Work done by reactions

Question

Consider the following changes.

1. H₂O(g) → H₂O(s)
2. H₂O₂(l) → H₂O₂(g)
3. H₂(g) + F₂(g) → 2 HF(g)
4. 2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l)
5. Mg₃P₂(s) + 6 H₂O(l) → 3 Mg(OH)₂(s) + 2 PH₃(g)

(a)

At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? (Select all that apply.)

(b)

In which of these changes is work done by the surroundings on the system? (Select all that apply.)

(c)

In which of these changes is no work done? (Select all that apply.)

Answer 1

Under these conditions, work = - pressure x change in volume or w = -P∆V

The convention is that if w is negative, work is done by the system and if w is positive, work is done on the system. So, for the reactions that follow, we need to find if the volume increases or decreases or stays the same (P is constant)

1.  H₂O(g) → H₂O(s) ∆V is negative, w is positive. Work is done on the system
2. H₂O₂(l) → H₂O₂(g) ∆V is positive, w is negative. Work is done by the system
3. H₂(g) + F₂(g) → 2 HF(g) ∆V is zero. No work is done
4. 2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l) ∆V is negative, w is positive. Work is done on system
5. Mg₃P₂(s) + 6 H₂O(l) → 3 Mg(OH)₂(s) + 2 PH₃(g) ∆V is positive, w is negative. Work is done by system