Report

Hess’s Law (4-4) • Hess suggested that the sum of the enthalpies (ΔH) of the steps of a reaction will equal the enthalpy of the overall reaction. 1 Why Does It Work? • If you turn an equation around, you change the sign: • If H2(g) + 1/2 O2(g) H2O(g) DH=-241.8 kJ • then, H2O(g) H2(g) + 1/2 O2(g) DH =+241.8 kJ • also, • If you multiply the equation by a number, you multiply the heat by that number: • 2 H2O(g) H2(g) + O2(g) DH =+483.6 kJ 2 • Find the heat of formation (∆H) of: 2 Al (s) + 3 CuO (s) → 3 Cu (s) + Al2O3 (s) Using two equations from the sheet: 2Al (s) + 3/2 O2 (g) → Al2O3 (s) ∆H=-1676kJ Cu(s) + 1/2 O2 (g) → CuO (s) ∆H = -157.3kJ Need to flip the second equation and change the sign on ∆H and multiply it by 3 2Al (s) + 3/2 O2 (g) → Al2O3 (s) ∆H=-1676kJ 3CuO (s)→3Cu(s) + 3/2 O2 (g) ∆H=(3)157.3kJ ___________________________________ 2 Al (s) + 3 CuO (s) → 3 Cu (s) + Al2O3 (s) ∆H = - 1204.1 kJ ( oxygen can be cancelled because it exists on both sides of the reactions) • Calculating Heats of Reaction using Hess's Law • 1) Write the overall equation for the reaction if not given. • 2) Manipulate the given equations for the steps of the reaction so they add up to the overall equation. • 3) Add up the equations canceling common substances in reactant and product. • 4) Add up the heats of the steps = heat of overall reaction. H2O(g) + C(s) → CO(g) +H2(g) ∆H=? These are the equations chosen from the sheet 1) H2(g) + 1/2O2(g) →H2O(g) ∆H = -242.kJ 2) C(s) +1/2 O2(g) → CO(g) ∆H = -110. kJ Note that the H2O (g) is a reactant and in step #1 it is a product. Thus step one needs to be reversed and the sign of the heat. H2O(g) → H2(g) + ½ O2(g) ∆H = 242.kJ C(s) + ½ O2(g) → CO(g) ∆H = -110. kJ _________________________________ H2O(g) + C(s) → CO(g) +H2(g) ∆H=132 Calculate the heat of reaction for the following equation C3H8(g) + 5 O2 (g) ---> 3 CO2(g)+4H2O (g) given the following steps in the reaction mechanism. 1) 3C (s)+ 4 H2 (g) -------> C3H8 (g) 2) H2 (g) + ½ O2 (g) -------> H2O (g) 3) C (s) + O2 (g) --------> CO2 (g) • We can manipulate the equations by: a) Reversing equation #1 b) Multiplying equation #2 by 4 c) Multiplying equation #3 by 3 ∆H C3H8 (g) -------> 3C (s)+ 4 H2 (g) 103.8 4 H2 (g) + 2O2 (g) -----> 4H2O (g) -967.2 3C (s) + 3O2 (g) -----> 3CO2 (g) -1180.5 C3H8(g) + 5 O2 (g) ---> 3 CO2(g)+4H2O (g) ∆H = - 2043.9 kJ