Thermodynamics The First Law Work, Heat, Energy

Transcription

1 The Basic Concepts: the system Thermodynamics The First Law Work, Heat, Energy System Matter Energy System Matter Energy Thermodynamics is the study of the transformations of energy. Surroundings (a) open System Matter Surroundings (b) closed Oxtoby, Chapter 10 ( ) Energy Surroundings (c) isolated The Basic Concepts: state and path fns A state function, X, is a property that depends only on the current state of the system, and not on how it was prepared. Changes in a state function depend only on the start and end points of an experiment DX = X final X initial e.g. the duration of this lecture depends only when I start and when I finish A path function, Y, is a property that depends on the history of the system. e.g. the boredom/interest factor for this lecture depends on a lot more than just my first and last sentence The Conservation of Energy: Energy can neither be created not destroyed (experimental observation) fl Energy is a state function The principle of conservation of energy can be used to assess the energy changes that accompany physical and chemical processes. ICE SOLID + heat + heat Melting Liquid water Liquid Work, Heat and Energy Work, w, is done when an object is moved against an opposing force C + O 2 CO 2 + heat Work is a path function (how strong is gravity?) Expanding gas pushes out piston against (e.g, atmospheric) pressure 1

2 Work, Heat and Energy The Energy, E of a system is its capacity to do work. Increasing its capacity to work When spring unwinds, work will be done When piston moves out against higher pressure, work will be done Heat, q When the energy of a system changes as a result of a temperature difference between it and its surroundings, it is said that energy has been transferred as Heat, q Processes that release energy as heat are exothermic; A A + heat B + heat Processes that absorb energy as heat are endothermic; B Combustion reactions Melting, Vaporization of Water The First Law In thermodynamics, the total energy of a system is called the internal energy, E. Experiments measure the change in energy between start and finish: E = E f - E i The change in internal energy E is the sum of work done on a system and the energy transferred as heat to a system according to; E = w + q Heat must be a path function: the system can gain/lose energy either as heat or as work. SI unit for work, heat and energy is the Joule (or J/mol, kj/mol, ) Hence, for an isolated system: System Matter Energy Surroundings and a closed system: System Matter Surroundings Energy E universe = 0 E = w + q = = 0 the change in internal energy of a closed system is equal to the energy that passed through its boundary either as heat or as work. Work, w Work and pressure In general: w = F. d work distance Force along path More specifically, for thermodynamics: w = - F. d Pressure - Volume Work: Area A External pressure, p h w = - F. ext d w = - F ext h We focus on the force the system has to push against w = - p ext A h Work is done on system by the surroundings w = - p ext V 2

3 Work, and pressure w = - p ext V For p ext = 0 Expansion V > 0 Compression V < 0 +/-? The Sign Convention Work done by the system on its surroundings is lost by the system (a negative w). Work done on the system by the surrounding increases the internal energy (positive w) energy is stored in the system w < 0, hence system performs work on surroundings w > 0, hence surroundings perform work on system A positive heat flow (positive q) corresponds to heat gained by the system. Hence, a heat loss is to be understood as a negative gain (negative q). Thermal Equilibrium is established when two bodies have the same temperature. Example: Calculate the work done when 25g of iron reacts with Hydrochloric Acid in (a) a closed cell, (b) in a open beaker at 25 o C? (a) V = 0 w = 0 (b) Fe s + 2 HCl (aq) FeCl 2(aq) + H 2(g) w = - p ext V V = V f -V i =V gas pv = nrt n= m/m 25g mol w = - p ext V = - p ext V gas = - n R T g = 0.45 mol w = mol x J x 298 K = kj K mol Recall: & E = w + q w = - p ext V If the system is kept at constant volume: E = q v We measure E usually using Calorimetry (the measurement of amounts of heat flowing into or out of a system and the accompanying temperature changes) Calorimetry Adiabatic Bomb Calorimeter Adiabatic means no heat is transferred from calorimeter to surroundings The change in temperature, T, of the calorimeter is proportional to the heat that the reaction releases or absorbs. q = C cal T Calorimeter Constant Calibrated using a process of known energy output (eg, burning of a substance of known mass) or from an electrical current, I, of known potential, V q = V I t Examples (a) When 100 g of Napthalene was burned in a bomb calorimeter releasing heat of 3.91 kj, the temperature rose by 3.01 K. What is the calorimeter constant? q = C cal T 3.91 kj = C cal µ 3.01 K C cal = 1.30 kj / K (b) 10.0 A from a 12 V battery were passed through the heater of a calorimeter for 150 s. The temperature rose by 5 K. What is the calorimeter constant? q = (10.0 A) µ (12 V) µ (150 s)=1.8 µ 10 4 A V s = 18 kj 18 kj = C cal µ 5 K C cal = 3.6 kj / K 3

