The document discusses enthalpy change (ΔH) which is the amount of heat released or absorbed during a chemical reaction under constant pressure. ΔH is calculated as the enthalpy of products minus reactants. Standard enthalpy change (ΔH°) is measured under standard temperature and pressure conditions. Some key enthalpy terms discussed include standard enthalpy of formation (ΔHf°), enthalpy of combustion (ΔHc°), enthalpy of solution (ΔHsoln), and enthalpy of neutralization. Practice problems are provided to calculate enthalpy changes using experimental heat and temperature change data.

1. Enthalpy Change(ΔH)

2. Enthalpy ChangeKey ConceptsThe heat content of a chemical reaction is called the enthalpy (symbol: H)

3. The enthalpy change (ΔH) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.

4. ΔH = H(products) - H(reactants)

5. The units are usually given as kJ mol-1 (kJ/mol) or sometimes as kcal mol-1 (kcal/mol)1 calorie (1 cal) = 4.184 joules (4.184 J)

6. Standard enthalpy change(ΔHѲ);Energy changes for any reaction which are measured under standard conditions of temperature and pressure25oC (298K) - 101.3 kPa (1 atmosphere) - 1 mol dm-3Manipulating the Enthalpy Change TermN2(g) + 3H2(g) -----> 2NH3(g) ΔH = - 92.4 kJ/mol 92.4 kJ of energy is released for every 1 mole of N2(g)

7. 92.4 kJ of energy is released for every 3 moles of H2(g)

8. 92.4 kJ of energy is released for every 2 moles of NH3(g) produced. How much energy is released if only 1 mole of ammonia (NH3) gas is produced?

9. a) Standard Enthalpy of formation (∆Hf0) It is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions of temperature and pressureEg: C(s) + O2 (g) -> CO2 (g) ∆Hf0 =-393.5 KJ practice question N2(g) + 3H2(g) -----> 2NH3(g) ΔH0 = - 92.4 kJ/molFind enthalpy of formation ∆Hf0of NH3 from the equation (remember enthalpy of formation ∆Hf0 is always for 1 mole)∆Hf0 of a free element in its standard state is taken as zero.Eg: ∆Hf0 = 0 for H2(g) , N2 (g), O2(g) , F2(g), Cl2(g), Br2(l), I2 (s) , C(s,graphite)How to find ∆H0 from std. enthalpy of formation ∆H0 = Σ ∆Hf0 (products) − Σ ∆Hf0 (reactants)

10. Calculate enthalpy change for the reactionCH4 +2O2(g) ->CO2(g) +2H2O(g) ∆Hf0of CH4 , CO2(g) , H2O(g, are -74.8, -393.5 and -285.8 kjmol-1 respectively