Thermochemistry

Energy, Work, and the First Law

Energy is the capacity to do work or supply heat. Kinetic energy is energy of motion (including thermal energy, the random motion of particles); potential energy is stored energy of position or chemical bonding. The system is the part of the universe under study, the surroundings is everything else, and the first law states that energy is conserved: any energy the system loses the surroundings gains.

• •ΔE = q + w, where q is heat transferred and w is work done.
• •Sign convention (system-centered): q > 0 means heat flows INTO the system (endothermic); q < 0 means heat leaves (exothermic).
• •w = −PΔV: gas expanding against pressure does work on the surroundings, so w is negative.
• •Energy units: 1 cal = 4.184 J exactly; 1 Cal (food calorie) = 1000 cal = 4.184 kJ.
• •Internal energy E is a state function — its change depends only on initial and final states, not the path. q and w individually are NOT state functions.

Heat Capacity and q = mCΔT

Heat capacity is the amount of heat needed to raise an object's temperature by 1 °C. Specific heat capacity C is that quantity per gram of substance; molar heat capacity is per mole. Water's unusually large specific heat (4.184 J/g·°C) comes from extensive hydrogen bonding and is why lakes moderate local climate and why water is the standard calorimetry fluid.

• •q = m C ΔT with ΔT = T_final − T_initial (positive when warming, negative when cooling).
• •A large C means the substance resists temperature change; metals have small C and heat up fast.
• •When two objects reach thermal equilibrium: q_lost = −q_gained, i.e. m₁C₁ΔT₁ = −m₂C₂ΔT₂.
• •Watch which mass goes with which C — the most common lost point in calorimetry.

Calorimetry: Constant Pressure vs Constant Volume

A coffee-cup calorimeter is open to the atmosphere, so measurements there occur at constant pressure and give ΔH directly. A bomb calorimeter is a sealed rigid vessel, so its volume is fixed, no PV work is possible, and the measured heat equals ΔE rather than ΔH.

• •Coffee cup: q_rxn = −(m_solution × C_solution × ΔT); divide by moles of limiting reactant for ΔH per mole.
• •Bomb: q_rxn = −C_cal × ΔT, where C_cal is the calorimeter constant in kJ/°C (heat capacity of the whole assembly).
• •A temperature RISE in the solution means the reaction released heat, so ΔH is negative.
• •Assume dilute aqueous solutions have the density and specific heat of water unless told otherwise.

Enthalpy and Thermochemical Equations

Enthalpy H = E + PV is defined so that ΔH equals the heat exchanged at constant pressure, which matches the conditions of nearly all lab chemistry. A thermochemical equation pairs a balanced equation with its ΔH, and the ΔH value is tied to the exact stoichiometric coefficients written.

• •ΔH < 0: exothermic, products lower in enthalpy, container feels warm.
• •ΔH > 0: endothermic, products higher in enthalpy, container feels cold.
• •Doubling all coefficients doubles ΔH; reversing the reaction flips the sign of ΔH.
• •ΔH is extensive — it scales with amount, so it is a legitimate conversion factor: (−890 kJ / 1 mol CH₄).
• •ΔH ≈ ΔE for reactions with no gas-mole change; they differ by Δn_gas RT when gases are produced or consumed.

Hess's Law

Because enthalpy is a state function, the enthalpy change of an overall reaction equals the sum of enthalpy changes of any set of steps that add up to it. This lets you compute ΔH for reactions that are impossible or impractical to measure directly, such as the formation of CO from graphite without also forming CO₂.

• •Reverse a given equation → change the sign of its ΔH.
• •Multiply an equation by a factor → multiply its ΔH by the same factor.
• •Add the manipulated equations; species appearing on both sides cancel.
• •Strategy: start with the compound that appears in only one given equation and place it correctly first.

Standard Enthalpies of Formation

The standard enthalpy of formation ΔH°f is the enthalpy change when exactly one mole of a compound forms from its constituent elements in their standard states at 1 bar (and usually 25 °C). By definition, an element in its standard state has ΔH°f = 0, which anchors the whole table.

