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22/03/2015 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). Thermodynamics part 2 THERMODYNAMIC STANDARD STATE Examples of elements in their standard states at 25°C are: hydrogen, gaseous diatomic molecules, H2(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). THERMODYNAMIC STANDARD STATE 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: 1.000 atm b. temperature: 0.00 oC; pressure: 1.000 atm c. temperature: 273.15 K; pressure: 1.000 Pa d. temperature: 298.15 K; pressure: 1.000 atm e. temperature: 298.15 K; pressure: 1.000 Pa f. temperature: 25oC; pressure: 1 atm Quiz 1 22/03/2015 • • • 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; in a mixture of gases, its partial pressure must be one atmosphere. 3. For a substance in solution, the standard state refers to one-molar concentration. TSS - RULES For ease of comparison and tabulation, we often refer to thermochemical or thermodynamic changes “at standard states” or, more simply, to a standard change. To indicate a change at standard pressure, we add a superscript zero; e.g. ∆H0 . STANDARD CHANGE Choose the answer/s which present TSS conditions: Answer Solute/ mixture Condition a) NaOH(aq) C = 0.1 M/L b) Mixture of O2 and N2 Po2 = 0.7atm; PN2 = 0.3atm c) He and Ar Po2 = 0.8 atm; PN2 = 1.3atm d) HClaq C = 1 M/L Quiz The standard enthalpy change, ∆H0 rxn, for reaction: reactants products refers to the H when the specified number of moles of reactants, all at standard states, are converted completely to the specified number of moles of products, all at standard states. Standard enthalpy change, ∆H0 rxn, for reaction 2 22/03/2015 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. T, P = T,P It is not possible 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. STANDARD MOLAR ENTHALPIES OF FORMATION, ∆H 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. At constant T and P The standard molar enthalpy of formation, ∆Hf0 , 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 ∆Hf0 value for any element in its standard state is zero. Standard molar enthalpy of formation 3 22/03/2015 The standard molar enthalpy of formation of ethanol, C2H5OH(l), is - 277.7 kJ/mol. Write the thermochemical equation for the reaction for which Hf0 rxn - 277.7 kJ/mol rxn. • • • Interpretation of Hf0 2 C(graphite) + 2H2(g) + ½ O2(g) → C2H5OH(l) ΔHf0 = -277.7 kJ/mol rxn Solution The definition of Hf0 of a substance refers to a reaction in which one mole of the substance is formed. We put one mole of C2H5OH(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. Plan Q.3. The standard molar enthalpy of formation of aluminium oxide, Al2O3(s), is -1675.7 kJ/mol. Write the thermochemical equation for the reaction for which Hf0 rxn -1675.7 kJ/mol rxn. Example to solve 4 22/03/2015 Q. 3. 2Al(s) + 3/2 O2(g) → Al2O3(s) ΔHf0 = -1675.7 kJ/ mol Solution 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 Hf0 rxn 26.48 kJ/mol rxn. Quiz 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. Q.4. ½ H2(g) + ½ I2(g) → HI(g) ΔHf0 = 26.48 kJ/mol Solution HESS’S LAW 5 22/03/2015 As an analogy, consider traveling from Kansas City (elevation 884 ft above sea level) to Denver (elevation 5280 ft). The change in elevation is (5280 - 884) ft = 4396 ft, regardless of the route taken. INTERPRETATION Hess’s law lets us calculate enthalpy changes for reactions for which the changes could be measured only with difficulty, if at all. In general terms, Hess’s Law of heat summation may be represented as: (next slide) Hess’s Law 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 equlation, minus the corresponding sum of the standard molar enthalpies of formation of the reactants. Useful form of Hess’s Law. A schematic representation of Hess’s Law 6 22/03/2015 C Consider the following reaction. (graphite) ∆H0 + 1/2 O2(g) CO(g) 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 O2(g), some CO2(g) is always produced as well. C(graphite) + O2(g) → CO2(g) ΔH0rxn = -393.5 kJ/mol rxn (1) CO(g) + ½ O2(g) → CO2(g) ΔH0rxn = -283.0 kJ/mol rxn (2) EXAMPLE Revise equation (2) to give (-2) thus we obtain the CO(g) from carbon dioxide, when the equation is revised, the sign of ΔH0 is changed because the reversed of exothermic reaction is endothermic reaction, then add to eq.