4 Heat Capacity Heat Capacity q Generally: C = Units: J K T -1, cal K -1 1 cal = J Molar Heat Capacity (J K -1 mol -1 or cal K -1 mol -1 ) q C c = = Heat capacity of a sample divided by the T n n chemical amount of substance, n The heat capacity C of a substance is the heat required to change its temperature by one Kelvin, and has units of energy per Kelvin. The heat capacity is an extensive variable: the quantity is proportional to the amount of matter present. Specific Heat Capacity (in J K -1 kg -1 or cal K -1 kg -1 ) q C c s = = Heat capacity of a sample divided by the T m m mass of substance, m c = M c s Heat Capacity measured at constant pressure c p = q p T n C p = q p T C v = C = p molar c n v = measured at constant volume (eg, conduct experiments in closed container) q v T q v T n If reaction contains only solids and/or liquids, c p c v = C v n Example: A new calorimeter is filled with 200g of room temperature water. Adding 1212 J of heat to the contents, raises the interior temperature by 1.42 K. Calculate the calorimeter constant given that the molar heat capacity of water, c p (H 2 O) is J K -1 mol J = q water + q calorimeter = C cal T + c p (H 2 O) n(h 2 O) T = C cal T + c p (H 2 O) m(h 2 O) T M(H 2 O) 1212 J J K -1 mol g 1.42 K C cal = 1.42 K 1.42 K 18 g mol -1 = 16 J K -1 Recall: & E = w + q w = - p ext V Enthalpy If the system is kept at constant volume: E = q v But if it is not kept at constant volume? E = q v Another type of energy, the ENTHALPY, is useful in these circumstances 4

5 The Enthalpy, H For a system kept at constant pressure: H = q p where H = H f - H i The Enthalpy, like the internal Energy, is a state function Example: Relating H and U The internal energy change when mol calcite (ρ c =2.71 g cm -3 ) converts to aragonite (ρ a =2.93 gcm -3 ) is kj. Calculate the difference between the changes in enthalpy and in internal energy at a pressure of 1 atm. [calcite and aragonite are different forms of CaCO 3 ] H = U + pv H = (U+pV)=H f - H i = (U+pV) a -(U+pV) c = U a - U c + (pv) a -(pv) c = U + p V as p = const. The Enthalpy is related to the Energy by: H = U + pv = U + nrt for an ideal gas H- U = p V H- U = J V a = m a ρ a M a = M c = 100 g/mol V a = 17.0 cm 3 m a = m c = 50.0 g V c = 18.5 cm 3 Enthalpy of Reaction - H r The Reaction Enthalpy, H r Example 1) A B In General: H r = H products - H reactants H A H B Enthalpy of Reaction: H r = H B - H A Example 2) 2 C + D E + 3 F 2 H C H D H E 3 H F Enthalpy of Reaction: H r = H E + 3 H F - H D -2 H C Example: H r = Σ ν H products - Σ ν H reactants Sum over all standard molar enthalpies taking into account their stochiometric factors, ν 2 C 2 H O 2 4 CO H 2 O ν= 2 ν= 7 ν= 4 ν= 6 H r = 6H(H 2 O) + 4H(CO 2 ) -7H(O 2 ) -2H(C 2 H 6 ) Standard Enthalpies Enthalpy are normally tabulated for substances in their standard state; these are called the standard enthalpy, H o. Denotes standard condition The standard state of a substance at a specified temperature is its pure form at 1 atm (should now be 1 bar). For dissolved species, the standard state is the concentration of 1M under a pressure of 1 atm at a specified temperature. Standard enthalpies at T = 298 K are often denoted Hʅ Enthalpies of phase change Enthalpy changes also occur when a substance melts/freezes or condenses/evaporates Fusion (melting): Vaporisation (boiling): H 2 O (s) H 2 O (l) H 2 O (l) H 2 O (g) H o fus (273)=+6.0 kj mol H o vap (373)=+40.7 kj mol endothermic! Hence, the standard enthalpy of reaction is H ro = Σ ν H o products - Σ ν H o reactants Sublimation H 2 O (s) (direct conversion from solid to gas) H 2 O (g) H o kj sub (298.15)=+46.7 mol 5

