Thermodynamics part 2

Thermodynamics part 2 The thermodynamic standard state of a substance is its most stable pure form under standard pressure (one atmosphere) and at some specific temperature (25 C or 298 K unless otherwise specified). THERMODYNAMIC STANDARD STATE 1 2 Examples of solutes in their standard states at 25 C are: hydrogen, gaseous diatomic molecules, H 2(g) ; mercury, a silver-coloured liquid metal, Hg (l) ; sodium, a silvery white solid metal, Na (s) ; carbon, a grayish black solid called graphite, C (graphite). Sodium chloride, a white crystals, NaCl (s) THERMODYNAMIC STANDARD STATE 1. For a pure substance in the liquid or solid phase, the standard state is the pure liquid or solid. 2. For a gas, the standard state is the gas at a pressure of one atmosphere; 3. For a mixture of gases, its partial pressure must be one atmosphere. 4. For a substance in solution, the standard state refers to one-molar concentration. TSS - RULES 3 4 1

Q.1.Could you listen other elements in their standard state (conditions): Q.2. The values associated with the term, "standard reference conditions or shortly standard condition in a thermochemical meaning refer to: a. temperature: 0.00 K; pressure: 1.00 atm b. temperature: 0.00 o C; pressure: 1.00 atm c. temperature: 273.15 K; pressure: 1.00 Pa d. temperature: 298.15 K; pressure: 1.00 atm e. temperature: 298.15 K; pressure: 1.00 Pa f. temperature: 25 o C; pressure: 1 atm Quiz 5 Choose the answer/s which present TSS conditions: Answer Quiz Solute/ mixture Condition a) NaOH (aq) C = 0.1 M b) Mixture of O 2 and N 2 P O2 = 0.7atm; P N2 = 0.3atm c) He and Ar P O2 = 0.8 atm; P N2 = 1.3atm d) HCl aq C = 1 M 6 For ease of comparison and tabulation the thermodynamic or thermochemical changes performed at standard states are known as standard changes. To indicate a change at standard pressure, we add a superscript zero in H 0. The standard enthalpy change, H 0 rxn, for reaction: reactants products refers to the H when the specified number of moles of reactants, all at their standard states, are converted completely to the specified number of moles of products, all at their standard states. STANDARD CHANGE 7 Standard enthalpy change, H 0 rxn, for reaction 8 2

When we describe a process as taking place at constant T and P, we mean that the initial and final conditions are the same. Because we are dealing with changes in state functions, the net change is the same as the change we would have obtained hypothetically with T and P actually held constant. We allow a reaction to take place, with changes in temperature or pressure if necessary. When the reaction is complete, we return the products to the same conditions of temperature and pressure that we started with, keeping track of energy or enthalpy changes as we do so. At constant T and P T, P = T,P 9 10 It is known that it is impossible to determine the total enthalpy content of a substance on an absolute scale. We need to describe only changes in this state function, however, so we can define an arbitrary scale as follows. The standard molar enthalpy of formation, H f0, of a substance is the enthalpy change for the reaction in which one mole of the substance in a specified state is formed from its elements in their standard states. By convention, the H f0 value for any element in its standard state is zero. STANDARD MOLAR ENTHALPY OF FORMATION, H 0 f 11 12 3

The standard molar enthalpy of formation of ethanol, C 2 H 5 OH (l), is - 277.7 kj/mol. Write the thermochemical equation for the reaction for which H f0 rxn - 277.7 kj/mol rxn. The definition of H f0 of a substance refers to a reaction in which one mole of the substance is formed. We put one mole of C 2 H 5 OH (l) on the right side of the chemical equation and put the appropriate elements in their standard states on the left. We balance the equation without changing the coefficient of the product, even if we must use fractional coefficients on the left. Interpretation of H f 0 Plan 13 14 2 C (graphite) + 2H 2(g) + ½ O 2(g) C 2 H 5 OH (l) ΔH f0 = -277.7 kj/mol rxn Q.3. The standard molar enthalpy of formation of aluminium oxide, Al 2 O 3 (s), is -1675.7 kj/mol. Write the thermochemical equation for the reaction for which H f0 rxn -1675.7 kj/mol rxn. 2Al(s) + 3/2 O 2 (g) Al 2 O 3 (s) ΔH f0 = -1675.7 kj/ mol Solution Example to solve 15 16 4

