Problem 96 - Chemistry An Atoms First Approach

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?

Step-by-Step Solution

Answer

In summary: - Work is done by the system on the surroundings in reactions \(c\) and \(e\). - Work is done by the surroundings on the system in reactions \(a\) and \(d\). - No work is done in reaction \(b\).

Step 1: Reaction (a) - N2(g) -> N2(l)

As we only need to consider gaseous substances, we can see that the number of gas moles changes from 1 to 0. The volume decreases, so the surroundings do work on the system.

Step 2: Reaction (b) - CO(g) + H2O(g) -> H2(g) + CO2(g)

In this reaction, we have 1 mole of CO(g) and 1 mole of H2O(g) on the reactants' side, and 1 mole of H2(g) and 1 mole of CO2(g) on the products' side. Both reactants and products have 2 moles of gases. Therefore, there is no change in a number of gas moles, and no work is done in this reaction.

Step 3: Reaction (c) - Ca3P2(s) + 6H2O(l) -> 3Ca(OH)2(s) + 2PH3(g)

Here, only the product PH3 is a gaseous substance. The change in total moles goes from 0 to 2 moles of gases. As the volume increases, work is done by the system on the surroundings.

Step 4: Reaction (d) - 2CH3OH(l) + 3O2(g) -> 2CO2(g) + 4H2O(l)

In this reaction, we have 3 moles of O2(g) on the reactants' side and 2 moles of CO2(g) on the products' side. So, there is a decrease in the number of gas moles from 3 to 2. The volume decreases, and the surroundings do work on the system.

Step 5: Reaction (e) - I2(s) -> I2(g)

For this reaction, the change in the number of moles goes from 0 to 1 mole of gas. Thus, the volume increases, and work is done by the system on the surroundings. In summary: - Work is done by the system on the surroundings in reactions (c) and (e). - Work is done by the surroundings on the system in reactions (a) and (d). - No work is done in reaction (b).