6 Rules for enthalpy changes (1) Enthalpy is a state function fl can derive some rules In general: (1) H o vap (373 K)= kj mol 1 H 2 O (l) vaporisation condensation H 2 O (g) H o cond (373 K)= kj mol 1 H o (A B) = - H o (B A) the enthalpy of a forward process differs from that of the backwards process only in sign. Rules for enthalpy changes (2) A change in enthalpy is independent of the path taken between two states: Sublimation Initial state H 2 O (s) solid solid H o fus H o sub H o vap H o sub = H o fus + H o vap Final state H 2 O (g) gas gas One could also have written Hess s Law H 2 O (s) H 2 O (l) H 2 O (l) H 2 O (g) H o fus = +6.0 kj mol 1 H o vap = kj mol 1 The enthalpy of an overall reaction is the sum of the enthalpies of the individual reactions into which the reaction can be divided. Overall: H 2 O (s) H 2 O (g) H o sub = kj mol 1 = kj mol 1 Because enthalpy is a state function, these rules apply to every type of reaction or change Example:Calculate the enthalpy of combustion of benzene (C 6 H 6 ) from its enthalpy of hydrogenation (-205 kj/mol) to cyclohexane, and the enthalpy of combustion of cyclohexane ( H o (C 6 H 12 ) = 3920 kj/mol)). The enthalpy of the combustion for H 2 is -286 kj/mol. Enthalpy of Combustion of Benzene We want to know: C 6 H O 2 6 CO H 2 O We do know: C 6 H H 2 C 6 H kj/mol C 6 H O 2 6 CO H 2 O kj/mol 3 H 2 O 3H O 2 -(3 µ -286) kj/mol Standard Molar Enthalpies of Formation, H o f The enthalpy of formation is the enthalpy change when a compound is formed from its elements, and those elements are in their most stable form under the prevailing conditions. When the prevailing conditions are the standard state, this is called the standard enthalpy of formation, H o f H 2(g) O 2(g) H 2 O (l) H o f = kj/mol C 6 H O 2 6 CO H 2 O kj/mol 6 C (s, graphite) + 3 H 2(g) C 6 H 6(l) H o f = + 49 kj/mol The standard enthalpies of elements in their reference states are zero at all temperatures (graphite is the reference state of carbon!). 6

