Chapter 9. Applications of Thermodynamics
9.1 Introduction to Entropy
9.1.1
9.1.2
Melting of ice: Entropy increases because matter becomes more dispersed as the rigid structure of ice transitions to the more disordered liquid phase.
Formation of a precipitate: Entropy decreases because the ions become less dispersed as they transition from the aqueous phase to the solid precipitate.
Heating of a copper block: Entropy increases because the thermal energy becomes more dispersed across the microstates of the copper atoms.
Mixing of two ideal gases: Entropy increases because the mixing of gases disperses the particles over a larger number of configurations.
Decomposition of hydrogen peroxide: Entropy increases because the formation of O₂(g) adds a highly dispersed gas phase to the system.
Isothermal compression of an ideal gas: Entropy decreases because the gas particles are confined to a smaller volume, reducing their freedom and the number of possible configurations.
Detailed explanation here.
PART A.
According to the kinetic molecular theory, temperature is directly proportional to the average kinetic energy and thus the speed of gas particles. As temperature increases, both the average speed of gas particles and the range of their speeds increase. This is visually represented by the Maxwell-Boltzmann distribution curve shifting to the right and becoming flatter, indicating a broader distribution of particle speeds.
Detailed explanation here.
PART B.
The entropy of the gas mixture increases with temperature. This is because at higher temperatures, the energy within the system becomes more dispersed as gas particles exhibit a greater variety of kinetic energies and velocities. This increased energy dispersion corresponds to a larger number of possible microstates for the system, which is the fundamental reason for the increase in entropy.
Detailed explanation here.
9.2 Absolute Entropy and Entropy Change
9.2.1
PART B.
ΔS°vap = S°(H₂O(g)) - S°(H₂O(l)) ΔS°vap = (188.8 J/mol·K) - (70.0 J/mol·K) ΔS°vap = +118.8 J/mol·K
Detailed explanation here.
PART C.
The entropy change (ΔS°) for the vaporization of water is positive (+118.8 J/mol·K). This is because the gaseous phase (vapor) is significantly more disordered than the liquid phase.
Detailed explanation here.
9.3 Gibbs Free Energy and Thermodynamic Favorability
9.3.1
9.3.2
9.3.3
9.3.4
9.3.5
The scenarios that describe a system under standard state conditions are:
B) Aqueous hydrochloric acid solution with a concentration of 1.0 M
D) Oxygen gas at 1.0 bar and 25°C
Detailed explanation here.
A negative ΔG° indicates a thermodynamically favorable reaction, meaning it can occur spontaneously under standard conditions. However, this does not imply that the reaction will proceed at a noticeable rate at room temperature, as a high activation energy might still prevent spontaneous ignition.
Detailed explanation here.
the standard Gibbs free energy change (ΔG°) for the decomposition of calcium carbonate is positive (+131.1 kJ/mol), the reaction is not thermodynamically favored under standard conditions (298.15 K and 1 atm pressure). This means that energy input is required for this reaction to proceed spontaneously.
Detailed explanation here.
PART B.
The entropy increase (more disorder as ions spread out in water) offsets the endothermic enthalpy, making ΔG negative and the process spontaneous.
Detailed explanation here.
PART C.
The process becomes more favorable at higher temperatures (ΔG becomes more negative).
Detailed explanation here.
PART A.
The student is incorrect. A process with ΔH° > 0 and ΔS° < 0 is never spontaneous, regardless of temperature.
Increasing temperature actually makes ΔG° more positive in this case.
Detailed explanation here.
PART B.
But when both ΔH° and ΔS° have the same sign, you must consider temperature, because the effect of the TΔS term can either outweigh or fail to outweigh ΔH° depending on the value of T.
Detailed explanation here.
9.4 Thermodynamic and Kinetic Control
9.4.1
PART B.
MnO₂ reduces the activation energy, allowing the thermodynamically favored reaction to proceed at a measurable rate.
Detailed explanation here.
