10.E: Thermodynamics (Exercises)
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What is a spontaneous reaction?
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A spontaneous reaction is one that occurs without the continual input of energy from an external source.
Indicate whether the following processes are spontaneous or nonspontaneous.
- Liquid water freezing below 0∘C (the freezing point of water)
- Liquid water freezing above 0∘C (the freezing point of water)
- A ball thrown upward into the air
- A raindrop falling to the ground
- Iron rusting when exposed to air and moisture
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a. Spontaneous – Water freezes below its freezing point without added energy.
b. Nonspontaneous – Water does not freeze above its freezing point under normal conditions because freezing requires heat to be removed.
c. Nonspontaneous – The ball requires added energy (a push) to rise; it won’t go up on its own.
d. Spontaneous – Gravity causes the raindrop to fall naturally.
e. Spontaneous – Rusting occurs without added energy when iron is exposed to air and moisture.
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
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Although the oxidation of plastics is thermodynamically spontaneous, the reaction occurs very slowly. Plastics are kinetically stable, meaning the reaction has a high activation energy and proceeds at an extremely slow rate. This means the process is spontaneous but extremely slow, so it doesn’t occur noticeably in the environment.
When you add a sugar cube to hot coffee, it dissolves and disperses in the coffee.
a. Is this process spontaneous?
b. Is the reverse process, the sugar crystallizing, spontaneous or nonspontaneous?
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a. Yes, the sugar dissolving and dispersing in hot coffee is spontaneous; it occurs without the continual input of energy.
b. The reverse process, sugar crystallizing out of the coffee, is nonspontaneous under these conditions. A process that is spontaneous in one direction is nonspontaneous in the reverse direction unless the conditions change.
Can a nonspontaneous process occur?
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Yes, but it requires continuous input of energy. For example, water does not flow uphill on its own, but it can be pumped using electricity.
Entropy
At room temperature, iodine (\(\ce{I2}\)) is a solid, bromine (\(\ce{Br2}\)) is a liquid, and chlorine (\(\ce{Cl2}\)) is a gas. Place the three halogens in order of increasing entropy.
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\(\ce{I2(s) < Br2 (l) < Cl2 (g)}\)
Entropy is strongly affected by physical state. Gases have the highest entropy, followed by liquids, then solids.
For each pair, indicate which substance has the higher entropy. Briefly explain your reasoning.
- \(\ce{C2H5OH(l)}\) or \(\ce{C3H7OH(l)}\)
- \(\ce{C2H5OH(l)}\) or \(\ce{C2H5OH(g)}\)
- 2 moles of \(\ce{H2(g)}\) or 1 mole of \(\ce{H2(g)}\)
- \(\ce{N2H4(g)}\) or \(\ce{N2(g)}\)
- \(\ce{H2(g)}\) or \(\ce{Na2CO3(s)}\)
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a. \(\ce{C3H7OH(l)}\) – It is a larger and more complex molecule, so it has more possible motions and microstates, resulting in a higher entropy.
b. \(\ce{C2H5OH(g)}\) – Gases have much higher entropy than liquids due to greater freedom of motion and dispersal.
c. 2 moles of \(\ce{H2(g)}\) – Entropy is an extensive property, so doubling the amount of substance doubles the entropy.
d. \(\ce{N2H4(g)}\) - It is a more complex molecule with more atoms and has more internal motions, giving it higher entropy.
e. \(\ce{H2(g)}\) – Gases have far more freedom of movement and more accessible microstates, so \(\ce{H2(g)}\) has a greater entropy even though \(\ce{Na2CO3(s)}\) is a more complex molecule.
Predict the sign of the entropy change (\(\Delta S\)) for each of the following processes. Give a reason for your prediction.
- \(\ce{Pb^2+}(aq) + \ce{S^2-}(aq) \rightarrow \ce{PbS}(s)\)
- \(\ce{4Fe}(s) + \ce{3O2}(g) \rightarrow \ce{2Fe2O3}(s)\)
- \(\ce{2C6H14}(l) + \ce{19O2}(g) \rightarrow \ce{14H2O}(g) + \ce{12CO2}(g)\)
- \(\ce{Ca}(s) + \ce{2H2O}(l) \rightarrow \ce{Ca(OH)2}(aq) + \ce{H2}(g)\)
- \(\ce{C2H2}(g) + \ce{2H2}(g) \rightarrow \ce{C2H6}(g)\)
- \(\ce{2HCl}(aq) + \ce{K2CO3}(aq) \rightarrow \ce{2KCl}(aq) + \ce{H2O}(l) +\ce{CO2}(g)\)
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- Negative. A solid precipitate forms from dissolved ions, decreasing the disorder in the system.
- Negative. Gaseous oxygen is consumed and converted into a solid, resulting in a net decrease in entropy.
- Positive. The number of gas molecules increases significantly, leading to greater energy dispersal and higher entropy.
- Positive – A gas is produced from non-gaseous reactants, increasing the dispersal of energy.
- Negative – The total number of gas molecules decreases from 3 to 1, reducing entropy.
- Positive – A gas is produced from aqueous reactants, increasing the number of microstates and entropy.
In the example of four gas molecules in a two-bulb container, the most probable arrangement is two molecules in each bulb.
a. Why is this arrangement more probable than having all four molecules in one bulb?
b. How is this related to the concept of entropy?
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a. There are more ways (microstates) to arrange the molecules so that two are in each bulb than there are ways for all four molecules to be in one bulb. For example, there is only one microstate where all four molecules are on one side, but there are six different microstates where two molecules are on each side. The arrangement with the most microstates is the most probable.
b. Entropy is a measure of the number of microstates available to a system. Arrangements with more microstates have higher entropy and are more probable. As the molecules spread out into both bulbs, the number of possible microstates increases, so the system's entropy increases. This makes the distribution with two molecules in each bulb more likely than all the molecules being in one bulb.
The Second and Third Law of Thermodynamics
Consider the following reaction:
\(\ce{2CO(g) + O2 (g) <=> 2CO2(g) \quad \Delta H = -566 kJ/mol}\)
a. What is the sign of \(\Delta S\) for the reaction?
b. Is it possible for this reaction to be spontaneous at any temperature?
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- \( \Delta S_{\text{system}} < 0 \). The number of gas molecules decreases from 3 to 2, so the system becomes more ordered and entropy decreases.
- Yes. Even though \( \Delta S_{\text{system}} \) is negative, the reaction releases a large amount of heat (\( \Delta H < 0 \)), which increases the entropy of the surroundings. If this increase in \( \Delta S_{\text{surroundings}} \) is greater than the decrease in \( \Delta S_{\text{system}} \), then the total entropy of the universe increases and the forward reaction is spontaneous.
What is the sign of \( \Delta S_{\text{surr}} \) for:
- an endothermic reaction?
- an exothermic reaction?
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- \( \Delta S_{\text{surr}} < 0 \). In an endothermic reaction, heat is absorbed from the surroundings, decreasing the entropy of the surroundings.
