In which of the following systems is(are) work done by the surroundings on the system? Assume pressure and temperature are constant. a. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) b. \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) c. \(4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCO}(s)+\mathrm{CO}_{2}(g)\)

Short Answer

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The short answer is: Reactions (a) and (c) have work done by the surroundings on the system.

Step by step solution

01

Reaction (a) analysis

For the first reaction, look at the change in moles of gas: 2 SO₂(g) + O₂(g) → 2 SO₃(g) On the left side (reactants) there are 2 + 1 = 3 moles of gas, and on the right side (products) there are 2 moles of gas. The change in moles of gas is negative, which means the surroundings are doing work on the system.
02

Reaction (b) analysis

For the second reaction, look at the change in moles of gas: CO₂(s) → CO₂(g) The reactant is in the solid phase and the product is in the gaseous phase. As there is an increase in the number of moles of gas, the system is doing work on the surroundings, not the other way around.
03

Reaction (c) analysis

For the third reaction, look at the change in moles of gas: 4 NH₃(g) + 7 O₂(g) → 4 NO₂(g) + 6 H₂O(g) On the left side, there are 4 + 7 = 11 moles of gas, and on the right side, there are 4 + 6 = 10 moles of gas. The change in moles of gas is negative, which means the surroundings are doing work on the system.
04

Reaction (d) analysis

For the fourth reaction, look at the change in moles of gas: N₂O₄(g) → 2 NO₂(g) On the left side, there is 1 mole, and on the right side, there are 2 moles of gas. The change in moles of gas is positive, meaning the system is doing work on the surroundings.
05

Reaction (e) analysis

For the fifth and last reaction, look at the change in moles of gas: CaCO₃(s) → CaO(s) + CO₂(g) The reactant (CaCO₃) is in the solid phase, while one product (CO₂) is in the gaseous phase, increasing the number of moles of gas, so the system is doing work on the surroundings. From the analysis above, reactions (a) and (c) have work done by the surroundings on the system.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions and Work
Understanding how chemical reactions involve work requires a grasp on the basics of thermochemistry. In the context of chemical reactions, work is often related to changes in volume against an external pressure. When a reaction occurs in a container at constant pressure, the system either expands, doing work on the surroundings, or contracts, with the surroundings doing work on the system.

Take for example the sublimation of solid carbon dioxide. We can represent this phase change as a chemical reaction: CO₂(s) → CO₂(g). The carbon dioxide starts as a solid (dry ice) and ends as a gas, expanding into the available volume which shows the system doing work on the surroundings by pushing the atmospheric pressure back. Conversely, reactions that lead to a reduction in the volume of gas, such as the synthesis of sulfur trioxide, 2 SO₂(g) + O₂(g) → 2 SO₃(g), imply that the surroundings are compressing the system, thus doing work on it.

Grasping this concept is pivotal, especially when dealing with the idea that a negative change in gas moles suggests work being done on the system, while a positive change indicates the system is doing work on the surroundings. This phenomenon is tied to the conservation of energy and the first law of thermodynamics, where the work can either be an input or output of the system.
Gas Moles Change in Reactions
The change in moles of gas during a chemical reaction is a critical component that affects work. When the number of moles of gaseous products is less than the number of moles of gaseous reactants, the overall volume tends to decrease, as seen in the reaction where ammonia and oxygen form nitrogen dioxide and water vapor: 4 NH₃(g) + 7 O₂(g) → 4 NO₂(g) + 6 H₂O(g).

It's important to note that only gases are considered in this calculation, mainly because gases are compressible and can expand to fill their containers, which is not the case with solids or liquids. An understanding of the ideal gas law, PV=nRT, also ties into these concepts, as 'n' (the number of moles of gas) directly influences the pressure (P) and volume (V) of the system at a given temperature (T) and gas constant (R).

For students, the challenge is often in counting the moles correctly and keeping in mind that while solids and liquids do undergo volume changes, these are generally negligible in the context of work done by or on the system under constant pressure conditions.
System and Surroundings in Thermochemistry
In thermochemistry, the system refers to the part of the universe we are focusing on, typically the reactants and products of a chemical reaction. The surroundings are everything outside the system, which could include the rest of the reaction vessel, the room, or even the entire universe. The interaction between a system and its surroundings often involves energy exchange, which can take the form of work or heat.

When considering our system, it is crucial to determine whether it is open, closed, or isolated. These terms describe whether the system can exchange energy or matter with its surroundings: open systems exchange both, closed systems exchange only energy, and isolated systems exchange neither. Most chemical reactions studied in classrooms are assumed to occur in closed systems at constant pressure.

For instance, the decomposition of calcium carbonate into calcium oxide and carbon dioxide gas, CaCO₃(s) → CaO(s) + CO₂(g), shows a system producing more gas moles, therefore pushing against the surroundings. This concept is vital in predicting the direction in which work is performed and is a fundamental consideration in chemical thermodynamics that gives us insights into energy requirements or releases during reactions.

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Most popular questions from this chapter

A system undergoes a process consisting of the following two steps: Step \(1:\) The system absorbs \(72 \mathrm{~J}\) of heat while \(35 \mathrm{~J}\) of work is done on it. Step 2: The system absorbs \(35 \mathrm{~J}\) of heat while performing \(72 \mathrm{~J}\) of work. Calculate \(\Delta E\) for the overall process.

Consider a balloon filled with helium at the following conditions. $$ \begin{array}{l} 313 \text { g He } \\ 1.00 \text { atm } \\ \text { 1910. L } \\ \text { Molar Heat Capacity }=20.8 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{mol} \end{array} $$ The temperature of this balloon is decreased by \(41.6^{\circ} \mathrm{C}\) as the volume decreases to \(1643 \mathrm{~L}\), with the pressure remaining constant. Determine \(q, w\), and \(\Delta E\) (in \(\mathrm{kJ}\) ) for the compression of the balloon.

The enthalpy change for the reaction $$ \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-891 \mathrm{~kJ}\) for the reaction as written. a. What quantity of heat is released for each mole of water formed? b. What quantity of heat is released for each mole of oxygen reacted?

In a coffee-cup calorimeter, \(150.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) is added to \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) to make \(200.0 \mathrm{~g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\). If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{~kJ} / \mathrm{mol}\), calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is \(4.184 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and assume no heat loss to the surroundings.

Why is it a good idea to rinse your thermos bottle with hot water before filling it with hot coffee?

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