Chapter 5. Recommended Homework Problems 5, 7, 11, 19, 31, 45, 49, 53, 59, 61, 63, 69, 73, 77, 83

Transcription

1 Recommended Homework Problems 5, 7, 11, 19, 31, 45, 49, 53, 59, 61, 63, 69, 73, 77, 83

2 Potential and kinetic energy. Energy of position and energy of motion. Law of conservation of energy the total energy of an isolated body is constant.

3 Heat once thought to be a substance like a gas. Related to molecular motion. The transfer of energy from one substance to another as the result of a difference in temperature. heat high low temperature

4 Work force through distance. SYSTEM and SURROUNDINGS convenient to keep track of heat flow.

5 EXOTHERMIC AND ENDOTHERMIC (p. 236) surroundings heat heat system system

6 Calorimetry - measures heat flow. Heat capacity of a substance the amount of heat needed to raise a given amount of the substance a given temperature C(joules or cals) = q (joules or cals) ΔT( o C) Specific heat = Heat capacity divided by mass Specific heat of water the amount of heat needed to raise 1 g of water 1 o C Specific heat of water = 1 cal = 4.18 joules.

7 1045 joules raises the temperature of 50 g of Fe 46.7 o C. What is the specific heat of Fe? How many joules needed to raise temp. of 20 g HOH 15 o C?

8 Have 50 g of water at 25 o C. In another container, heat 20 g of metal to 100 o C. Drop the metal into the water. Final temperature of the water (and metal) is 31 o C. Assume the walls of the calorimeter absorb no heat all is absorbed by the water in the calorimeter. Find the SH of the metal.

9 Heat lost = heat gained SH metal x wt x ΔT = SH water x wt x ΔT

10 The total energy of an isolated body is constant. P. 255 State functions vs non state functions. P. 256 State functions independent of how it was achieved depends upon starting and final points. Work in not a state function. Depends upon the route.

11 We keep track of energy changes. The change in internal energy is given by: ΔE = E final - E initial because internal energy is a state function The internal energy of a system can change by doing work on the system: ΔE = w and the internal energy of a system can change by supplying heat to the system: ΔE = q

12 Changing the internal energy of the system by both doing work and adding heat gives: ΔE = q + w Very important both q and w have signs: + means that heat or work is added to the system and - means that heat or work is removed from the system.

13 ENTHALPY Symbol is H. Book (p. 257) The change in enthalpy of a system is equal to the heat released or absorbed at constant pressure. Difference between ΔH and ΔE: ΔH = ΔE + P ΔV We are going to call ΔH heat content

14 Exothermic vs endothermic in terms of enthalpy reactants products ΔH Heat ΔH Heat products reactants

15 Hess's Law Takes advantage of the fact that H is a state function ΔH = ΔH(products) - ΔH(reactants) Standard enthalpy of formation: ΔH f o Define one mol of the substance formed from its elements at standard conditions p. 263

16 Do all of these describe the ΔH fo of the product? S(s) + O 2 (g) --> SO 2 (g) (note - exothermic) C(s) + O 2 (g) --> CO 2 (g) -393 Pb(s) + O 2 (g) --> PbO 2 (s) -277 H 2 (g) + 1/2O 2 (g) --> H 2 O(g) kj/mol 1/2 N 2 (g) + 1/2 O 2 (g) --> NO CH 2 (g) + H 2 (g) --> CH 4 (g) -35 We define the ΔH fo for any free element as ZERO

17 Laws of Thermochemistry 1. ΔH is proportional to the amount of substance that reacts or is produced. 2. Reverse the sign and reverse the heat flow. 3. A reaction may be the sum of two or more reactions. Then ΔH is the sum of the ΔH's for the individual reactions.

18 Examples: Suppose we have: kj H 2 (g) + O 2 (g) --> H 2 O 2 (l) H 2 (g) + 1/2 O 2 (g) --> H 2 O(l) Find ΔH for 2 H 2 O 2 (l) --> 2 H 2 O(l) + O 2 (g)

19 Given: (all gases) 1/2 N 2 + 3/2 H 2 --> NH H 2 + 1/2 O 2 --> H 2 O /2 N 2 + 1/2 O 2 --> NO Find ΔH for 4 NH O 2 --> 6 H 2 O + 4 NO

20 Do the two problems again, using just Hess's Law: For 2 H 2 O 2 --> 2 H 2 O + O 2 H 2 (g) + O 2 (g) --> H 2 O 2 (l) H 2 (g) + 1/2 O 2 (g) --> H 2 O(l)

21 4 NH O 2 --> 6 H 2 O + 4 NO 1/2 N 2 + 3/2 H 2 --> NH H 2 + 1/2 O 2 --> H 2 O /2 N 2 + 1/2 O 2 --> NO +90.3

22 Find ΔH fo for WC Given: 2 W + 3 O 2 > 2 WO C + O 2 > CO WC + 5 O 2 > 2 WO CO Need: W + C > WC

23 HEAT OUTPUT OF REACTIONS. The heat absorbed or given off by a reaction can be treated like a product or reactant in stoichometric reaction. 2 C 2 H O 2 --> 4 CO H 2 O ΔH = kj How much heat released when 80.0 g of C 2 H 6 used? 2 C 2 H O 2 --> 4 CO H 2 O + heat

24 C 3 H O 2 --> 3 CO H 2 O ΔH = kj Calculate the mass of propane needed to produce kj of heat. C 3 H O 2 --> 3 CO H 2 O + heat

25 For 2 C 3 H 5 (NO 3 ) 3 > 3 N 2 (g) + ½ O 2 (g) + 6 CO 2 (g) + 5 H 2 O(g) Calculate ΔH when 10.0 g of nitroglycerine is exploded. Given:ΔH f : Nitroglycerine: -364 kj/mol; CO 2 (g): kj/mol; H 2 O(g): kj/mol First need to calc. ΔH rxn.

26 2 C 3 H 5 (NO 3 ) 3 > 3 N 2 (g) + ½ O 2 (g) + 6 CO 2 (g) + 5 H 2 O(g) ΔH rxn = 6[ΔH f CO 2 (g)] + 5[ΔH f H 2 O(g)] -2([ΔH f C 3 H 5 (NO 3 ) 3 ]