Practice Problems Thermodynamics

Answer

The second law states that a process is spontaneous if the system and the surroundings have an increase in entropy. Thus, even is a given system has a decrease in entropy (suggesting nonspontaneity), if there is sufficient increase in the entropy of the surroundings then the process can be spontaneous.

Answer

The thermodynamic equilibrium constant K_eq is formally written in terms of activities, a unitless quantity. Activities are approximated by using solution concentrations in units of molarity (divided by 1 M to remove the units) and gas partial pressures in units of atm (divided by 1 atm to remove the units). K_c is written solely in terms of molarity and K_p is written solely in terms of partial pressures. K_eq can mix "units".

Answer

(a) CH₃OH(l) → CH₃OH(g)

The entropy, S, increases because there are more moles of gaseous products.

(b) N₂O₄(g) → 2 NO₂(g)

The entropy, S, increases because there are more moles of gaseous products.

(c) CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Can not predict the entropy change because there are the same number of moles of gaseous products and reactants.

(d) 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

The entropy, S, increases because there are more moles of gaseous products.

(e) CH₃COOH(l)→CH₃COOH(s)

ΔS° < 0 (decrease in entropy) since the reaction goes from a liquid to a solid.

(f) N₂(g) + O₂(g)→2 NO(g)

ΔS° ~ 0 (little change in entropy) since the reaction goes from 2 moles of gases to 2 moles of gases.

(g) N₂H₄(l)→N₂(g) + 2 H₂(g)

ΔS° > 0 (increase in entropy) since the reaction goes from a liquid to a 3 moles of gases.

(h) 2 NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)

ΔS° < 0 (decrease in entropy) since the reaction goes from a liquid + 2 moles of gases to an all liquid state.

Answer

Pollution is caused by random and widespread disposal of items, a high entropy situation. To eliminate pollution, the reverse process must be invoked, so that the entropy of the system now becomes negative, which requires and large positive entropy change for the surroundings and a large energy input.

Answer

(a) PCl₃(g) + Cl₂(g) → PCl₅(g) ΔH° = –87.9 kJ

ΔS° is predicted to be negative since there are fewer moles of gaseous products than reactants. Thus, at low temperatures enthalpy will dominate (TΔS° is small) and the reaction will be spontaneous (ΔH° < 0) while at high temperature the entropy dominates and the reaction becomes nonspontaneous (TΔS° becomes large and negative).

(b) 2 NH₃(g) → N₂(g) + 3 H₂(g) ΔH° = +92.2 kJ

ΔS° is predicted to be positive since there are more moles of gaseous products than reactants. Thus, at low temperatures enthalpy will dominate (TΔS° is small) and the reaction will be nonspontaneous (ΔH° > 0) while at high temperature the entropy dominates and the reaction becomes spontaneous (TΔS° becomes large and positive).

(c) 2 N₂O(g) → 2 N₂(g) + O₂(g) ΔH° = –164.1 kJ

ΔS° is predicted to be positive since there are more moles of gaseous products than reactants. Thus, at low temperatures enthalpy will dominate (TΔS° is small) and the reaction will be spontaneous (ΔH° < 0) while at high temperature the entropy dominates and the reaction is also spontaneous (TΔS° becomes large and positive). Therefore, the reaction is expected to be spontaneous over all temperatures.

(d) H₂O(g) + ½O₂(g) → H₂O₂(g) ΔH° = +105.5 kJ

ΔS° will be negative (1.5 moles gases in reactants but only 1 mole of gases in the product) and ΔH° is positive so ΔG° will be positive at all temperatures. Thus, the reaction is never spontaneous.

(e) CH₄(g) + O₂(g) → CO₂(g) + 2 H₂O(g) ΔH° = –802.3 kJ

ΔS° will be positive (2 moles gases in reactants but 3 moles of gases in the product) and ΔH° is negative so ΔG° will be negative at all temperatures. Thus, the reaction is spontaneous at all temperatures.

(f) 2 CO(g) + O₂(g) → 2 CO₂(g) ΔH° = –566.0 kJ

ΔS° will be negative (3 moles gases in reactants but only 2 moles of gases in the product) and ΔH° is negative so ΔG° will be negative at low temperatures and positive at high temperatures. Thus, the reaction is spontaneous only at low temperature.

