Calculating equilibrium pressures

Question

Suppose 1.09 atm of CH4(g), 2.22 atm of C2H6(g), and 14.21 atm of O2(g) are placed in a flask at a given temperature. The reactions are given below.

CH4(g) 2 O2(g) CO2(g) 2 H2O(g) KP = 1.0x10^4

2 C2H6(g) 7 O2(g) 4 CO2(g) 6 H2O(g) KP = 1.0x10^8

Calculate the equilibrium pressures of CH4 and C2H6.

I made an ICE table and came up with

Kp = ([5x]^5[8x]^8) / ([1.14-x][2.43-2x]^2[14.67-9x]^9) = 1.0 x 10^12

The top part I simplified to 5.243 x 10^10, but how do I multiply/simplify the bottom part of the equation?

Answer 1

Let us drop (g) label since every chemical here is a gas:

CH4 + 2O2 ==> CO2 + 2H2O, Kp = 1.0x10^4

2C2H6 + 7O2 ==> 4CO2 + 6H2O, Kp = 1.0x10^8

Now, let X (atm) and 2Y(atm) are the REDUCED pressures of CH4 and C2H6.  That is:

CH4 . . + . . 2O2 . ==> . CO2 . + . 2H2O . . . . . (1)

1.09-X . . 14.21-2X-7Y . . X+4Y . . . 2X+6Y

2C2H6 . . + . . 7O2 . ==> . 4CO2 . + . 6H2O . . . (2)

2.22-2Y . . 14.21-2X-7Y . . . X+4Y . . . 2X+6Y

Notice that the concentrations [O2], [CO2] and [H2O] in both (1) and (2) are the same, since they are in the same reactor.  Now, applying the reaction constant:

1.0x10^4 = (X+4Y)*(2X+6Y)^2 /{(1.09-X)* (14.21-2X-7Y)^2} . . . (3)

and

1.0x10^8 = (X+4Y)^4*(2X+6Y)^6 /{(2.22-2Y)^2* (14.21-2X-7Y)^7} . (4)

The order of Eq.(4) is very high.  We may simplify it by (4)/(3)^3:

1.0x10^-4 = (X+4Y)*(1.09-X)^3 /{(2.22-2Y)^2* (14.21-2X-7Y)} . . . . (5)

Solving the coupled Eq.(3) and (5) for X and Y. I cannot get analytical solution since they are 4th order equations.  You have to get numerical solutions.