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Flashcards · AP® Chemistry · Unit 9

Unit 9: Thermodynamics and Electrochemistry flashcards

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9.1 Introduction to Entropy

What is entropy?
Entropy, S, measures how dispersed the matter and the energy of a system are. ΔS > 0 means more dispersal; ΔS negative means less. Name what spreads out (particles, their energy, or both), not just "disorder".
What are matter dispersal and energy dispersal?
Matter dispersal: particles spread over more space or become freer to move (melting, boiling, a gas expanding). Energy dispersal: particles' kinetic energies spread over a wider range of values, which happens when temperature rises.
How does entropy change with phase changes?
Solid → liquid: small increase (particles can move past each other). Liquid → gas: large increase (particles separate and fill a far larger volume). Boiling gives a much larger increase than melting. Reverse changes have negative ΔS (CO₂(g) → CO₂(s) is negative).
How do you predict the sign of ΔS for a reaction involving gases?
Count moles of gas only (not solids or liquids). More moles of gas in products: usually positive (2N₂O(g) → 2N₂(g) + O₂(g) is positive). Fewer: usually negative. Equal moles: small, and the sign is not predictable by counting (do not say ΔS = 0).
Does heating a sealed container of gas change its entropy?
Yes, S increases. Even with no phase change or volume change, the energy spreads over a wider range of kinetic energies.
Is the sign of ΔS linked to the sign of ΔH?
No. They are independent. Entropy explains why some endothermic changes, such as ice melting on a warm day, still happen.

9.2 Absolute Entropy and Entropy Change

What is absolute entropy, and what is its zero?
Entropy measured from a true zero: a perfect crystal at 0 K. Every substance at 298 K therefore has a positive entropy. Standard molar entropy, S°, is the absolute entropy of 1 mol at 1 bar, usually listed at 298 K, in J K⁻¹ mol⁻¹.
Is S° zero for elements in their standard states?
No. Only ΔH°f (and ΔG°f) are zero for elements. O₂(g) has S° = 205.2 J K⁻¹ mol⁻¹. A negative S° from a calculation means a sign slip.
How do S° values compare for the phases of one substance?
S°(g) > S°(l) > S°(s); gases have much larger values. H₂O(l) is 70.0 and H₂O(g) is 188.8 J K⁻¹ mol⁻¹, so the state symbol matters.
How do you calculate ΔS° for a reaction?
ΔS° = ΣnS°(products) − ΣnS°(reactants), multiplying each S° by its coefficient. 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l): (427.6 + 280.0) − (253.6 + 615.6) = −161.6 J K⁻¹ mol⁻¹.
What unit conversion is needed before combining ΔS° with ΔH°?
S° is in joules: divide ΔS° by 1000 to get kJ K⁻¹ mol⁻¹. −161.6 J K⁻¹ mol⁻¹ is −0.1616 kJ K⁻¹ mol⁻¹.
How does ΔS° change when an equation is reversed or doubled?
Reversing changes the sign; doubling doubles it. ΔS° = −332.3 J K⁻¹ mol⁻¹ for CO + 2H₂ → CH₃OH(l) gives +664.6 for 2CH₃OH(l) → 2CO + 4H₂.
What sign check should you do on a calculated ΔS°?
Compare with the change in moles of gas. If the sign disagrees (for example, gas moles decrease but ΔS° comes out positive), recheck the arithmetic and states.

9.3 Gibbs Free Energy and Thermodynamic Favorability

What does the sign of ΔG° tell you?
ΔG° negative: the process is thermodynamically favoured under standard conditions. ΔG° > 0: the reverse is favoured. ΔG° says which direction is favoured, never how fast.
What are standard states for ΔG°?
Pure solids and liquids, 1 M solutions, and gases at 1 bar (about 1 atm). Different concentrations or pressures change ΔG.
How do you calculate ΔG° from ΔH° and ΔS°?
ΔG° = ΔH° − TΔS°, with ΔH° in kJ mol⁻¹, ΔS° divided by 1000 to kJ K⁻¹ mol⁻¹, and T in K. ΔH° = +30.0 kJ mol⁻¹, ΔS° = +80.0 J K⁻¹ mol⁻¹ at 500 K: 30.0 − 500 × 0.0800 = −10.0 kJ mol⁻¹, favoured.
How do the signs of ΔH° and ΔS° decide when a process is favoured?
ΔH° −, ΔS° +: favoured at all T. ΔH° +, ΔS° −: favoured at no T. Both +: favoured at high T. Both −: favoured at low T.
How do you find the temperature at which a process becomes favoured?
Set ΔG° = 0: T = ΔH° ÷ ΔS°, with the same energy units top and bottom. ΔH° = +45.0 kJ mol⁻¹ and ΔS° = +150 J K⁻¹ mol⁻¹: 45 000 ÷ 150 = 300 K. This assumes ΔH° and ΔS° are constant with temperature.
How do you calculate ΔG° from free energies of formation?
ΔG° = ΣnΔG°f(products) − ΣnΔG°f(reactants), in kJ mol⁻¹; ΔG°f = 0 for elements in standard states. Multiply each by its coefficient.
Is every exothermic process thermodynamically favoured?
No. Check ΔS° too. Freezing water is exothermic, but with negative ΔS° it is not favoured above 273 K.