Q.4. The standard molar enthalpy of formation of hydrogen iodide, HI(g), is 26.48 kj/mol. Write the thermochemical equation for the reaction for which H f 0 rxn 26.48 kj/mol rxn. In 1840, G. H. Hess (1802 1850) published his law of heat summation, which he derived on the basis of numerous thermochemical observations. The enthalpy change for a reaction is the same whether it occurs by one step or by any series of steps. Quiz HESS S LAW 17 18 As an analogy, consider traveling from Kansas City (elevation 269.64m (884 ft) above sea level) to Denver (elevation 1609.34m (5280 ft )). The change in elevation is (1609.34 269.64) m = 1339.7 m, regardless of the route taken. The value of ΔH 0 for any reaction that can be written in steps equals the sum of the values of ΔH 0 of each of the individual steps INTERPRETATION Hess law 19 20 5

Hess s law lets us calculate enthalpy changes for reactions for which the changes could be measured only with difficulty, if at all. A schematic representation of Hess s Law 21 Hess s Law application 22 Consider the following reaction. C (graphite) + 1/2 O 2(g) CO (g) H 0 rxn? The enthalpy change for this reaction cannot be measured directly. Even though CO (g) is the predominant product of the reaction of graphite with a limited amount of O 2(g), some CO 2(g) is always produced as well. EXAMPLE C (graphite) + O 2(g) CO 2(g) ΔH 0 rxn = -393.5 kj/mol rxn (1) CO (g) + ½ O 2(g) CO 2(g) ΔH 0 rxn = -283.0 kj/mol rxn (2) 25 26 6

Revise equation (2) to give (-2) thus we obtain the CO (g) from carbon dioxide, when the equation is revised, the sign of ΔH 0 is changed because the reversed of exothermic reaction is endothermic reaction, then add to eq.(1). C (graphite) + O 2(g) CO 2(g) ΔH 0 rxn = -393.5 kj/mol rxn (1) CO 2(g) CO (g) + ½ O 2(g) ΔH 0 rxn = -(-283.0 kj/mol rxn) (2) C (graphite) + ½ O 2(g) CO (g) ΔH rxn0 = - 110.5 kj/mol rxn Plan 27 EXAMPLE 28 Calculate heat of reaction at 298K. C 2 H 4(g) + H 2 O (l) C 2 H 5 OH (l) use the following thermochemical equations ΔH 0 C 2 H 5 OH (l) + 3O 2(g) 2CO 2(g) + 3H 2 O (l) -1367kJ/mol (1) C 2 H 4(g) + 3O 2(g) 2CO 2(g) + 2H 2 O (l) - 1411kJ/mol (2) INTERPRETATION Example 29 30 7

Revise equation (1) to give (-1) when the equation is revised, the sign of ΔH 0 is changed because the reversed of exothermic reaction is endothermic reaction, then add to eq.(2). ΔH 0 2CO 2(g) + 3H 2 O (l) C 2 H 5 OH (l) + 3O 2(g) +1367kJ/mol (-1) C 2 H 4(g) + 3O 2(g) 2CO 2(g) + 2H 2 O (l) - 1411kJ/mol (2) C 2 H 4(g) + H 2 O (l) C 2 H 5 OH (l) ΔH 0 = -44 kj/mol rxn Plan Solution 31 32 What is the value for ΔH for the following reaction? CS 2(l) + 3O 2(g) CO 2(g) + 2 SO 2(g) Given: C (s) + O 2(g) CO 2(g) ; ΔH f = -393.5 kj/mol S (s) + O 2(g) SO 2(g) ; ΔH f = -296.8 kj/mol C (s) + 2 S (s) CS 2(l) ; ΔH f = 87.9 kj/mol Calculate the standard enthalpy of combustion, ΔH f0, for burning β-dglucose, C 6 H 12 O 6. The required values of ΔH f 0 are presented above: β-d-glucose = -1268 kj/mol CO 2(g) = -393 kj/mol H 2 O (l) = -285,5 kj/mol O 2(g) = 0 kj/mol Quiz 33 34 8

