# When 1.50 mol CO_2 and 1.50 mol H_2 are placed in a 0.750-L container at 395^0 C, `CO_2(g) + H_2(g)harrCO(g) + H_2O(g)` If K_c = 0.802, What are the concentrations of each substance in...

When 1.50 mol CO_2 and 1.50 mol H_2 are placed in a 0.750-L container at 395^0 C,

`CO_2(g) + H_2(g)harrCO(g) + H_2O(g)`

If K_c = 0.802, What are the concentrations of each substance in equilibrium mixture?

### 1 Answer | Add Yours

`CO_2+H_2 harr CO+H_2O`

Initial mol 1.5 1.5 - -

Reacted mol -x -x x x

Equilibrium 1.5-x 1.5-x x x

`K_c = {[CO][H_2O]}/{[CO_2][H_2]}`

For the equilibrium solution;

`[CO_2] [H_2] = (1.5-x)/0.75` mol/l `= (3(1.5-x))/4` (mol)/l

`[CO] = [H_2O] = x/0.75 ` mol/l `= (3x)/4` (mol)/l

Only liquids and solids are omitted in concentration. But here we have to take `H_2O ` since it is a gas.

`K_c = {[CO][H_2O]}/{[CO_2][H_2]}`

`K_c = ((3x)/4)^2/((3(1.5-x))/4)^2`

`K_c = x^2/(1.5-x)^2`

`0.802 = x^2/(1.5-x)^2`

`sqrt0.802 = +-x/(1.5-x)`

`sqrt0.802 = +x/(1.5-x)`

`sqrt0.802 = -x/(1.5-x)`

Solving the above will give you x = 0.66759 or x = -6.076

Since x>0 then x = 0.6679

*So the final concentration would be;*

`[CO_2] [H_2] = (1.5-x)/0.75 = 1.109(mol)/l`

`[CO] = [H_2O] = x/0.75 = 0.89 (mol)/l`

**Sources:**