Answer

(a) C₂H₄(g) + H₂(g) → C₂H₆(g)

ΔG_f°(C₂H₆(g)) = –32.89 kJ/mol

ΔG_f°(H₂(g)) = 0.00 kJ/mol

ΔG_f°(C₂H₄(g)) = 68.12 kJ/mol

so ΔG° = [–32.89] – [68.12 + 0.00] = –101.01 kJ

(b) SO₃(g) + CaO(s) → CaSO₄(s)

ΔG_f°(CaSO₄(s)) = –1322 kJ/mol

ΔG_f°(CaO(s)) = –604.0 kJ/mol

ΔG_f°(SO₃(g)) = –371.1 kJ/mol

so ΔG° = [–1322] – [–604.0 + –371.1] = –347 kJ

Answer

The enthalpy of reaction is found from the enthalpies of formation:

ΔH_f° (CS₂(l)) = 89.70 kJ/mol

ΔH_f° (O₂(g)) = 0 kJ/mol

ΔH_f° (CO₂(g)) = –393.5 kJ/mol

ΔH_f° (SO₂(g)) = –296.8 kJ/mol

ΔH° = [(–393.5) + 2(–296.8)] – [(89.70) + 3(0)] = –1076.8 kJ

The entropy of reaction is found from the standard entropies:

S° (CS₂(l)) = 151.3 J/mol·K

S° (O₂(g)) = 205.0 J/mol·K

S° (CO₂(g)) = 213.6 J/mol·K

S° (SO₂(g)) = 248.1 J/mol·K

ΔS° = [(213.6) + 2(248.1)] – [(151.3) + 3(205.0)] = –56.5 J/K

One way to find ΔG° is to use the free energies of formation:

ΔG_f° (CS₂(l)) = 65.27 kJ/mol

ΔG_f° (O₂(g)) = 0 kJ/mol

ΔG_f° (CO₂(g)) = –394.4 kJ/mol

ΔG_f° (SO₂(g)) = –300.2 kJ/mol

ΔG° = [(–394.4) + 2(–300.2)] – [(65.27) + 3(0)] = –1060.1 kJ

Another way to find the free energy of reaction is to use the enthalpy and entropy changes:

ΔG° = ΔH° – TΔS° = –1076.8 – (298)( –0.0565) = –1060.0 kJ

As required, the Gibb's free energies are the same, within experimental error.

Answer

C(graphite)

ΔH_f° = 0.00 kJ/mol

S° = 5.74 J/mol·K

ΔG_f° = 0.00 kJ/mol

H₂O(g)

ΔH_f° = –241.8 kJ/mol

S° = 188.7 J/mol·K

ΔG_f° = –228.6 kJ/mol

CO(g)

ΔH_f° = –110.5 kJ/mol

S° = 197.6 J/mol·K

ΔG_f° = –137.2 kJ/mol

H₂(g)

ΔH_f° = 0.00 kJ/mol

S° = 130.6 J/mol·K

ΔG_f° = 0.00 kJ/mol

ΔH° = [–110.5 + 0.0] – [0.0 + –241.8] = 131.3 kJ

ΔS° = [197.6 + 130.6] – [5.74 + 188.7] = 133.8 J/K = 0.1338 kJ/K

ΔG° = [–137.2 + 0.0] – [0.0 + –228.6] = 91.4 kJ

ΔG° = 131.3 – (298)(0.1338) = 91.4 kJ

Both methods give the same ΔG°, as indeed they should.

Answer

Use Trouton's Rule: ΔS°_vpn = ΔH°_vpn/T_bp ≈ 87 J/mol·K

so T_bp = ΔH°_vapn/(87 J/mol·K) = (31690 J/mol)/(87 J/mol·K) = 364 K = 91 °C.

Answer

Trouton's Rule suggests that ΔH°_vapn = 87 J/mol·K×T_bp = (87)(273.15 + 59.47) = 29000 J/mol = 29 kJ/mol.

To use the data from the Table of Thermodynamic Quantities, use the reaction:

Br₂(l) → Br₂(g)

with ΔH_f° (Br₂(l)) = 0 kJ/mol and ΔH_f° (Br₂(l)) = 30.91 kJ/mol so ΔH° = ΔH°_vapn = [30.91] – [0] = 30.91 kJ/mol.

