CH10007/87. Thermodynamics. Dr Toby Jenkins

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1 CH10007/87 Thermodynamics Dr Toby Jenkins 1

2 Objectives To introduce the basic concepts of thermodynamics To apply them to chemical systems To develop competence in thermodynamics calculations 2

3 Equilibrium Kinetic equilibrium A B At equilibrium rates of forward and reverse reactions are equal.. f [ ] = [ ] k A k B b Thermodynamic equilibrium K = [ B] [ A] and so. K = [ k ] f [ k ] b Linking Kinetics and Thermodynamics! 3

4 Basic concepts and units : force, energy and work Velocity m s -1 Acceleration m s -1 s -1 = m s kg 1 m Lifting a weight - requires work to be done against gravity 4

5 10 kg what is the downward force due to gravity?? Newton's 2 nd Law force = mass acceleration f = m g = 10 kg x 9.81 m s -2 = 98.1 kg m s -2 = 98.1 N (Newtons) 1 N = 1 kg m s -2 (definition of the unit of force) 5

6 10 kg Lift. force upwards = force downwards Work = force x distance = 98.1 N x 1 m = 98.1 Nm = 98.1 J 1 J = 1 Nm (definition of unit of work or energy change) 6

7 10 kg potential energy difference 10 kg = work done p.e. of the mass is 98.1 J 10 kg Drop 10 kg p.e. goes back to zero or initial state 10 kg 7

8 Where does the energy go when we drop the weight? Energy is conserved. potential energy is transformed into kinetic energy k.e. = 1/2 mv 2 p.e. total k.e. total energy is conserved 1 0 m distance above floor 8

9 Electrical energy I V motor 10 kg electrical energy = charge passed x voltage 1 C = 1 A s Coulombs charge = current x time Volts 1 J = 1 CV (relationship between electrical and mechanical work) 9

10 Energy output from "ideal" electrical motor: = V x I x t 1 J = 1 V A s Also heat output from an electrical heater (bomb calorimeter experiment) 10

11 work done?? h w = 80 kg x 9.81 ms -2 x 10m = 7850 J Chemical energy - running up to the second floor.. Where does the energy come from? C 6 H 12 O O 2 6 CO H 2 O i.e. from the "heat of combustion" of glucose 11

12 We can also produce heat from work and vice versa heat work BUT Unfortunately, heat work is never 100% efficient! (Second Law of Thermodynamics) 12

13 Take 1 mol of solid glucose at 298 K. Burn to form 6 moles of CO 2 gas and 6 moles of H 2 O liquid C 6 H 12 O 6(s) + 6O 2(g) 6CO 2(g) + 6H 2 O (l) H comb = x10 6 J mol -1 If energy work conversion efficiency is 10%, how much glucose is burnt in climbing the stairs? work = 7850 J heat needed = 7850 x 10 = J (conversion to work only 10% efficient) 13

14 We get 2.808x10 6 J for each mol of glucose (molar enthalpy of combustion), so we need to burn: 7.85x10 4 mol = 2.8x10-2 mol of glucose 2.808x10 6 Since 1 mol glucose (C 6 H 12 O 6 ) 180g g = 5.04g of glucose is burned Turning work into heat Heat of friction (e.g. rubbing hands together to keep warm, mechanical work) 14

15 Heat from electrical work: e.g. electric fire, heater in bomb calorimeter Electrical work done in time t is = I x V x t = 240 V x 1A x 1s i.e. 240 Joules in 1 second 1J s -1 = 1 Watt (W) (unit of power) So heat output (power) from fire is 240 J s -1 = 240 W 15

16 Internal energy and the First Law of Thermodynamics For an isolated system, the change in internal energy is the sum of heat, q, and work, w, changes. U = q + w or du = dq + dw The total energy of an isolated system is constant Principle of Conservation of energy U U+ U 16

17 To change the energy of a system, we either introduce heat into the system 1 mol Ar gas 298 K fixed volume 1 mol Ar gas 398 K internal energy = kinetic energy Energy = ½ mv ½ mv ½ mv

18 = ½ m(v v v ) Number atoms speed V of with Mean speed increased Total kinetic energy increased Maxwell Boltzmann distribution of molecular speeds V do work on the system 18

19 do work (w syst = -w surr ) squeeze Change in kinetic energy = work done in compression Change in internal energy (J) U = Q + W Heat IN (J) Work ON (J) The 1 st Law U = w if Q =0 (i.e. if insulated as shown) 19

20 EXAMPLE The temperature of 1 mol of a liquid is raised by heating it with 750 J of energy. It expands and does the equivalent of 200 J of work. Calculate the change in internal energy. Since heat is passed TO the system, q = J (note PLUS sign). In an expansion, the system does work on the surroundings so that w = J (note the MINUSs sign). U = q + w = (+750) + (-200) = J. 20