(1). Plan C(graphite) + O2(g) → CO2(g) ΔH0rxn = -393.5 kJ/mol rxn (1) CO2(g) → CO (g) + ½ O2(g) ΔH0rxn = -(-283.0 kJ/mol rxn) (2) C(graphite) + ½ O2(g) → CO(g) ΔHrxn0 = - 110.5 kJ/mol rxn EXAMPLE 7 22/03/2015 Calculate heat of reaction at 298K. C2H4(g) + H2O(l) → C2H5OH(l) use the following thermochemical equations ΔH0 C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) -1367kJ/mol (1) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) - 1411kJ/mol (2) Example INTERPRETATION ΔH0 2CO2(g) + 3H2O(l) → C2H5OH(l) + 3O2(g) +1367kJ/mol (-1) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) - 1411kJ/mol (2) Revise equation (1) to give (-1) when the equation is revised, the sign of ΔH0 is changed because the reversed of exothermic reaction is endothermic reaction, then add to eq.(2). Plan C2H4(g) + H2O(l) → C2H5OH(l) ΔH0 = -44 kJ/mol rxn Solution 8 22/03/2015 What is the value for ΔH for the following reaction? CS2(l) + 2 O2(g) → CO2(g) + 2 SO2(g) Given: C(s) + O2(g) → CO2(g); ΔHf = -393.5 kJ/mol S(s) + O2(g) → SO2(g); ΔHf = -296.8 kJ/mol C(s) + 2 S(s) → CS2(l); ΔHf = 87.9 kJ/mol Calculate the standard enthalpy of combustion, ΔHf0, for burning β-Dglucose, C6H12O6. The required values of ΔHf0 are presented above: β-D-glucose = -1268 kJ/mol CO2(g) = -393 kJ/mol H2O(l) = -285,5 kJ/mol O2(g) = 0 kJ/mol Quiz The internal energy, E, of a specific amount of a substance represents all 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 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 CHANGES IN INTERNAL ENERGY, E 9 22/03/2015 Δ E = Efinal - Einitial - Eproducts – Ereactants - q + w CHANGES IN INTERNAL ENERGY, E Sign conventions for q and w. 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. Enthalpy, free energy, spontaneous and non-spontaneous reaction What was the character of the process? Answer - spontaneous You will learn about the conditions that will produce a spontaneous chemical reaction. Entropy and Free Energy 10 22/03/2015 • What are two characteristics of spontaneous reactions? • A spontaneous reaction occurs naturally and favours the formation of products at the specified conditions. Any process which occurs without 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 spontaneous. Spontaneous reaction 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 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 = ∆Hvap of heat is removed. To reverse the process, q = ∆Hvap must be added to the 1 mol of ice at 0°C and 1 atm to form 1 mol of water at 0 °C Spontaneous reactions 11 22/03/2015 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 The major concern of thermodynamics is predicting whether a particular process can occur under specified conditions to give predominantly products. Q. 6. Non-spontaneous changes are: rusting of an iron burning a piece of paper metling of ice at room temperature electrolysis photosynthesis 2H2O→2H2 +O2 producing the aniline from nitrobenzene Quiz 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. Product favored 12 22/03/2015 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. 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. Reactant- favored 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. 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 very slowly. The rate at which a spontaneous reaction occurs is addressed by kinetics (see in future lectures). Spontaneus change 13 22/03/2015 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 Not all exothermic changes are spontaneous, however, nor are all spontaneous changes exothermic. As an example, 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. Enthaply of exothermic reaction 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. Another factor, related to the disorder of reactants and products, also plays a role in determining spontaneity. The dissolution of ammonium nitrate, NH4NO3, in water is spontaneous, although it is endothermic reaction. 14 22/03/2015 The system (consisting of the water, the Nevertheless, the process is spontaneous solid NH4NO3, and the resulting hydrated because the system becomes more NH4 and NO3 ions) absorbs heat from disordered as the regularly arranged ions the surroundings as the endothermic of crystalline ammonium nitrate become process occurs. 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. Two factors affect the spontaneity of any physical or chemical change: 1. Spontaneity is favoured when heat is released during the change (exothermic). 2. Spontaneity is favoured when the change causes an increase in disorder. Factors affect the spontaneity 15 22/03/2015 In spontaneous changes, the universe tends toward a state of greater disorder. The Second Law of Thermodynamics is based on our experiences. State of entropy of the entire universe, 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. Second law of thermodynamics introduction 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. 