PART C.
It’s incorrect because the system is simply not reacting, not because it has reached equilibrium. It is stuck under kinetic control, not thermodynamic balance.
Detailed explanation here.
9.5 Free Energy and Equilibrium
9.5.1
9.5.2
9.5.3
9.5.4
Reactions C and D are thermodynamically favored because they have K > 1, meaning products are favored at equilibrium. Reactions A and B are not thermodynamically favored because K < 1.
Detailed explanation here.
Reaction A has the larger equilibrium constant because its ΔG° is more negative. This means it is more thermodynamically favored and shifts more toward products at equilibrium.
Detailed explanation here.
Since ΔG° is small and close to zero, the equilibrium constant K is close to 1, meaning neither products nor reactants are heavily favored at equilibrium.
Detailed explanation here.
Final Answer: Options A and C are consistent with the relationship between ΔG° and K.
Option B is not consistent because a positive ΔG° should not correspond with a large K.
Detailed explanation here.
9.6 Coupled Reactions
9.6.1
9.6.2
Charging a battery is thermodynamically unfavorable (ΔG° > 0), so it requires external electrical energy to force the reverse redox reaction. This confirms that an unfavorable process can be made to occur using an outside energy source.
Detailed explanation here.
The overall ΔG° is –8.0 kJ/mol, so the coupled reaction is thermodynamically favorable. Coupling the unfavorable A → B process to a more favorable C → D reaction makes the combined process spontaneous.
Detailed explanation here.
9.7 Galvanic (Voltaic) and Electrolytic Cells
9.7.1
9.7.2
9.7.3
PART A. Electrons flow from the magnesium electrode (anode) to the silver electrode (cathode) in the external circuit.
Detailed explanation here.
PART B. The magnesium electrode loses mass, and the silver electrode gains mass as the cell operates.
Detailed explanation here.
PART C. The salt bridge maintains electrical neutrality. Na⁺ ions flow to the cathode half-cell, and NO₃⁻ ions flow to the anode half-cell.
Detailed explanation here.
PART D. The cell will still function because the redox reaction involving Ag⁺ ions does not depend on the silver electrode itself. The platinum electrode serves as a surface for the reduction reaction.
Detailed explanation here.
PART E. In the anode solution, Mg²⁺ concentration increases, and NO₃⁻ ions enter to balance charge. In the cathode solution, Ag⁺ concentration decreases, and Na⁺ ions enter to balance charge.
Detailed explanation here.
PART F. If the solutions are mixed into a single container, the redox reaction will occur directly, and the cell’s voltage will drop to zero because the circuit is bypassed.
Detailed explanation here.
PART A.
Anode: Fe(s)
Cathode: Ni(s)
Electron flow: From iron to nickel.
Reaction is thermodynamically favored.
Detailed explanation here.
PART B.
Anode: Ni(s)
Cathode: Fe(s)
Electron flow: From nickel to iron.
An external power source is required to drive the reaction because it is thermodynamically unfavored.
Detailed explanation here.
PART C.
In the galvanic cell, Ni²⁺ ions are reduced, decreasing their concentration.
In the electrolytic cell, Ni²⁺ ions are produced, increasing their concentration.
Detailed explanation here.
PART D.
In the galvanic cell, the Fe(s) electrode loses mass as Fe(s) is oxidized to Fe²⁺.
In the electrolytic cell, the Fe(s) electrode gains mass as Fe²⁺ is reduced to Fe(s).
Detailed explanation here.
PART E.
In the galvanic cell, the salt bridge balances ion flow: K⁺ ions flow to the cathode, and Cl⁻ ions flow to the anode.
In the electrolytic cell, a salt bridge is not required because the external power source provides the necessary driving force.
Detailed explanation here.
The anode is always the site of oxidation, so its designation doesn't change between galvanic and electrolytic cells; in the provided reactions, Fe(s) and H₂O(l) are oxidized, identifying them as anodes in their respective cells.