Given the approximate nature of Trouton's Rule, the estimate and the experimental value are remarkably similar.

Answer

(a) 2 NO(g) + O₂(g) → ← 2 NO₂(g)

(b) MgSO₃(s) → ← MgO(s) + SO₂(g)

(c) HCN(aq) + H₂O(l) → ← H₃O⁺(aq) + CN^–(aq)

Answer

(a) 2 SO₂(g) + O₂(g) → ← 2 SO₃(g)

ΔG_f° (SO₃(g)) = –371.1 kJ/mol

ΔG_f° (SO₂(g)) = –300.2 kJ/mol

ΔG_f° (O₂(g)) = 0 kJ/mol

ΔG° = [2(–371.1)] – [2(–300.2) + (0)] = –141.8 kJ

ΔG° = –RTlnK_p

–141800 = –(8.314)(298)lnK_p

K_p = e^57.23 = 7.2×10²⁴

(b) CH₄(g) + 2 H₂O(g) → ← CO₂(g) + 4 H₂(g)

ΔG_f° (CO₂(g)) = –394.4 kJ/mol

ΔG_f° (H₂(g)) = 0 kJ/mol

ΔG_f° (CH₄(g)) = –50.75 kJ/mol

ΔG_f° (H₂O(g)) = –228.6 kJ/mol

ΔG° = [(–394.4) + 4(0)] – [(–50.75) + 2(–228.6)] = 113.6 kJ

ΔG° = –RTlnK_p

113600 = –(8.314)(298)lnK_p

K_p = e^–45.85 = 1.2×10^–20

Answer

(a) 2 N₂O(g) + O₂(g) → ← 4 NO(g)

ΔG°_f(NO(g)) = 86.57 kJ/mol

ΔG°_f(O₂(g)) = 0.00 kJ/mol

ΔG°_f(N₂O(g)) = 104.2 kJ/mol

ΔG° = [4(86.57)] – [2(104.2) + 0.0] = 137.9 kJ

137900 = –(8.314)(298)lnK_p

lnK_p = –55.7

K_p = e^–55.7 = 6×10^–25

(b) 2 NH₃(g) + 2 O₂(g) → ← N₂O(g) + 3 H₂O(g)

ΔG°_f(N₂O(g)) = 104.2 kJ/mol

ΔG°_f(H₂O(g)) = –228.6 kJ/mol

ΔG°_f(NH₃(g)) = –16.48 kJ/mol

ΔG°_f(O₂(g)) = 0.00 kJ/mol

ΔG° = [104.2 + 3(–228.6)] – [2(–16.48) + 2(0.00)] = –548.6 kJ

–548600 = –(8.314)(298)lnK_p

lnK_p = 221.4

K_p = e^221.4 = 1×10⁹⁶

Answer

Find the enthalpy of reaction at 298 K:

ΔH_f°(CF₄(g)) = –925 kJ/mol

ΔH_f°(SO₂(g)) = –296.8 kJ/mol

ΔH_f°(SF₄(g)) = –763 kJ/mol

ΔH_f°(CO₂(g)) = –393.5 kJ/mol

ΔH° = [(–925) + (–296.8)] – [(–393.5) + (–763)] = –65 kJ

Find the entropy of reaction at 298 K:

S°(CF₄(g)) = 261.6 J/mol·K

S°(SO₂(g)) = 248.1 J/mol·K

S°(SF₄(g)) = 299.6 J/mol·K

S°(CO₂(g)) = 213.6 J/mol·K

ΔS° = [(261.6) + (248.1)] – [(213.6) + (299.6)] = –3.5 J/K

Assuming that ΔH° and ΔS° do not change significantly between 298 K and 318 K (45 °C):

