Preview

Thermochemistry & Spontaneity

Q: Burning 1 mol of propane (C₃H₈) releases 2220 kJ. How much heat is released burning 44 g of propane? (M = 44 g/mol).

44 g / 44 g/mol = 1 mol → 2220 kJ released.

Q: Using bond energies: H₂ + Cl₂ → 2HCl. H-H = 432, Cl-Cl = 243, H-Cl = 431 kJ/mol. Calculate ΔH.

Bonds broken: H-H (432) + Cl-Cl (243) = 675. Bonds formed: 2×H-Cl = 2×431 = 862. ΔH = 675 - 862 = -187 kJ.

Q: Dissolving 8 g of NH₄NO₃ in water cools 100 g of water from 25°C to 19°C. Calculate ΔH in kJ/mol. (c_water = 4.18 J/g°C, M(NH₄NO₃) = 80 g/mol).

Q_water = 100×4.18×(-6) = -2508 J (water lost heat). ΔH_rxn = +2508 J for 0.1 mol → +25080 J/mol = +25.1 kJ/mol (endothermic).

Q: Use Hess's Law: C(s)+O₂(g)→CO₂(g), ΔH₁=-393. CO(g)+½O₂(g)→CO₂(g), ΔH₂=-283. Find ΔH for C(s)+½O₂(g)→CO(g).

Target = Eq1 - Eq2. ΔH = ΔH₁ - ΔH₂ = -393 - (-283) = -110 kJ/mol.

Gibbs free energy, entropy, and the ΔG = ΔH − TΔS verdict

Some reactions just go. Others never do, no matter how long you wait. The referee isn't heat alone — an endothermic reaction CAN run by itself, whenever nature gets something it wants even more than low energy: disorder. The score is one number, ΔG, and temperature is the wildcard that can flip the verdict mid-game. The referee's formula is ΔG = ΔH − TΔS: enthalpy minus temperature times entropy. One unit trap decides every calculation — ΔH is counted in kJ/mol while ΔS is counted in J/(mol·K), a factor of 1000 apart, so ΔS must be divided by 1000 before it can stand against ΔH. When ΔG < 0 the verdict is spontaneous; when ΔG > 0 the reaction refuses; and when ΔH and ΔS share a sign the flip happens at exactly T = ΔH/ΔS — when they have opposite signs, no finite positive-temperature flip exists. Today you learn to read the referee — and to find the exact degree where an 'impossible' reaction becomes inevitable.

What you'll be able to do

Judge spontaneity from the signs of ΔH and ΔS using ΔG = ΔH − TΔS
Compute the crossover temperature where a temperature-dependent sign combination (ΔH>0/ΔS>0 or ΔH<0/ΔS<0) flips its verdict
Distinguish thermodynamic spontaneity (ΔG < 0) from kinetic speed (activation energy)

Formulas

\Delta G = \Delta H - T\Delta S

Gibbs free energy decides spontaneity: ΔG < 0 spontaneous, ΔG > 0 non-spontaneous

\Delta S\,[\mathrm{kJ/(mol\cdot K)}] = \Delta S\,[\mathrm{J/(mol\cdot K)}] \,/\, 1000

Unit bridge: ΔS is given in J/(mol·K), ΔH in kJ/mol — divide ΔS by 1000 before combining

T_{\text{crossover}} = \frac{\Delta H}{\Delta S}

Temperature where the verdict flips; exists only when ΔH and ΔS share a sign

\Delta H \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})

Bond-energy estimate of reaction enthalpy: breaking bonds costs energy, forming them releases it

q = m \, c \, \Delta T

Calorimetry: heat exchanged by the surroundings; ΔH_rxn = −q divided by moles of reactant

\Delta H_{\text{rxn}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})

Hess's law: reaction enthalpy is path-independent — add or subtract known step enthalpies

Make a prediction

A reaction has ΔH = +100 kJ/mol (endothermic) and ΔS = +150 J/(mol·K). At 298 K it refuses to run. You start heating it. What happens?

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Your prediction

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Answer: Above about 667 K it runs by itself — the TΔS term finally outweighs ΔH

ΔG = ΔH − TΔS. At 298 K: ΔG = 100 − 298×0.150 ≈ +55 kJ/mol — non-spontaneous, correct. But ΔS is positive, so every degree of heating subtracts a little more. The crossover is T = ΔH/ΔS = 100/0.150 ≈ 667 K: past it, ΔG turns negative and the 'impossible' reaction runs on its own. Endothermic does NOT mean non-spontaneous — spontaneity is a fight between enthalpy and entropy, and temperature picks the winner.

Quiz (0/4)

Burning 1 mol of propane (C₃H₈) releases 2220 kJ. How much heat is released burning 44 g of propane? (M = 44 g/mol)

Using bond energies: H₂ + Cl₂ → 2HCl. H-H = 432, Cl-Cl = 243, H-Cl = 431 kJ/mol. Calculate ΔH.

Dissolving 8 g of NH₄NO₃ in water cools 100 g of water from 25°C to 19°C. Calculate ΔH in kJ/mol. (c_water = 4.18 J/g°C, M(NH₄NO₃) = 80 g/mol)

Use Hess's Law: C(s)+O₂(g)→CO₂(g), ΔH₁=-393. CO(g)+½O₂(g)→CO₂(g), ΔH₂=-283. Find ΔH for C(s)+½O₂(g)→CO(g).

You can now

Judge spontaneity from the signs of ΔH and ΔS using ΔG = ΔH − TΔS
Compute the crossover temperature where a temperature-dependent sign combination (ΔH>0/ΔS>0 or ΔH<0/ΔS<0) flips its verdict
Distinguish thermodynamic spontaneity (ΔG < 0) from kinetic speed (activation energy)