Energy and Thermochemistry

Transcription

1 Energy and Thermochemistry 1

2 Energy The ability to do work 2 types Potential: stored energy Kinetic: energy in motion 2

3 Thermochemistry Changes of heat content and heat transfer Follow Law of Conservation of Energy Or, 1 st Law of Thermodynamics Energy can neither be created nor destroyed 3

4 Temperature & Heat Heat not same as temperature Heat = energy transferred to one system by another due to temperature difference Temperature = measure of heat energy content & ability to transfer heat Thermometer Higher thermal energy, greater motion of constituents Sum of individual energies of constituents = total thermal energy 4

5 Systems and Surroundings System = the object in question Surrounding(s) ) = everything outside the system When both system and surrounding at same temperature thermal equilibrium When not Heat transfer to surrounding = exothermic (you feel the heat) hot metal! Heat transfer to system = endothermic (you feel cold) cold metal! 5

6 Math! Joules (J) used for energy quantities But usually kj (1000 J) used Ye Royal Olde School used calorie (cal) cal = amt of heat required to raise the temperature of 1.00 g of water by 1 C1 1 cal = J (SI-unit) But Calorie (Cal) = 1000 cal Used in nutrition science and on food labels Joule (J) = 1 kg m s 2 2 6

7 Heat Capacity Specific heat capacity Quantity of heat required to raise the temp of 1 gram of any substance by 1 K C = g J K J Molar heat capacity Quantity of heat required to raise the temp of 1 mole of any substance by 1 K J c = mol K specific heat capacity of water = g K 75.4 J molar heat capacity of water = mol K 7

8 Calculating heat transfer Q = C m T Q = transferred heat, m = mass of substance, T = temperature change FYI Specific heat capacity of metals is very low < J/(g K) What does this tell us about heat transfer in metals? 8

9 Let s s do an example In your backyard, you have a swimming pool that contains 5.19 x 10 3 kg of water. How many kj are required to raise the temperature of this water from 7.2 C C to 25.0 C? 9

10 Example solved J g K Q = C m T = (4.184 ) (5.19 x 10 g) (298.2 K K) = J = kj Trick: T in K = T T in C 10

11 Practice How many kj are required to raise the temperature of 25.8 g of quicksilver from 22.5 C C to 28.0 C? C Hg = J/(g K) 11

12 Solution T = 28.0 C-22.5 C = 5.5 C J kj = = = = g K 1000J 3 Q C m T ( ) 25.8g 5.5 C 20.J kj 12

13 But what if there s s a change of state? Temperature constant throughout change of state Added energy overcomes inter-molecular forces 13

14 Change of state What do the flat areas represent? 14

15 Change of state q tot = q s + q s l + q l + q l g + q g q s l = heat h of fusion Heat required to convert solid at melting pt. to liq Ice = 333 J/g q l g = heat of vaporization Heat required to convert liq. at boiling pt. to gas Water = 2256 J/g 15

16 Practice How much heat is required to vaporize g of ice at C C to C? Given: Specific heat capacity of ice = 2.06 J/g K Specific heat capacity of water = J/g K Specific heat capacity of steam = 1.92 J/g K Let s s do this 16

17 Solution J g K 4 Q s=250.0g 2.06 (0.0 C 25.0 C)= J J Q s l =250.0g 333 = 83, 250J g J Q l=250.0g (100.0 C-0.0 C)=104,600J g K J g 5 Q l g=250.0g 2256 = J J J 3 Q g =250.0g 1.92 (110.0 C C)= J g K Qtot = J + 83, 250J + 104,600J J J = 769,650J 17

18 Calorimetry The process of measuring heat transfer in chemical/physical process q rxn + q soln = 0 q rxn = -q soln Rxn = system Soln = surrounding What you ll do in lab Heat given off by rxn Measured by thermometer Figure out q rxn indirectly 18