21 Chemical Energy energy A.B locked up in bonds x AB x the energy of formation can be defined from the elements for any compound e.g. C (s) + O 2(g) CO 2(g) 21

22 Types of energy change Only two to worry about so far more later internal energy change U - process at constant volume enthalpy change H - process at constant pressure EXAMPLE Energy change for burning acetylene (ethyne) C 2 H 2 + 5/2 O 2 2CO 2 + H 2 O C 2 H 2(g) + 5/2 O 2(g) 2CO 2(g) + H 2 O (l) 3½ moles gas 2 moles gas (Note: the water (liquid) occupies a negligible volume compared with the gas). 22

23 CONSTANT VOLUME The bomb calorimeter U = Q + W constant volume no work done, nothing moves so U = Q 23

24 CONSTANT PRESSURE p = 1 atm 3 ½ moles gas 2 moles gas change = - 1½ moles (note minus sign!) So the piston moves in The surroundings do work on the system. How much work is done by the surroundings on the system? x pv = n RT p = force/area Nm -2 force = pressure x area = pa work done = force distance = pax Ideal = p V gas law 24

25 p is the pressure when n moles of a gas occupy a volume V at temperature T (measured in Kelvin). R is the gas constant - (in S.I. units) J K -1 mol -1. Since p = nrt V = (nrt)/p V start = 7/2 (RT)/p and V final = 2 (RT)/p V = -3/2 (RT)/p P V = -3/2 (RT) = work due to change in volume U = Q - p V change in internal energy (1 st Law) 25

26 U = H - p V H = U + p V in this case P V = -3/2(RT) so, H < U enthalpy change change in internal energy H = U + nrt where n is the change in number of moles of gas 26

27 The bomb calorimeter burn material in excess O 2 allow heat to pass to bath & reach constant T measure temp. rise T 1 calibrate by passing IVt T 2 electrical energy = IVt T 1 = chemical energy = U T 2 = electrical energy IVt Calibration experiment : determines the heat capacity of the calorimeter. 27

28 Heat capacity heat required to change temperature by 1 K so if IVt T 2 C calorimeter = IVt T 2 So heat capacity has units of JK -1 J K In Chemistry usually JK -1 mol -1 of substance. EXAMPLE Complete combustion of 2.00 g of sucrose in a bomb calorimeter gave a T 1 of K. Calibration run at 10V, 2A, 2min gave T 2 of K. 28

29 Heat capacity = 10 x 2 x 120 = 1140 JK U = heat released = 1140 JK -1 x K = kj No of moles of sucrose (C 12 H 22 O 12 ) = x 10-3 mol U = kj mol -1 (Note the negative sign, heat is released, i.e. exothermic). U U = Q + W heat out negative = 0 29

30 "system" C 12 H 22 O 11(s) + 12 O 2(g) 12 CO 2(g) + 11H 2 O (l) 12 moles gas 12 moles gas i.e. no change in the No of moles of gas in this case, H = U Simple Constant pressure Calorimeters "coffee cup" "thermos" or "dewar flask" silvered 30

31 State functions The internal energy U is an example of a state function. Value depends only on the present state of the system. No dependence on how system achieved its present state Change in value depends only on the final and initial conditions. U = U(final) U(initial) The enthalpy H = U + pv is also a state function H = H(final) H(initial) = U + (pv) H is useful for reactions at constant pressure 31

32 State functions Changes in state functions U - internal energy U o H enthalpy H o standard conditions Changes in state functions are independent of the number of steps taken, or the route taken, to reach an end point. 32

33 Hess's Law "The total enthalpy change for a reaction is independent of the path by which the reaction occurs". Initial state H = H final - H initial Final state Enthalpy changes - Hess Law The total enthalpy change for a reaction is independent of the path by which the reaction occurs. 33

34 Hess Law allows us to treat the chemical equations and the enthalpy changes in the same way as algebraic equations. EXAMPLE (1) CO (g) + ½ O 2(g) CO 2(g) H 1 = kj mol -1 (2) C (graph.) + ½ O 2(g) CO (g) H 2 = kj mol -1 (3) C (graph.) + O 2(g) CO 2(g) H 3 = kj mol -1 (3) - (2) {C (graph.) + O 2(g) } - {C (graph.) + ½ O 2(g) } {CO 2(g) } - {CO (g) } CO (g) + ½ O 2(g) CO 2(g) H = (-393.5) - ( ) = kj mol -1 34

35 Graphical diagram of the energy levels. C (gra p h ) + O 2 H 2, C O + ½ O 2 EXAMPLE H H Calculate the H ethene to give ethane. C O 2 for the hydrogenation of 35

