Question

Tetrafluoroethene is made from chlorodifluoromethane in this reversible reaction. 2 CHCIF₂(g) = C₂F4(g) + 2HCl(g) AH = +128 kJ mol¹ A 2.00 mol sample of CHCIF₂ is placed in a container of volume 23.2 dm³ and heated. When equilibrium is reached, the mixture contains 0.270 mol of CHCIF2 Calculate the amount, in moles, of C₂F4 and of HCI in the equilibrium mixture.​

Answer 1

Let the amount of C₂F4 and HCl produced at equilibrium be x mol.

Using the stoichiometry of the balanced chemical equation, we can write the equilibrium expression for the reaction as follows:

Kc = ([C₂F4][HCl]²) / [CHCIF₂]²

where Kc is the equilibrium constant, [C₂F4], [HCl], and [CHCIF₂] are the equilibrium concentrations of each species in mol dm⁻³.

At equilibrium, the concentration of CHCIF₂ is 0.270 mol / 23.2 dm³ = 0.0116 mol dm⁻³.

We can use the equilibrium expression and the given equilibrium constant to solve for the concentrations of C₂F4 and HCl at equilibrium:

Kc = ([C₂F4][HCl]²) / [CHCIF₂]²

128 × 10³ = (x)(2x²) / (0.0116)²

Solving for x, we get:

x = 0.153 mol

Therefore, at equilibrium, the amount of C₂F4 produced is 0.153 mol and the amount of HCl produced is 0.306 mol (twice the amount of C₂F4, according to the stoichiometry of the balanced chemical equation).