Problem 41 - Chemistry The Molecular Science

Question

What is the sign of \(w\) for these processes if they occur at constant pressure? Consider only \(P \Delta V\) work from gases. (a) \(\mathrm{Fe}_{2} \mathrm{~S}_{3}(\mathrm{~s})+6 \mathrm{HNO}_{3}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

(a) Negative, (b) Positive

Step 1: Understand the Concept of Work

In thermodynamics, work done by or on a system at constant pressure is determined by the expression \(w = - P\Delta V\), where \(\Delta V\) is the change in volume. The sign of \(w\) depends upon whether the process involves expansion or compression.

Step 2: Determine Initial and Final Moles of Gas for Reaction (a)

For the reaction \(\mathrm{Fe}_{2} \mathrm{S}_{3}(s) + 6 \mathrm{HNO}_{3}(aq) \rightarrow 2 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(aq) + 3 \mathrm{H}_{2} \mathrm{S}(g)\), we only consider gases. Initially, there are 0 moles of gas, and finally, there are 3 moles of \(\mathrm{H}_{2} \mathrm{S}\, \text{(g)}\). So, \(\Delta V > 0\) due to the production of gas.

Step 3: Determine the Sign of \(w\) for Reaction (a)

Since \(\Delta V > 0\), the system does work on the surroundings (expansion), thus \(w = - P\Delta V < 0\). The work is negative.

Step 4: Determine Initial and Final Moles of Gas for Reaction (b)

For the reaction \(\mathrm{C}_{3} \mathrm{H}_{8}(g) + 5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}(\ell)\), initially there are 6 moles of gas (1 mole of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and 5 moles of \(\mathrm{O}_{2}\)). Finally, there are 3 moles of \(\mathrm{CO}_{2}\). So, \(\Delta V < 0\).

Step 5: Determine the Sign of \(w\) for Reaction (b)

Since \(\Delta V < 0\), the surroundings do work on the system (compression), thus \(w = - P\Delta V > 0\). The work is positive.

Key Concepts

Pressure-volume work

In the world of thermodynamics, understanding how work is calculated in terms of pressure and volume can unlock insights into various physical processes. When a gas expands or compresses, it performs work, often calculated with the equation \(w = - P\Delta V\). The symbol \(w\) represents work, \(P\) stands for the constant external pressure applied, and \(\Delta V\) is the change in volume of the system.

• If the volume increases (\(\Delta V > 0\)), the gas is applying work as it expands. Consequently, the sign of \(w\) is negative because the system uses its energy to do work on the surroundings.
• Conversely, if the volume decreases (\(\Delta V < 0\)), the system is undergoing compression. Here, the surroundings are doing work on the system, and \(w\) becomes positive, reflecting energy going into the system.
This fundamental principle is crucial, especially when working with chemical reactions involving gases under constant pressure. It helps predict whether a process results in energy being absorbed or released by the system.

Expansion and compression

When thinking about expansion and compression of gases, it's important to note the directions of work and energy shifts in these processes. During **expansion**, a gas occupies more volume, spreading its molecules further apart. This involves using energy from the system to work against atmospheric pressure, often observed as a release of gas during a reaction. If we take the reaction \(\mathrm{Fe}_{2} \mathrm{~S}_{3} (s) + 6 \mathrm{HNO}_{3}(aq) \rightarrow 2 \mathrm{Fe} ext{(NO}_3\text{)}_{3}(aq) + 3 \mathrm{H}_{2} \mathrm{~S} (g)\), initially, there are zero moles of gas while post-reaction yields 3 moles of \(\text{H}_2\text{S}\). This increase in gaseous moles signifies an expansion.In **compression**, such as in the second reaction \(\mathrm{C}_{3} \mathrm{H}_{8}(g) + 5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}( ext{l})\), there are fewer gas molecules after the reaction, meaning a decrease in volume (from 6 moles of gas to 3 moles of \(\text{CO}_2\)), indicating a compression. The decreased volume suggests the surrounding pressure is performing work on the system, rather than energy being released externally.

Chemical reactions at constant pressure

Chemical reactions behaving at constant pressure allow us to examine the energetic changes during the process using the concept of thermodynamic work. At constant pressure, we can understand energy transfer, particularly heat and work, by focusing on gas-producing reactions.For instance, the reaction involving \(\mathrm{Fe}_{2} \mathrm{~S}_{3} (s) + 6 \mathrm{HNO}_{3}(aq) \rightarrow 2 \mathrm{Fe} ext{(NO}_3\text{)}_{3}(aq) + 3 \mathrm{H}_{2} \mathrm{~S} (g)\) at constant pressure demonstrates expansion. Here, the formation of gaseous hydrogen sulfide from solid and aqueous reactants indicates a movement of energy outwards, with the system doing work, represented by \(w < 0\).Conversely, in the reaction \(\mathrm{C}_{3} \mathrm{H}_{8}(g) + 5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_2(g) + 4 \mathrm{H}_{2} \mathrm{O}(\ell)\), the shift from gas reactants to fewer gas products (formulating liquid water also) signals compression, where energy goes into the system, making the work positive \(w > 0\).Understanding these energy dynamics at constant pressure is key for analyzing reaction energetics and predicting the direction of energy flow in a reaction.