9.4 Thermodynamic and Kinetic Control

What is the difference between thermodynamic and kinetic control?
Thermodynamics (sign of ΔG°) says whether a process is favoured; kinetics (Eₐ, rate) says how fast. A process that is favoured (ΔG° negative) but does not occur at a measurable rate is under kinetic control.
Give examples of favoured processes that are extremely slow.
Paper burning in air, hydrogen and oxygen forming water, diamond becoming graphite. Each has ΔG° negative but a high activation energy.
Why is a favoured reaction under kinetic control so slow?
Usually because of a high activation energy: very few collisions have enough energy to reach the transition state. The size of ΔG° and the size of Eₐ are unrelated.
Is a mixture that shows no change for a year at equilibrium?
Not necessarily. At equilibrium the forward and reverse rates are equal and Q = K. Under kinetic control both rates are nearly zero and Q ≪ K. Compare Q with K.
What does adding a catalyst do to a system under kinetic control or at equilibrium?
Under kinetic control, the reaction starts, because Eₐ is lowered. At equilibrium, nothing changes in amounts, because both rates rise equally. A catalyst never changes ΔG°, ΔH° or K.
How do you describe a reaction with ΔG° negative that is not observed?
Say it is "thermodynamically favoured but under kinetic control", not "not favoured". Explain with particles: few collisions have enough energy to overcome Eₐ.

9.5 Free Energy and Equilibrium

How are ΔG° and K related?
ΔG° = −RT ln K, with ΔG° in J mol⁻¹, R = 8.314 J mol⁻¹ K⁻¹ and T in K. Rearranged: K = e^(−ΔG°/RT). Use ln, not log, and convert kJ to J.
What does the sign of ΔG° tell you about K?
ΔG° negative: K > 1, products favoured at equilibrium. ΔG° > 0: K below 1, reactants favoured, but some product still forms (K is never zero). ΔG° = 0: K = 1. K = 1.0 × 10⁻³ at 298 K gives ΔG° = +17.1 kJ mol⁻¹.
How can you judge whether K is close to 1 from ΔG°?
Compare ΔG° with RT, about 2.48 kJ mol⁻¹ at 298 K. If ΔG° is about the size of RT or smaller, K is near 1; if it is many times RT, K is very far from 1.
How can you estimate K from ΔG° without a calculator at 298 K?
Every 5.7 kJ mol⁻¹ of ΔG° is a factor of about 10 in K (from RT ln 10). ΔG° = −11.4 kJ mol⁻¹ gives K ≈ 10² (exactly 99.6).
How does temperature change K?
Mainly by changing ΔG° = ΔH° − TΔS°; then K = e^(−ΔG°/RT) at the new temperature, assuming ΔH° and ΔS° barely change.
What sign check catches errors in ΔG° = −RT ln K?
K below 1 must give ΔG° > 0, and K > 1 must give ΔG° negative. If not, a minus sign was lost.

