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

Expert verified
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 by step solution

01

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.
02

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.
03

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.
04

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.
05

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One way to lose weight is to exercise! Walking briskly at 4.0 miles per hour for an hour consumes about 400 kcal of energy. How many hours would you have to walk at 4.0 miles per hour to lose one pound of body fat? One gram of body fat is equivalent to 7.7 kcal of energy. There are 454 g in 1 lb.

Consider the reaction $$2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \Delta H=-118 \mathrm{kJ}$$ Calculate the heat when \(100.0 \mathrm{mL}\) of \(0.500 \mathrm{M}\) HCl is mixed with \(300.0 \mathrm{mL}\) of \(0.100 M \mathrm{Ba}(\mathrm{OH})_{2} .\) Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of \(400.0 \mathrm{g}\) and a specific heat capacity of \(4.18 \mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g},\) calculate the final temperature of the mixture.

Explain why oceanfront areas generally have smaller temperature fluctuations than inland areas.

The specific heat capacity of silver is \(0.24 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\). a. Calculate the energy required to raise the temperature of \(150.0 \mathrm{g}\) Ag from \(273 \mathrm{K}\) to \(298 \mathrm{K}\). b. Calculate the energy required to raise the temperature of 1.0 mole of Ag by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes \(1.25 \mathrm{kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\). Calculate the mass of the sample silver.

For the reaction \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2} \mathrm{O}_{2}(g), \Delta H=+90.7 \mathrm{kJ}\). a. What quantity of heat is required to produce 1 mole of mercury by this reaction? b. What quantity of heat is required to produce 1 mole of oxygen gas by this reaction? c. What quantity of heat would be released in the following reaction as written? $$2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s)$$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free