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. Spontaneous vs. nonspontaneous 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 16 22/03/2015 Entropy increases when a substance is divided into parts. INCREASING ENTROPY Entropy tends to increase when temperature increases. As the temperature increases, the molecules move faster and faster, which increases the disorder. INCREASING ENTROPY Entropy tends to increase in chemical reactions in which the total number of product molecules is greater than the total number of reactant molecules. ENTROPY IN CHEMICAL REACTION Q. Which ones of the processes below are not accompanied by an increase in entropy? a. Cooking the soup b. decomposition of SO3 2SO3→ 2SO2 + 2O2 d. Cooling the tea in cup e. restocking a canned goods shelf display in a supermarket f. Forming a snowball Entropy quiz 17 22/03/2015 Q. Which system has the greater enthropy: A) granulated sugar or icing sugar B) wall of bricks or several bricks C) desk of pedant person or desk of Spontaneous reactions produce substantial amounts of products at equilibrium and release free energy. Free energy is energy that is available to do work. messy person D) the lecture room before or after the lecture Quiz • 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. ENTROPY Free Energy and Spontaneous Reactions • An increase in entropy favours the spontaneous chemical reaction. • A decrease favours the nonspontaneous reaction. ENTROPY 18 22/03/2015 • 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 Combustion of acetylene C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g) Enthalpies of formation ∆Horxn = -1238 kJ Standard molar entropies ∆Sorxn = -97.4 J/K or -0.0974 kJ/K ∆Gorxn = -1238 kJ - (298 K)(-0.0974 J/K) = -1209 kJ Reaction is product-favored in spite of negative ∆Sorxn. Reaction is “enthalpy driven” Gibbs energy in questions • The Gibbs free-energy change (ΔG) is the maximum amount of energy that can be coupled to another process to do useful work. 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 NH4NO3(s) + heat ---> NH4NO3(aq) From tables of thermodynamic data we find ∆Horxn = +25.7 kJ ∆Sorxn = +108.7 J/K or +0.1087 kJ/K ∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ Reaction is product-favored in spite of positive ∆Horxn. Reaction is “entropy driven” Calculating ∆Gorxn 19 22/03/2015 : In the Haber process for the manufacture of ammonia N2 + 3H2→ 2 NH3 At what temperature will the reaction above become spontaneous? If the enthalpy and entropy of this proces 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 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 and T = 93000/198 Kelvin therefore the reaction becomes spontaneous when T = 469 K (196 ºC) below this temperature the reaction is spontaneous. Influence of enthalpy and entropy changes on reaction spontaneity Enthalpy change Entropy Spontaneity reaction Decreases (exothermic) Increases (more disorder in products than in reactants) Yes Increases (endothermic) Increases Only if unfavourable enthalpy change is offset by favourable entropy change Decreases (exothermic) Decreases (more disorder in reactants than in products) Only if unfavourable entropy change is offset by favourable enthalpy change Increases (endothermic) Decreases No 20 22/03/2015 21 22/03/2015 True and False The equation, Horeaction = Hof (products) - Hof (reactants) is a statement of the second law of thermodynamics. ___ Quiz 1. Free energy from a reaction is the amount of True/False: ΔG > 0, the process is spontaneous True/False: A nonspontaneous process cannot occur with external intervention. energy that is absorbed by an entropy decrease. equal to the enthalpy change. wasted as heat. avaiable to do work QUIZ 22 22/03/2015 2. Free energy is always available from 3. Choose the correct words for the spaces: reactions that are Spontaneous reactions produce ________ endothermic. and substantial amounts of _________ at nonspontaneous. equilibrium. at equilibrium. free energy, products spontaneous. no free energy, reactants free energy, reactants QUIZ no free energy, products QUIZ 4. Which of the following involves a decrease in entropy? Natural gas burns. A liquid freezes. Dry ice sublimes. Water evaporates. 6. A reaction is spontaneous if enthalpy enthalpy enthalpy enthalpy decreases and entropy increases. increases and entropy increases. decreases and entropy decreases. increases and entropy decreases 5. Both the enthalpy and the entropy are combined to calculate the a. reaction time b. free energy c. bound energy QUIZ QUIZ 23 22/03/2015 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 QUIZ 24