Calorimetry and Heat Capacity

Calorimetry and Heat Capacity

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Termokimia tracks the heat (enthalpy, H) exchanged during a chemical reaction or physical change. The must-know formula for calorimetry is q = m · c · ΔT, where m is mass (g), c is specific heat capacity (J g⁻¹ K⁻¹), and ΔT is the temperature change (K or °C). For a coffee-cup calorimeter, q_reaction = −q_solution. Two sign conventions to memorise: exothermic reactions release heat (ΔH < 0, system loses energy); endothermic reactions absorb heat (ΔH > 0). Hess’s Law lets you calculate ΔH by adding intermediate reactions — flip a reaction reverses the sign, multiply coefficients multiplies ΔH. For UTBK, expect 1–2 questions: one calorimetry numerical plus one Hess’s Law multi-step reaction. Watch the sign on the energy-of-bonds formula: ΔH = Σ bonds broken (reactants) − Σ bonds formed (products). Always include physical states (s, l, g, aq) in thermochemical equations.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Calorimetry and Heat Capacity

A calorimeter isolates a reaction so that q_system + q_surroundings = 0. In a simple (open) calorimeter, the surroundings are the solution, so q_reaction = −(m · c · ΔT)_solution. Use c_water = 4.18 J g⁻¹ K⁻¹. When the calorimeter itself absorbs heat, use the heat capacity of the calorimeter (C_cal, J K⁻¹) and write q_reaction = −(m·c·ΔT + C_cal·ΔT). A bomb calorimeter operates at constant volume, so the measured heat equals the change in internal energy (q_v = ΔU), not ΔH — important for combustion of gases where Δn_gas ≠ 0.

Hess’s Law

Hess’s Law states that ΔH is a state function: it depends only on the initial and final states. To use it:

1. Write the target reaction.
2. Manipulate given reactions — reverse a reaction flips the sign of its ΔH; multiply coefficients by n multiplies ΔH by n.
3. Add them so intermediates cancel.

Standard Enthalpies

• ΔH°f = enthalpy of formation of 1 mol of compound from its elements in standard states (298 K, 1 atm); ΔH°f of any element in its standard state is 0.
• ΔH°c = enthalpy of combustion of 1 mol of a substance with excess O₂.
• Calculation: ΔH°_rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants), with each ΔH°f multiplied by its stoichiometric coefficient.

Bond Energies

ΔH_rxn = Σ BE(reactants) − Σ BE(products), where BE is the average bond energy (kJ mol⁻¹). Bond breaking is endothermic (+), bond forming is exothermic (−); the net gives the sign of ΔH.

Quantity	Formula	Sign convention
Heat absorbed by solution	q = m·c·ΔT	+ if T rises
Reaction enthalpy (open calorimeter)	q_rxn = −q_solution	− for exothermic
From formation enthalpies	Σ ΔH°f(prod) − Σ ΔH°f(react)	follows sign directly
From bond energies	Σ BE(react) − Σ BE(prod)	− for exothermic

Typical UTBK Question Patterns

• Numerical: 50 mL of 1.0 M HCl mixed with 50 mL of 1.0 M NaOH; ΔT = 6.8 °C. Compute q, then ΔH per mole of water formed.
• Conceptual: which reaction has the largest −ΔH°f? (Most stable product from its elements.)
• Multi-step Hess: given three reactions with ΔH values, calculate ΔH of a target.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Subtleties

ΔH vs ΔU. For reactions in solution (ΔV ≈ 0) or in an open calorimeter, ΔH ≈ q_p. Only a bomb calorimeter gives ΔU, and the two are related by ΔH = ΔU + Δn_gas·RT. UTBK rarely tests this conversion, but recognising the distinction prevents losing a “why is q_v not equal to ΔH?” conceptual question.

Why physical state matters. H₂O(l) and H₂O(g) have ΔH°f of −285.8 and −241.8 kJ mol⁻¹ respectively. Using the gaseous value in a problem where water condenses would give a ΔH answer off by 44 kJ mol⁻¹ — a classic trap.

Lavosier–Laplace Law. The enthalpy change of a forward reaction equals minus the enthalpy change of the reverse reaction at the same temperature; this is a special case of Hess’s Law and is sometimes quoted separately in UTBK passages.

Common Mistakes to Avoid

1. Forgetting to flip the sign of ΔH when reversing a reaction in Hess cycles.
2. Mixing up Σ ΔH°f(prod) − Σ ΔH°f(react) with Σ ΔH°c(prod) − Σ ΔH°c(react). The combustion version uses reactants minus products because combustion is the reverse of forming the compound from CO₂ and H₂O.
3. Treating the calorimeter as having zero heat capacity when a value for C_cal is provided.
4. Dividing q by mass of solute instead of moles of reaction when asked for “ΔH per mole of reaction”.

Connections to Other Topics

Termokimia links to chemical equilibrium (van ‘t Hoff equation: d(ln K)/dT = ΔH°/RT²), electrochemistry (ΔG° = −nFE°, and ΔG° = ΔH° − TΔS°), and kinetics (activation energy E_a is the kinetic analogue of the enthalpy barrier). UTBK sometimes embeds thermochemistry inside a longer passage that also tests entropy or Gibbs free energy — read for sign conventions consistently.

Worked Micro-Example

Combustion of methane: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), ΔH° = −890 kJ mol⁻¹. Using ΔH°f values: ΔH° = [ΔH°f(CO₂) + 2·ΔH°f(H₂O,l)] − [ΔH°f(CH₄) + 2·ΔH°f(O₂)] = [−393.5 + 2(−285.8)] − [−74.8 + 0] = −965.1 + 74.8 = −890.3 kJ mol⁻¹ ✓.

Practice Prompts

1. Given ΔH°f of CS₂(l) = +89.7 kJ mol⁻¹, CO₂(g) = −393.5 kJ mol⁻¹, SO₂(g) = −296.8 kJ mol⁻¹, calculate ΔH° for CS₂(l) + 3 O₂(g) → CO₂(g) + 2 SO₂(g).
2. A 1.20 g sample of benzoic acid (M = 122 g mol⁻¹) is burned in a bomb calorimeter (C_cal = 8.92 kJ K⁻¹) and T rises by 2.45 K. Find the internal energy change ΔU per mole of benzoic acid.

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