Question

H₂(g) + I₂(g) <-> 2HI(g)

K_p= 0.456 at 398K

Assume that initially there are n moles of HI in a varaible volume priston and no moles of H₂ and I₂ are present.

a) Find the partial pressures of HI, HI₂, and I₂ when the piston is set to give 1.00 bar total pressure.

b) Find Δ_rG^o

Answer 1

1.00 bar = P HI + P I₂ + P H₂

Reaction:________ 2HI ______ ⇌ H₂____+___I₂
Initial pressure____1.00atm_____0 atm______0 atm
Change__________-2x_________+x________...
Final pressure____1.00 - 2x_____x_________x
Using Kp and the final pressures we obtain:

Kp = P_I₂ * P_H₂ / (P_HI) ^2
0.456 = x * x / (1.0 - 2x) ^2

0.456 = x^2 / (1.0 - 2x) ^2

0.675 = x/1.0-2x

x= 0.287

Now the partial pressures of

HI =1.0-2x = 1.0-2*0.287 = 0.426 bar,

HI₂, = I2 = 0.287 bar,

b) Find Δ_rG^o

Δ_rG^o = –RT ln(K_p)

Δ_rG^o = –RT ln(K_p)

Here K_p= 0.456 at 398K and R = 8.3145 J / K-mol

Δ_rG^o = –8.3145 J / K-mol * 398 K ln(0.456)

Δ_rG^o= 2598.6 J / mol or 2.6 KJ /mol