Question

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Short answer

In summary: - In reactions (c) and (e), the system does work on the surroundings. - In reaction (d), the surroundings do work on the system. - In reactions (a) and (b), there is no work done.

Step-by-step solution

a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\)

: In this reaction, nitrogen gas is transforming into nitrogen liquid. Since it is undergoing a phase transition from gas to liquid, volume decreases, and there is no change in the number of moles. The surroundings are doing work on the system as it contracts.

b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\)

: For this reaction, we need to count the number of gas moles on both sides: Reactant side: 1 (CO) + 1 (H2O) = 2 moles of gas Product side: 1 (H2) + 1 (CO2) = 2 moles of gas The number of gas moles stays constant, and therefore, no work is done in this reaction.

c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\)

: For this reaction, we need to count the number of gas moles on both sides: Reactant side: 0 moles of gas (as none of the reactants is in the gaseous state) Product side: 2 (PH3) = 2 moles of gas As there is an increase in the number of gas moles, the system expands, and therefore, the system does work on the surroundings.

d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

: For this reaction, we need to count the number of gas moles on both sides: Reactant side: 3 (O2) = 3 moles of gas Product side: 2 (CO2) = 2 moles of gas As there is a decrease in the number of gas moles, the system contracts, and the surroundings do work on the system.

e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\)

: In this reaction, iodine is transforming from solid to gas. Since it is undergoing a phase transition, the number of moles remains constant, but the volume increases. The system expands, and therefore, the system does work on the surroundings. In summary, - In reactions (c) and (e), the system does work on the surroundings. - In reaction (d), the surroundings do work on the system. - In reactions (a) and (b), there is no work done.