36 Information provided enthalpies of formation Ethene: 2 C (graph) + 2 H 2(g) = C 2 H 4(g) H= kj mol -1 (A) Ethane: 2 C (graph) + 3 H 2(g) = C 2 H 6(g) H= kj mol -1 (B) The reaction of interest is C 2 H 4(g) + H 2(g) = C 2 H 6(g) H=? Subtract reaction A from reaction B {2 C (graph) + 3 H 2(g) } - {2 C (graph) + 2 H 2(g) } = {C 2 H 6(g) - C 2 H 4(g) } H 2(g) = C 2 H 6(g) - C 2 H 4(g) Add C 2 H 4 (g) to both sides 36

37 C 2 H 4(g) + H 2(g) = C 2 H 6(g) i.e. H? = H (B) - H (A) H? = (+52.26) kj mol -1 = kj mol -1 standard conditions Enthalpy changes in chemical reactions Standard temperature: 25 C, K. Standard pressure: 1 bar, Pa Standard state: pure component at 1 bar pressure 37

38 standard enthalpy of reaction H 298 enthalpy change at 1 bar at 25 C all components in their standard states Thermochemical equations N 2(g) + 3 H 2(g) 2 NH 3(g) H 298 = kj mol -1 Translated when 1 mol of nitrogen reacts with three moles hydrogen to form 2 moles of ammonia at 25 C and 1 bar pressure, kj is evolved NB: the enthalpy of reaction is kj mol -1 is ambiguous. ½ N 2(g) + 1½ H 2(g) NH 3(g) H 298 = kj mol -1 38

39 Standard enthalpy of formation H f is the enthalpy change when 1 mole of a compound is formed under standard conditions from its constituent elements in their standard states. Values are quoted at the standard temperature of 25 C or K. Tables of H f,298 are available in text books. For elements in their standard states, H f,298 = 0 1/2N 2(g) + 3/2H 2(g) NH 3(g) H 298 = H f,298 (NH 3 ) = kj mol -1 The standard state is the one with lowest energy 39

40 EXAMPLE Combustion of graphite C (s) + O 2(g) kj mol CO (g) + O2(g) 2 kj mol kj mol -1 CO 2(g) Hess's Law Sum of H values must = 0 H 1 B A C H 4 D H 2 H 3 40

41 EXAMPLE H for C 2 H 4(g) + H 2(g) = C 2 H 6(g) using standard heats of formation of compounds from the elements. C 2 H 4(g) + H 2(g) H (formation) -( ) kj mol -1 = kj mol -1 2C (s) + 3H 2(g) H (formation) kj mol -1 C 2 H 6(g) 41

42 EXAMPLE Balancing an equation using Hess's Law. Find H for C 3 H 6(g) + 9/2 O 2(g) = 3 CO 2(g) + 3 H 2 O (l) Given H C 3 H 6(g) + H 2(g) = C 3 H 8(g) -124 kj C 3 H 8(g) + 5 O 2(g ) = 3CO 2(g) + 4 H 2 O (l) kj H C 3 H 6 + H 2 = C 3 H kj C 3 H O 2 = 3 CO H 2 O kj H 2 O (l) = H 2(g) + ½ O 2(g) +286 kj Adding: C 3 H 6(g) O2(g) = 3CO 2(g) + 3 H 2 O (l) kj heat of combustion of C 3 H 6 is kj mol -1 Types of enthalpy changes to be found in tables 42

43 Enthalpies of formation H (formation) Enthalpies of combustion H (combustion) bond enthalpies - get an estimate of enthalpy changes by looking at: bonds broken (energy in) bonds formed (energy out) Example CH 4 CH 3 + H 432 kj mol -1 CH 3 + H CH 2 + H 469 kj mol -1 CH 2 + H CH + H 422 kj mol -1 CH + H C + H 339 kj mol -1 "bond enthalpy for C-H bond" 416 kj mol -1 average 43

44 Applications of enthalpies of formation e.g. 3 C 2 H 2(g) C 6 H 6(g) o [ ] g Ha Hb 3 C2 H2( g) 6 C (graph) + 3 H2( ) C6H 6( g) o H 298 (reaction) = H a + H b. H 298 (reaction) = (3 x ) = kj mol -1 general way of writing the result H 298 (Rxn) = sum H f, 298 (Prods) sum H f, 298 (Reacts) 44

45 EXAMPLE Calculate the enthalpy of reaction at 25 C for 3 Fe 2 O 3 (s) + 2 NH 3 (g) 6 FeO (s) + 3 H 2 O (l) + N 2 (g) H = [6 H f,298 (FeO (s) ) + 3 H f,298 (H 2 O (l) ) + H f,298 (N 2 (g) )] - [3 H f,298 (Fe 2 O 3 (s) ) + 2 H f,298 (NH 3 (g) ) ] = [6 (-266.3) + 3 (-285.8) + 0 ] - [3 (-824.2) + 2 (-46.1) ] = [ ] - [ ] = kj mol -1 45

The Nature of Energy. Chapter Six: Kinetic vs. Potential Energy. Energy and Work. Temperature vs. Heat

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