Detailed explanation here.
9.8 Cell Potential and Free Energy
9.8.1
9.8.2
PART B.
Since the E° cell for the electrolytic reaction is negative (-0.54V), the reaction is not thermodynamically favored (it is non-spontaneous).
Detailed explanation here.
PART C.
Because the E° cell for the electrolysis reaction is negative, the process is non-spontaneous. Non-spontaneous reactions require an input of energy to proceed. In an electrolytic cell, this energy is supplied by an external power source (a battery or power supply). The power source provides the electrical potential necessary to overcome the non-spontaneity and drive the reaction forward.
Detailed explanation here.
PART C.
ΔG° = +786 kJ (or +785.6 kJ if maintaining more significant figures)
Detailed explanation here.
PART D.
Since ΔG° is positive (+786 kJ), the reaction is not thermodynamically favored (it is non-spontaneous) as written. A positive ΔG° indicates that the reaction will not proceed spontaneously in the forward direction. The negative E°<sub>cell</sub> (-4.07 V) also confirms this; a negative cell potential means the reaction is non-spontaneous. This is expected for an electrolytic cell, which requires an external power source to drive the non-spontaneous reaction.
Detailed explanation here.
9.9 Cell Potential Under Nonstandard Conditions
9.9.1
9.9.2
9.9.3
9.9.4
PART A. The cell potential will decrease compared to E°cell because the concentrations favor the reverse reaction, bringing the system closer to equilibrium.
Detailed explanation here.
PART B. The reaction quotient is Q = 160.
Detailed explanation here.
PART C. The calculated Q = 160 decreases the cell potential, as the system favors the reverse reaction and is farther from equilibrium in the forward direction.
Detailed explanation here.
PART D. Increasing [Ag⁺] to 1.00 M decreases Q, which increases the cell potential, as the reaction is farther from equilibrium in the forward direction.
Detailed explanation here.
no question for 9.9.2
PART A. The cell potential will decrease compared to E°cell.
Detailed explanation here.
PART B.
Scenario A:
Q = 1, so the reaction is at standard conditions.
The system is at standard conditions, and the driving force is at its standard value.
Scenario B:
Q = 0.010, so the system favors product formation and is farther from equilibrium in the reverse direction.
The low concentration of product (Mg²⁺) and high concentration of reactant (Fe²⁺) provides a strong driving force for the forward reaction, increasing the cell potential.
Scenario C:
Q = 100, so the system favors reactant formation and is farther from equilibrium in the forward direction.
The high concentration of product (Mg²⁺) and low concentration of reactant (Fe²⁺) decreases the driving force for the forward reaction, lowering the cell potential.
Detailed explanation here.
PART C.
Increasing [Mg²⁺] decreases the cell potential by reducing the driving force of the forward reaction.
Detailed explanation here.
PART D.
Scenario A is the closest to equilibrium, as Q is equal to 1, meaning the reaction is at standard conditions, which is a reference point for comparing how far other conditions are from equilibrium.
Detailed explanation here.
PART A.
Scenario A: Ecell = E°cell = +1.10 V.
Scenario B: The cell potential will increase compared to E°cell.
Scenario C: The cell potential will decrease compared to E°cell.
Detailed explanation here.
PART B.
The cell potential will increase compared to E°cell because the system is far from equilibrium in the reverse direction, strongly favoring the forward reaction.
Detailed explanation here.
PART C.
The cell potential will decrease compared to E°cell because the system favors the reverse reaction and is farther from equilibrium in the forward direction, reducing the driving force for the forward reaction.
Detailed explanation here.
PART D.
Scenario A is at standard conditions (Q=1). Without knowing the value of K, we cannot determine which scenario is closest to equilibrium.
Detailed explanation here.
PART E.
Decreasing [Zn²⁺] in Scenario C will decrease Q, increasing the cell potential by moving the system farther from equilibrium in the reverse direction.
Detailed explanation here.