- \( \Delta S_{\text{surr}} > 0 \). In an exothermic reaction, heat is transferred from the system to the surroundings, increasing the entropy of the surroundings.
According to the Third Law of Thermodynamics, the entropy of a perfect crystal is zero at 0 K. What does this imply about the motion and arrangement of particles in a perfect crystal at absolute zero?
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At 0 K, a perfect crystal would have no particle motion and a perfectly ordered arrangement of atoms. Because there is only one possible microstate, the entropy is zero.
Predict the sign of \(ΔS^\circ\) for each process. Then calculate \(ΔS^\circ\) at 298 K, using the values for the standard molar entropies in the reference table.
Reactions:
- \(\ce{CS2}(g) \rightarrow \ce{CS2}(l)\)
- \(\ce{Cu}(s) \rightarrow \ce{Cu}(g)\)
- \(\ce{2H2}(g) + \ce{O2}(g) \rightarrow \ce{2H2O}(l)\)
- \(\ce{2S}(s) + \ce{3O2}(g) \rightarrow \ce{2SO3}(g)\)
- \(\ce{SO3}(g) + \ce{H2O}(l) \rightarrow \ce{H2SO4}(aq)\)
- \(\ce{NH4NO3}(s) \rightarrow \ce{N2O}(g) + \ce{2H2O}(g)\)
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\[ \Delta S^\circ = \sum m\, S^\circ(\text{products}) - \sum n\, S^\circ(\text{reactants}) \nonumber \]
a. \( \Delta S^\circ \) is negative because a gas condenses to a liquid.
\[ \Delta S^\circ = S^\circ(\ce{CS2}(l)) - S^\circ(\ce{CS2}(g)) = 151.3\ \mathrm{J/(mol \cdot K)} - 237.8\ \mathrm{J/(mol \cdot K)} = -86.5\ \mathrm{J/(mol \cdot K)} \nonumber \]
b. \( \Delta S^\circ \) is positive because a gas is formed from a solid.
\[ \Delta S^\circ = S^\circ(\ce{Cu}(g)) - S^\circ(\ce{Cu}(s)) = 166.4\ \mathrm{J/(mol \cdot K)} - 33.2\ \mathrm{J/(mol \cdot K)} = 133.2\ \mathrm{J/(mol \cdot K)} \nonumber\]
c. \( \Delta S^\circ \) is negative because gases react to form a liquid.
\[ \begin{align*} \Delta S^\circ &= 2S^\circ(\ce{H2O}(l)) - [2S^\circ(\ce{H2}(g)) + S^\circ(\ce{O2}(g))] \\ &= 2(70.0) - [2(130.7) + 205.2] \\ &= 140.0 - 466.6 = -326.6\ \mathrm{J/(mol \cdot K)} \end{align*} \]
d. \( \Delta S^\circ \) is negative because 3 moles of gas form only 2 moles of gas.
\[ \begin{align*} \Delta S^\circ &= 2S^\circ(\ce{SO3}(g)) - [2S^\circ(\ce{S}(s)) + 3S^\circ(\ce{O2}(g))] \\ &= 2(256.8) - [2(32.1) + 3(205.2)] \\ &= 513.6 - [64.2 + 615.6] = 513.6 - 679.8 = -166.2\ \mathrm{J/(mol \cdot K)} \end{align*} \]
e. \( \Delta S^\circ \) is negative because a gas is consumed and no gas is formed.
\[ \begin{align*} \Delta S^\circ &= S^\circ(\ce{H2SO4}(aq)) - [S^\circ(\ce{SO3}(g)) + S^\circ(\ce{H2O}(l))] \\ &= 20.1 - [256.8 + 70.0] = 20.1 - 326.8 = -306.7\ \mathrm{J/(mol \cdot K)} \end{align*} \]
f. \( \Delta S^\circ \) is positive because gases are formed from a solid.
\[ \begin{align*} \Delta S^\circ &= S^\circ(\ce{N2O}(g)) + 2S^\circ(\ce{H2O}(g)) - S^\circ(\ce{NH4NO3}(s)) \\ &= 220.0 + 2(188.8) - 151.1 \\ &= 220.0 + 377.6 - 151.1 = 446.5\ \mathrm{J/(mol \cdot K)} \end{align*} \]
Free Energy
Calculate the standard free energy change at 298 K using the values for the standard free energies of formation in the reference table. Are the reactions spontaneous under standard state conditions?
- \(\ce{N2}(g)+\ce{3H2}(g) \rightleftharpoons\ce{2NH3}(g)\)
- \(\ce{H2}(g)+\ce{Br2}(l)\rightleftharpoons\ce{2HBr}(g)\)
- \(\ce{MnO2}(s)\rightleftharpoons\ce{Mn}(s)+\ce{O2}(g)\)
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\[ \Delta G^\circ = \sum m\, \Delta G_f^\circ(\text{products}) - \sum n\, \Delta G_f^\circ(\text{reactants}) \nonumber \]
a.\[ \begin{align*} \Delta G^\circ &= 2\Delta G_f^\circ(\ce{NH3}(g)) - [\Delta G_f^\circ(\ce{N2}(g)) + 3\Delta G_f^\circ(\ce{H2}(g))] \\ &= 2(-16.4~\text{kJ/mol}) - [0 + 3(0)] \\ &= -32.8~\text{kJ/mol} \end{align*} \]
Spontaneous (\( \Delta G < 0 \))
b.\[ \begin{align*} \Delta G^\circ &= 2\Delta G_f^\circ(\ce{HBr}(g)) - [\Delta G_f^\circ(\ce{H2}(g)) + \Delta G_f^\circ(\ce{Br2}(l))] \\ &= 2(-53.4) - [0 + 0] \\ &= -106.8~\text{kJ/mol} \end{align*} \]
Spontaneous (\( \Delta G < 0 \))
c.\[ \begin{align*} \Delta G^\circ &= \Delta G_f^\circ(\ce{Mn}(s)) + \Delta G_f^\circ(\ce{O2}(g)) - \Delta G_f^\circ(\ce{MnO2}(s)) \\ &= 0 + 0 - (-465.1) \\ &= 465.1~\text{kJ/mol} \end{align*} \]
Nonspontaneous (\( \Delta G > 0 \))
Consider the following reaction:
\(\ce{CaCO3}(s) \rightleftharpoons \ce{CaO}(s) + \ce{CO2 (g)}\)
What is the standard free energy change (\( \Delta G^\circ \)) at
(a) 298 K
(b) 1225 K?
Thermodynamic data at 298 K:
| \( \Delta H_f^\circ \) (\(\mathrm{kJ/mol}\)) | \( \Delta G_f^\circ \) (\(\mathrm{kJ/mol}\)) | \(S^\circ \) (\(\mathrm{J/(mol \cdot K)}\)) | |
| \(\ce{CaCO3}(s) \) | -1207.6 | -1129.1 | 91.7 |
| \( \ce{CaO}(s)\) | -634.9 | -603.3 | 38.1 |
| \( \ce{CO2 (g)}\) | -393.5 | -394.4 | 213.8 |
Assume \( \Delta H^\circ \) and \( \Delta S^\circ \) do not change significantly with temperature.
