Select whether there is no work done by the system, work done by the system or work done by the surroundings for each system described below:

A. H2O (l) ⟶ H2O (g) ΔH = 44.0 kJ/mol B. 2 NO (g) + O2 (g) ⟶ 2 NO2 (g) ΔH = - 114.1 kJ/mol C. Cl (g) + O3 (g) ⟶ ClO (g) + O2 (g) ΔH = - 163 kJ/mol D. CaCO3 (s) ⟶ CaO (s) + CO2 (g) ΔH = 110.1 kJ/mol E. 4 NH3 (g) + 5 O2 (g) ⟶ 4 NO (g) + 6 H2O (l) ΔH = - 906 kJ/mol

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.