Question

Calculate the value of the equilibrium constant, Kc , for the reaction below, if 0.1908 moles of CO2, 0.0908 moles of H2, 0.0092 moles of CO, and 0.0092 moles of H2O vapor were present in a 2.00-liter reaction vessel at equilibrium.
CO2 (g) + H2 (g) 2-way arrow CO (g) + H2O (g)


Keq = 4.89 x 10-3

Keq = 1.13

Keq = 8.87 x 10-1

Keq = 205

Answer 1

C = N/V

calculate the concentration for each compound

given that V = 2 liters

N(Co2) = 0.0908 , and so on

then use the formula

Kc = [CO] * [H2O] / [CO2] * [H2]

Answer 2

I'm not entirely sure how this type of problem works since math is more my specialty (not chem) but this page may help

http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_Equilibrium_Constants.htm

Answer 3

@jim_thompson5910, Actually he is too lazy to use the formulas, I wrote.

[CO2] = 0.1908 / 2 = 0.0954 mol/l
[H2] = 0.0908 /2 = 0.0454 mol/4
[CO] = 0.0092 / 2= 4.6 x 10^-3 mol/l
[[H20] = 0.0092 / 2 = 4.6 x 10^-3 mol/l


Kc = ( 4.6 x 10^-3 ) * ( 4.6 x 10^-3) / ( 0.0954 * 0.0454)
= 4.89 x 10^-3