• •ΔH°rxn = Σ n ΔH°f(products) − Σ n ΔH°f(reactants) — products minus reactants, each weighted by its coefficient.
• •Standard states: O₂(g), N₂(g), Br₂(l), I₂(s), Hg(l), C(graphite, not diamond).
• •Highly negative ΔH°f indicates a thermodynamically stable compound relative to its elements.
• •Watch the physical state: ΔH°f of H₂O(l) and H₂O(g) differ by the enthalpy of vaporization.

Heating Curves and Phase Changes

Heating a substance from solid to gas produces a curve with sloped segments (temperature rising within one phase, governed by q = mCΔT) and flat plateaus (phase change at constant temperature, governed by q = nΔH_fus or nΔH_vap). During a plateau, all added energy goes into overcoming intermolecular forces rather than increasing kinetic energy.

• •Total heat for a multi-segment problem is the SUM of every segment; never skip the plateaus.
• •ΔH_vap is always larger than ΔH_fus — melting only loosens the lattice, vaporizing separates particles completely.
• •For water: ΔH_fus = 6.02 kJ/mol, ΔH_vap = 40.7 kJ/mol; C_ice ≈ 2.09, C_water = 4.184, C_steam ≈ 2.01 J/g·°C.
• •Condensation and freezing release the same magnitude of heat with the opposite sign.

Bond Energies as an Estimate

Breaking bonds always costs energy and forming bonds always releases it. Using tabulated average bond energies gives an approximate ΔH for gas-phase reactions, and it explains at the molecular level why combustion reactions are so exothermic: weak C–H and O=O bonds are replaced by very strong C=O and O–H bonds.

• •ΔH ≈ Σ BE(bonds broken) − Σ BE(bonds formed).
• •The estimate is approximate because tabulated values are averages across many molecules.
• •Only valid for gas-phase species; it ignores intermolecular contributions.

A 55.0 g piece of copper (C = 0.385 J/g·°C) at 99.8 °C is dropped into 125.0 g of water at 22.0 °C. What is the final temperature?

1. 1Heat lost by copper equals heat gained by water: −(m C ΔT)_Cu = (m C ΔT)_water.
2. 2−(55.0)(0.385)(T_f − 99.8) = (125.0)(4.184)(T_f − 22.0).
3. 3−21.18(T_f − 99.8) = 523.0(T_f − 22.0).
4. 4−21.18 T_f + 2113 = 523.0 T_f − 11506.
5. 513619 = 544.2 T_f, so T_f = 25.0 °C.

T_f = 25.0 °C — the water barely warms because its heat capacity and mass dominate.

50.0 mL of 1.00 M HCl is mixed with 50.0 mL of 1.00 M NaOH in a coffee-cup calorimeter. The temperature rises from 21.0 °C to 27.5 °C. Find ΔH per mole of water formed. (Assume d = 1.00 g/mL, C = 4.184 J/g·°C.)

1. 1Total solution mass = 100.0 mL × 1.00 g/mL = 100.0 g; ΔT = 6.5 °C.
2. 2q_solution = (100.0)(4.184)(6.5) = 2720 J = 2.72 kJ absorbed by the solution.
3. 3q_rxn = −2.72 kJ (the reaction released this heat).
4. 4Moles of water formed = 0.0500 L × 1.00 M = 0.0500 mol (both reactants exactly consumed).
5. 5ΔH = −2.72 kJ / 0.0500 mol = −54 kJ/mol.

ΔH ≈ −54 kJ/mol, close to the accepted −55.8 kJ/mol for strong acid–strong base neutralization.

Use Hess's law to find ΔH for C(s) + ½O₂(g) → CO(g), given C(s) + O₂ → CO₂, ΔH = −393.5 kJ, and CO(g) + ½O₂ → CO₂, ΔH = −283.0 kJ.

1. 1Keep equation 1 as written: C + O₂ → CO₂, ΔH = −393.5 kJ.
2. 2Reverse equation 2 so CO appears as a product: CO₂ → CO + ½O₂, ΔH = +283.0 kJ.
3. 3Add: C + O₂ + CO₂ → CO₂ + CO + ½O₂; cancel CO₂ from both sides and ½O₂ from the left.
4. 4Net: C(s) + ½O₂(g) → CO(g).
5. 5ΔH = −393.5 + 283.0 = −110.5 kJ.

