Question

Consider these changes.

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.