Sodium metal (Na) is reacted with H2O in a rigid cylinder with a piston that can move up and down depending on the pressure in the cylinder and the external pressure. The equation for the reaction is:2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) ∆H° = –386.6 kJ mol–1 What will happen to the work (w) and internal energy (∆U) of the system in the cylinder after the reaction reaches completion?单项选择题
A
a. w is positive and ∆U is unchanged
B
b. w is negative and ∆U is negative
C
c. w is positive and ∆U is negative
D
d. w is positive and ∆U is positive

Question

Sodium metal (Na) is reacted with H2O in a rigid cylinder with a piston that can move up and down depending on the pressure in the cylinder and the external pressure. The equation for the reaction is:2 Na (s) + 2 H2O (l) → 2 NaOH (aq) + H2 (g) ∆H° = –386.6 kJ mol–1 What will happen to the work (w) and internal energy (∆U) of the system in the cylinder after the reaction reaches completion?单项选择题

A

a. w is positive and ∆U is unchanged

B

b. w is negative and ∆U is negative

C

c. w is positive and ∆U is negative

D

d. w is positive and ∆U is positive

BlackTom AI · Accepted Answer

Question restatement: A stoichiometric reaction occurs in a piston-cylinder setup where Na(s) reacts with H2O(l) to form NaOH(aq) and H2(g) with an enthalpy change of ΔH° = –386.6 kJ per reaction as written. The system is not fixed in volume because the piston can move in response to pressures. The prompt asks what happens to the reaction work w and the internal energy ΔU of the system once the reaction finishes.

Option by option analysis:

Option a: w is positive and ΔU is unchanged
- A positive w would mean the system is doing work on the surroundings (expansion). In this setup, the reaction produces a gas (H2) and the piston is free to move, so expansion work is possible. However, the internal energy change ΔU being unchanged (zero) would imply no net energy transfer as heat or work, which contradicts the exothermic nature of the reaction (ΔH° negative) and the fact that there is gas evolution that typically leads to PV work. Therefore this option is inconsistent with the thermodynamics of an exothermic reaction with gas evolution in a piston system.

Option b: w is negative and ΔU is negative
- Negative w indicates that the system does work on the surroundings (expansion). This aligns with gas production (H2) generating pressure that can push the piston outward, performing PV work on the surroundings. ΔU being negative implies that the internal energy of the system decreases overall. For an exothermic reaction, heat is released to the surroundings (q is negative at constant pressure, q = ΔH for a constant-pressure process), and some of the released energy goes into doing work (PV work). Since ΔU = q + w, if q is negative (approximately the enthalpy change adjusted for the PV work) and w is negative, ΔU is also negative. This option is consistent with the physics of the system: heat is released and the gas expansion does work, both contributing to a decrease in internal energy.

Option c: w is positive and ΔU is negative
- A positive w would mean the system is doing work on the surroundings? No, positive work typically denotes work done on the system by the surroundings in some conventions, but standard chemistry convention uses w < 0 for expansion (system doing work on surroundings). If we interpret w as the work done by the system (typical convention), positive w would suggest compression, which contradicts the production of a gas and piston expansion. Even if one adopted a different sign convention, the overall scenario involves expansion and energy release as heat; having w positive while the system expands would be inconsistent with the standard sign convention used in thermodynamics and with the physical picture of gas formation driving piston movement outward.
- Additionally, ΔU being negative while w is positive would require q to be sufficiently negative to satisfy ΔU = q + w; while possible in principle, it conflicts with the usual arrangement for this reaction where the gas evolution and heat release accompany expansion and a net decrease in internal energy. This combination is therefore unlikely.

Option d: w is positive and ΔU is positive
- This would imply the system both gains energy and does work on the surroundings, which is incompatible with the exothermic reaction (which releases energy as heat) and with gas formation driving expansion. A positive ΔU would mean the internal energy increases, contrary to the fact that the reaction releases enthalpy and energy tends to leave the system as heat or as PV work, not accumulate in the system. This option does not reflect the energetics of a spontaneous exothermic reaction with gas evolution in a piston setup.

Synthesis of reasoning across options: The reaction releases heat (negative ΔH), and gas production tends to push the piston outward, performing work on the surroundings (negative w in the standard convention where w is work done by the system). With q (heat exchange) being negative at constant pressure for an exothermic process, and w being negative due to expansion, the internal energy ΔU = q + w becomes negative as well. Therefore, the combination 'work negative (expansion) and internal energy negative' is consistent with both the heat release and the PV work done by the system during gas evolution.

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