(c) Why is the reaction spontaneous at higher temperatures?
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(a) At 298 K, \( \Delta G^\circ \) can be found either by using \( \Delta G_f^\circ \) values:
\[ \begin{align*} \Delta G^\circ &= \Delta G_f^\circ(\ce{CaO}(s)) + \Delta G_f^\circ(\ce{CO2}(g)) - \Delta G_f^\circ(\ce{CaCO3}(s)) \\ &= (-603.3) + (-394.4) - (-1129.1) \\ &= 131.4~\text{kJ/mol} \end{align*} \]
Or by finding \( \Delta H^\circ \) and \( \Delta S^\circ \):
\[ \begin{align*} \Delta H^\circ &= \Delta H_f^\circ(\ce{CaO}(s)) + \Delta H_f^\circ(\ce{CO2}(g)) - \Delta H_f^\circ(\ce{CaCO3}(s)) \\ &= (-634.9) + (-393.5) - (-1207.6) \\ &= 179.2~\text{kJ/mol} \end{align*} \]
\[ \begin{align*} \Delta S^\circ &= S^\circ(\ce{CaO}(s)) + S^\circ(\ce{CO2}(g)) - S^\circ(\ce{CaCO3}(s)) \\ &= 38.1 + 213.8 - 91.7 \\ &= 160.2\ \mathrm{J/(mol \cdot K)} \end{align*} \]
\[\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=179.2~\text{kJ/mol}-298 \, \mathrm{K}(160.2\ \mathrm{J/(mol \cdot K)})\left(\frac{\text{kJ}}{1000~\text{J}}\right) =131.5~\text{kJ/mol}\nonumber \]
(b) Since \( \Delta G^\circ \) is temperature-dependent, \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\) must be used to find \( \Delta G^\circ \) at 1225 K:
\[\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=179.2~\text{kJ/mol}-1225 \, \mathrm{K}(160.2\ \mathrm{J/(mol \cdot K)})\left(\frac{\text{kJ}}{1000~\text{J}}\right) =-17.0~\text{kJ/mol}\nonumber \]
(c) For this reaction, both \( \Delta H^\circ \) and \( \Delta S^\circ \) are positive. Using the Gibbs free energy equation:
\( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)
At low temperatures, the \( \Delta H^\circ \) term dominates, so \( \Delta G^\circ \) is positive and the reaction is nonspontaneous.
At higher temperatures, the \( T\Delta S^\circ \) term becomes large enough to outweigh \( \Delta H^\circ \). As a result, \( \Delta G^\circ \) becomes negative, and the reaction becomes spontaneous.
In summary, the reaction becomes spontaneous at higher temperatures because it is entropy-driven: the positive entropy change favours the reaction as temperature increases.
For the previous reaction in Exercise \(\PageIndex{15}\), at what temperature will the reaction switch from being nonspontaneous to spontaneous under standard state conditions? Use the data from the previous question.
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The reaction switches from nonspontaneous to spontaneous when \( \Delta G^\circ = 0 \):
\[\Delta G^{\circ}=0 = \Delta H^{\circ}-T \Delta S^{\circ} \\
T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{179.2~\text{kJ/mol}\left(\frac{1000~\text{J}}{\text{kJ}}\right)}{160.2\ \mathrm{J/(mol \cdot K)}}=1119~\text{K}\nonumber \]At temperatures above 1119 K, the reaction becomes spontaneous under standard state conditions.
Consider the following reaction:
\(\ce{CaO(s) + 3C(s) <=> CaC2 (s) + CO (g)} \quad \Delta H^\circ = 464.6\;\text{kJ/mol}\)
Is the reaction spontaneous at all temperatures, only at lower temperatures, only at higher temperatures, or nonspontaneous at all temperatures? Explain your reasoning.
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\( \Delta S^\circ \) is positive because a gas is produced from solid reactants. The reaction is endothermic (\( \Delta H^\circ > 0 \)). Therefore, the reaction will be spontaneous only at higher temperatures, where the \( T\Delta S^\circ \) term is large enough to overcome the positive enthalpy.
Consider the following molecular scene:
(a) Predict the sign of \( \Delta S^\circ \) and \( \Delta H^\circ \).
(b) Is the reaction spontaneous at all temperatures, only at lower temperatures, only at higher temperatures, or nonspontaneous at all temperatures? Explain your reasoning.
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a) \(\Delta S^\circ < 0\) because the number of particles decreases — six separate atoms become three molecules, leading to a more ordered system.
\(\Delta H^\circ < 0\) because bonds are being formed, which releases energy.(b) The reaction is spontaneous only at lower temperatures. Since \(\Delta H^\circ < 0\) (exothermic) and \(\Delta S^\circ < 0\) , the enthalpy change favours spontaneity, but the entropy change does not. According to the Gibbs free energy equation, this combination results in spontaneity only when the temperature is low enough that the \(T \Delta S^\circ\) term doesn't dominate.
Consider the following reaction:
\(\ce{2Na(s) + 2H2O (l) <=> 2NaOH (aq) + H2 (g)} \quad \Delta H^\circ = -368.4\;\text{kJ/mol}\)
Is the reaction spontaneous at all temperatures, only at lower temperatures, only at higher temperatures, or nonspontaneous at all temperatures? Explain your reasoning.
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\(\Delta S^\circ > 0\) because a gas is formed. The reaction is also exothermic (\(\Delta H^\circ < 0\)). Since both \(\Delta H^\circ < 0\) and \(\Delta S^\circ > 0\), \(\Delta G^\circ < 0\) at all temperatures. Therefore, the reaction is spontaneous at all temperatures.
For the following reaction:
\(\ce{2Al(s) + 3Cl2 (g) \rightleftharpoons 2AlCl3(s)}\)
(a) What is the standard free energy change (\( \Delta G^\circ \)) at 25 °C?
(b) At what temperature will the reaction switch from being spontaneous to nonspontaneous under standard state conditions?
Thermodynamic data at 298 K:
| \( \Delta H_f^\circ \) (kJ/mol) | \(S^\circ \) \((\mathrm{J/(mol \cdot K)})\) | |
| \(\ce{Al}(s) \) | 0 | 28.3 |
| \( \ce{Cl2}(g)\) | 0 | 223.1 |
| \( \ce{AlCl3 (s)}\) | -704.2 | 109.3 |
Assume \( \Delta H^\circ \) and \( \Delta S^\circ \) do not change significantly with temperature.