7 Bond enthalpies A B (g) A (g) + B (g) H rxn = energy of the A B bond But bond enthalpies are affected by the neighbouring bonds CH 4(g) CH 3(g) + H (g) H = 439 kj mol 1 C 2 H 6(g) C 2 H 5(g) + H (g) H = 410 kj mol 1 CHCl 3(g) CCl 3(g ) + H (g) H = 380 kj mol 1 Applications of bond enthalpies Using average bond enthalpies, estimate the enthalpy change for C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O (g) Step 1: what bonds are in the reactants (draw Lewis structures) C 3 H 8(g) H H H H C C C H fl usually tabulate average bond enthalpies (determined from A B bond enthalpies in many different A B containing molecules) Average bond enthalpies can be used to estimate the enthalpy of a compound: just count the number and type of bonds involved. nb: angles are not 90 o H H H 1 µ (8 C-H bonds), 1 µ (2 C-C bonds) O 2 O O 5 µ (1 O=O double bond) Step 2: what bonds are in the products (draw Lewis structures) 3 CO 2 O C O O H H 4 H 2 O 3 µ (2 µ CO double bonds) 4 µ (2 µ OH single bonds) Step 3: calculate the net breakage / formation of bonds: energy is absorbed in breaking reactant bonds, and released it in making them DH rxn º 8H(C H) + 2H(C C) + 5H(O=O) 6H(C=O) 8H(O H) Finally DH rxn º 8 µ µ µ 497 6µ µ 463) kj mol 1 = 1685 kj mol 1 Value from calorimetry experiments: H co = kj/mol Bond enthalpies appear to give a 30% error! Works better when not all bonds are being broken (1 st, 2 nd, 3 rd and 4 th C H bonds in methane have very different strengths) better to use thermochemical groups bonded to at least two other atoms rather than individual bonds Spontaneous Changes: Entropy Why do some processes happen spontaneously? Why does a hot body get cooler (rather than hotter) when surrounded by a cooler medium? Why does a gas expand into all available volume of a container rather than contract? The driving force for spontaneous change (change that happens without intervention doing work or heating) is described in the second law of Thermodynamics Chapter 11 7

8 No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. Kinetic energy converted into thermal motion Energy is not accumulated in ball and thermal motion is not directional Entropy, S, and the second law These spontaneous changes happen because they increase the randomness with which energy is spread through an isolated system The Entropy, S, a thermodynamic state function, is a measure of molecular disorder, or freedom of movement molecules have, and helps us to define the direction of spontaneous change The Entropy of an isolated system increases in the course of a spontaneous change S system + S surroundings = S total > 0 Hence, in a spontaneous process: S universe > 0 Entropy and Equilibrium Nothing changes when a system is at equilibrium, including the entropy of the system So, for any process S system + S surroundings = S universe r 0 But, equilibrium is dynamic at a microscopic level reactants Ý products This introduces the idea that changes can be reversible, i.e. a change can be made, and then exactly undone ( reversed ), so S universe = 0 for a reversible process In practice, reversible changes are an idealised limit in which changes happen infinitely slowly via a series of imperceptible shifts in the equilibrium Entropy and Heat Entropy measures dispersal of energy in a system Heat changes kinetic energy of molecules, i.e. disperses energy by increasing the velocities of all the molecules fl heat and entropy are related? For a reversible process DS = q rev / T For an irreversible process DS > q rev / T = DH / T (at constant pressure) Entropy and Disorder The Entropy of the System But Water freezes (spontaneously) on a cold night! Entropy increases?? Solid Liquid Gas Entropy of system increases Liquid (water molecules able to move) Solid (molecules fixed in crystal) 8

9 Spontaneous Freezing Water freezing is an exothermic process Entropy of Phase Changes System at transition temperature H 2 O (l) H 2 O (s) H = kj/mol So heat is put into the surroundings H system = + H surroundings = q p,surroundings b S surroundings decreases increases S system + S surroundings = S universe > 0 Entropy is not the final determinant of spontaneity see tomorrow Solid similarly, for vaporization S Sys fus = H fus T Fus S Sys vap = H vap T vap Liquid Entropy of Reaction As entropy is a state function, the entropy of reaction is defined (by analogy to Hess law for the enthalpy) as the difference between the molar entropies for the pure, separated products and the pure, separated reactants: Trends in entropy Increasing Temperature S r = ΣνS o products - ΣνSo reactants Where o as before indicates that all substances are in their standard states at the specified temperature. translation rotation vibration More modes to distribute energy become available ENTROPY INCREASES Entropy and Disorder: the microscopic view Increase in Entropy Example: mixing of two gases (pure O 2 and pure N 2 ) To quantify S, see microstate considerations below The Statistical Entropy Let s say you are tossing a coin with your mates to check who will have to go to the next lecture to take notes (if handouts are not provided). You decide on head and your mate on tails. As you toss the coin for the first time, the likelihood that of head:tail is 50:50, or 1/2:1/2. Tail Head # of combinations 1 1 The chance that you win heads twice is ¼, as it is to win tails twice; the mixed result has a probability of ½ # Tail/Tail Tail/Head or Head/Tail Head

