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)...
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!!
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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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