Question

A mixture containing 3.9 moles of NO and 0.88 mole of CO₂ was allowed to react in a flask at a certain temperature according to the equation:

NO(g) + CO₂(g) ⇌ NO₂(g) + CO(g)

At equilibrium, 0.11 mole of CO₂ was present. Calculate the equilibrium constant K_c of the reaction.

Answer 1

       NO(g) + CO2(g) ⇌ NO2(g) + CO(g)

Initial        3.9      0.88     0        0
Change         -x       -x       x        x
Equilibrium    (3.9-x) (0.88-x) (x)   (x)

Given equilibrium CO2 = 0.88-x= 0.11
So x = 0.77 moles
Let the volume = V liter
Equilibrium concentrations are given by
[NO]eq=(3.9-X)/V=(3.9-0.77)/V = 3.13/V
[CO2]eq=0.11/V
[NO2]eq=x/V =0.77/V
[CO]eq=x/V=0.77/V

Kc = ([NO]eq*[CO2]eq)/([NO2]eq*[CO]eq) =(3.13/V*0.11/V)/(0.77*0.77)=0.58