The standard enthalpy change of a reaction is equal to the sum of the standard molar enthalpies of formation of the products, each multiplied by its coefficient,n, in the balanced equation, minus the corresponding sum of the standard molar enthalpies of formation of the reactants. Hess Law and standard enthalpies of formation. The equation for the decomposition of NaHCO 3 is following 2NaHCO 3(s) Na 2 CO 3(s) + H 2 O (l) +CO 2(g) Use the standard heats of formation to calculate ΔH 0 for this reaction in kj. Standard molar enthalpies of formation of NaHCO 3(s), Na 2 CO 3(s), H 2 O (l) and CO 2(g) are as follows: -1131kJ; - 947.7 kj; -285.9 kj and 393.5 kj. Example 35 36 ΔH 0 r = [1 mol Na 2 CO 3(s) X ΔH 0 f Na 2 CO 3(s) + 1 mol H 2 O (l) X ΔH 0 f H 2 O (l) + 1mol CO 2(g) X ΔH 0 f CO 2(g) ] [2 mol NaHCO 3(s) X ΔH 0 f NaHCO 3(s) ] = = [(-1131 kj) + (-285.9 kj) + (-393.5 kj)] - [2x (-947.7 kj)] = +85 kj The internal energy, E, of a specific amount of a substance represents all types of the energy contained within the substance. It includes such forms as: kinetic energies of the molecules; energies of attraction, repulsion among subatomic particles, atoms, ions, or molecules; other forms of energy. CHANGES IN INTERNAL ENERGY, E 37 38 9

The internal energy of a collection of molecules is a state function. The difference between the internal energy of the products and the internal energy of the reactants of a chemical reaction or physical change, E, is given by the equation Δ E = E final - E initial - E products E reactants - q + w CHANGES IN INTERNAL ENERGY, E CHANGES IN INTERNAL ENERGY, E 39 40 Sign conventions for q and w. Entropy, free energy, spontaneous and non-spontaneous reaction 41 42 10

Inside a pile of oily rags or a stack of hay that has not been thoroughly dried, decomposition causes heat to build up. When heat cannot escape, the temperature can become high enough to cause a fire. What was the character of the process? Answer - spontaneous You will learn about the conditions that will produce a spontaneous chemical reaction. What are two characteristics of spontaneous reactions? A spontaneous reaction occurs naturally and favours the formation of products at the specified conditions. Spontaneous reaction 43 Spontaneous reaction 44 Any process which occurs without an outside intervention is spontaneous. When two eggs are dropped they spontaneously break. The reverse reaction (two eggs leaping into your hand with their shells back intact) is not We can conclude that a spontaneous process has a direction. A process that is spontaneous in one direction is not spontaneous in the opposite direction. The direction of a spontaneous process can depend on temperature. Ice turning to water is spontaneous at T > 0 C. Water turning to ice is spontaneous at T < 0 C spontaneous. Direction of spontaneous process 45 46 11

A reversible process is one which can go back and forth between states along the same path. When 1 mol of water is frozen at 1 atm at 0 C to form 1 mol of ice, q = H sol of heat is removed. To reverse the process, q = H melt must be added to the 1 mol of ice at 0 C and 1 atm to form 1 mol of water at 0 C. ΔH melt = ΔH sol = 6.008 kj/mol Reversible process 47 A nonspontaneous reaction is a reaction that does not favour the formation of products at the specified conditions. Photosynthesis is a nonspontaneous reaction that requires an input of energy. Nonspontaneous reaction 48 Q. Non-spontaneous changes are (choose proper answer): rusting of an iron burning a piece of paper melting of ice at room temperature electrolysis photosynthesis 2H 2 O 2H 2 +O 2 The major concern of thermodynamics is predicting whether a particular process can occur under specified conditions to give predominantly products. Quiz 49 50 12

A change for which the collection of products is thermodynamically more stable than the collection of reactants under the given conditions is said to be product favoured, or spontaneous, under those conditions. A change for which the products are thermodynamically less stable than the reactants under the given conditions is described as reactant-favoured, or nonspontaneous, under those conditions. Product favoured, spontaneous Reactant- favoured, nonspontaneous 51 52 The concept of spontaneity has a very specific interpretation in thermodynamics. A spontaneous chemical reaction or physical change is one that can happen without any continuing outside influence. Any spontaneous change has a natural direction, like: the rusting of a piece of iron, the burning of a piece of paper, or the melting of ice at room temperature. We can think of a spontaneous process as one for which products are favoured over reactants at the specified conditions. Spontaneus change 53 54 13