10 # For three throw T/T/T TTH,THT,HHT HHT,HHT,THH H/H/H Similarly for particles in boxes Imagine you have four balls and can put them into two containers (T or H) then you can distribute them as follows;. and for four throws T/T/T/T TTTH,TTHT, THTT, HTTT Most likely outcome TTHH, THHT, THTH, HHTT, HTHT, HTTH HHHT,HHTH, HTHH, THHH H/H/H/H T/T/T/T # TTTH,TTHT, THTT, HTTT TTHH,THHT,HTTH, THTH, HTHT,HTTH HHHT,HHTH, HTHH, THHH H/H/H/H # Even split most favored T H T H T H T H T H The Boltzmann Formula S = k ln W Third Law of Thermodynamics The entropy change for any equilibrium between pure substance becomes zero as the temperature approaches absolute zero The Entropy of any crystalline substance approaches zero as T 0 K. W = weight of most probable configuration of the system! k = Boltzmann constant = R/N 0 = x J K -1 Walther Nernst Nobel Laureate in chemistry in recognition of his work in thermochemistry The third law S = k ln W As T 0 K, every atoms is located at the perfect crystal location, so W = 1 But Removing every defect is difficult (if not impossible) QM ensures atoms still have zero-point energy, and so some entropy persists (see last week) fl can t actually achieve absolute zero Temperatures < K have been achieved Let s remember when processes are spontaneous S uni > 0 S uni = 0 S uni < 0 Spontaneous Equilibrium; also true for idealised changes that are so slow they never disturb equilibrium (reversible) Does not happen (reverse reaction will be spontaneous) 10

11 The Gibbs Free Energy, G S uni = S system + S surroundings Gibbs Free Energy, G Requires knowledge of both system and surroundings At constant pressure and temperature, and for reversible changes: T S surroundings = q p,rev = H system Define a new form of the energy function G = H TS Called the Gibbs free energy. G is a state function The Gibbs Free Energy: criterion for spontaneity Criterion for Spontaneity At constant T: G sys = (H sys - T sys S sys ) = H sys - T S sys G = H - T S S uni > 0 S uni = 0 Spontaneous Equilibrium G sys < 0 G sys = 0 But also: H sys T = - S surr S uni < 0 Not spontaneous (reverse reaction is spontaneous) G sys > 0 Resulting in: G sys = - T S surr -T S sys = - T ( S surr + S sys ) and as S uni > 0 = - T ( S uni ) < 0 for spontaneous pr. N.B. left hand describes the whole universe, but right hand is just the system. We can drive the system in the wrong direction (hence can have reactions where DG is positive, but cannot force the universe to show negative entropy Examples Remember G sys < 0 for spontaneous processes G = H -T S G = H -T S G < 0 for spontaneous 2) An endothermic process where entropy decreases 1) An exothermic process which also increase entropy H = + ve (>0) H = - ve (<0) S = + ve ; -T S = - ve (<0) G = -ve (<0) spontaneous S = - ve ; -T S = + ve (>0) G = +ve (>0) NOT spontaneous (reverse reaction would be spontaneous as G is a state function) 11