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(a) \[ \begin{align*} \Delta H^\circ &= 2\Delta H_f^\circ(\ce{AlCl3}(s)) - [2\Delta H_f^\circ(\ce{Al}(s)) + 3\Delta H_f^\circ(\ce{Cl2}(g))] \\ &= 2(-704.2\;\text{kJ/mol}) \\ &= -1408.4~\text{kJ/mol} \end{align*} \]
\[ \begin{align*} \Delta S^\circ &= 2S^\circ(\ce{AlCl3}(s)) - [2S^\circ(\ce{Al}(s)) + 3S^\circ(\ce{Cl2}(g))] \\ &= 2(109.3 \ \mathrm{J/(mol \cdot K)}) -[2(28.3\ \mathrm{J/(mol \cdot K)})+3(223.1\ \mathrm{J/(mol \cdot K)})] \\ &= -507.3\ \mathrm{J/(mol \cdot K)} \end{align*} \]
\[\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=-1408.4~\text{kJ/mol}-298K(-507.3\ \mathrm{J/(mol \cdot K)})\left(\frac{\text{kJ}}{1000~\text{J}}\right) =-1257.2~\text{kJ/mol}\nonumber \]
The reaction is spontaneous at 298 K.
(b) The reaction switches from spontaneous to nonspontaneous when \( \Delta G^\circ = 0 \):
\[\Delta G^{\circ}=0 = \Delta H^{\circ}-T \Delta S^{\circ} \\
T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{-1408.4~\text{kJ/mol}\left(\frac{1000~\text{J}}{\text{kJ}}\right)}{-507.3\ \mathrm{J/(mol \cdot K)}}=2776~\text{K}\nonumber \]At temperatures above 2776 K, the reaction becomes nonspontaneous under standard state conditions.
Two chemical reactions are shown below:
1. Conversion of methane to ethane and hydrogen:
\(\ce{2CH4 (g) -> C2H6(g) + H2 (g) } \quad \Delta G^\circ = +68.1 \ \text{kJ/mol}\)
2. Formation of water:
\(\ce{H2(g) + \dfrac{1}{2} O2 (g) -> H2O (g) } \quad \Delta G^\circ = -228.6 \ \text{kJ/mol}\)
(a) Write the overall chemical equation that results from coupling these two processes to give the reaction of methane with oxygen gas.
(b) Calculate the overall standard free energy change, \(\Delta G^\circ\), for the coupled process.
(c) Is the overall coupled reaction spontaneous under standard conditions? Explain.
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(a) Add the two reactions and cancel \( \ce{H2(g)} \):
\[\ce{2CH4 (g) + \dfrac{1}{2}O2(g) -> C2H6 (g) + H2O(g)} \nonumber\](b) \[ \Delta G^\circ = +68.1 \ \text{kJ/mol} + (-228.6 \ \text{kJ/mol}) = -160.5\ \text{kJ/mol} \nonumber\]
(c) Yes, the overall reaction is spontaneous under standard conditions because the total \( \Delta G^\circ \) is negative. The formation of water provides enough free energy to drive the otherwise nonspontaneous conversion of methane.
A reaction has \(ΔH^\circ_{298}=\textrm{100 kJ/mol}\) and \(ΔS^\circ_{298}=\textrm{250 J/mol⋅K}\).
(a) Is the reaction spontaneous at room temperature under standard state conditions?
(b) If not, under what temperature conditions will it become spontaneous? (Assume \(ΔH^\circ\) and \(ΔS^\circ\) do not change significantly with temperature.)
(c) What is the standard free energy change (\(ΔG^\circ\)) at 650 K?
- Answer
-
(a) \[\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=100~\text{kJ/mol}-298\,K(250\ \mathrm{J/(mol \cdot K)})\left(\frac{\text{kJ}}{1000~\text{J}}\right) =25.5~\text{kJ/mol}\nonumber \]
At 298 K, the reaction is nonspontaneous.
(b) \[\Delta G^{\circ}=0 = \Delta H^{\circ}-T \Delta S^{\circ} \\
T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{100~\text{kJ/mol}\left(\frac{1000~\text{J}}{\text{kJ}}\right)}{250\ \mathrm{J/(mol \cdot K)}}=400~\text{K}\nonumber \]The reaction is spontaneous above 400 K.
(c) \[\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=100~\text{kJ/mol}-650\,K(250\ \mathrm{J/(mol \cdot K)})\left(\frac{\text{kJ}}{1000~\text{J}}\right) =-62.5~\text{kJ/mol}\nonumber \]
At 650 K, the reaction is spontaneous.
Free Energy, Equilibrium, and Non-Standard State Conditions
If \(\Delta G^\circ < 0\) at 298 K, is the reaction always spontaneous at 298 K?
- Answer
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No, \(\Delta G^\circ\) refers to standard conditions. A reaction with \(\Delta G^\circ < 0\) is nonspontaneous (\(\Delta G > 0\)) under nonstandard conditions when \(Q > K\).
What is the relationship between \(Q\) and \(K\) and the sign of \(\Delta G\) when:
(a) the reaction is spontaneous in the forward direction?
(b) the reaction is at equilibrium?
(c) the reaction is nonspontaneous in the forward direction?
- Answer
-
(a) \(Q < K, \Delta G < 0\)
(b) \(Q = K, \Delta G = 0\)
(c) \(Q > K, \Delta G > 0\)
Why might a reaction with a very large \(K\) not proceed spontaneously?
- Answer
-
A large \(K\) value means that the reaction mixture contains mostly products. However, if the reaction mixture already has more products than it would at equilibrium (\(Q > K\)), the forward reaction will not be spontaneous (\(\Delta G > 0\)). The reverse reaction will proceed to reach equilibrium.
For the following reaction at 298 K:
\[\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g) \hspace{20px} ΔG^\circ=\mathrm{−457.18\: kJ/mol} \nonumber\]
what is \(\Delta G\) at 298 K when \(P_{\ce{H2}}\) = 2.0 atm; \(P_{\ce{O2}}\) = 2.0 atm; \(P_{\ce{H2O}}\) = 0.010 atm?
- Answer
-
\[\Delta G =\Delta G^{\circ}+R T \ln Q \\ Q = \frac{P_{\ce{H2O}}^2}{P_{\ce{H2}}^2 P_{\ce{O2}}} \\[4pt]
\Delta G = -457.18 \frac{ \text{kJ} }{ \text{mol} }+\left(8.314 \frac{ \text{J} }{ \text{mol K} }\right)\left(\frac{\text{kJ}}{1000\;\text{J}}\right)(298\;\text{K})\ln\left( \frac{(0.010)^2}{(2.0)^2(2.0)}\right) = -485 ~\text{kJ/mol} \nonumber \]
For the following reaction at 298 K:
\[\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g) \hspace{20px} ΔG°=\mathrm{−79\: kJ/mol}\nonumber\]
what is \(\Delta G\) at 298 K?
(a) when \(P_{\ce{CO2}}\) = 2.0 atm; \(P_{\ce{H2O}}\) = 0.10 atm; with excess \(\ce{LiOH(s)}\) and \(\ce{Li2CO3 (s)}\) present?
(b) when \(P_{\ce{CO2}}\) = 0.00010 atm; \(P_{\ce{H2O}}\) = 2.0 atm; with excess \(\ce{LiOH(s)}\) and \(\ce{Li2CO3 (s)}\) present?
