Hess' Law Learning Goals: Will be able to apply Hess’s Law to determine the enthalpy change of chemical equations. Will be able to write target equations from word equations

Hess’ Law Start Finish Path independent Both lines accomplished the same result, they went from start to finish. Net result = same.

Hess’s Law states…. The enthalpy change (ΔH) in a chemical reaction – from reactants to products – is the same whether the conversion occurs in one step or several steps. ie. ΔH is independent of the path taken

Hess’s Law Rules The value of the ΔH for any reaction can be written in steps equals the sum of the values of ΔH for each of the individual reactions. ∆Htarget = ∑ ∆Hunknowns If a chemical reaction is reversed then the sign of ΔH changes. If changes are made to the coefficients of a chemical equation (i.e. to balance it) then the value of ΔH is altered in the same way.

Example 1: What is the enthalpy change for the formation of nitrogen monoxide from its elements? N2(g) +O2(g) → NO(g) ΔH=? Given the following known reactions: ½N2(g) +O2(g) → NO2(g) ΔH= + 34kJ NO(g) + ½ O2(g) → NO2(g) ΔH= - 56kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the H values must be treated accordingly.

Example 2: Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: 2CO2 (g) +H2O(g) →C2H2 (g) + 5/2 O2 (g) C2H2 (g) +2H2 (g) →C2H6 (g) ΔH=-94.5kJ H2O(g)  H2 (g) + ½ O2 (g) ΔH=71.2kJ C2H6 (g) +7/2O2 (g) →2CO2 (g) +3H2O(g) ΔH=-283kJ

Example 3: How much energy can be obtained from the roasting of 50.0kg of zinc sulphide ore? ZnS (s) + 3/2 O2 (g)  ZnO(s) + SO2(g) ZnO(s)  Zn(s) + ½ O2(g) ΔH= 350.5kJ S(s) + O2(g)  SO2(g) ΔH=-296.8kJ ZnS(s)  Zn(s) + S (s) ΔH= 206.0kJ

Found in more than one place, SKIP IT (its hard). Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: (1) N2(g) + O2(g)  2NO(g) H = 180.6 kJ (2) N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ (3) 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: (2)(Reverse and x 2) Found in more than one place, SKIP IT (its hard). O2 : NO: (1) (Same x2) 2N2 + 2O2  4NO H = 361.2 kJ H2O: (3)(Same x3) 6H2 + 3O2  6H2O H = -1451.1 kJ

Found in more than one place, SKIP IT. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse and x2 Found in more than one place, SKIP IT. O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ 6H2 + 3O2  6H2O H = -1451.1 kJ H2O: x3 Cancel terms and take sum. H = 183.6 kJ + 361.2 kJ + (-1451kJ) + 5O2 + 6H2O H = -906.3 kJ 4NH3  4NO Is the reaction endothermic or exothermic?

Consult your neighbor if necessary. Determine the heat of reaction for the reaction: TARGET  C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: (1) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ (2) C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ (3) H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ Consult your neighbor if necessary.

Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: (1) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ (2) C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ (3) H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l)  C2H6(g) + 7/2O2(g) H = +1550 kJ C2H4(g) + H2(g)  C2H6(g) H = -137 kJ

Used for reactions that Summary: Used for reactions that cannot be determined experimentally Summary: H is independent of the path taken