ΔHsol = +711 + (−322) + (−381) = +8 kJ mol−1[1]
Mg2+. Mg2+ has a higher charge and a smaller ionic radius than Na+, so it has a higher charge density.[1]
Hence, Mg2+ forms stronger ion–dipole interactions with water molecules.[1]
ΔG = −92 − 298(−0.199) = −92 + 59.3 = −32.7 kJ mol−1[1]
Since ΔG is negative, the reaction is spontaneous at 298 K.[1]
At 298 K, ΔG = 130 − 298(0.335) = +30.2 kJ mol−1, not spontaneous[1]
T > 1300.335 = 388 K[1]
NoteLikewise MgCO3 decomposes above 576 K but BaCO3 only above 1566 K: ΔH(MgCO3) is less positive, ΔS similar.
AIt is spontaneous only at low temperature.[1]
ΔS < 0 due to a decrease in the number of moles of gas (4 mol to 2 mol). Hence −TΔS > 0.
Since ΔH < 0 and ΔG = ΔH − TΔS, ΔG < 0 when |−TΔS| < |ΔH|. Hence the reaction is spontaneous only at low temperature.
D+436 J K−1[1]
Amount of H2O = 7218.0 = 4 mol
At the boiling point, ΔG = 0, and ΔG = ΔH − TΔS
0 = (40.7 × 103 × 4) − (100 + 273)ΔS
ΔS = +436 J K−1
Energy cycle: MgCl2(s) + aq → Mg2+(aq) + 2Cl−(aq) (ΔHsol), with both sides formed from Mg(s) + Cl2(g): −641 for MgCl2(s), and ΔHf = −467 + 2(−167) for the ions.[1]
By Hess' Law, ΔHsol = 641 + (−467) + 2(−167) = −160 kJ mol−1[1]
Since ΔHsol is highly exothermic, MgCl2 is very soluble in water.[1]
NoteThe magnitude of ΔS is often much smaller than that of ΔH. For most reactions, ΔS is less than 100 J K−1 mol−1. Hence when ΔH is very negative, ΔG is likely to be negative and the reaction is spontaneous.
ΔHsoln[LiCl(s)] = −lattice energy + ΔHhyd[Li+(g)] + ΔHhyd[Cl−(g)]
= +843 + (−883) = −40 kJ mol−1[1]
Heat absorbed = mcΔT = 50.0 × 4.20 × 3.3 = 693 J[1]
ΔHsoln[KCl(s)] = +693 ÷ 3.0074.6 = +17.2 kJ mol−1 (3 s.f.)[1]
Since |LE| is inversely proportional to the interionic distance, the lattice energy of LiCl is slightly more exothermic than that of KCl.[1]
However, since hydration energy is proportional to charge density, Li+ will have a much more exothermic hydration enthalpy than K+. Therefore, the enthalpy change of solution of LiCl is more exothermic than that of potassium chloride.[1]
ΔGdecomposition = ΔHdecomposition − TΔS
= −1423 − [298 × (+208 ÷ 1000)] = −1485 kJ mol−1[1]
Since ΔG is negative, the reaction is spontaneous at 298 K.[1]
The spontaneity is not affected by changes in temperature because ΔHdecomposition is negative, T is always positive and ΔS is positive (or −TΔS is always negative), i.e. ΔG = ΔH − TΔS = negative − [(positive) × (positive)] = negative.
Yes, the decomposition of nitroglycerin is spontaneous at all temperatures.[1]
ΔG⦵ = ΔH⦵ − TΔS⦵. ΔG⦵ (+172.6 kJ mol−1) is positive at 298 K, hence reaction I is not spontaneous at room temperature.[1]
The high temperature in the car engine causes TΔS⦵ to become more positive (or −TΔS⦵ to become more negative). This leads to a more negative ΔG⦵, and hence reaction I becomes spontaneous.[1]
The amount of gases decreases from 3 mol to 2 mol, hence there are fewer ways to arrange the gaseous particles and their energy in the system.[1]
ΔS⦵ is negative.[1]
ΔHsoln = −(lattice energy) + ΔHhyd
For NaOH: −44 = 896 + (−390) + ΔHhyd(OH−), so ΔHhyd(OH−) = −550 kJ mol−1[1]
For Mg(OH)2: x = 2995 − 1890 + 2(−550) = +5.00 kJ mol−1[1]