Question

At constant pressure, in which of the reactions is work done by the system on the surroundings?

A. Hg(l) ---> Hg(s)
B. 3 O₂(g) ---> 2 O₃(g)
C. CuSO₄ x 5 H₂O(s) ---> CuSO₄(s) + 5 H₂O(g)
D. H₂(g) + F₂(g) ---> 2 HF(g)

Answer 1

Work is done by the system on the surroundings in reaction C, where CuSO4 × 5 H2O(s) turns into CuSO4(s) and H2O(g), leading to an expansion and the system performing PV work.

Step-by-step explanation:

At constant pressure, work done by the system on the surroundings occurs when there is an expansion of gases, as the system performs pressure-volume work (PV work). Based on the given options, the reaction where work is done by the system on the surroundings is option C, CuSO₄ × 5 H₂O(s) → CuSO₄(s) + 5 H₂O(g). Here, a solid is converting to gas, increasing the number of moles of gas and therefore, the volume in the system. The increase in volume will push the piston up, performing work on the surroundings as per the equation w = -PΔV, where the work (w) done is negative, indicating energy loss by the system. Options A, B, and D either do not involve a change in the number of moles of gas or result in a decrease in moles of gas, so these would not result in the system doing work on the surroundings.