ΔH = −110.5 kJ, which is also ΔH°f of CO(g).

Calculate ΔH°rxn for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), given ΔH°f: CH₄ = −74.6, CO₂ = −393.5, H₂O(l) = −285.8 kJ/mol.

1. 1ΔH°rxn = Σ n ΔH°f(products) − Σ n ΔH°f(reactants).
2. 2Products: (1)(−393.5) + (2)(−285.8) = −393.5 − 571.6 = −965.1 kJ.
3. 3Reactants: (1)(−74.6) + (2)(0) = −74.6 kJ (O₂ is an element in its standard state).
4. 4ΔH°rxn = −965.1 − (−74.6) = −890.5 kJ.

ΔH°rxn = −890.5 kJ per mole of methane burned — strongly exothermic.

How much heat is needed to convert 25.0 g of ice at −15.0 °C to steam at 115.0 °C? (C_ice = 2.09, C_water = 4.184, C_steam = 2.01 J/g·°C; ΔH_fus = 6.02, ΔH_vap = 40.7 kJ/mol.)

1. 1Moles of water = 25.0 g / 18.02 g/mol = 1.387 mol.
2. 2Warm ice −15.0 → 0 °C: q = (25.0)(2.09)(15.0) = 784 J = 0.784 kJ.
3. 3Melt at 0 °C: q = (1.387)(6.02) = 8.35 kJ.
4. 4Warm water 0 → 100 °C: q = (25.0)(4.184)(100.0) = 10460 J = 10.46 kJ.
5. 5Vaporize at 100 °C: q = (1.387)(40.7) = 56.5 kJ.
6. 6Warm steam 100 → 115 °C: q = (25.0)(2.01)(15.0) = 754 J = 0.754 kJ.
7. 7Total = 0.784 + 8.35 + 10.46 + 56.5 + 0.754 = 76.8 kJ.

≈ 76.8 kJ, with vaporization alone accounting for about 74% of the total.

A gas expands from 2.00 L to 6.00 L against a constant external pressure of 1.50 atm while absorbing 850 J of heat. Find ΔE. (1 L·atm = 101.3 J)

1. 1w = −PΔV = −(1.50 atm)(6.00 − 2.00 L) = −6.00 L·atm.
2. 2Convert: −6.00 L·atm × 101.3 J/L·atm = −608 J (system does work on surroundings).
3. 3q = +850 J (heat absorbed).
4. 4ΔE = q + w = 850 + (−608) = +242 J.

ΔE = +242 J — internal energy rises even though the gas did work, because it absorbed more heat than it spent.

Q1. Which sign combination describes an exothermic reaction in which the system also does work on the surroundings?

Q2. Which quantity is NOT a state function?

Q3. How much heat is required to warm 150.0 g of water from 20.0 °C to 65.0 °C?

Q4. In a coffee-cup calorimeter, the solution temperature rises. What is true of the reaction?

Q5. Which substance has ΔH°f = 0 by definition?

Q6. For 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −571.6 kJ. What is ΔH for H₂O(l) → H₂(g) + ½O₂(g)?

Q7. A gas is compressed from 8.0 L to 3.0 L at a constant external pressure of 2.0 atm. What is w?

Q8. During the flat plateau of a heating curve, added energy is used to:

Q9. Given ΔH°f values in kJ/mol: NH₃(g) = −45.9. What is ΔH° for N₂(g) + 3H₂(g) → 2NH₃(g)?

Q10. A bomb calorimeter with C_cal = 5.65 kJ/°C shows a temperature rise of 2.30 °C. How much heat did the reaction release?

Q11. Which statement about ΔH_vap and ΔH_fus for the same substance is correct?

Q12. Why is ΔH the more useful quantity than ΔE for typical laboratory chemistry?

Q13. Using bond energies, why is combustion of hydrocarbons strongly exothermic?

Q14. 125 g of a metal at 95.0 °C is added to 100.0 g of water at 22.0 °C; the final temperature is 28.0 °C. What is the metal's specific heat?

Q15. Which process has a positive ΔH?