Although a spontaneous reaction might occur rapidly, thermodynamic spontaneity is not related to speed. The fact that a process is spontaneous does not mean that it will occur at an observable rate. It may occur rapidly, at a moderate rate, or even very slowly. The rate at which a spontaneous reaction occurs is addressed by kinetics (see in future lectures). Spontaneity and rate of reaction 55 Many product-favoured reactions are exothermic. The combustion (burning)reactions of hydrocarbons such as methane and octane are all exothermic and highly productfavoured (spontaneous). The enthalpy contents of the products are lower than those of the reactants Enthaply of exothermic reaction 56 Not all exothermic changes are spontaneous and vice versa not all spontaneous changes are exothermic. As an example, let consider the freezing of water, which is an exothermic process (heat is released). This process is spontaneous at temperatures below 0 C, but it certainly is not spontaneous at temperatures above 0 C. Likewise, we can find conditions at which the melting of ice, an endothermic process, is spontaneous. Spontaneity is favoured but not required when heat is released during a chemical reaction or a physical change. 57 58 14

Another factor, related to the disorder of reactants and products, also plays a role in determining spontaneity. For example: the dissolution of ammonium nitrate, NH 4 NO 3, in water is spontaneous, although it is endothermic reaction. The system (consisting of the water, the solid NH 4 NO 3, and the resulting hydrated NH + 4 and NO 3- ions) absorbs heat from the surroundings as the endothermic process occurs. Spontaneity and disorder 59 60 Nevertheless, the process is spontaneous because the system becomes more disordered as the regularly arranged ions of crystalline ammonium nitrate become more randomly distributed hydrated ions in solution. An increase in disorder in the system favours the spontaneity of a reaction. In this particular case, the increase in disorder overrides the effect of endothermicity. Disorder and spontaneity 61 62 15

Two factors affect the spontaneity of any physical or chemical change. Spontaneity is favoured when 1. heat is released during the change (exothermic). 2. the change causes an increase in disorder. The reverse of any spontaneous change is nonspontaneous, because if it did occur, the universe would tend toward a state of greater order. This is contrary to our experience. Factors affect the spontaneity Spontaneous vs. nonspontaneous 63 64 The thermodynamic state function entropy, S, is a measure of the disorder of the system. The greater the disorder of a system, the higher is its entropy. For a given substance, the entropy of the gas is greater than the entropy of the liquid or the solid. Similarly, the entropy of the liquid is greater than that of the solid. Now lets show some examples of systems of increasing entropy. Entropy ENTROPY - SOLID-LIQUID-GAS 65 66 16

Entropy increases when a substance is divided into parts. Entropy tends to increase in chemical reactions in which the total number of product molecules is greater than the total number of reactant molecules. INCREASING ENTROPY ENTROPY IN CHEMICAL REACTION 67 68 Entropy tends to increase when temperature increases. As the temperature increases, the molecules move faster and faster, which increases the disorder. Q. Which ones of the processes below are not accompanied by an increase in entropy? a. Cooking the soup b. Decomposition of SO 3 according to the reaction 2SO 3 2SO 2 + 2O 2 c. Cooling the tea in cup d. Forming a snowball INCREASING ENTROPY Entropy quiz 69 70 17

Q. Which system has the greater entropy: A) granulated sugar or icing sugar B) wall of bricks or several bricks C) desk of pedant person or desk of messy person D) the lecture room before or after the lecture Quiz Based on the relationship between entropy and disorder, indicate which, if any, of the following changes represents an increase in the entropy of the system. a. the freezing of acetic acid b. the sublimation of the moth repellent para-dichlorobenzene (C 6 H 4 Cl 2 ) c. the burning of gasoline Quiz 71 72 In which of the following reactions will the entropy of the system increase? a. CH 3 OH(l) CH 3 OH(g) b. N 2 O 4 (g) 2 NO 2 (g) c. 2 KClO 3 (s) 2 KCl(s) + 3O 2 (g) d. 2SO 2(g) + O 2(g) 2SO 3(g) e. Ca(OH) 2 Ca 2+ aq + 2OH - aq f. 2H 2(g) + O 2(g) 2H 2 O (l) Quiz S o = ns o products - ns o reactants ENTROPY CHANGES IN CHEMICAL REACTIONS 73 74 18

