Homework Answers
The equilibrium reaction is given as
H2 (g) + CO2 (g) <=======> H2O (g) + CO (g)
The initial concentrations are obtained by dividing the number of moles of the gaseous reactants by the volume of the container.
[H2]ini = (0.492 moles)/(6.00 L) = 0.082 mol/L = 0.082 M
[CO2]ini = (0.492 moles)/(6.00 L) = 0.082 mol/L = 0.082 M
Set up the ICE chart as below.
H2 (g) + CO (g) <=========> H2O (g) + CO (g)
Initial (M) 0.082 0.082 - -
Change (M) -x -x +x +x
Equilibrium (M) (0.082 – x)(0.082 – x) x x
Write down the expression for the equilibrium constant as below.
Kc = [H2O][CO]/[H2][CO2]
where the square braces denote molar concentrations at equilibrium.
Plug in the values and get
0.798 = (x)(x)/(0.082 – x)(0.082 – x)
=======> 0.798 = x2/(0.082 – x)2
Take the positive square root of both the sides and get
0.8933 = x/(0.082 – x)
=======> 0.8933*(0.082 – x) = x
=======> 0.0732506 – 0.8933x = x
=======> 1.8933x = 0.0732506
=======> x = 0.0732506/1.8933 = 0.03869 ≈ 0.0387
The equilibrium concentration of H2 is (B) 0.0387 M (ans).
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