# Exactly 2.00 mol of each gas was placed in a 200.0 L flask and the mixture was allowed to react. Find the equilibrium concentration of each gas.The reaction is: CO(g) + H2O(g) <--> CO2(g) +...

Exactly 2.00 mol of each gas was placed in a 200.0 L flask and the mixture was allowed to react. Find the equilibrium concentration of each gas.

The reaction is:

CO(g) + H2O(g) <--> CO2(g) + H2(g)

Keq = 0.36

Ok so I converted the moles into molarity, so all the gases have a concentration of 0.01. But I don't know what to do next. :( How do I find the equilibrium concentrations if the change in concentration isn't given? Help please!!

### 1 Answer | Add Yours

So far what you’re doing is correct. The next step is to establish the equilibrium equation. First we rewrite the equation:

CO + H2O <--> CO2 + H2

Then we draw the ICE table.

CO + H2O <--> CO2 + H2

I 0.01 0.01 0 0

C -x -x x x

-------------------------------------------

E 0.01-x 0.01-x x x

The equilibrium expression for this reaction is:

keq = [CO2][H2]/[CO][H2O]

keq = [x][x]/[0.01-x][0.01-x] = 0.36

Next, simplify the equation.

0.36 = x^2 / (0.01-x)^2

0.36 = x^2 / 0.0001 - 0.02x + x^2

0.36 (0.0001 - 0.02x + x^2) = x^2

3.6x10^-5 - 7.2x10^-3 + 0.36x^2 = x^2

Simplify the equation in quadratic form.

x^2 - 0.36x^2 + 7.2x10^-3 -3.6x10^-5 = 0

using the quadratic equation, the values for x are:

**x = 3.75x10^-3**

x = -1.5x10^-2

We take the positive x value.

**x = 3.75X10^-3 = [CO2] = [H2]**

[CO] = [H2O] = 0.01-x = 0.01 - 3.75x10^-3

**[CO] = [H2O] = 6.25x10^-3**