A brief note on Enthalpy and Hess's Law . Read it before a test. Try each question at the end before you open its answer.
Enthalpy H = U + p V H = U + pV H = U + p V , a state function. At constant pressure, the heat taken in is Δ H \Delta H Δ H . Δ H = Δ U + Δ n g R T \Delta H = \Delta U + \Delta n_g R T Δ H = Δ U + Δ n g R T . Count gases only. R = 8.314 J K − 1 mol − 1 R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} R = 8.314 J K − 1 mol − 1 . Exothermic: Δ H < 0 \Delta H < 0 Δ H < 0 . Endothermic: Δ H > 0 \Delta H > 0 Δ H > 0 . q = C Δ T = m c Δ T q = C\Delta T = m c \Delta T q = C Δ T = m c Δ T . For an ideal gas, C p − C v = R C_p - C_v = R C p − C v = R . Bomb calorimeter: constant volume, gives Δ U \Delta U Δ U . Coffee-cup calorimeter: constant pressure, gives Δ H \Delta H Δ H . Standard enthalpies Standard state: the pure substance at 1 bar, at the stated temperature. Reverse an equation: change the sign of Δ H \Delta H Δ H . Multiply it: multiply Δ H \Delta H Δ H . Δ s u b H = Δ f u s H + Δ v a p H \Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H Δ s u b H = Δ f u s H + Δ v a p H . Δ f H ⊖ \Delta_f H^\ominus Δ f H ⊖ : one mole of a compound from its elements. Elements in their standard states have zero. Δ r H ⊖ = ∑ a i Δ f H ⊖ ( products ) − ∑ b i Δ f H ⊖ ( reactants ) \Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) - \sum b_i \Delta_f H^\ominus(\text{reactants}) Δ r H ⊖ = ∑ a i Δ f H ⊖ ( products ) − ∑ b i Δ f H ⊖ ( reactants ) . Combustion is always negative, atomisation always positive. Δ s o l H = Δ l a t t i c e H + Δ h y d H \Delta_{sol}H = \Delta_{lattice}H + \Delta_{hyd}H Δ so l H = Δ l a tt i ce H + Δ h y d H . Hess's law and bond enthalpies The enthalpy change is the same by any route. Add equations like algebra. For gases, Δ r H ≈ ∑ ( bonds broken ) − ∑ ( bonds formed ) \Delta_r H \approx \sum(\text{bonds broken}) - \sum(\text{bonds formed}) Δ r H ≈ ∑ ( bonds broken ) − ∑ ( bonds formed ) . Mean bond enthalpies are averages, so the answer is an estimate. A worked example Find Δ f H ⊖ \Delta_f H^\ominus Δ f H ⊖ of CO from two combustions: carbon gives − 393.5 kJ mol − 1 -393.5\ \text{kJ mol}^{-1} − 393.5 kJ mol − 1 , and CO gives − 283.0 kJ mol − 1 -283.0\ \text{kJ mol}^{-1} − 283.0 kJ mol − 1 .
Target. C + 1 2 O 2 → CO \text{C} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO} C + 2 1 O 2 → CO . Steps. Burn the carbon, then reverse the burning of CO: − 393.5 -393.5 − 393.5 and + 283.0 +283.0 + 283.0 . Add. Δ f H ⊖ = − 393.5 + 283.0 = − 110.5 kJ mol − 1 \Delta_f H^\ominus = -393.5 + 283.0 = -110.5\ \text{kJ mol}^{-1} Δ f H ⊖ = − 393.5 + 283.0 = − 110.5 kJ mol − 1 . Practice with new numbers
A program to try The program writes each equation as a dictionary and adds them with their enthalpy changes.
Run it. Which substances cancel in the sum?
1. Change the factors to (1, *burn_C) and (1, *burn_CO). What does the sum describe now?
Mistakes to avoid Liquids in Δ n g \Delta n_g Δ n g . Count gases only. Joules with kilojoules. R T RT R T comes out in joules. Reactants minus products. Formation enthalpies go products minus reactants. Formed minus broken. Bond enthalpies go broken minus formed. A reversed equation with the old sign. Change the sign. Check yourself What is Δ f H ⊖ \Delta_f H^\ominus Δ f H ⊖ of graphite? Stuck? Show a hint Show the answerWhich calorimeter measures Δ H \Delta H Δ H directly? Stuck? Show a hint Show the answerFor CaCO 3 (s) → CaO(s) + CO 2 (g) \text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)} CaCO 3 (s) → CaO(s) + CO 2 (g) , is Δ H \Delta H Δ H larger or smaller than Δ U \Delta U Δ U ? Stuck? Show a hint Show the answerWhy is breaking a bond always endothermic? Stuck? Show a hint Show the answerIf X → Y \text{X} \rightarrow \text{Y} X → Y has Δ H = + 30 kJ \Delta H = +30\ \text{kJ} Δ H = + 30 kJ and Y → Z \text{Y} \rightarrow \text{Z} Y → Z has Δ H = − 50 kJ \Delta H = -50\ \text{kJ} Δ H = − 50 kJ , what is Δ H \Delta H Δ H for X → Z \text{X} \rightarrow \text{Z} X → Z ? Stuck? Show a hint Show the answer