Hess’s Law Start Finish A State Function: Path independent.

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1 Hesss Law Start Finish A State Function: Path independent. Both lines accomplished the same result, they went from start to finish. Net result = same.

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

and.. the H values must be treated accordingly. Determine the heat of reaction for the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the H values must be treated accordingly.

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: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse and x 2 Found in more than one place, SKIP IT (its hard). O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ H2O: 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 H2O: x3 6H2 + 3O2  6H2O H = -1451.1 kJ Cancel terms and take sum. + 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: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ 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: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ 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