1. Ch 5: Hess ’ s Law

2. Hess ’ s Law states.. “ the enthalpy change for a chem rxn is the same whether the rxn takes place in one step or several steps ” ΔH rxn = sum of enthalpy changes in each step Use the following procedure to calculate an enthalpy change for a rxn, having been given the known thermochemical equations:

3. Calculating ΔH rxn 1. Write the equation for desired chem process 2. Write several equations that when added together give the desired chem process 3. If the desired reactant or product is on the wrong side of the equation, reverse the equation (and change the sign of ΔH) 4. Multiply (or divide) the known equation by appropriate coefficients so that the react. and prod. have same coeff. as desired equation 5. Add equations, cancel species that appear on react. and prod. side, make sure everything stays balanced 6. Add ΔH ’ s to determine ΔH rxn

4. Ex#1 2S(s) + 3O 2 (g)  2SO 3 (g) Calculate ΔH rxn for the following reaction: (1) 2S(s) + 3O 2 (g)  2SO 3 (g) … given, (2) ΔH for related reactions: (a) S(s) + O 2 (g)  SO 2 (g) = -296.9 kJ (b) 2SO 2 (g) + O 2 (g)  2SO 3 (g) = -196.6 kJ (3) react. and prod. are on the correct side of equation (4) coeff. in eqn. (a) needs to be x2 to match desired eqn. 2S(s) + 2O 2 (g)  2SO 2 (g) = 2(-296.9 kJ)

5. Ex#1 2S(s) + 3O 2 (g)  2SO 3 (g) (5) Adding (a) and (b) you get: 2S(s) + 2O 2 (g) + O 2 (g) +2SO 2 (g)  2SO 2 (g) + 2SO 3 (g) Combine the O 2 and cancel the SO 2 to get the desired equation (6) – 593.8 kJ + (-196.6 kJ) = -790.4kJ

6. Ex#2 Calc ΔH rxn for: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (l) (2) given the following data: (a) C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l) … -845.2 kJ (b) 2H 2 (g) + O 2 (g)  2H 2 O(l) ………………… -571.7 kJ (c) 2C 2 H 6 (l) + 7O 2 (g)  4CO 2 (g) + 6H 2 O(l) … -1987.4 kJ (3) Flip eqn. (c) so that C 2 H 6 (l) appears on product side (and change to + ΔH) (4) Divide coeff in (b) by 2 to get correct # of H 2 Divide coeff in (c) by 2 to get correct # of C 2 H 6

7. (5) You now have … (a) C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l) … -845.2 kJ (b) H 2 (g) + 0.5 O 2 (g)  H 2 O(l) ………………… -289.5 kJ (c) 2CO 2 (g) + 3H 2 O(l)  C 2 H 6 (l) + 3.5 O 2 (g)..+993.7 kJ Combine the equations to get: C 2 H 4 (g) + 3.5 O 2 (g) + H 2 (g) + 2CO 2 (g) + 3H 2 O(l)  2CO 2 (g) + 3H 2 O(l) + C 2 H 6 (l) + 3.5 O 2 (g) The O 2 and H 2 O on cancel to give the desired equation (6) Add the ΔH ’ s you get: -845.2 kJ + (-289.5 kJ) + 993.7kJ = -137.4kJ

8. For HW Page 205 #59, 61-64 Tomorrow ’ s lesson is on “ Enthalpies of Formation ” so read pages 188-192 Savannah says, “ Shaylah, I heard what you did to me on the SmartBoard ”