Question

I got 1.11 x 10^-6 for K.

Here is my work:

x= 0.004

-2x= -0.008 (change for CH4)

Equilibrium for CH4 is 0.087 - 0.008= 0.079

Change in eq for C2H2 is 0.004 and eq is 0.004 (=x)

Change in eq for H2 is 0.012

Eq for H2 is 0.012

K= [C2H2][H2]^3/[CH4]^2
K= (0.004)(0.012)^3/(0.079)^2
K= 1.11 x 10^-6




The reaction of CH4 in

2CH4(g) ---->/<---- C2H2(g) + 3 H2 (g)

is carried out at a different temperature with an initial concentration of [CH4]=0.087M. At equilibrium, the concentration of H2 is 0.012 M. Find the equilibrium constant at this temperature.
Chemistry - DrBob222, Monday, February 29, 2016 at 10:59am
.........2CH4 ==> C2H2 + 3H2
I ......0.087......0......0
C........-2x.......x......3x
E.....0.087-2x.....x......3x

The problem tells you that 3x = 0.012 which allows you to calculate x and 0.087-2x (as well as 3x). Substitute those values into the Keq expression and evaluate Keq.

Answer 1

That looks ok to me.