Enthalpy (H)

Enthalpy: Heat content at constant pressure

$\Delta H = q_p$

At constant pressure (most reactions): Heat change = enthalpy change

Exothermic reaction:

• ΔH < 0 (negative)
• Releases heat to surroundings
• Products lower energy than reactants
• Feels warm

Endothermic reaction:

• ΔH > 0 (positive)
• Absorbs heat from surroundings
• Products higher energy than reactants
• Feels cold

Calorimetry

Calorimetry: Measuring heat changes

Heat capacity equation:

$q = mc\Delta T$

Where:

• q = heat (J)
• m = mass (g)
• c = specific heat capacity (J/g·°C)
• ΔT = T_{final} - T_{initial}

Common specific heats:

• Water: 4.18 J/(g·°C)
• Metals: 0.1-1 J/(g·°C)

Coffee Cup Calorimeter

For solution reactions:

• Constant pressure (open to atmosphere)
• Measures q_p = ΔH
• Simple styrofoam cup

Assumption: All heat goes to solution (usually aqueous)

$q_{\text{solution}} = -q_{\text{reaction}}$

Bomb Calorimeter

For combustion:

• Constant volume
• Measures ΔE (not ΔH)
• More precise, sealed container

$q = C_{\text{cal}}\Delta T$

C_{cal} = calorimeter constant (J/°C)

Hess's Law

Hess's Law: ΔH is path independent - only depends on initial and final states

Use: Calculate ΔH for reaction from known ΔH values

Rules:

1. Reverse reaction → change sign of ΔH
2. Multiply reaction by n → multiply ΔH by n
3. Add reactions → add ΔH values

Example approach:

• Manipulate given equations to match target
• Add them to get target equation
• Add corresponding ΔH values

Standard Enthalpy of Formation (ΔH°_f)

ΔH°_f: Enthalpy change to form 1 mole from elements in standard states

Standard conditions:

• 25°C (298 K)
• 1 atm pressure
• 1 M concentration

Key rules:

• ΔH°_f of element in standard state = 0
• ΔH°_f values tabulated for compounds

Examples:

• C(graphite): ΔH°_f = 0
• O₂(g): ΔH°_f = 0
• H₂O(l): ΔH°_f = -285.8 kJ/mol
• CO₂(g): ΔH°_f = -393.5 kJ/mol

Calculating ΔH°_{rxn}

From ΔH°_f values:

$\Delta H°_{\text{rxn}} = \sum n\Delta H°_f(\text{products}) - \sum n\Delta H°_f(\text{reactants})$

Steps:

1. Sum (coefficient × ΔH°_f) for products
2. Sum (coefficient × ΔH°_f) for reactants
3. Subtract: products - reactants

Shortcut: Products minus reactants (with coefficients)

Bond Enthalpies

Bond enthalpy: Energy to break 1 mole of bonds (always positive)

Estimating ΔH_{rxn}:

$\Delta H_{\text{rxn}} = \sum \text{bonds broken} - \sum \text{bonds formed}$

Energy required to break bonds (positive) Energy released forming bonds (negative)

Note: Bond enthalpies are averages, less accurate than ΔH°_f

ΔT = 27.5 - 21.0 = 6.5°C

$q_{\text{solution}} = mc\Delta T$ $q_{\text{solution}} = (100.0)(4.18)(6.5)$ $q_{\text{solution}} = 2717 \text{ J} = 2.72 \text{ kJ}$

Step 2: Calculate heat of reaction

$q_{\text{reaction}} = -q_{\text{solution}} = -2.72 \text{ kJ}$

(Negative because reaction releases heat - exothermic)

Step 3: Calculate moles reacted

Moles HCl = (1.0 M)(0.050 L) = 0.050 mol Moles NaOH = (1.0 M)(0.050 L) = 0.050 mol

Limiting reactant: Both 0.050 mol (1:1 ratio) → 0.050 mol reacts

Step 4: Calculate ΔH per mole

$\Delta H = \frac{q_{\text{reaction}}}{\text{moles}} = \frac{-2.72 \text{ kJ}}{0.050 \text{ mol}}$

$\Delta H = -54.4 \text{ kJ/mol}$

Answer: ΔH = -54 kJ/mol (exothermic)

Note: Literature value is -57.1 kJ/mol - our answer is close!