2. First Law of Thermodynamics Enthalpy -  H, HfHf Second Law Third Law Entropy  S, Free Energy GG Spontaneity n Chap 18. Thermodynamics and Equilibrium

3. Thermochemistry n Heat changes during chemical reactions Thermochemical equation. eg. H 2 (g) + O 2 (g) ---> 2H 2 O(l)  H =- 256 kJ;  is called the enthalpy of reaction. if  H is + reaction is called endothermic if  H is - reaction is called exothermic

4. Why is it necessary to divide Universe into System and Surrounding n Universe = System + Surrounding

5. Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding

6. What is the internal energy change (  U) (  U) of a system?  U is associated with changes in atoms, molecules and subatomic particles E total = E ke + E pe +  U  U = heat (q) + w (work)  U = q + w  U = q -P  V; w =- P  V

7. What forms of energy are found in the Universe? n mechanical n thermal n electrical n nuclear n mass: E = mc 2 n others yet to discover

8. What is 1 st Law of Thermodynamics n Eenergy is conserved in the universe n All forms of energy are inter-convertible and conserved n Energy is neither created nor destroyed.

9. What exactly is  H? n Heat measured at constant pressure q p n Chemical reactions exposed to atmosphere and are held at a constant pressure. Volume of materials or gases produced can change. ie: work = -P  V  U = q p + w;  U = q p -P  V q p =  U + P  V;w = -P  V  H =  U + P  V;q p =  H(enthalpy )

10. How do you measure  U? Heat measured at constant volume q v Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -P  V= 0  U = q v + w q v =  U + o;w = 0  U = q v =  U(internal energy )

11. What is Hess's Law of Summation of Heat? n To heat of reaction for new reactions. n Two methods? 1st method: new  H is calculated by adding  Hs of other reactions. 2nd method: Where  H f (  H of formation) of reactants and products are used to calculate  H of a reaction.

12. Method 1: Calculate  H for the reaction: SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) -----> H 2 SO 4 (l)  H = ? Other reactions: SO 2 (g) ------> S(s) + O 2 (g)  H = 297kJ H 2 SO 4 (l)------> H 2 (g) + S(s) + 2O 2 (g)  H = 814 kJ H 2 (g) +1/2O 2 (g) -----> H 2 O(g)  H = -242 kJ

13. SO 2 (g) ------> S(s) + O 2 (g);  H 1 = 297 kJ - 1 H 2 (g) + S(s) + 2O 2 (g) ------> H 2 SO 4 (l)  H 2 = -814 kJ - 2 H 2 O(g) ----->H 2 (g) + 1/2 O 2 (g)  H 3 = +242 kJ - 3 ______________________________________ SO 2 (g) + 1/2 O 2 (g) + H 2 O(g) -----> H 2 SO 4 (l) ª H =  H 1 +  H 2 +  H 3 ª H = +297 - 814 + 242 ª H = -275 kJ

14. Calculate Heat (enthalpy) of Combustion: 2 nd method C 7 H 16 (l) + 11 O 2 (g) -----> 7 CO 2 (g) + 8 H 2 O(g) ;  H = ?  H f (C 7 H 16 ) = -198.8 kJ/mol  H f (CO 2 ) = -393.5 kJ/mol  H f (H 2 O) = -285.9 kJ/mol  H f O 2 (g) = 0 (zero) n What method? n 2 nd method

15. ª H = [ 3 n ( ª H o f ) Products] - [ 3 n ( ª H o f ) reactants] ª H = [ 7(-393.5 + 8 (-285.9)] - [-198.8 + 11 (0)] n = [-2754.5 - 2287.2] - [-198.8] n = -5041.7 + 198.8 n = 4842.9 kJ = 4843 kJ

16. Why is ª H o f of elements is zero? ª H o f, Heat formations are for compounds Note: ª H o f of elements is zero

17. What is 2 nd Law of Thermodynamics Entropy (  S) of the Universe increases during spontaneous process. n What is entropy Entropy (  S) : A measure of randomness or disorder of a system. n Spontaneous Process: A process that occurs without outside intervention.

18. Definitions n The Universe: The sum of all parts under consideration. n System: Part of the Universe we are interested in and a change is taking place. n Surrounding: Part of the Universe we are not interested in or can not observe.

19. Entropy  S  S univ = entropy of the Universe  S sys = entropy of the System  S surr = entropy of the Surrounding  S univ =  S sys +  S surr

20.  S univ  S sys  S surr n + + + + +(  S sys >  S surr) - + - + (  S surr >  S sys)

21. Entropy Change Entropy (  S) normally increase (+) for the following changes: n i) Solid ---> liquid (melting) n ii) Liquid ---> gas n iii) Solid ----> gas n iv) Increase in temperature n v) Increasing in pressure n vi) Increase in volume ( at constant temperature and pressure)

22. What is  G ? n Free Energy  G = - T  S uni.  G =  H - T  S.

23. How do you get:  G =  H - T  S.  s univ =  s sys +  s surr  s surr = -  H sys /T)  S univ =  S sys -  H sys /T  S univ x T = T  S sys -  H sys : x T -  S univ x T = -T  S sys +  H sys : x -1 -  S univ x T =  H sys -T  S sys -  S univ x T =  G;  G =  H sys -T  S sys or  G =  H - T  S.

24. What  G means If  G is - for a change it will take place n spontaneously If  G is + for a change it will not take place If  G is 0 for a change it will be in equilibrium

25. Qualitative prediction of  S of Chemical Reactions n Look for (l) or (s) --> (g) n If all are gases: calculate  n = 3 n(gaseous prod.) - 3 n(gaseous reac.) n N 2 (g) + 3 H 2 (g) --------> 2 NH 3 (g)  n = 2 - 4 = -2 If  n is -  S is negative (decrease in S) If  n is +  S is positive (increase in S)

26. Predict  S! n 2 C 2 H 6 (g) + 7 O 2 (g)--> 4 CO 2 (g) + 6H 2 0(g). n 2 CO(g) + O 2 (g)-->2 CO 2 (g). n HCl(g) + NH 3 (g)-->NH 4 Cl(s). n H 2 (g) + Br 2 (l) --> 2 HBr(g).

27. Calculating  S of reactions n Based on Hess’s Law second method:  S o = 3  S o (prod.) - 3  S o (react.)

28. How do you calculate  G There are two ways to calculate  G n for chemical reactions. i)  G =  H - T  S. ii)  G o = 3  G o f (products) - 3  G o f (reactants)

29. How do you calculate  G at different T and P  G =  G o + RT ln Q n Q = reaction quotient at equilibrium  G =   =  G o + RT ln K  G o = - RT ln K If you know  G o you could calculate K