- Answer
-
(a) \[\Delta G =\Delta G^{\circ}+R T \ln Q \\ Q = \frac{P_{\ce{H2O}}}{P_{\ce{CO2}}} \\[4pt]
\Delta G =-79 \frac{ \text{kJ} }{ \text{mol} }+\left(8.314 \frac{ \text{J} }{ \text{mol K} }\right)\left(\frac{\text{kJ}}{1000\;\text{J}}\right)(298\text{ K})\ln\left( \frac{0.10}{2.0}\right) = -86 ~\text{kJ/mol} \nonumber \]The reaction is spontaneous under these conditions.
(b) \[\Delta G =\Delta G^{\circ}+R T \ln Q \\ Q = \frac{P_{\ce{H2O}}}{P_{\ce{CO2}}} \\[4pt]
\Delta G =-79 \frac{ \text{kJ} }{ \text{mol} }+\left(8.314 \frac{ \text{J} }{ \text{mol K} }\right)\left(\frac{\text{kJ}}{1000\;\text{J}}\right)(298\;\text{K})\ln\left( \frac{2.0}{0.00010}\right) = -54 ~\text{kJ/mol} \nonumber \]The reaction is spontaneous under these conditions. The smaller negative value of \(\Delta G\) indicates that the reaction is closer to equilibrium than in part (a).
Calculate the equilibrium constant at 25 °C for each of the following reactions:
(a) \(\ce{O2}(g)+\ce{2F2}(g)⟶\ce{2OF2}(g) \hspace{20px} ΔG°=\mathrm{−9.2\: kJ/mol}\)
(b) \(\ce{I2}(s)+\ce{Br2}(l)⟶\ce{2IBr}(g) \hspace{20px} ΔG°=\mathrm{7.3\: kJ/mol}\)
(c) \(\ce{2LiOH}(s)+\ce{CO2}(g)⟶\ce{Li2CO3}(s)+\ce{H2O}(g) \hspace{20px} ΔG°=\mathrm{−79\: kJ/mol}\)
- Answer
-
Use the relationship:
\[ \Delta G^\circ = -RT \ln K \nonumber\]
Rearranged:
\[ K = e^{-\dfrac{\Delta G^\circ}{RT}}\nonumber \]
(a) \[ K = e^{-\dfrac{-9.2\:kJ/mol}{\left(8.314\frac{J}{mol \cdot K}\right)(298\,K)}\left(\dfrac{1000\,J}{kJ}\right)}=41\nonumber \]
(b) \[ K = e^{-\dfrac{7.3\:kJ/mol}{\left(8.314\frac{J}{mol \cdot K}\right)(298\,K)}\left(\dfrac{1000\,J}{kJ}\right)}=0.053\nonumber \]
(c) \[ K = e^{-\dfrac{-79\:kJ/mol}{\left(8.314\frac{J}{mol \cdot K}\right)(298\,K)}\left(\dfrac{1000\,J}{kJ}\right)}=7.0 \times 10^{13}\nonumber \]
Calculate \(K_p\) for the following reaction at 298 K using the provided thermodynamic data:
\(\ce{Fe2O3}(s)+\ce{3CO}(g)⟶\ce{2Fe}(s)+\ce{3CO2}(g)\)
Thermodynamic data at 298 K:
| \( \Delta H_f^\circ \mathrm{(kJ/mol)} \) | \(S^\circ \mathrm{(J/(mol \cdot K))}\) | |
| \(\ce{Fe2O3}(s) \) | -842.2 | 87.4 |
| \( \ce{CO}(g)\) | -110.5 | 197.7 |
| \( \ce{Fe (s)}\) | 0 | 27.3 |
| \( \ce{CO2}(g)\) | -393.5 | 213.8 |
- Answer
-
Step 1: Calculate \(\Delta H^\circ\):
\[ \begin{align*} \Delta H^\circ &= 2\Delta H_f^\circ(\ce{Fe}(s))+3\Delta H_f^\circ(\ce{CO2}(g)) - [\Delta H_f^\circ(\ce{Fe2O3}(s)) + 3\Delta H_f^\circ(\ce{CO}(g))] \\ &= 2(0\;\text{kJ/mol}) + 3(-393.5\;\text{kJ/mol})- [1(-842.2\;\text{kJ/mol})+3(-110.5\;\text{kJ/mol})] \\ &= -6.8~\text{kJ/mol} \end{align*} \]
Step 2: Calculate \(\Delta S^\circ\):
\[ \begin{align*} \Delta S^\circ &= 2S^\circ(\ce{Fe}(s))+ 3S^\circ(\ce{CO2}(g))- [S^\circ(\ce{Fe2O3}(s)) + 3S^\circ(\ce{CO}(g))] \\ &= 2(27.3 \ \mathrm{J/(mol \cdot K)})+3(213.8 \ \mathrm{J/(mol \cdot K)}) -[1(87.4\ \mathrm{J/(mol \cdot K)})+3(197.7\ \mathrm{J/(mol \cdot K)})] \\ &= 15.5\ \mathrm{J/(mol \cdot K)} \end{align*} \]
Step 3: Calculate \(\Delta G^\circ\) at 298 K:
\[\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=-6.8~\text{kJ/mol}-298\text{ K}(15.5\ \mathrm{J/(mol \cdot K)})\left(\frac{\text{kJ}}{1000~\text{J}}\right) =-11.4~\text{kJ/mol}\nonumber \]
Step 4: Calculate \(K\):
\[ K = e^{-\dfrac{-11.4\:kJ/mol}{\left(8.314\frac{J}{mol \cdot K}\right)(298K)}\left(\dfrac{1000J}{kJ}\right)}=100\nonumber \]
Calculate \(\Delta G^\circ\) for each of the following reactions from the equilibrium constant at the temperature given.
(a) \(\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g) \hspace{20px} \mathrm{T=2000\:°C} \hspace{20px} K_p=4.1×10^{−4}\)
(b) \(\ce{H2}(g)+\ce{I2}(g)⟶\ce{2HI}(g) \hspace{20px} \mathrm{T=400\:°C} \hspace{20px} K_p=50.0\)
(c) \(\ce{CO2}(g)+\ce{H2}(g)⟶\ce{CO}(g)+\ce{H2O}(g) \hspace{20px} \mathrm{T=980\:°C} \hspace{20px} K_p=1.67\)
(d) \(\ce{AgBr}(s)⟶\ce{Ag+}(aq)+\ce{Br-}(aq) \hspace{20px} \mathrm{T=25\:°C} \hspace{20px} K_p=3.3×10^{−13}\)
- Answer
-
\[ \Delta G^\circ = -RT \ln K \nonumber\]
(a) \[ \Delta G^\circ = -\left(8.314 \frac{J}{mol \cdot K}\right)(2273\;K)\left(\frac{kJ}{1000\;J}\right) \ln(4.1\times10^{-4}) =147 \; kJ/mol\nonumber\]
(b) \[ \Delta G^\circ = -\left(8.314 \frac{J}{mol \cdot K}\right)(673\;K)\left(\frac{kJ}{1000\;J}\right) \ln(50) =-21.9 \; kJ/mol\nonumber\]
(c) \[ \Delta G^\circ = -\left(8.314 \frac{J}{mol \cdot K}\right)(1253\;K)\left(\frac{kJ}{1000\;J}\right) \ln(1.67) =-5.34 \; kJ/mol\nonumber\]
(d) \[ \Delta G^\circ = -\left(8.314 \frac{J}{mol \cdot K}\right)(298\;K)\left(\frac{kJ}{1000\;J}\right) \ln(3.3\times10^{-13}) =71 \; kJ/mol\nonumber\]
Consider the following reaction:
\(\ce{CaCO3}(s) \rightleftharpoons \ce{CaO}(s) + \ce{CO2 (g)}\)
At 298 K, \(\Delta H^\circ\) = 179.2 kJ/mol and \(\Delta S^\circ\) = 160.2 J/(mol K).
