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Chapter 19: Problem 97

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1 .\) (c) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. $$ \begin{array}{l}{\text { (i) } 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)} \\ {\text { (ii) } 2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{K}(g)+\mathrm{I}_{2}(g)} \\ {\text { (iii) } \mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)} \\ {\text { (iv) } 2 \mathrm{V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{V}(s)+5 \mathrm{O}_{2}(g)}\end{array} $$

Short Answer

Expert verified
The key findings of the analysis are: (i) Mg(s) + O₂(g) ↔ 2MgO(s): ΔH⁰ < 0, ΔS⁰ < 0, K > 1 at low temperatures, K decreases as temperature increases. (ii) 2KI(s) ↔ 2K(g) + I₂(g): ΔH⁰ > 0, ΔS⁰ > 0, K > 1 at high temperatures, K increases as temperature increases. (iii) Na₂(g) ↔ 2Na(g): ΔH⁰ > 0, ΔS⁰ > 0, K > 1 at high temperatures, K increases as temperature increases. (iv) 2V₂O₅(s) ↔ 4V(s) + 5O₂(g): ΔH⁰ > 0, ΔS⁰ > 0, K > 1 at high temperatures, K increases as temperature increases.

Step by step solution

01

(i) Reaction 1: Predicting ΔH⁰ and ΔS⁰

Mg(s) reacts with O₂(g) to form MgO(s). Since the product MgO is more stable than the reactants, the reaction is likely exothermic, meaning that ΔH⁰ will be negative. The reaction also decreases the number of gas molecules, implying a decrease in entropy (ΔS⁰ < 0).
02

(i) Reaction 1: Predicting K > 1 and the Temperature Dependence of K

Since both ΔH⁰ and ΔS⁰ are negative for this reaction, its spontaneity is favored at lower temperatures (K > 1), and K decreases as the temperature increases.
03

(ii) Reaction 2: Predicting ΔH⁰ and ΔS⁰

In this reaction, solid potassium iodide dissociates into gaseous potassium and iodine molecules. As this process requires breaking ionic bonds, it is endothermic (ΔH⁰ > 0). Moreover, the reaction results in an increase in the number of gas molecules, leading to an increase in entropy (ΔS⁰ > 0).
04

(ii) Reaction 2: Predicting K > 1 and the Temperature Dependence of K

With both ΔH⁰ and ΔS⁰ being positive, the reaction is more spontaneous at higher temperatures, which means K > 1 at high temperatures. K increases with increasing temperature.
05

(iii) Reaction 3: Predicting ΔH⁰ and ΔS⁰

In this reaction, Na₂(g) dissociates into 2Na(g). The reaction requires bond breaking, so it's endothermic (ΔH⁰ > 0). Additionally, since the reaction increases the number of gas molecules, there is an increase in entropy (ΔS⁰ > 0).
06

(iii) Reaction 3: Predicting K > 1 and the Temperature Dependence of K

Both ΔH⁰ and ΔS⁰ are positive for this reaction, which means the reaction is more spontaneous at higher temperatures, and K increases with an increase in temperature.
07

(iv) Reaction 4: Predicting ΔH⁰ and ΔS⁰

In this reaction, 2V₂O₅(s) decomposes into 4V(s) and 5O₂(g). The reaction requires bond breaking, so it's endothermic (ΔH⁰ > 0). Furthermore, the reaction leads to an increase in the number of gas molecules, suggesting an increase in entropy (ΔS⁰ > 0).
08

(iv) Reaction 4: Predicting K > 1 and the Temperature Dependence of K

As both ΔH⁰ and ΔS⁰ are positive for this reaction, the reaction is more spontaneous at higher temperatures (K > 1), and K increases with increasing temperature.

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Most popular questions from this chapter

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on!). What do you observe? Are your observations consistent with your answer to part (b)?

A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{F}_{2} .\) An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of q for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at \(31^{\circ} \mathrm{C}\) . Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\) . Is this process spontaneous or nonspontaneous?

Would each of the following changes increase, decrease, or have no effect on the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas?

For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0,\) and \(w=0 .\) (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) What is the "driving force" for the expansion of the gas: enthalpy or entropy?

The normal freezing point of \(n\) -octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is \(-57^{\circ} \mathrm{C}\) . (a) Is the freezing of \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the freezing of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and solid \(n\) -octane are in equilibrium? Explain.
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