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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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Presentation on theme: "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."— Presentation transcript:

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2 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.

3 2 Determine the heat of reaction for the reaction: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Using the following sets of reactions: N 2 (g) + O 2 (g) 2NO(g) H = kJ N 2 (g) + 3H 2 (g) 2NH 3 (g) H = kJ 2H 2 (g) + O 2 (g) 2H 2 O(g) H = kJ Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction. and.. the H values must be treated accordingly.

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

5 4 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?

6 5 Determine the heat of reaction for the reaction: C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) Use the following reactions: C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) H = kJ C 2 H 6 (g) + 7/2O 2 (g) 2CO 2 (g) + 3H 2 O(l) H = kJ H 2 (g) + 1/2O 2 (g) H 2 O(l) H = -286 kJ Consult your neighbor if necessary.

7 6 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: C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(l) H = kJ C 2 H 6 (g) + 7/2O 2 (g) 2CO 2 (g) + 3H 2 O(l) H = kJ 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

8 7 Summary: enthalpy is a state function and is path independent.

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10 9 Standard Enthalpies of formation:

11 10 Thermodynamic Quantities of Selected K


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