(a) Does the spontaneity of the reaction under standard state conditions change with temperature?
(b) What is \(\Delta G^\circ\) at 1225 K, assuming \(\Delta H^\circ\) and \(\Delta S^\circ\) do not change significantly with temperature?
(c) Given this value of \(\Delta G^\circ\), is K = 1, K > 1, K < 1 at 1225 K?
(d) What is the value of K at 1225 K?
(e) The reaction takes place in a closed system containing only solid CaCO₃ and CaO at equilibrium. What is the equilibrium pressure of \(\ce{CO2}\) at 1225 K?
- Answer
-
(a) Yes. Since the reaction is endothermic (\( \Delta H^\circ > 0 \)) with increasing entropy (\( \Delta S^\circ > 0 \)), the reaction is nonspontaneous at lower temperatures but becomes spontaneous at higher temperatures. Increasing temperature increases the value of \( T \Delta S^\circ \), which favours spontaneity.
(b)\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ = 179.2\ \text{kJ/mol} - (1225\ \text{K}) \left(160.2\ \text{J/(mol·K)}\right)\left(\frac{\text{kJ}}{1000\ \text{J}}\right) = -17.0\ \text{kJ/mol} \nonumber\]
(c) Since \( \Delta G^\circ < 0 \), the reaction is product-favoured at 1225 K. Therefore, \( K > 1 \).
(d)\[
\Delta G^\circ = -RT \ln K \quad \Rightarrow \quad \ln K = -\frac{\Delta G^\circ}{RT}
\nonumber\]
\[
\ln K = -\frac{-17.0\ \text{kJ/mol}(1000\ \text{J/kJ})}{(8.314\ \text{J/mol·K})(1225\ \text{K})}
= \frac{17000\ \text{J/mol}}{(10184.65\ \text{J/mol})}
= 1.67
\nonumber \]
\[
K = e^{1.67} = 5.3 \nonumber \](e) Since the only gas in the equilibrium expression is \( \ce{CO2} \), and the solids are not included in \( K_p \), we have: \[ K_p = P_{\ce{CO2}} = 5.3\ \text{atm} \nonumber\]
Phase Transitions and Phase Diagrams
What are the signs of \(\Delta H\), \(\Delta S\), and \(\Delta G\) for the spontaneous melting of a solid?
- Answer
-
\(\Delta G\) < 0 because the process is spontaneous.
\(\Delta H\) > 0 because energy is required to overcome intermolecular forces as the solid melts.
\(\Delta S\) > 0 because particles in a liquid have more freedom of motion than in a solid, resulting in an increase in entropy.
Consider the melting of sodium chloride:
\[\ce{NaCl(s) <=> NaCl(l)} \quad \Delta H_{\text{fusion}}^\circ =28.16\;\text{kJ/mol} \quad \Delta S_{\text{fusion}}^\circ = 26.22\;\text{J/(mol·K)}\nonumber\]
Assuming \(\Delta H^\circ\) and \(\Delta S^\circ\) do not change with temperature:
(a) Is melting spontaneous at 575 K?
(b) Calculate the melting point of NaCl(s).
- Answer
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(a) \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ = 28.16\ \text{kJ/mol} - (575\ \text{K}) \left(26.22\ \text{J/(mol·K)}\right)\left(\frac{\text{kJ}}{1000\ \text{J}}\right) = 13.08\ \text{kJ/mol} \nonumber\]
Since \(\Delta G^\circ\) > 0, melting is nonspontaneous at 575 K.
(b) At the melting point, the solid and liquid are in equilibrium, so \( \Delta G^\circ = 0 \):
\[\Delta G^{\circ}=0 = \Delta H^{\circ}-T \Delta S^{\circ} \\
T_{\text{melt}} = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{28.16~\text{kJ/mol}\left(\frac{1000~\text{J}}{\text{kJ}}\right)}{26.22\ \mathrm{J/(mol \cdot K)}}=1074~\text{K}\nonumber \]The melting point of NaCl is 1074 K. Above 1074 K, NaCl melts spontaneously.
Find the normal boiling point of \(\ce{Br2}\) using the thermodyamic data below:
\[\ce{Br2(l)<=> Br2(g) }\nonumber\]
Assume \(\Delta H^\circ\) and \(\Delta S^\circ\) do not change with temperature.
Thermodynamic data at 298 K:
| \( \Delta H_f^\circ \) (kJ/mol) | \(S^\circ \) (\(\mathrm{J/(mol \cdot K)}\)) | |
| \(\ce{Br2}(l) \) | 0 | 152.2 |
| \( \ce{Br2}(g)\) | 30.9 | 245.5 |
- Answer
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Step 1: Calculate \(\Delta H^\circ\):
\[ \begin{align*} \Delta H^\circ &= 1\Delta H_f^\circ(\ce{Br2(g)})-1\Delta H_f^\circ(\ce{Br2(l)}) \\ &= 1(30.9\;\text{kJ/mol}) - 1(0\;\text{kJ/mol}) \\ &= 30.9~\text{kJ/mol} \end{align*} \]
Step 2: Calculate \(\Delta S^\circ\):
\[ \begin{align*} \Delta S^\circ &= 1S^\circ(\ce{Br2(g)})- 1S^\circ(\ce{Br2(l)}) \\ &= 1(245.5 \ \mathrm{J/(mol \cdot K)})-1(152.2 \ \mathrm{J/(mol \cdot K)}) \\ &= 93.3 \mathrm{~J/(mol \cdot K)} \end{align*} \]
Step 3: At the boiling point, the vapour and liquid are in equilibrium, so \( \Delta G^\circ = 0 \):
\[\Delta G^{\circ}=0 = \Delta H^{\circ}-T \Delta S^{\circ} \\
T_{\text{boil}} = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{30.9~\text{kJ/mol}\left(\frac{1000~\text{J}}{\text{kJ}}\right)}{93.3\ \mathrm{J/(mol \cdot K)}}=331~\text{K}\nonumber \]
What is the relationship between vapour pressure and temperature? Explain your answer.
