Answer:
2.3 M
Explanation:
First, we have to calculate the initial molarities of the substances.
[tex][H_{2}O]=\frac{0.50mol}{0.250L} =2.0M[/tex]
[tex][CO_{2}]=\frac{1.2mol}{0.250L} =4.8M[/tex]
[tex][H_{2}]=\frac{1.3mol}{0.250L} =5.2M[/tex]
In the beginning there is no CO, so the reaction must proceed to the left to attain equilibrium. In order to know the concentrations at equilibrium we will use an ICE chart. "I" stands for Initial, "C" stands for Change and "E" stands for equilibrium. We complete each row with the concentration or change in concentration in that stage.
CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)
I 0 2.0 4.8 5.2
C +x +x -x -x
E x 2.0 + x 4.8 - x 5.2 - x
The equilibrium constant K is:
[tex]K=0.289=\frac{[CO_{2}].[H_{2}]}{[CO].[H_{2}O]} =\frac{(4.8-x).(5.2-x)}{x.(2.0+x)} \\0.289=\frac{25-10x+x^{2} }{2.0x+x^{2} } \\0.289x^{2} +0.58x=25-10x+x^{2}\\0.711x^{2} -10.58x+25=0[/tex]
The answers are x₁ = 11.9 and x₂=2.9. Since x₁ would make concentrations negative (which is not possible) we choose x₂.
The molarity of H₂ is:
5.2 - x = 5.2 -2.9 = 2.3 M
