The equation for another reaction used in industry isco(g) + h₂o(g) h₂(g) + co₂(g), = -42 kj

(i) under certain conditions of temperature and pressure, 2.0 mol of carbon monoxide and 3.2 mol of steam were left to reach equilibrium. at equilibrium, 1.6 mol of both hydrogen and carbon dioxide were present. calculate the amounts of carbon monoxide and steam at equilibrium and the value of kc.
(ii) under the same conditions of temperature and pressure, 2.0 mol of carbon monoxide and 2.0 mol of steam were left to reach equilibrium. calculate the amounts of each reactant and product at equilibrium. (if you were unable to calculate a value for kc in (i) use the value 9.0, although this is not the correct value.)

(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              Initial

-x              -x                +x        +x            Reacts (stoichiometry is 1: 1: 1: 1)

2.0-x       3.2-x            x           x             Equilibrium

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                 Initial

-x              -x             +x         +x               Reacts (stoichiometry is 1: 1: 1: 1)

2.0-x        2.0-x         x           x                Equilibrium

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.

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >