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. ∆H target = ∑ ∆H unknowns 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? N 2(g) +O 2(g) → NO (g) ΔH=? Given the following known reactions: ½N 2(g) +O 2(g) → NO 2(g) ΔH= + 34kJ NO (g) + ½ O 2(g) → NO 2(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: 2CO 2 (g) +H 2 O (g) →C 2 H 2 (g) + 5/2 O 2 (g) C 2 H 2 (g) +2H 2 (g) →C 2 H 6 (g) ΔH=-94.5kJ H 2 O (g)  H 2 (g) + ½ O 2 (g) ΔH=71.2kJ C 2 H 6 (g) +7/2O 2 (g) →2CO 2 (g) +3H 2 O (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 O 2 (g)  ZnO (s) + SO 2(g) ZnO (s)  Zn (s) + ½ O 2(g) ΔH= 350.5kJ S (s) + O 2(g)  SO 2(g) ΔH=-296.8kJ ZnS (s)  Zn (s) + S (s) ΔH= 206.0kJ

4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Using the following sets of reactions: (1) N 2 (g) + O 2 (g)  2NO(g)  H = kJ (2) N 2 (g) + 3H 2 (g)  2NH 3 (g)  H = kJ (3) 2H 2 (g) + O 2 (g)  2H 2 O(g)  H = kJ Goal: NH 3 : O2 O2 : NO: H 2 O: (2)( Reverse and x 2) 4NH 3  2N 2 + 6H 2  H = kJ Found in more than one place, SKIP IT (its hard). (1) ( Same x2) 2N 2 + 2O 2  4NO  H = kJ (3)( Same x3 ) 6H 2 + 3O 2  6H 2 O  H = kJ

4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) Goal: NH 3 : O 2 : NO: H 2 O: Reverse and x2 4NH 3  2N 2 + 6H 2  H = kJ Found in more than one place, SKIP IT. x2 2N 2 + 2O 2  4NO  H = kJ x3 6H 2 + 3O 2  6H 2 O  H = kJ Cancel terms and take sum. 4NH 3 + 5O 2  4NO + 6H 2 O  H = kJ Is the reaction endothermic or exothermic?  H = kJ kJ + (-1451kJ)

Determine the heat of reaction for the reaction: TARGET  C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = ? Use the following reactions: (1) C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ (2) C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = kJ (3) H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ Consult your neighbor if necessary.

Determine the heat of reaction for the reaction: Goal: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = ? Use the following reactions: (1) C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ (2) C 2 H 6 (g) + 7/2O 2 (g)  2CO 2 (g) + 3H 2 O(l)  H = kJ (3) H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 4 (g) :use 1 as is C 2 H 4 (g) + 3O 2 (g)  2CO 2 (g) + 2H 2 O(l)  H = kJ H 2 (g) :# 3 as is H 2 (g) + 1/2O 2 (g)  H 2 O(l)  H = -286 kJ C 2 H 6 (g) : rev #2 2CO 2 (g) + 3H 2 O(l)  C 2 H 6 (g) + 7/2O 2 (g)  H = kJ C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = -137 kJ

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

Worksheet Hess’ Law – under homework on class site