19 Enthalpy = heat content at constant pressure If H H = +,, process endothermic If H H = -,, process exothermic Enthalpy change dependent on states of matter and molar quantities For example: Is vaporizing ice an exothermic or endothermic process? Thus, will H H be + or -? 19

20 Hess s s Law If a rxn is the sum of 2 or more other reactions, H H = sum of H s s for those rxns So, H tot = H 1 + H 2 + H H n 20

21 Let s s solve a problem C (s) + 2S (s) CS 2(l) ; H H =? Given: C (s) + O 2(g) CO 2(g) ; H H = kj/mol S (s) + O 2(g) SO 2(g) ; H H = kj/mol CS 2(l) + 3O 2(g) CO 2(g) + 2SO 2(g) ; H H = kj/mol How do we manipulate the 3 rxns to achieve the necessary net rxn? Does H H change if the rxns are reversed and/or their mole ratios are changed? Let s s talk about this on the next slide 21

22 Let s s work it out 1. Switch this rxn: : CS 2(l) + 3O 2(g) CO 2(g) + 2SO 2(g) ; H H = kj Thus, CO 2(g) + 2SO 2(g) CS 2(l) + 3O 2(g) ; H H = kj Thus, - H fwd = + H+ rev 2. Double this rxn: S (s) + O 2(g) SO 2(g) ; H H = kj Thus, 2S 2 (s) + 2O 2(g) 2SO 2(g) ; H H = ( kj) x 2 = kj Since H H is per mole, changing the stoichiometric ratios entails an equivalent change in H 3. Keep this rxn: C (s) + O 2(g) CO 2(g) ; H H = kj 4. Add those on same side of rxns/eliminate those on opposite sides of rxn: CO 2, 2SO 2, 3O 2 5. Net rxn: C (s) + 2S (s) CS 2(l) H H = kj kj kj = kj Is it an exo- or endothermic rxn? 22

23 Practice Given: CH 4(g) C (s) + 2H 2(g) ; H H = 74.6 kj/mol C (s) + O 2(g) CO 2(g) ; H H = kj/mol H 2(g) + ½ O 2(g) H 2 O (g) ; H= kj/mol CH 4(g) + 2O 2(g) CO 2(g) + 2H 2 O (g) ; H rxn =? 23

24 Standard Energies of Formation Standard molar enthalpies of formation = H f = enthalpy change for formation of 1 mol of cmpd directly from component elements in standard states Standard state = most stable form of substance in physical state that 1 bar pressure & a specific temp., usually 0 C 0 (273K) 1 bar = 100kPa kpa = 1 atm So 1 bar 1 atm (an SI unit) Example C (s) + O 2(g) CO 2(g) ; H H = H f = kj H f = 0 for elements in standard state 24

25 Enthalpy Change for a Rxn Must know all standard molar enthalpies H rxn = H f prods - H f reactants Given to you in a table in the back of the book Again, keep in mind the mole ratios for each species involved! 25

26 Practice Determine H rxn for: 4NH 3(g) + 5O 2(g) 4NO (g) + 6H 2 O (g) Given: NH 3(g) = kj/mol NO (g) = 91.3 H 2 O (g) =

27 Solution Determine H for: rxn 4NH + 5O 4NO + 6H O 3(g) 2(g) (g) 2 (g) kj kj kj kj H rxn = [4mol mol ] [4mol mol 0 ] = 902kJ mol mol mol mol 27

28 Example H rxn for 10.0 g of nitroglycerin? 2C 3 H 5 (NO 3 ) 3(l) 3N 2(g) + ½O 2(g) + 6CO 2(g) + 5H 2 O (g) C 3 H 5 (NO 3 ) 3(l) = -364 kj/mol CO 2(g) = kj/mol H 2 O (g) = kj/mol 28

29 Solution kj 1 kj kj kj 1) H rxn = [3 mol 0 + mol mol mol ] mol 2 mol mol mol kj - [2 mol -364 ] = kj mol 1mol -2842kJ 2)10.0 g = 62.6kJ 227.2g 2mol 29