Enthalpy Notes

Enthalpy

  • Enthalpy is the heat (q) value divided by moles or grams of a reactant.

  • SHSH = enthalpy.

  • Enthalpy of reaction: the heat per mole or per gram of a reactant.

  • Enthalpy of combustion: the heat per mole or gram of a burned substance.

  • Standard combustion equation for a hydrocarbon:

    Hydrocarbon + O<em>2O<em>2 → CO</em>2CO</em>2 + H2OH_2O

Problem 1

  • Combustion of 0.92 grams of octane (C<em>8H</em>18C<em>8H</em>{18}) in a calorimeter holding 74 mL of H2OH_2O.

  • Initial temperature of water: 22.4°C.

  • Final temperature of water: 46.1°C.

a. Energy Released (in Joules and kJ)

  • Formula used: q=mCΔTq = mCΔT

  • Assuming 1g/mL density for water, mass of water = 74 g.

  • q=(74 g)×(4.184 J/g°C)×(46.1°C−22.4°C)q = (74 \text{ g}) \times (4.184 \text{ J/g°C}) \times (46.1°C - 22.4°C)

  • q=(74 g)×(4.184 J/g°C)×(23.7°C)q = (74 \text{ g}) \times (4.184 \text{ J/g°C}) \times (23.7°C)

  • q=7.3×103 Jq = 7.3 \times 10^3 \text{ J}

  • In kJ: 7.3×103 J÷1000 J/kJ=7.3 kJ7.3 \times 10^3 \text{ J} ÷ 1000 \text{ J/kJ} = 7.3 \text{ kJ}

b. Heat of Combustion (kJ/g) for Octane

  • 7.3 kJ0.92 g=7.9 kJ/g\frac{7.3 \text{ kJ}}{0.92 \text{ g}} = 7.9 \text{ kJ/g}

c. Molar Heat of Combustion (kJ/mol) for Octane

  • Molar mass of octane (C<em>8H</em>18C<em>8H</em>{18}) = 114.24 g/mol.

  • 7.9 kJg×114.24 g1 mol=9.0×102 kJ/mol\frac{7.9 \text{ kJ}}{\text{g}} \times \frac{114.24 \text{ g}}{1 \text{ mol}} = 9.0 \times 10^2 \text{ kJ/mol}

d. Balanced Chemical Equation

  • 2C<em>8H</em>18+25O<em>2→16CO</em>2+18H2O+1800 kJ2 C<em>8H</em>{18} + 25 O<em>2 → 16 CO</em>2 + 18 H_2O + 1800 \text{ kJ}

  • Note: The energy value corresponds to 2 moles of octane.

Problem 2

  • Decomposition of 1.45 grams of calcium carbonate (CaCO<em>3CaCO<em>3) in a calorimeter holding 100.2 mL of H</em>2OH</em>2O.

  • Initial temperature of water: 20.6°C.

  • Final temperature of water: 12.4°C.

  • Reaction: CaCO<em>3→CO</em>2+CaOCaCO<em>3 → CO</em>2 + CaO

a. Energy Absorbed (in Joules and kJ)

  • q=mCΔTq = mCΔT

  • q=(100.2 g)×(4.184 J/g°C)×(12.4°C−20.6°C)q = (100.2 \text{ g}) \times (4.184 \text{ J/g°C}) \times (12.4°C - 20.6°C)

  • q=(100.2 g)×(4.184 J/g°C)×(−8.2°C)q = (100.2 \text{ g}) \times (4.184 \text{ J/g°C}) \times (-8.2°C)

  • q=−3.4×103 Jq = -3.4 \times 10^3 \text{ J}

  • In kJ: −3.4×103 J÷1000 J/kJ=−3.4 kJ-3.4 \times 10^3 \text{ J} ÷ 1000 \text{ J/kJ} = -3.4 \text{ kJ}

b. Heat of Combustion (kJ/g) for Calcium Carbonate

  • −3.4 kJ1.45 g=−2.3 kJ/g\frac{-3.4 \text{ kJ}}{1.45 \text{ g}} = -2.3 \text{ kJ/g}

c. Molar Heat of Combustion (kJ/mol) for Calcium Carbonate

  • Molar mass of CaCO3CaCO_3 = 100.09 g/mol.

  • −2.3 kJg×100.09 g1 mol=−230 kJ/mol\frac{-2.3 \text{ kJ}}{\text{g}} \times \frac{100.09 \text{ g}}{1 \text{ mol}} = -230 \text{ kJ/mol}