9.6 Free Energy of Dissolution

What three factors make up the free energy of dissolving a salt?
1. Breaking up the solid: ion–ion attractions overcome; ΔH > 0 (large), ΔS > 0. 2. Reorganizing the solvent: some water–water attractions broken and water ordered around ions; ΔH usually > 0, ΔS usually negative. 3. Ion–solvent (ion–dipole) attractions form: ΔH negative (large).
Why is it hard to predict whether a salt will dissolve?
ΔG° of dissolution is a small difference of large opposing terms. The three-factor model explains results better than it predicts them; a full explanation compares the opposing contributions.
How are ΔG° of dissolution and K_sp related?
ΔG° = −RT ln K_sp. ΔG° negative ⇔ K_sp > 1 (very soluble); ΔG° > 0 ⇔ K_sp below 1. ΔH° = +8.0 kJ mol⁻¹, ΔS° = +20 J mol⁻¹ K⁻¹ at 298 K: ΔG° = +2.04 kJ mol⁻¹, K_sp = 0.44, not favoured.
Why can dissolving decrease entropy?
Small or highly charged ions order water molecules into shells around them, which lowers entropy. So ΔS° of dissolving can be small or even negative.
How does temperature affect K_sp?
Endothermic dissolving (ΔH° > 0): K_sp rises with T. Exothermic dissolving (ΔH° negative): K_sp falls with T.
Is an endothermic dissolving process always unfavourable?
No. A large positive ΔS° can make ΔG° negative. Likewise, an exothermic one can be unfavourable if ΔS° is strongly negative.

9.7 Coupled Reactions

How can a thermodynamically unfavourable process (ΔG° > 0) be made to happen?
By supplying free energy: from an outside source (electrical energy in an electrolytic cell or recharging a battery; light in photosynthesis), or by coupling it to a favourable reaction.
What is a coupled reaction?
A favourable reaction (large negative ΔG°) linked to an unfavourable one through a common intermediate that cancels when the equations are added. The unfavourable step keeps its positive ΔG°; only the overall reaction is favourable.
How do you find ΔG° and K for a coupled pair?
ΔG°(overall) = ΔG°₁ + ΔG°₂; K(overall) = K₁ × K₂. +12.0 and −20.0 kJ mol⁻¹ give −8.0 kJ mol⁻¹ (favourable). K₁ = 2.0 × 10⁻³ and K₂ = 4.0 × 10⁴ give K = 80.
What role does ATP play in living cells?
Hydrolysis of ATP to ADP (ΔG° about −30 kJ mol⁻¹) is often the favourable partner coupled to unfavourable reactions. Using 2 ATP multiplies its ΔG° by 2. Questions give the value to use.
Are two reactions in separate beakers coupled?
No. Coupling requires a shared intermediate passed directly from one step to the other.
Does a catalyst drive an unfavourable reaction?
No. Catalysts change rate, not ΔG°. A negative overall ΔG° also says products are favoured, not that the reaction is fast.

9.8 Galvanic (Voltaic) and Electrolytic Cells

What is the difference between a galvanic and an electrolytic cell?
A galvanic (voltaic) cell runs a thermodynamically favourable redox reaction (ΔG° negative) and produces electrical energy. An electrolytic cell uses a power supply to drive an unfavourable reaction (ΔG° > 0), turning electrical energy into chemical energy.
What happens at the anode and cathode in every electrochemical cell?
Oxidation at the anode; reduction at the cathode. Electrons move through the external wire from anode to cathode, in both galvanic and electrolytic cells. Rely on anode = oxidation, not on electrode signs.
What does the salt bridge do?
It completes the circuit with moving ions, keeps each half-cell electrically neutral and keeps the solutions apart. Only ions move in it, never electrons. Without it, charge builds up and the current stops.
Which way do ions move in a salt bridge?
Anions move towards the anode; cations towards the cathode. In a KNO₃ bridge, K⁺ moves to the cathode half-cell and NO₃⁻ to the anode half-cell.
How do you identify the anode and cathode from observations?
An electrode that loses mass is the anode (metal oxidized, e.g. a zinc electrode getting thinner). One that gains mass is the cathode (metal ions reduced and deposited). Gas may form, e.g. Cl₂ at the anode from Cl⁻.
How do you link mass changes at the two electrodes?
Electrons lost at the anode = electrons gained at the cathode. 0.0200 mol e⁻ reducing Cu²⁺ gives 0.0100 mol Cu = 0.636 g. Balance electrons before comparing moles.
Do bigger electrodes raise a cell's voltage?
No. Electrode size does not change the voltage; it only lets the cell run longer.

