Question

A mixture of 0.01484 mol of H2O, 0.07019 mol of CH4, 0.01819 mol of CO, and 0.09015 mol of H2 is placed in a 1.0-L steel pressure vessel at 1239 K. The following equilibrium is established:

1 H2O(g) + 1 CH4(g)--> 1 CO(g) + 3 H2(g)

At equilibrium 0.01097 mol of H2O is found in the reaction mixture.

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

Peq(H2O) =

Peq(CH4) =

Peq(CO) = .

Peq(H2) = .

(b) Calculate KP for this reaction.

KP =

Answer 1

initial concentrations:

[H2O] = 0.01484

[CH4] = 0.07019

[CO] = 0.01819

[H2] = 0.09015

in equilibrium:

[H2O] = 0.01484 - x

[CH4] = 0.07019 -x

[CO] = 0.01819 + x

[H2] = 0.09015 +3x

then, we know tht

[H2O] = 0.01097 in equilibrium

so get x:

[H2O] = 0.01484 - x = 0.01097

x = 0.01484 -0.01097 = 0.00387 M

so...

[H2O] = 0.01484 - 0.00387 = 0.01097

[CH4] = 0.07019 -0.00387 = 0.06632

[CO] = 0.01819 + 0.00387 = 0.02206

[H2] = 0.09015 +3*0.00387 = 0.10176

so...

PV = nRT

P = n/V * RT and n/V = M

so

P = m*RT

P-H2O = 0.01097*0.082*1239 = 1.1145

P-CH4 = 0.06632*0.082*1239 = 6.737

P-CO = 0.02206*0.082*1239 = 2.2412

P-H2 = 0.10176*0.082*1239 = 10.338

so...

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

Kp = 2.2412 *(10.338^3)/ ( 1.1145*6.737)

Kp = 329.79