Answer:
(i) CO = 0.4 mol; H₂O = 1.6 mol; Kc = 4
(ii) CO = 0.67 mol; H₂O = 0.67 mol; CO₂ = 1.33 mol; H₂ = 1.33 mol
Explanation:
(i) For the equation given let's make a table of the concentrations for equilibrium (the volume is constant, so, we can do it with moles number)
CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)
2.0 mol 3.2 mol 0 0 <em>Initial</em>
-x -x +x +x <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)
2.0-x 3.2-x x x <em>Equilibrium</em>
In the equilibrum, the moles number of hydrogen and carbon dioxide are 1.6 mol, so x = 1.6 mol
The amounts of CO and H₂O are:
CO = 2.0 - 1.6 = 0.4 mol
H₂O = 3.2 - 1.6 = 1.6 mol
The constant of the equilibrium is the multiplications of the concentrations of products divided by the multiplication of the concentration of the reactants (all the concentrations elevated to the coefficient). So:
Kc = (1.6x1.6)/(0.4x1.6)
Kc = 1.6/0.4
Kc = 4
(ii) Kc must remais constant (it only changes with the temperature), so let's construct a new table of equilibrium:
CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)
2.0 mol 2.0 mol 0 0 <em>Initial</em>
-x -x +x +x <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)
2.0-x 2.0-x x x <em>Equilibrium</em>
Kc = (x*x)/((2.0-x)*(2.0-x))
4 = x²/(4 - 4x + x²)
16 - 16x + 4x² = x²
3x² - 16x + 16 = 0
Using Baskhara's equation:
Δ =(-16)² - 4x3x16
Δ = 256 - 192
Δ = 64
x = (-(-16) +/- √64)/(2*3)
x' = (16 + 8)/6 = 4
x'' = (16 - 8)/6 = 1.33
x must be small than 2.0, so x = 1.33 mol, which is the amount of hydrogen and carbon dioxide at equilibrium. The both reactants has 2.0 - 1.33 = 0.67 mol at equilibrium.