9.9 Cell Potential and Free Energy

What is a standard reduction potential?
E° measures how easily a species is reduced, relative to the hydrogen half-cell (0.00 V by definition). More positive E° = more easily reduced = stronger oxidizing agent.
How do you calculate a standard cell potential?
The half-reaction with the higher E° is the reduction (cathode); the lower one runs as oxidation (anode). E°cell = E°(cathode) − E°(anode), using reduction potentials. Mg + 2Ag⁺: 0.80 − (−2.37) = +3.17 V.
Should E° be multiplied by coefficients when balancing a cell equation?
No. Balance electrons for the equation and for n, but E° stays the same. Multiplying the equation by k multiplies n and ΔG° by k, not E°.
How are ΔG° and E° related?
ΔG° = −nFE°, with n = mol e⁻ per mol reaction (from the balanced equation), F = 96,485 C mol⁻¹, E° in V, ΔG° in J mol⁻¹. n = 1 and E° = +0.50 V gives −48.2 kJ mol⁻¹. They always have opposite signs.
What does the sign of E°cell tell you?
E°cell > 0: favoured, galvanic, positive voltage. E°cell negative: unfavoured, needs an externally applied potential (electrolytic). Reversing the reaction changes the sign of both E° and ΔG°. A favourable E° says nothing about rate.
Is Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g) favoured under standard conditions?
Yes. E° = 0.00 − (−0.76) = +0.76 V, and ΔG° = −2 × 96,485 × 0.76 = −147 kJ mol⁻¹.
What sign error is common when calculating E°cell?
Changing the sign twice. Either subtract the anode's reduction potential or add its oxidation potential, never both.

9.10 Cell Potential Under Nonstandard Conditions

Why does a real cell's potential differ from E°?
E° is for standard conditions, where Q = 1. A real cell usually has Q ≠ 1. The cell potential is a driving force towards equilibrium: the further from equilibrium, the larger E. Q below 1 → E > E°; Q > 1 → E below E°.
What is the Nernst equation, and what does it show?
E = E° − (RT/nF) ln Q. With RT/F = 0.0257 V at 298 K, a tenfold change in Q shifts E by only a few hundredths of a volt. Explain the change in Q first; the number only supports it.
Why does a galvanic cell's voltage fall as it runs?
As it runs, Q rises towards K, so E falls. At Q = K, E = 0: the cell is "dead", meaning at equilibrium, not necessarily out of reactant.
How should you explain a change in E when a concentration changes?
Use Q, not Le Châtelier language, because a working cell is not at equilibrium. For example: "Q decreases, so the cell is further from equilibrium, so E increases." Concentrations change E, never E°.
What is a concentration cell?
A cell with the same half-reaction on both sides, so E° = 0, which still gives a voltage while concentrations differ. The dilute side is the anode; electrons flow from dilute to concentrated until concentrations are equal. Ni in 0.10 M and 1.0 M Ni²⁺: E ≈ (0.0257/2) ln 10 ≈ 0.030 V.
How are E° and K related?
E° = (RT/nF) ln K. Positive E° ⇔ K > 1.

9.11 Electrolysis and Faraday’s Law

How do you find the charge passed from current and time?
q = I × t, with I in amperes, t in seconds and q in coulombs. Convert minutes or hours to seconds first. 1.00 A for 1.00 h passes 3600 C.
What is the Faraday constant, and how is it used?
F = 96,485 C mol⁻¹, the charge on one mole of electrons. n(e⁻) = q / F. 3600 C is 0.0373 mol e⁻. Coulombs are not moles.
How do you calculate the mass of metal deposited in electrolysis?
n(e⁻) = It/F; n(metal) = n(e⁻) ÷ electrons per ion (the ion's charge); m = n × M. 0.100 mol e⁻ with Cu²⁺ gives 0.0500 mol Cu = 3.18 g. Al³⁺ needs 3 mol e⁻ per mol Al.
How many electrons are needed per molecule of gas formed in electrolysis?
2 e⁻ per Cl₂ or H₂, but 4 e⁻ per O₂ from water. Take the count from the half-reaction.
How do you find the charge on an unknown metal ion from electrolysis data?
Electrons per ion = n(e⁻) ÷ n(metal), using n(e⁻) from It/F and n(metal) from the measured mass ÷ molar mass.
Why is a Faraday's law answer a maximum?
It assumes all the charge goes into the reaction calculated, with a steady current and no side reactions such as hydrogen forming at the cathode. Concentration and electrode size do not change the yield for a fixed charge.
Do cells in series give equal masses at their electrodes?
They pass the same charge, so equal moles of electrons, not equal masses. Mass depends on each ion's charge and molar mass.

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