1. Q-Hess’s Law Prem Sattsangi Copyright 2007

2. #2-Hess’s Law Study Please have your pencil and paper and BLB text book as you attempt these problem. Write your response. Press enter to check the correct answer.

3. #3-Definitions BLB-Page 170 Please Define: SYSTEM: Portion of the Universe under study. SURROUNDINGS: Rest of the Universe.

4. #4-Definitions BLB-Page 174 Please Define: ENDOTHERMIC: Process in which the system Absorbs Heat. EXOTHERMIC: Process in which the system Evolves Heat. p. 175Fig. 5.8 p. 178Fig. 5.12

5. #5-Definitions BLB-Page 183 Please Define: HEAT CAPACITY: Amount of Heat required to raise the temperature of an object by 1 o C. SPECIFIC HEAT : HEAT CAPACITY per gram of a substance. OR Amount of Heat required to raise the temperature of 1g of a substance by 1 o C.

6. #6-Definitions Please Define: HESS’S LAW: (p. 189) If a reaction is carried out in a series of steps,  H for the overall reaction will equal the sum of enthalpy changes for individual steps. PRACTICE: SE 5.8 STANDARD STATE : (p. 191) Physical State (s, l, or g) of a substance under STANDARD conditions, 1 atm and 25 o C STANDARD ENTHALPY OF FORMATION:  Hform of 1 mol of a compound starting from its elements with all substances under STANDARD conditions.

7. #7 Hess’s Law calculations (p. 189) Calculate  H form for C 3 H 8 [A: -106 kJ] REACTANTS  PRODUCTS 3C(s) + 4H 2 (g)  C 3 H 8 (g) FINAL EQUATION GIVEN: (a) C(s) + O 2 (g)  CO 2 (g)-394 kJ (b) H 2 (g) + ½ O 2 (g)  H 2 O (l)-286 kJ (c) C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l)-2220.0 kJ Rewrite each equation to: (i) MATCH the REACTANTS and PRODUCTS in the FINAL EQUATION. (ii) If you reverse the equation, change the sign of  H. (a) 3C(s) + 3O 2 (g)  3 CO 2 (g) -394 kJ x 3 (to get 3C) (b) 4H 2 (g) + 2 O 2 (g)  4 H 2 O (l) -286 kJ x 4 (to get 4 H 2 ) (c) 3 CO 2 (g) + 4 H 2 O (l)  C 3 H 8 (g) + 5 O 2 (g) +2220.0 kJ Add the equations to obtain the FINAL EQUATION. Calculate the kJ.

8. #8- Calculating”Q/g”, p. 183, SE 5.5 Q = m(H 2 O) x s (H 2 O) x  T Eq. 5.22 [s (H 2 O) = 4.184 J/g o C, m (water only) ] Calculate  H soln. /g for the U.K.#1 Wt. Of sample =2.05 g Amt of water=100 mL (D = 1g/mL)  T = 1.65 o C Q = 100 g x 4.184 J/g o C x 1.65 o C = 690 J J/g sample = 690 J/2.05 g = 336.6 J/g

9. #9- Calculating”  H soln. /mol”  H soln. /g for the U.K.#1 was = 336.6 J/g Calculate  H soln. /mol in kJ, if the U.K.#1 was NH 4 NO 2. [Atomic weights: N(14.0), H(1.01), O(16.0)] FW: NH 4 NO 2 = 28.0 + 4.04 + 32.0 = 64.0 g  H soln. = 336.6 J x 64.0 g x 1kJ___ = 21.54 kJ/mol  mol g mol 1000J

10. #10-Calculating Q/mol, Heat of Neutralization of solutions On mixing 60.0 mL each of 0.50 M HCl(aq) and 0.50 M KOH(aq) the temperature rose by 7.25 o C. Calculate  H per mol KOH. Total volume of Aq. Solutions = 60.0 mL + 60.0 mL= 120.0 mL Hence: Q = 120 g x 7.25 o C x 4.184 J = 3,640.1 J g o C Mol KOH present in 60.0 mL of 0.50 M solution(aq): n = M x V or 0.50 mol x 60.0 mL x 1 L = 0.030 mol L 1x10 3 mL Q = 3,640.1 J x ( 1 kJ) = 121 kJ Mol 0.030 mol (1000J) mol

11. #11 Neutralization reactions Role of a LIMITING REAGENT If you used 50.0 mL each of the following solutions, Heat of Neutralization of sets (b) and (c) will be (I) Same as (a), (ii) more, (iii) Less LIMITING REAGENT (a) 0.5 M HCl + 0.5 M NaOH NONE (b) 0.4 M HCl + 1.0 M NaOH Reagent with lower “M” (c) 1.3 M HCl + 0.6 M NaOHReagent with lower “M” Consider the role of LIMITING REAGENT (c) 0.6 M NaOH is less than 1.3 M HCl. It is the LIMITING REAGENT Comparing it to “0.5 M NaOH in (a),” this is more. Heat of Neutralization of set (C) will be more.