Presentation on theme: "ENTHALPY (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."— Presentation transcript:
ENTHALPY (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
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 – W In a process carried out at constant pressure: ΔU = q p + W or (Equation 2) (Equation 1) = (Equation 2) ΔH = q p (at constant pressure) q p = ΔU - W
ONLY at constant Pressure: ΔH = Q p For this reason, the term heat of reaction and change in enthalpy are used interchangeably for rxns studied at constant pressure.
In summary At constant volume At constant pressure ΔU measures heat lost/g a ined. ΔH measures heat lost/gained. The difference between ΔU & ΔH is very small. ΔH is generally satisfactory to use.
Exercise Under 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?
For a chemical rxn: ΔH = H products –H reactants
When H products > H reactants -ΔH is (+). -Heat will be absorbed by the system. -Reactants are more stable than the products. -reactants have stronger bonds than the product s.
When H products < H reactants -ΔH is (-). -Heat will be released by the system. -Reactants are less stable than the products. -Products have stronger bonds than the reactants.
ENTHALPIES OF RXNS 1) Enthalpy is an extensive property. ΔH is directly proportional to the amounts of reactants consumed in chemical rxns. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l) ΔH = -890 kJ 2 CH4(g) + 4 O2(g) ---> 2 CO2(g) + 4 H2O(l) ΔH = kJ (890x2)
Exercise When 1 mole of methane is burned a t constant pressure, 890 kJ of energy is released as heat. Calculate ΔH for a process in which a 5.8 g sample of meth a ne is burned at constant pressure.
Solution: CH4: 16 g/mol When 16g of CH4 burns 890 kJ energy is released 5.8 g of CH4 ? ________________________________________ ? = kJ of energy
2) ΔH for a rxn is equal in magnitude but opposite in sign, to ΔH for the reverse rxn. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l) ΔH = -890 kJ CO2(g) + 2 H2O(l) ---> CH4(g) + 2 O2(g) ΔH = kJ
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) + 2 O2(g) ---> CO2(g) + 2 H2O(l) ΔH1 = -890 kJ - II. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(g) ΔH2 = ? kJ - Which one is greater in amount?