- Answer
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Vapour pressure increases with increasing temperature. As temperature rises, a greater fraction of molecules in the liquid have enough kinetic energy to overcome intermolecular forces and escape into the gas phase, increasing the number of gas-phase molecules and the resulting vapour pressure.
Which substance has the greater vapour pressure at 273 K:
(a) \(\ce{NH3}\) or \(\ce{PH3}\)
(b) \(\ce{CH3CH2OH}\) or \(\ce{CH3CH2Cl}\)
- Answer
-
(a) \(\ce{PH3}\) Although both molecules are polar, \(\ce{NH3}\) can form hydrogen bonds due to the presence of N–H bonds, while \(\ce{PH3}\) cannot. Hydrogen bonding gives \(\ce{NH3}\) stronger intermolecular forces, which lowers its vapour pressure relative to \(\ce{PH3}\).
(b) \(\ce{CH3CH2Cl}\)
\(\ce{CH3CH2OH}\) can hydrogen-bond, while \(\ce{CH3CH2Cl}\) cannot. Hydrogen bonding results in stronger intermolecular forces in \(\ce{CH3CH2OH}\), so fewer molecules escape into the gas phase, leading to a lower vapour pressure compared to \(\ce{CH3CH2Cl}\).
The following heating curve shows how the temperature of a pure substance changes as heat is added at a constant rate:
1. Identify which segment corresponds to:
(a) the solid being heated
(b) the liquid being heated
(c) the gas being heated
(d) melting
(e) boiling
2. Why does the temperature stay constant during the melting and boiling processes?
- Answer
-
1. (a) i (b) iii (c) v (d) ii (e) iv
2. During melting and boiling, the added heat is used to overcome intermolecular forces, not to increase kinetic energy. As a result, the temperature stays constant until the phase change is complete.
Krypton is an inert gas used in lighting. It has normal boiling and freezing points of 120.9 K and 116.6 K, respectively. The triple point of argon is 104.2 K at 0.18 atm. Use these data to sketch a phase diagram for krypton and label all the regions. (The diagram does not need to be to scale.) Is solid krypton more or less dense than liquid krypton?
- Answer
-
Solid krypton is more dense than liquid krypton because the melting curve has a positive slope. This indicates that increasing pressure favours the solid phase, meaning the solid occupies less volume than the same amount of liquid.
Use the unlabeled phase diagram for water to answer the following questions:
Triple point: 0.01 °C, 0.006 atm
Critical point: 374.4 °C, 217.7 atm
Normal melting point: 0 °C, Normal boiling point: 100 °C
(a) What is the maximum pressure at which ice and water vapour can coexist in equilibrium?
(b) What happens to a sample of water at 50 °C and 0.006 atm when the pressure is increased to 10 atm at 50 °C?
(c) Describe the phase changes that occur as a sample at 200 °C and 1.0 atm is cooled at constant pressure to –10 °C.
- Answer
-
(a) The maximum pressure at which ice and water vapour can coexist in equilibrium is at the triple point, which occurs at 0.006 atm. Above this pressure, any transition from solid to gas must pass through the liquid phase.
(b) At 50 °C and 0.006 atm, the water exists as vapour. As the pressure increases to 10 atm at constant temperature, the water vapour condenses to form liquid water.
(c) At 200 °C and 1.0 atm, the water is in the vapour phase. As it is cooled at constant pressure:
- It condenses to liquid at 100 °C (normal boiling point),
- It freezes to solid at 0 °C (normal melting point),
- It remains solid as the temperature decreases to –10 °C.
Additional Problems
What is the difference between \(\Delta G\), \(\Delta G^\circ\), and \(\Delta G^\circ\) at 500 K for a chemical reaction?
- Answer
-
\(\Delta G\) is the free energy change under any conditions. It depends on the temperature and the actual concentrations or partial pressures of reactants and products. A negative value indicates that the forward reaction is spontaneous under the current conditions.
\(\Delta G^\circ\) is the standard free energy change. It refers to the free energy change when all reactants and products are in their standard states (1 atm pressure for gases, 1 M concentration for solutions), usually at 298 K. It reflects the spontaneity of a reaction under standard conditions and is often used to calculate equilibrium constants.
\(\Delta G^\circ\) at 500 K still assumes standard states, but the temperature is different. Its value differs from \(\Delta G^\circ\) at 298 K because Gibbs free energy depends on temperature through the equation:
\[\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \nonumber\]
Even if \(\Delta H^\circ\) and \(\Delta S^\circ\) do not change significantly with temperature, the term \(T\Delta S^\circ\) can have a large effect. As a result, the sign or magnitude of \(\Delta G^\circ\) may change at higher or lower temperatures.
Why is \(\Delta H_{vap} >\Delta H_{fus}\) for a substance?
- Answer
-
Vaporization requires a complete separation of particles into the gas phase, which involves fully overcoming the intermolecular forces holding the particles together. In contrast, fusion (melting) only requires enough energy for particles to move past one another in the liquid phase, while still remaining relatively close. Because vaporization involves a greater disruption of intermolecular forces than fusion, the enthalpy change for vaporization is larger: \(\Delta H_{vap} >\Delta H_{fus}\).
Determine the standard free energy of formation, \( \Delta G^\circ_{\ce{f}} \), for \( \ce{S^{2-}(aq)} \), given the following data:
- \( \Delta G^\circ_{\ce{f}}(\ce{Ag^+}(aq)) = +77.1\ \text{kJ/mol} \)
- \( \Delta G^\circ_{\ce{f}}(\ce{Ag2S}(s)) = -39.5\ \text{kJ/mol} \)
- \( K_{\text{sp}}(\ce{Ag2S}) = 8 \times 10^{-51} \)
- Answer
-
The dissolution equilibrium is:
\[ \ce{Ag2S(s) <=> 2Ag^+(aq) + S^{2-}(aq)} \nonumber \]
Step 1: Use the relationship \( \Delta G^\circ = -RT \ln K \) to find the standard free energy change for the dissolution.
\[ \Delta G^\circ = - (8.314\ \text{J/mol·K})\left(\frac{1\ \text{kJ}}{1000\ \text{J}}\right)(298\ \text{K}) \ln(8 \times 10^{-51}) = 285.8\ \text{kJ/mol} \nonumber\]
Step 2: Relate the \( \Delta G^\circ\) for the dissolution to standard free energies of formation:
\[ \Delta G^\circ = \Delta G^\circ_{\ce{f}}(\ce{2Ag^+}) + \Delta G^\circ_{\ce{f}}(\ce{S^{2-}}) - \Delta G^\circ_{\ce{f}}(\ce{Ag2S(s)}) \nonumber\]
\[ 285.8 \ \text{kJ/mol} = 2(77.1 \ \text{kJ/mol}) + \Delta G^\circ_{\ce{f}}(\ce{S^{2-}}) - (-39.5\ \text{kJ/mol}) \nonumber \]
\[ \Delta G^\circ_{\ce{f}}(\ce{S^{2-}}) = 285.8 \ \text{kJ/mol} - 193.7 \ \text{kJ/mol}= 92.1\ \text{kJ/mol} \nonumber\]
Determine \(\Delta G^\circ\) for the decomposition of antimony pentachloride at 448 °C:
\[\ce{SbCl5}(g)⟶\ce{SbCl3}(g)+\ce{Cl2}(g) \nonumber\]
An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.43 g of SbCl5, 9.14 g of SbCl3, and 2.84 g of Cl2.
