AP Chemistry Unit 9 Practice

Thermodynamics and units

Use the values below and keep extra digits until the final answer. Entropies are in J/(mol·K); reaction enthalpies and free energies are in kJ/mol. Reaction quantities refer to the balanced equation as written.

ΔS°rxn = Σ S°(products) − Σ S°(reactants)

Multiply each S° by its balanced coefficient. S° of an element at room temperature is not zero, unlike ΔH°f of an element in its standard state. A perfect crystal at 0 K has zero absolute entropy.

ΔG° = ΔH° − TΔS°

ΔG° = −RT ln K

ΔG = ΔG° + RT ln Q = RT ln(Q/K)

R = 8.314 J/(mol·K)

J/kJ check: divide ΔS° in J/(mol·K) by 1000 before combining TΔS° with ΔH° in kJ/mol. With R in J/(mol·K), RT ln K is in J/mol; divide by 1000 for kJ/mol. Use T in K, never °C.

Keep the + or − sign on every entropy-change, free-energy and cell-potential answer. The opposite sign is incorrect. Positive values may be typed with or without +. Precision notes are advisory. The unit selector does not convert values.

Favorability, equilibrium and coupling

ΔG° < 0: standard forward direction is thermodynamically favorable and K > 1. ΔG° > 0: it is unfavorable and K < 1. At actual equilibrium Q = K and ΔG = 0.

For constant ΔH° and ΔS°: −/+ favors all positive T; +/− favors none; −/− favors low T; +/+ favors high T. The boundary is T = ΔH°/ΔS° with matching energy units.

A negative free-energy change does not imply a fast reaction. A high activation barrier can make a favorable reaction slow. A catalyst changes rates, not ΔG° or K at fixed T.

Coupled reactions must transfer energy through a physical mechanism. Add their ΔG° values with the balanced equations. Reverse the sign when reversing a step; multiply ΔG° when scaling a step. Light or electrical work can drive an unfavorable chemical process.

Thermodynamic data

Rounded classroom values at 298 K. S°: J/(mol·K). ΔH°f: kJ/mol. Assume the given ΔH° and ΔS° remain approximately constant when a problem changes temperature.

Electrochemical cells

ΔG° = −nFE°

E°cell = E°(cathode) − E°(anode)

E = E° − (RT/nF) ln Q

F = 96,485 C/mol e⁻

With E in V and F in C/mol, nFE is in J/mol. Divide by 1000 for kJ/mol. Standard reduction potentials are never multiplied by stoichiometric coefficients. The balanced charge transfer determines n.

Oxidation is at the anode; reduction is at the cathode. A galvanic (voltaic) cell produces electrical work. An electrolytic cell requires external electrical work. Electrode signs change between these cell types, but oxidation and reduction roles do not.

For Q > 1, E < E°. For Q < 1, E > E°. At equilibrium Q = K and E = 0. Use the natural logarithm and ideal dilute activities. For Zn/Cu, Q is the zinc-ion activity divided by copper-ion activity; pure metals are omitted.

Electrolysis and Faraday’s law

I = q/t

q = It

n(e⁻) = q/F

Use current in A, charge in C and time in s. Divide mol of transferred charge carriers by the balanced half-reaction coefficient to find mol of metal. Multiply by molar mass in g/mol for deposited mass. Calculations assume 100% current efficiency unless stated otherwise.

Atomic masses (g/mol)

H

1.01

C

12.01

N

14.01

O

16.00

Na

22.99

Mg

24.31

Al

26.98

S

32.06

Cl

35.45

K

39.10

Ca

40.08

Fe

55.85

Cu

63.55

Zn

65.38

Ag

107.87