What is the value ΔS rxn for the dissolution of sodium sulphate if S 0 salt = 149.6 J mol -1 K S 0 sodium ion = 59.0 J mol -1 K S 0 sulphate ion = 20.0 J mol -1 K Spontaneous reactions produce substantial amounts of products at equilibrium and release free energy. Free energy is energy that is available to do work. ΔS = 2 59 + 20-149.6 = - 12 J/mol K Free Energy and Spontaneous Reactions 75 76 What part does entropy play in chemical reactions? Entropy is a measure of the disorder of a system. Physical and chemical systems attain the lowest possible energy. The law of disorder states that the natural tendency is for systems to move in the direction of maximum disorder or randomness. Such system possesses the lowest energy. An increase in entropy favours the spontaneous chemical reaction. A decrease favours the nonspontaneous reaction. In spontaneous changes, the universe tends toward a state of greater disorder. ENTROPY ENTROPY vs. Spontaneous 77 78 19

The Second Law of Thermodynamics is based on our experiences. State of entropy of the entire universe (total), as a closed isolated system, will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative. ΔS total > 0. Second law of thermodynamics 79 80 81 82 20

ΔS surroundings = ΔH T [ Enthalpy change [J mol 1 ] ] Temperature [K] Enthalpy and entropy 84 83 ΔS = Q T Reversible process ΔS > Q T Irreversible process, spontaneous Reversible and irreversible processes Entropy change Process Process kind ΔS > 0 Process happened irreversible ΔS = 0 ΔS < 0 Process is at equilibrium and it is impossible to change its direction Process is not happened reversible 21

If protein undergoes denaturation at 80 o C, the heat of this process is equal to 500 kj/mol. Calculate entropy of this process. ΔS rxn = Q T 500 000 J mol = 273.14+80 K = 1.42 J/mol K Change of enthalpy of carbon burning in oxygen is equal to -393,5 kj/mol. Calculate the change of entropy of surroundings for this reaction. The temperature of the reaction is 298 K. ΔS surr = ΔH/T ΔS surr = ( 393.5 kj/mol)/(298 K) ΔS = 1.32kJ/mol K QUIZ 88 Calculate total entropy ΔS total of reaction which undergoes at 25 o C 2 CuO 2Cu (S) + O 2(g) 1. Find ΔS rxn (entropy of system) using molar entropy values Cu, ΔS = 33.15 J/molK CuO, ΔS = 42.63 J/molK O 2, ΔS = 205.14 J/molK S o = ns 0 products - ns 0 reactants= 186.18 J/K 2. Find ΔH rxn 2 CuO 2Cu (S) + O 2(g) Using molar enthalpy values Cu, ΔH = 0 J/mol CuO, ΔH = -157.3 kj/mol O 2, ΔH = 0 J/mol H 0 = nh 0 products- nh 0 reactants ΔH 0 = [(2 x 0) + (1x0)] [- (2 mol x157.3 kj/mol) ] = 314.6 kj 90 22

3. Calculate ΔS surr using ΔH 0 f and T 314600 kj ΔS surr = - = - 1055.7 J/K 298 K ΔS tot = ΔS system + ΔS surroundings S tot = 186.18 J/ K + (- 1055.7 J/K) ΔS tot = -869.52 J/K ΔS tot is negative so reaction is non-spontaneous What two factors determine the spontaneity of a reaction? The size and direction of enthalpy changes and entropy changes together determine whether a reaction is spontaneous that is, whether it favours products and releases free energy. Enthalpy, Entropy, and Free Energy 92 The Gibbs free-energy change (ΔG) is the maximum amount of energy that can be coupled to another process to do useful work. The following data are given for the sublimation of naphthalene [C 10 H 8(s) C 10 H 8(g) ] Is the Gibbs free-energy change positive or negative in a spontaneous process? The numerical value of ΔG is negative in spontaneous processes because the system loses free energy. Gibbs free-energy at 25 C; ΔS 0 = 168.7 J. K -1. mol -1 ; ΔH 0 = 73.6 kj. mol -1. Determine ΔG 0 for this process at this temperature. 93 94 23

