Question

Initially, only H2S is present at a pressure of 0.256 atm in a closed container. What is the total pressure in the container at equilibrium?

Answer 1

Assume:

H2S <-> H2 + S

K = ?

we can calculate K with

dG° = -RT*lnK

K = exp((dG°/(-RT))

dG° = must be calculated as H - TS

R = 8.314 and T = 298 K (we assume this is STP since no other data is givne)

then

dG° = H - TS

H = (H2 + S) - (H2S) = −20.6 kJ/mol = -20600 J/mol

T = 298

S = (H2 + S) - (H2S) = (130.6 + 0.5*31.8 ) - (205.7 ) = -59.2 J/mol

dG° = -20600 - 298*(-59.2) = -2958.4

Then

K = exp((-2958.4/(-0.082*298)) = 3.7918*10^52

Since K >>> 1; expect the equilibrium to shift toward products

that is

H2 and S

then

1 mol of H2S = 1 mol of H2

1 mol of H2S = 1 mol of S

therefore

PH2 = 0.256 atm

PS = 0.256 atm

total pressure

P = 0.256+0.256 = 0.512 atm