For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(g),\) which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

Short Answer

Expert verified
The given process A(l) → A(g) is favored by both energy probability and positional probability as it requires sufficient energy to break the intermolecular forces, and transitioning from liquid to gas increases the positional probability. To favor this process, the temperature of the system should be raised, as it will provide the necessary energy to increase the energy probability for the forward reaction.

Step by step solution

01

Understand the process

We are given a process in which substance A changes its state from liquid (l) to gas (g). We need to determine whether this process is favored by energy probability and positional probability.
02

Analyze energy probability

Energy probability refers to the likelihood of a system adopting a certain energy state. For a system to transition from one state to another, it needs to overcome the potential energy barrier between those states. In the case of a liquid-to-gas transition, the system needs to overcome the intermolecular forces that hold the molecules together in the liquid state. When the energy probability favors the forward reaction (A(l) → A(g)), it means that the system has enough thermal energy to break the intermolecular forces of the liquid state. As a result, the molecules gain freedom to move as a gas. In this case, the forward reaction (liquid to gas) is favored by energy probability. Therefore, energy probability favors the direction from liquid to gas.
03

Analyze positional probability

Positional probability refers to the likelihood of a particle being found in a certain location or state. For a substance to transition from a more ordered to a less ordered state, the positional probability favors the process. When going from liquid to gas, the substance becomes less ordered and the particles have more freedom to move. As a result, the positional probability favors the forward reaction (A(l) → A(g)) as it allows the particles to occupy more positions in space, increasing the entropy of the system. Therefore, positional probability favors the direction from liquid to gas.
04

Determine the effect of temperature on the process

To favor the given process (A(l) → A(g)), we need to ensure that the energy probability and positional probability are favored. As we concluded in steps 2 and 3, both energy and positional probabilities favor the forward reaction. However, we need to know how temperature affects this process. As the forward reaction (A(l) → A(g)) requires energy to break the intermolecular forces, increasing the temperature provides this energy and increases the energy probability for the forward reaction. Therefore, we should raise the temperature of the system to favor the process as written.

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

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\) . Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\) . Consider the process $\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)$ a. Determine the sign of \(\Delta S, \Delta S_{\text { sum }}\) and $\Delta S_{\text { univ }}$ for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text { sum }},\) and $\Delta S_{\text { univ }}\( for the process in vessel \)2 .$ (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is (are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K\) is greater than 1.00 .

As \(\mathrm{O}_{2}(l)\) is cooled at \(1 \mathrm{atm},\) it freezes at 54.5 \(\mathrm{K}\) to form solid I. At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the $I \rightarrow\( II phase transition is \)-743.1 \mathrm{J} / \mathrm{mol}$ , and \(\Delta S\) for the same transition is \(-17.0 \mathrm{J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

Describe how the following changes affect the positional probability of a substance. a. increase in volume of a gas at constant T b. increase in temperature of a gas at constant V c. increase in pressure of a gas at constant T

At \(100 .^{\circ} \mathrm{C}\) and $1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\( for the vaporization of water. Estimate \)\Delta G^{\circ}$ for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C}\) . Assume $\Delta H^{\circ}\( and \)\Delta S^{\circ}\( at \)100 .^{\circ} \mathrm{C}$ and 1.00 \(\mathrm{atm}\) do not depend on temperature.

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