Question

The Equilibrium constant Kc for the reaction H2(g) + CO2(g) -> H2O(g) + CO(g) is 4.2 at 1650 deg C. Initially .74 mol H2 and .74 mol CO2 are injected into a 4.6-L flask. Calculate the concentration of each species at equilibrium.

H2=

CO2 =

H2O=

CO=

Answer 1

Kc = 4.2

Kc = H2O*CO / (CO2*H2)

initially

CO2 = 0.74/4.6 -x = 0.160869 - x

H2 =  0.74/4.6 -x = 0.160869 - x

H2O = 0 + x

CO = 0 +x

then

Kc = H2O*CO / (CO2*H2)

4.2 = (x*x)/(0.160869 - x)^2

solve for x

sqrt(4.2) = x/(0.160869 - x)

2.04939(0.160869 - x) = x

0.32968331 -2.04939x = x

3.04939x = 0.32968331

x = 0.32968331 /3.04939 = 0.108114

then

CO2 =0.160869 - x = 0.160869 -0.108114 = 0.052755

H2 =0.160869 - x = 0.160869 -0.108114 = 0.052755

H2O = 0 + x = 0.108114

CO = 0 +x =0.108114