1. Thermochemistry Standard Enthalpies of Formation

2. Standard Enthalpy of Formation  ∆H f °  change in enthalpy that accompanies the formation of one mole of a compound from its elements in standard states  ° means that the process happened under standard conditions so we can compare more easily

3. Standard States  For a COMPOUND:  for gas: P = 1 atm  pure liquid or solid state  in solution: concentration is 1 M  For an ELEMENT:  form that it exists in at 1 atm and 25°C O: O 2 (g)K: K(s)Br: Br 2 (l)

4. Writing Formation Equations  always write equation where 1 mole of compound is formed (even if you must use non-integer coefficients) NO 2 (g): ½N 2 (g) + O 2 (g)  NO 2 (g) ∆H f °= 34 kJ/mol CH 3 OH(l): C(s) + 2H 2 (g) + ½O 2 (g)  CH 3 OH(l) ∆H f °= -239 kJ/mol

5. Using Standard Enthalpies of Formation where  n = number of moles of products/reactants  ∑ means “sum of”  ∆H f ° is the standard enthalpy of formation for reactants or products  ∆H f ° for any element in standard state is zero so elements are not included in the summation

6. Using Standard Enthalpies of Formation  since ∆H is a state function, we can use any pathway to calculate it  one convenient pathway is to break reactants into elements and then recombine them into products

7. Using Standard Enthalpies of Formation

9. Example 1  Calculate the standard enthalpy change for the reaction that occurs when ammonia is burned in air to make nitrogen dioxide and water 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(l)  break them apart into elements and then recombine them into products

10. Example 1

11.  can be solved using Hess’ Law: (1) 4NH 3 (g)  2N 2 (g) + 6H 2 (g) -4 ∆H f ° NH3 (2) 7O 2 (g)  7O 2 (g) 0 (3) 2N 2 (g) + 4O 2 (g)  4NO 2 (g) 4 ∆H f ° NO2 (4) 6H 2 (g) + 3O 2 (g)  6H 2 O(l) 6 ∆H f ° H2O 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(l) Look up: -46 kJ/mol 34 kJ/mol -286 kJ/mol values are in Appendix 3: p. A8-A12

12. Example 1  can also be solved using enthalpy of formation equation:

13. Example 2  Calculate the standard enthalpy change for the following reaction: 2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s) 2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s) = 0 kJ/mol -826 kJ/mol -1676 kJ/mol 0 kJ/mol

14. Example 3  Compare the standard enthalpy of combustion per gram of methanol with per gram of gasoline (it is C 8 H 18 ).  Write equations: 2CH 3 OH(l) + 3O 2 (g)  2CO 2 (g) + 4H 2 O(l) 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O(l)

15. Example 3  Calculate the enthalpy of combustion per mole:

16. Example 3  Convert to per gram using molar mass: so octane is about 2x more effective