12 G = H -T S G < 0 for spontaneous 3) An exothermic process in which entropy decreases (freeze water) G = H -T S G < 0 for spontaneous 4) An endothermic process in which entropy increases (melting ice) H = - ve (<0) S = - ve ; -T S = + ve (>0) Depends on relative magnitudes of H and -T S H = + ve (>0) S = + ve ; -T S = - ve (<0) Depends on relative magnitudes of H and -T S H < T S H = T S H > T S H < T S H = T S H > T S G > 0 G = 0 G < 0 G < 0 G = 0 G > 0 Not spontaneous Equilibrium Spontaneous Spontaneous Equilibrium Not spontaneous For fixed H and S, spontaneity depends on T For fixed H and S, depends on T Driving forces in chemistry Two driving forces underpin Chemistry: Systems tend to a state of minimum enthalpy Systems tend to a state of maximum entropy The Gibbs free energy expresses the balance between these two driving forces DG = DH T DS (b 0?) Gibbs Energy for Phase Transitions P = 1atm H 2 O (s) H 2 O (l) H o fus (273.15)= kj mol H 2 O (l) H 2 O (s) H o kj frze (273.15)= 6.0 mol G = H - T S S frze, = H frze, T frze= G = H T frze= Equilibrium S = 0 If H and S are independent of temperature (a good, but not perfect, approximation), then: T < T frze G = H - T S Gibbs Energy for Chemical Reaction As the Gibbs energy is a state function, the (standard) Gibbs energy of reaction is defined (in analogy to Hess law for the enthalpy) as: G < 0 Spontaneous Freezing G ro = Σ ν G o products - Σ ν Go reactants T > T frze G = H - T S G > 0 Spontaneous Melting of Ice Clearly, the spontaneity of freezing (negative S and negative H) depends on a balance of the entropy and enthalpy Where o as before indicates that all substances are in their standard states at the specified temperature. G ro = H ro - T S r o 12

13 Standard Molar Gibbs Energy of Formation Example: By analogy with the definition of the standard molar Enthalpy of formation we define Pt (NH 3 ) 2 I 2, cis Pt (NH 3 ) 2 I 2, trans G fo = H fo -T S f o as the standard molar Gibbs energy of formation when 1 mole of a substance forms in a standard state at a specified temperature from the most stable forms of its constituent elements in standard states at the same temperature. H fo (298.15K) G fo (298.15K) cis kj/mol kj/mol trans kj/mol kj/mol Calculate the standard entropy, S o, of both compounds at K. S fo = H fo - G f o Pt (s) + 2 NH 3,(g) + I 2,(s) Pt(NH 3 ) 2 I 2(s) cis trans S fo = kj/mol K S fo = ( ) kj/mol K = kj/(k mol) = kj/(k mol) T Gibbs Free Energy & Equilibrium Constants aa g + bb g cc g + dd g For the standard reaction all reactants and products in standard states, but in general they are related by the reaction quotient. G r = G ro + RT lnq S fo = S o (Pt(NH 3 ) 2 I 2 ) - S o (Pt (s) ) - 2 S o (NH 3,(g) ) - S o (I 2,(s) ) kj/(k mol) S o (Pt(NH 3 ) 2 I 2 ) cis = 17.9 J/(K mol) S o (Pt(NH 3 ) 2 I 2 ) trans = 21.3 J/(K mol) N.B. both G and Q can be used to determine the direction of change: G 0 Q K Q = (P C ) c (P D ) d (P A ) a (P B ) b Now remember, at equilibrium, G r = 0 and Q = K Spontaneous to products G<0 Q<K G reactants G r = G ro + RT lnq G ro = - RT lnk G r Equilibrium G=0 Q=K = - RT lnk + RT lnq = RT ln Q/K G>0 Q>K products Not spontaneous to reactants Temperature and the Equilibrium Constant Assume S and H ro don t change with temperature! RT1lnK1 = H T1 S RT2lnK2 = H T2 S K 1 H T1 S H T2 S ln = lnk1 lnk2 = K2 RT1 RT2 The van t Hoff Equation ln K 2 K 1 = H r o R ( ) 1 1 T 1 T Nobel Prize in Chemistry,

14 Similarly, for liquid-vapour equilibrium The Clausius-Clapeyron Equation ln P 2 P 1 = H o vap R 1 1 T 1 ( ) T 2 assuming S & H ro independent of temperature This expression is particularly useful when you want to determine the boiling points of substances at different pressures. Example: Water boils at K at an atmospheric pressure of 1.0 atm. Calculate the boiling temperature of water on top of Mt Everest (assuming a pressure of about 0.36 atm). The enthalpy of vaporization of water can be assumed to be 40.7 kj/mol. ln P 2 P J ln = H o vap R 0.36 atm 1 atm K mol J mol T 1 ( ) T 2 P 1 =1.0 atm T 1 = K P 2 = 0.36atm H vapo = 40.7 kj/mol = = K T K 14