) + S(s) Kp =1.34x10-5 @60°

ID: 1026933 • Letter: C

Question

Consider the equilibrium below.
H2S(g) + I2 (s) 2HI(g) + S(s) Kp =1.34x10-5 @60°C

P(H2S)initial = 0.010 atm P(HI)initial = 0.00010 atm

What is the algebraic form of Qp for the reaction above? (note: use partial pressures, not concentrations).

What is the numerical value of Qp using the initial values of the partial pressures provided above?

How does Qp compare to Kp? Is it less than, greater than, or equal to Kp ?

Which direction will the reaction proceed to reach equilibrium?

What are the equilibrium partial pressures of the gases involved in the reaction above?

If the volume of the container is increased, how will this shift the equilibrium once it has been established?

What is the value of DG0 for this reaction?

Draw the free energy vs. reaction progress diagram for this reaction.

Part a

algebraic form of Qp = PHI2 /( PH2S)

Part b

numerical value of Qp = (0.00010)2 / (0.010)

= 10^-6

Part C

Kp > Qp

1.34x10^-5 > 10^-6

Kp is greater than Qp

Part d

The reaction shift to the right

Increase PHI and decrease PH2S

PART E

H2S(g) + I2 (s) 2HI(g) + S(s)

I 0.010 0.00010

C - x +2x

E 0.010-x 0.00010+2x

Kp =1.34x10-5 = (0.00010+2x)2 /(0.010-x)

0.000000134 - 0.0000134x = 10^8 + 4x2 + 0.0004x

x = 0.00013181970

Equilibrium partial pressure

PHI = 0.00010+2*0.00013181970 = 0.0003636394

= 0.0003636

PH2S = 0.010-0.00013181970 = 0.0098681803

= 0.009868