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Chapter 16: Problem 1

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
In summary, energy probability favors the reverse direction (A(g) -> A(l)) since substances generally have lower energy in the liquid state compared to the gaseous state. However, positional probability favors the process as written (A(l) -> A(g)) since particles have more available positions in the gaseous state. To favor the process A(l) -> A(g), we would need to increase the temperature of the system, as it provides particles with more energy to transition from the liquid to the gaseous state.

Step by step solution

01

Understanding energy probability

Energy probability refers to the likelihood of a system being in a specific energy state. In general, a system tends to be more energetically favorable when it is in a lower energy state. In this case, comparing the energy probability of A in the liquid state to A in the gaseous state will help us determine which direction the reaction is more likely to proceed.
02

Understanding positional probability

Positional probability refers to the likelihood of a particle being found at a specific location in a system. A substance in the gaseous state has a higher positional probability than the same substance in the liquid state because the particles in a gas have more available positions and are more spread out than in a liquid. Comparing the positional probability of A in the liquid state to A in the gaseous state will help us determine which direction the reaction is more likely to proceed based on this factor.
03

Determining favored direction based on energy probability and positional probability

Based on energy probability alone, the process is more likely to be favored in the reverse direction (A(g) -> A(l)), as substances generally have lower energy in the liquid state compared to the gaseous state. However, based on positional probability alone, the process as written (A(l) -> A(g)) is more likely to be favored, since there are more possible positions for the particles in the gaseous state compared to the liquid state.
04

Understanding the effect of temperature on the process

Temperature plays a significant role in determining the direction of a reaction. As the temperature of a system increases, the energy of the particles also increases. This increased energy makes it more likely for a substance to transition from a lower-energy state (such as a liquid) to a higher-energy state (such as a gas). Conversely, a decrease in temperature would typically favor the opposite direction, with particles more likely to transition from a higher-energy state to a lower-energy state.
05

Determining whether to increase or decrease the temperature to favor the process

Since we want to favor the process as it is written (A(l) -> A(g)), we would want to increase the temperature of the system. Doing so would provide the particles in the liquid state of A with more energy, making them more likely to transition into the gaseous state, and increasing the energy probability in favor of the process.

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Most popular questions from this chapter

As \(\mathrm{O}_{2}(l)\) is cooled at 1 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 \(\mathrm{I} \rightarrow\) 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?

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

For the reaction at \(298 \mathrm{K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ}\) and \(-176.6 \mathrm{J} / \mathrm{K},\) respectively. What is the value of \(\Delta G^{\circ}\) at 298 K? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Given the following data: $$2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) \quad \Delta G^{\circ}=-51 \mathrm{kJ}$$ $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-474 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at $$25^{\circ} \mathrm{C}$$ a. Assuming that \(G_{A}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol}\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C}.\) c. Show by calculations that \(\Delta G=0\) at equilibrium.
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