# 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!! 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    ...

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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

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