Answer:
A and D: system does work
B and E: surroundings do work
C: No work done by system or surroundings
Explanation:
A system does work on the surroundings if the volume increases, because it is pushing back against the atmosphere.
The volumes of liquids don't change much during a reaction. However, if a gas is absorbed or released at constant pressure, the volume changes significantly.
Thus, we can ignore the solids and liquids and consider only whether there is a change in the number of moles of gas.
The formula for the work done is
w = -pΔV = -ΔnRT where
Δn = n₂ - n₁ = moles of product gases - moles of reacting gases
If Δn is +, w is -, and the system does work on the surroundings.
If Δn is -, w is +, and the surroundings do work on the system.
A. H₂O(ℓ) ⟶ H₂O(g); ΔH = 44.0 kJ/mol
Δn = 1 - 0 = 1
w = -1RT = -RT, so the system does work.
B. 2NO(g) + O₂(g) ⟶ 2NO₂(g); ΔH = -114.1 kJ/mol
Δn = 2 - 3 = -1
w = 1RT = RT, so the surroundings do work.
C. Cl(g) + O₃ (g) ⟶ ClO(g) + O₂(g); ΔH = -163 kJ/mol
Δn = 2 - 2 = 0
w = 0, so the system does no work.
D. CaCO₃(s) ⟶ CaO s) + CO₂(g); ΔH = 110.1 kJ/mol
Δn = 1 - 0 = 1
w = -1RT = -RT, so the system does work.
E. 4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(l); ΔH = -906 kJ/mol
Δn = 4 - 9 = -5
w = 5RT, so the surroundings do work.
