Question

10. A mixture of 0.08729 mol of H₂O, 0.1041 mol of CH₄, 0.08069 mol of CO, and 0.1098 mol of H₂ is placed in a 1.0-L steel pressure vessel at 1034 K. The following equilibrium is established:

1 H₂O(g) + 1 CH₄(g) 1 CO(g) + 3 H₂(g)

At equilibrium 0.06283 mol of CO is found in the reaction mixture.

(a) Calculate the equilibrium partial pressures of H₂O, CH₄, CO, and H₂.

P_eq(H₂O) = ____ .

P_eq(CH₄) = ____ .

P_eq(CO) = ____ .

P_eq(H₂) = ____ .

(b) Calculate K_P for this reaction.

K_P = ____ .

Answer 1

initially:

[H2O] = 0.08729

[CH4] = 0.1041

[CO] = 0.08069

[H2] = 0.1098

after equilibrium is formed:

[H2O] = 0.08729 - x

[CH4] = 0.1041 - x

[CO] = 0.08069 + x

[H2] = 0.1098 + 3x

so..

w ealso know that

[CO] = 0.08069 + x = 0.06283

solve for

x

x = 0.06283 -0.08069 = -0.01786 ; note that CO is actually being reacted and not produced!

so..

[H2O] = 0.08729 - x = 0.08729 + 0.01786 = 0.10515

[CH4] = 0.1041 - x = 0.1041 + 0.01786 = 0.12196

[CO] = 0.08069 + x = 0.08069 -0.01786 = 0.06283

[H2] = 0.1098 + 3x = 0.1098 -3*0.01786 = 0.05622

for total P:

PV = nRT

P = nRT/V = (0.10515+0.12196+0.06283+0.05622)(0.082)(1034)/(1)= 29.350 atm

then..

for partial pressure, find mol fractions

total mol = 0.34616

so..

x-H2O = 0.10515/0.34616= 0.30381

P-H2O = 0.30381*32.378 = 9.8367

x-CH4 = 0.12196/0.34616= 0.35232

P-CH4 = 0.35232*32.378 = 11.407

x-CO = 0.06283/0.34616= 0.181505

P-CO = 0.181505*32.378 = 5.8767

x-H2 = 0.09194/0.34616= 0.2655

P-H2 = 0.2655*32.378 = 8.596359

for Kp :

Kp = P-CO * P-H2^3 / (P-H2O * P-CH4)

Kp = 5.8767 *8.596359^3 / (9.8367 * 11.407)

Kp = 33.270