ΔG° = ΔH° – TΔS° = –65 – (318)(–0.0035) = –64 kJ = –64000 J

ΔG° = –RTlnK_eq

–64000 = –(8.314)(318)lnK_eq

K_eq = e^24.2 = 3×10¹⁰

Answer

ΔH°_f(SO₂(g)) = –296.8 kJ/mol

ΔH°_f(SO₃(g)) = –395.7 kJ/mol

ΔH°_f(SCl₂(g)) = –50.0 kJ/mol

ΔH°_f(OSCl₂(g)) = –245.6 kJ/mol

ΔH° = [–245.6 + –296.8] – [–395.7 + –50.0] = –96.7 kJ

S°(SO₂(g)) = 248.1 J/mol·K

S°_f(SO₃(g)) = 256.6 J/mol·K

S°_f(SCl₂(g)) = 184 J/mol·K

S°_f(OSCl₂(g)) = 121 J/mol·K

ΔS° = [121 + 248.1] – [256.6 + 184] = –72 J/K

–(8.314)Tln(1.0×10¹⁵) = –96700 – T(–72)

–287T = –96700 + 72T

–359T = –96700

T = 269 K (= –4 °C)

Answer

Since ΔG° = ΔH° – TΔS°, the Gibb's energy changes linearly with temperature. If ΔS° is a large value, then ΔG° can change significantly with temperature.

If the enthalpy change and the entropy change are both constant with respect to temperature and a reaction is nonspontaneous at standard-state conditions, the reaction can not become spontaneous both at higher and lower temperature. For example, if both the enthalpy change and the entropy change are negative, the reaction could be nonspontaneous at standard-state conditions and could become spontaneous at lower temperature; however, this scenario does not allow spontaneity at higher temperature (the Gibb's energy change is always positive after a certain temperature is attained).

Answer

The van't Hoff equation is so the slope of the ln K_eq versus 1/T plot equals –ΔH°/R.

The slope of the plot = 22700 so ΔH° = –R×slope = –(8.314)×(22700) = –189000 J/mol = –189 kJ/mol

Answer

ΔG_f° (Mg²⁺(aq)) = –454.8 kJ/mol

ΔG_f° (NH₃(aq)) = –26.57 kJ/mol

ΔG_f° (H₂O(l)) = –237.2 kJ/mol

ΔG_f° (NH₄⁺(aq)) = –79.31 kJ/mol

ΔG_f° (Mg(OH)₂(s)) = –833.9 kJ/mol

ΔG° = [(–454.8) + 2(–26.57) +2(–237.2)] – [(–833.9) + 2(–79.31)] = 10.2 kJ

ΔG° = –RTlnK_eq

10200 = –(8.314)(298)lnK_eq

K_eq = e^–4.12 = 1.6×10^–2

To use published equilibrium constants, a series of reactions must be used:

Mg(OH)₂(s) → ← Mg²⁺(aq) + 2 OH^–(aq)K_sp = 1.8×10^–11

NH₄⁺(aq) + H₂O(l) → ← H₃O⁺(aq) + NH₃(aq) K_a = 5.6×10^–10

NH₄⁺(aq) + H₂O(l) → ← H₃O⁺(aq) + NH₃(aq) K_a = 5.6×10^–10

H₃O⁺(aq) + OH^–(aq) → ← 2 H₂O(l) K_c = 1/K_w = 1/1.0×10^–14 = 1.0×10¹⁴

H₃O⁺(aq) + OH^–(aq) → ← 2 H₂O(l)K_c = 1/K_w = 1/1.0×10^–14 = 1.0×10¹⁴

Summing the reactions gives the desired reaction with

K_c = K_sp×K_a×K_a×1/K_w×1/K_w = (1.8×10^–11)(5.6×10^–10)(5.6×10^–10) (1.0×10¹⁴)(1.0×10¹⁴) = 5.6×10^–2

The two equilibrium constants are the same order of magnitude, but differ by about a factor of 3.5. This is not too bad.

Answer

The reaction is:

Ag₂SO₄(s) → ← 2 Ag⁺(aq) + SO₄^2–(aq)

ΔG°_f(SO₄^2–) = –744.5 kJ/mol

ΔG°_f(Ag⁺) = 77.11 kJ/mol

ΔG°_f(Ag₂SO₄) = –618.5 kJ/mol

ΔG° = [2(77.11) + –744.5] – [–618.5] = 28.2 kJ

ΔG° = –RTln K_eq = –RTlnK_sp

28200 = –(8.314)(298)ln K_sp

ln K_sp = –11.4

K_sp = e^–11.4 = 1×10^–5

From the Table, K_sp = 1.4×10^–5; the two values are the same, within error.