Plan 1. Convert o C into K 25 o C + 273 = 298 K 2. Present ΔH in J 73.6 kj. mol -1 = 73600 J mol -1 3. Use formula ΔG = 73600 J mol -1-298 K. 168.7 J. K -1. mol -1 ΔG = 73600 J mol -1 50272.6 J mol -1 ΔG = 23327.4 J mol -1 = 23.33 kj mol -1 95 Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Enthalpies of formation H 0 rxn = -1238 kj Standard molar entropies S 0 rxn = -97.4 J/K or -0.0974 kj/k G 0 rxn = -1238 kj - (298 K)(-0.0974 J/K) = -1209 kj Reaction is product-favored in spite of negative S 0 rxn. Reaction is enthalpy driven Gibbs energy in questions 96 NH 4 NO 3 (s) + heat ---> NH 4 NO 3 (aq) From tables of thermodynamic data we find H 0 rxn = +25.7 kj S 0 rxn = +108.7 J/K or +0.1087 kj/k G 0 rxn = +25.7 kj - (298 K)(+0.1087 kj/k) = -6.7 kj Reaction is product-favored in spite of positive H 0 rxn. Reaction is entropy driven Calculating G 0 rxn 97 : In the Haber process for the manufacture of ammonia N 2 + 3H 2 2 NH 3 At what temperature will the reaction above become spontaneous? If the enthalpy and entropy of this process ΔH = -93 kj/mol; ΔS= -198 J/ mol K The fact that both terms are negative means that the Gibbs free energy equation is balanced and temperature dependent: ΔG = ΔH - TΔS 98 24

ΔG = -93000 - (T x -198) note that the enthalpy is given in kilojoules if ΔG = 0 then the system is at the limit of reaction spontaneity When ΔG = 0 then (T x -198) = -93000 and T = 93000/198 Kelvin therefore the reaction becomes spontaneous when T = 469 K (196 ºC) below this temperature the reaction is spontaneous. ΔG < 0 The reaction is spontaneous in the forward direction (and nonspontaneous in the reverse direction). ΔG > 0 The reaction is nonspontaneous in the forward direction (and spontaneous in the reverse direction). ΔG = 0 The system is at equilibrium. Free Gibbs energy and spontaneity of reaction 99 100 Influence of enthalpy and entropy changes on reaction spontaneity Enthalpy change Decreases (exothermic) Increases (endothermic) Decreases (exothermic) Entropy Increases (more disorder in products than in reactants) Increases Decreases (more disorder in reactants than in products) Spontaneity reaction Yes Only if unfavourable enthalpy change is offset by favourable entropy change Only if unfavourable entropy change is offset by favourable enthalpy change Increases (endothermic) Decreases No 101 102 25

103 104 True and False The equation, H 0 reaction = H0 f (products) - H0 f (reactants) is a statement of the second law of thermodynamics. True/False: ΔG > 0, the process is spontaneous True/False: A nonspontaneous process cannot occur with external intervention. Quizzes 105 1. Free energy from a reaction is the amount of energy that is absorbed by an entropy decrease. equal to the enthalpy change. wasted as heat. available to do work QUIZ 106 26

2. Free energy is always available from reactions that are endothermic. nonspontaneous. at equilibrium. spontaneous. QUIZ 3. Choose the correct words for the spaces: Spontaneous reactions produce and substantial amounts of at equilibrium. free energy, products no free energy, reactants free energy, reactants no free energy, products 107 QUIZ 108 4. Which of the following involves a decrease in entropy? Natural gas burns. A liquid freezes. Dry ice sublimes. Water evaporates. 5. Both the enthalpy and the entropy are combined to calculate the a. reaction time b. free energy c. bound energy 6. A reaction is spontaneous if enthalpy decreases and entropy increases. enthalpy increases and entropy increases. enthalpy decreases and entropy decreases. enthalpy increases and entropy decreases QUIZ QUIZ 109 110 27

6. Choose the correct words for the spaces: Gibbs free-energy change is the amount of energy that can be another process to do useful work. maximum, coupled to maximum, duplicated by spontaneous, coupled to minimum, duplicated by Next lecture is 16 of April 2018 QUIZ 111 112 28