1. IIIIII Chapter 16 Hess’s Law

2. HESS’S LAW n If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.

3. Hess’s law can be used to determine the enthalpy change for a reaction that cannot be measured directly!

4.  H NET =  H 1 +  H 2 N 2(g) + O 2(g)  2NO (g) ΔH 1 = +181kJ 2NO (g) + O 2(g)  2NO 2(g) ΔH 2 = -113kJ ADD THEM UP ALEGBRAICALLY

5. N 2(g) + O 2(g)  2NO (g) ΔH 1 = +181kJ 2NO (g) + O 2(g)  2NO 2(g) ΔH 2 = -113kJ N 2(g) + 2O 2(g) First, add up the chemical equations. + 2NO (g)  2NO (g) + 2NO 2(g)

6. Notice that 2NO(g) is on both the reactants and products side and can be cancelled out. N 2(g) + O 2(g)  2NO (g) ΔH 1 = +181kJ 2NO (g) + O 2(g)  2NO 2(g) ΔH 2 = -113kJ N 2(g) + 2O 2(g) + 2NO (g)  2NO (g) + 2NO 2(g)

7. Write the net equation: N 2(g) + 2O 2(g) + 2NO (g)  2NO (g) + 2NO 2(g) N 2(g) + 2O 2(g)  2NO 2(g)

8.  H NET =  H 1 +  H 2 ΔH 1 = +181kJ ΔH 2 = -113kJ Apply Hess’s Law to calculate the enthalpy for the reaction.

9.  H NET =  H 1 +  H 2 ΔH NET = (+181kJ) + (-113kJ) ΔH NET = +68kJ Overall, the formation of NO 2 from N 2 and O 2 is an endothermic process, although one of the steps is exothermic.

10. ΔH Reaction Progress N 2(g) + 2O 2(g) 2NO (g) + O 2(g) 2NO 2(g) ΔH NET = +68kJ ΔH 1 = +181kJ ΔH 2 = -113kJ

11. RULES for Hess’s Law Problems 1.If the coefficients are multiplied by a factor, then the enthalpy value MUST also be multiplied by the same factor. 2.If an equation is reversed, the sign of ΔH MUST also be reversed.

12. C (s) + ½O 2(g)  CO (g) ΔH 1 = -110.5kJ CO (g) + ½O 2(g)  CO 2(g) ΔH 2 = -283.0kJ C (s) + O 2(g) + CO (g)  CO (g) + CO 2(g) Practice Problem: #1 C (s) + O 2(g)  CO 2(g)  H NET =  H 1 +  H 2  H NET = (-110.5kJ) + (-283.0kJ)  H NET = -393.5kJ Net Equation

13. C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (g) Practice Problem: #3 CH 3 OCH 3(l) + 3O 2(g)  2CO 2(g) + 3H 2 O (g) ΔH 1 = -1234.7kJ ΔH 2 = -1328.3kJ You have to REVERSE equation 2 to get the NET equation. DON’T forget to change the sign Of ΔH 2

14. C 2 H 5 OH (l) + 3O 2(g)  2CO 2(g) + 3H 2 O (g) Practice Problem: #3 2CO 2(g) + 3H 2 O (g)  CH 3 OCH 3(l) + 3O 2(g) ΔH 1 = -1234.7kJ ΔH 2 = +1328.3kJ C 2 H 5 OH (l) + 3O 2(g) + 2CO 2(g) + 3H 2 O (g)  2CO 2(g) + 3H 2 O (g) + CH 3 OCH 3(l) + 3O 2(g) Net Equation C 2 H 5 OH (l)  CH 3 OCH 3(l)

15. Net Equation C 2 H 5 OH (l)  CH 3 OCH 3(l)  H NET =  H 1 +  H 2  H NET = (-1234.7kJ) + (+1328.3kJ)  H NET = +93.6kJ

16. H 2(g) + F 2(g)  2HF (g) ΔH 1 = -542.2kJ 2H 2(g) + O 2(g)  2H 2 O (g) ΔH 2 = -571.6kJ Practice Problem: #5 You have to REVERSE equation 2 to get the NET equation. DON’T forget to change the sign Of ΔH 2

17. H 2(g) + F 2(g)  2HF (g) ΔH 1 = -542.2kJ 2H 2 O (g)  2H 2(g) + O 2(g) ΔH 2 = +571.6kJ Practice Problem: #5 You will need to multiply the first equation by 2. DON’T forget to multiply the ΔH by 2 also.

18. 2H 2(g) + 2F 2(g)  4HF (g) ΔH 1 = -1084.4kJ 2H 2 O (g)  2H 2(g) + O 2(g) ΔH 2 = +571.6kJ Practice Problem: #5 Net Equation 2H 2(g) + 2F 2(g) + 2H 2 O (g)  4HF (g) + 2H 2(g) + O 2(g) 2F 2(g) + 2H 2 O (g)  4HF (g) + O 2(g)  H NET =  H 1 +  H 2  H NET = (-1084.4kJ) + (+571.6kJ)  H NET = -512.8kJ

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