- Answer
-
Step 1: Convert masses to moles
\[ \begin{aligned} \text{Molar mass of } \ce{SbCl5} &= 299.0\ \text{g/mol} \quad \Rightarrow\quad n_{\ce{SbCl5}} = \frac{3.43\ \text{g}}{299.0\ \text{g/mol}} = 0.01147\ \text{mol} \\ \text{Molar mass of } \ce{SbCl3} &= 228.1\ \text{g/mol} \quad \Rightarrow\quad n_{\ce{SbCl3}} = \frac{9.14\ \text{g}}{228.1\ \text{g/mol}} = 0.04007\ \text{mol} \\ \text{Molar mass of } \ce{Cl2} &= 70.9\ \text{g/mol} \quad \Rightarrow\quad n_{\ce{Cl2}} = \frac{2.84\ \text{g}}{70.9\ \text{g/mol}} = 0.04006\ \text{mol} \end{aligned} \]
Step 2: Calculate partial pressures using the ideal gas law
Use \( P = \frac{nRT}{V} \), with:
- \( R = 0.08206\ \text{L·atm/(mol·K)} \)
- \( T = 448^\circ \text{C} = 721\ \text{K} \)
- \( V = 5.00\ \text{L} \)
\[ \begin{align*} P_{\ce{SbCl5}} &= \frac{(0.01147\ \text{mol})(0.08206\ \text{L·atm/mol·K})(721\ \text{K})}{5.00\ \text{L}} = 0.1357\ \text{atm} \\ P_{\ce{SbCl3}} &= \frac{(0.04007\ \text{mol})(0.08206\ \text{L·atm/mol·K})(721\ \text{K})}{5.00\ \text{L}} = 0.4742\ \text{atm} \\ P_{\ce{Cl2}} &= \frac{(0.04006\ \text{mol})(0.08206\ \text{L·atm/mol·K})(721\ \text{K})}{5.00\ \text{L}} = 0.4740\ \text{atm} \end{align*} \]
Step 3: Write the equilibrium expression and solve for \( K_p \)
\[ K_p = \frac{P_{\ce{SbCl3}} \cdot P_{\ce{Cl2}}}{P_{\ce{SbCl5}}} = \frac{(0.4742\ \text{atm})(0.4740\ \text{atm})}{0.1357\ \text{atm}} = 1.656 \nonumber\]
Step 4: Calculate \( \Delta G^\circ \) from \( K_p \)
Use \( \Delta G^\circ = -RT \ln K \), with:
- \( R = 8.314\ \text{J/(mol·K)} \)
- \( T = 721\ \text{K} \)
- \( K_p = 1.656 \)
\[ \begin{align*} \Delta G^\circ &= -(8.314\ \text{J/mol·K})(721\ \text{K}) \ln(1.656) \cdot \frac{1\ \text{kJ}}{1000\ \text{J}} \\ &= -3023.6\ \text{J/mol} = \boxed{-3.02\ \text{kJ/mol}} \end{align*} \]
The negative value of \( \Delta G^\circ \) indicates that the decomposition of \( \ce{SbCl5} \) is slightly spontaneous under standard conditions at 448 °C.
When ammonium chloride is added to water and stirred, it dissolves spontaneously, and the resulting solution feels cold. Without doing any calculations, deduce the signs of ΔG, ΔH, and ΔS for this process, and justify your choices.
- Answer
-
\(\Delta G\) is negative because the process is spontaneous.
\(\Delta H\) is positive because the solution feels cold, indicating that energy is absorbed from the surroundings (the process is endothermic).
\(\Delta S\) is positive. For the process to be spontaneous, it must occur with an increase in entropy since the process is endothermic.
An important source of copper is the ore chalcocite, a form of copper(I) sulfide. When heated, the Cu2S decomposes to form copper and sulfur, as described by the following equation:
\[ \ce{Cu2S}(s) \rightleftharpoons \ce{2Cu}(s) + \ce{S}(s) \nonumber\]
- Determine \( \Delta G^\circ \) for the decomposition of Cu2S(s) at 298 K.
- The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine \( \Delta G^\circ \) for the process at 298 K.
- The production of copper from chalcocite is performed by roasting the Cu2S in air to produce Cu. By combining the equations from parts (a) and (b), write the equation that describes the roasting of chalcocite, and explain why coupling these reactions makes for a more efficient copper production process.
Thermodynamic data at 298 K:
| Substance | \( \Delta G_f^\circ \) (kJ/mol) |
|---|---|
| \( \ce{Cu2S}(s) \) | -86.2 |
| \( \ce{Cu}(s) \) | 0 |
| \( \ce{S}(s) \) | 0 |
| \( \ce{O2}(g) \) | 0 |
| \( \ce{SO2}(g) \) | -300.1 |
- Answer
-
(a) \[ \begin{align*} \Delta G^\circ &= 2\Delta G_f^\circ(\ce{Cu}(s)) + 1\Delta G_f^\circ(\ce{S}(s))- \Delta G_f^\circ(\ce{Cu2S}(s)) \\ &= 0~\text{kJ/mol} + 0~\text{kJ/mol} - (-86.2 ~\text{kJ/mol}) \\ &= 86.2~\text{kJ/mol} \end{align*} \]
(b) \(\ce{S(s) + O2 (g) <=> SO2 (g)} \)
\[ \begin{align*} \Delta G^\circ &= 1\Delta G_f^\circ(\ce{SO2}(g)) - [1\Delta G_f^\circ(\ce{S}(s))+ 1\Delta G_f^\circ(\ce{O2}(s))] \\ &= -300.1~\text{kJ/mol} - [0~\text{kJ/mol} + (0 ~\text{kJ/mol})] \\ &= -300.1~\text{kJ/mol} \end{align*} \]
(c) \[\begin{align*} \ce{Cu2S(s) &<=>2Cu(s) +\cancel{S(s)}} \quad &\Delta G^\circ = 86.2~\text{kJ/mol} \\
\ce{\cancel{S(s)} + O2(g) &<=> SO2 (g)} &\Delta G^\circ = -300.1~\text{kJ/mol} \\
\hline
\ce{Cu2S(s) + O2(g) &<=>2Cu(s) +SO2(g)} &\Delta G^\circ = 86.2 + (-300.1) = -213.9~\text{kJ/mol}
\end{align*}\]The overall \( \Delta G^\circ \) is negative, so the coupled reaction proceeds spontaneously under standard state conditions. Coupling the decomposition of Cu2S with the oxidation of sulfur makes the overall process thermodynamically favourable, enabling efficient copper production.