2ENTHALPY (H)Accounts for heat flow in chemical reactions happening at constant pressure when work is performed due to P-V change.The enthalpy of a substance is its internal energy plus a term that takes into account the pressure & volume of the substance.H = U + PV
3ΔH = ΔU – W qp= ΔU - W In a process carried out at constant pressure: We cannot calculate the actual value of H. Instead, we can calculate the change in enthalpy.ΔH = ΔU + PΔV (since W = - PΔV)(Equation 1)ΔH = ΔU – WIn a process carried out at constant pressure:ΔU = qp + W or (Equation 2)(Equation 1) = (Equation 2)ΔH = qp (at constant pressure)qp= ΔU - W
4ONLY at constant Pressure: ΔH = QpFor this reason, the term “ heat of reaction” and “change in enthalpy” are used interchangeably for rxns studied at constant pressure.
5In summary ΔU measures heat lost/gained. ΔH measures heat lost/gained. At constant volumeAt constant pressureΔU measures heat lost/gained.ΔH measures heat lost/gained.The difference between ΔU & ΔH is very small. ΔH is generally satisfactory to use.
6ExerciseUnder standard conditions, 1 mol CO is burnt in a sealed flask w/ constant volume and kJ energy is released. The same amount of CO is burnt under the same conditions in a flask w/ frictionless movable piston, 283kJ of energy is released. What are the values of ΔH, ΔU, and w for both conditions?
8ΔH = Hproducts –Hreactants For a chemical rxn:ΔH = Hproducts –Hreactants
9When Hproducts > Hreactants ΔH is (+).Heat will be absorbed by the system.Reactants are more stable than the products.reactants have stronger bonds than the products.
10When Hproducts < Hreactants ΔH is (-).Heat will be released by the system.Reactants are less stable than the products.Products have stronger bonds than the reactants.
11ENTHALPIES OF RXNSEnthalpy is an extensive property.ΔH is directly proportional to the amounts of reactants consumed in chemical rxns.CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH = -890 kJ2CH4(g) + 4O2(g) --->2 CO2(g) + 4H2O(l) ΔH = kJ (890x2)
12ExerciseWhen 1 mole of methane is burned at constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which a 5.8 g sample of methane is burned at constant pressure.
13Solution:CH4: 16 g/molWhen 16g of CH4 burns 890 kJ energy is released5.8 g of CH ?________________________________________? = kJ of energy
14CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH = -890 kJ 2) ΔH for a rxn is equal in magnitude but opposite in sign, to ΔH for the reverse rxn.CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH = -890 kJCO2(g) + 2H2O(l) ---> CH4(g) + 2O2(g) ΔH = kJ
15Which one is greater in amount? 3) ΔH depends on the state (gas, liquid, solid, crystalline structure)of the reactants and products.Therefore, we should write the states of the substances in the rxn equation.I. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) ΔH1 = -890 kJII. CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(g) ΔH2 = ? kJWhich one is greater in amount?
174) ΔH depends on temperature & pressure of the rxn medium 4) ΔH depends on temperature & pressure of the rxn medium. - We will generally assume that the reactants & products are both at the same temperature, 25 ° C, unless otherwise stated.
185) ΔH is a state function. Therefore, it doesn’t depend on the rxn pathway.