Question

Consider these changes.

(a) Hg(l)->Hg(g)

(b) 3O2(g)->2O3(g)

(c) CuSo4*5H2O(s)->CuSO4(s)+5H2O(g)

(d) H2(g)+F2(g)->2HF(g)


At constant pressure, in which of the reactions is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Answer 1

Answer:

(a) The system does work on the surroundings.

(b) The surroundings do work on the system.

(c) The system does work on the surroundings.

(d) No work is done.

Explanation:

The work (W) done in a chemical reaction can be calculated using the following expression:

W = -R.T.Δn(g)

where,

R is the ideal gas constant

T is the absolute temperature

Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants

R and T are always positive.

• If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.

• If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.

• If Δn(g) = 0, W = 0, which means that no work is done.

(a) Hg(l) ⇒ Hg(g)

Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.

(b) 3 O₂(g) ⇒ 2 O₃(g)

Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.

(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g)

Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.

(d) H₂(g) + F₂(g) ⇒ 2 HF(g)

Δn(g) = 2 - 2 = 0. W = 0. No work is done.