# Chemistry 102 Chapter 17 THERMODYNAMICS

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1 THERMODYNAMICS Thermodynamics is concerned with the energy changes that accompany chemical and physical processes. Two conditions must be fulfilled in order to observe a chemical or physical change: The change must be possible (determined by Thermodynamics) The change must occur at a reasonable speed (determined by Kinetics) Important Terms in Thermodynamics: 1. Nature of Energy: Energy is the capacity to do work, and work is defined as force acting through a distance. There are many types of energy. Among them are kinetic (moving), potential (stored), thermal energy and chemical energy. Energy can be transformed from one type to another through work. 2. System and Surroundings: System : The part of the universe we happen to be studying Surroundings: Everything outside the system Three types of systems can be present: Open: exchange of energy and matter Closed: exchange of energy but not matter Isolated: No exchange of energy or matter 1

2 3. State Function THERMODYNAMICS A quantity whose value depends only on the current state of the system, and does not depend on the prior history of the system. Two example of state functions are temperature, internal energy 2

3 INTERNAL ENERGY Internal Energy (E) is the sum of the kinetic and potential energies of all of the particles that compose the system. Internal energy is a state function. Internal Energy Kinetic Energy of the particles making the system Potential Energy of the particles making the system Results from the energy of motion of Results from electrons atoms molecules chemical bonding between atoms attraction between molecules The 1 st Law of Thermodynamics is the Law of Conservation of Energy applied to thermodynamic systems. When a system changes from one state to another (ex: warming), its internal energy changes from one definite energy to another. Change in Internal Energy : E = E products E reactants where: E products = final value of the internal energy E reactants = initial value of the internal energy 3

4 CHANGES IN INTERNAL ENERGY C (s) + O 2 (g) CO 2 (g) CO 2 (g) C (s) + O 2 (g) SUMMARY: If the reactants have higher internal energy than products, E sys is negative and energy flows out of the system and into the surroundings. If the reactants have lower internal energy than products, E sys is positive and energy flows into the system from the surroundings. 4

5 1 ST LAW OF THERMODYNAMICS The changes in internal energy ( E) are more important in thermodynamics than the absolute values of Internal Energy The changes in IE are measured by noting the exchanges of energy between the system and the surroundings. These exchanges occur as one of two forms: Heat and Work Heat (q) = the energy that moves into or out of the system because of a temperature difference between the system and its surroundings. When heat is lost, internal energy (E) decreases and q is therefore assigned a negative sign. When heat is absorbed, internal energy (E) increases and q is therefore assigned a positive sign. Work (w) = the energy exchange that results when a force (F) moves an object through a distance (d). work = force x distance If work is done by the system, internal energy (E) decreases and w is therefore assigned a negative sign. If work is done on the system, internal energy (E) increases and w is therefore assigned a positive sign. According to the 1 st law of thermodynamics, the changes in the internal energy of the system ( E) are the sum of the heat transferred (q) and the work done (w). E = q + w Examples: Determine the sign of the work for each of the changes shown below (carried out at constant P): 1) 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) w = 2) CaCO 3 (s) CaO (s) + CO 2 (g) w = 3) CO 2 (g) + H 2 O (l) + CaCO 3 (s) Ca 2+ (aq) + 2 HCO 3 (aq) w = 5

6 1 ST LAW OF THERMODYNAMICS (ANALOGY) A rolling billiard ball has energy due to its motion. When it collides with the second ball, it does work, transferring energy to the second ball. The second ball now has now energy to roll away from the collision. The system can be defined as the first ball rolling across the table. The surroundings are the second ball and the pool table. Assume that when the first ball collides with the second ball, it loses all of its energy and remains still at the point of collision. After collision, the first ball loses all of its energy and therefore has E= 5.0 J. All this energy is gained by the surroundings. ( E sys = E surr ) In an actual situation, due to friction, some of the energy of the first ball is lost to the table as heat. As a result, energy change of the ball ( E sys ) is the sum of heat and work. On a table with a rough surface, more energy is lost to heat due to friction, leaving less available for work to transfer energy to the second ball. However, the sum of the heat and work still equals the change in energy of the first ball. Note that the amount of energy converted to heat and work depends on the details of the pool table and the path taken, while the change in internal energy of the rolling ball does not. This is because internal energy is a state function and only depends on the initial and final states of the ball. Heat and work are not state functions, and therefore are affected by the details of the ball s journey across the table. 6

7 Chemistry 102 HEAT AND WORK IN A PHYSICAL SYSTEM Case 1: A gas is enclosed in a vessel equipped with a movable piston, momentarily fixed in position. Heat passes from the surroundings to the vessel and the temperature is increased (q = J). In lifting the weight, the system does work equal to the energy gained by piston and the weight (w= -92 J). Heat Piston Gas E = q + w = (+165 J) + ( 92 J) = + 73 J Case 2: A gas is enclosed in a vessel equipped with a movable piston, momentarily fixed in position. Work is done on the system by pushing down the piston and thus compressing the gas (w= + 92 J). The gas pressure increases and causes the gas to warm up. This will cause heat to flow out of the system (q = 165 J). E = q + w = ( 165 J) + (+92 J) = 73 J Examples: 1. A cylinder and piston assembly is warmed by an external flame. As a result, the system absorbs 559 J of heat. The contents of the cylinder expand and perform 488 J of work during the expansion. What is the change in internal energy of the system? 2. A system releases 622 kj of heat and does 105 kj of work on the surroundings. What is the change in the internal energy of the system? 7

8 PRESSURE-VOLUME WORK To calculate the work associated with a chemical reaction, we limit discussion to those associated with a volume change and therefore called pressure-volume work. Pressure-volume work occurs when the force is caused by a volume change against an external pressure. Assume a cylinder, where the pressure exerted by the atmosphere is replaced with a piston, whose downward force of gravity, F, creates a pressure on the gas equivalent to that of the atmosphere. When the volume of the cylinder increases, it pushes against the external force. The work done by the system in expanding = (Force of Gravity) x (Distance the piston moves) w = F x h since V = A x h V F w = F x = x V A A Atmospheric pressure (P) Since the volume of cylinder is increasing, work is done by the system and therefore assigned a negative sign. w = P V 8

9 Examples: 1. In the reaction shown below, 1.00mol of Zn reacts with excess HCl to produce 1.00 mol of H 2 gas. At 25C and 1.00 atm, this amount of H 2 occupies 24.5 L. Calculate the change in Internal Energy ( U) for this reaction. (1 Latm = kj) Zn (s) + 2H 3 O + (aq) Zn 2+ (aq) + 2 H 2 O (l) + H 2 (g) q = kj/mol E = q + w q = kj w =? w (done by system) = PV = (1.00 atm)(24.5 L)(0.101 kj/1 Latm) = 2.47 kj E = q + w = ( kj ) + ( 2.47 kj) = kj 2. What is E when 1.00 mol of liquid water vaporizes at 100C? (H vap = kj/mol at 100 C) 3. Calculate the amount of work done in each of the following reactions. In each case, is the work done by or on the system? a) oxidation of HCl at 200 C 4 HCl (g) + O 2 (g) 2 Cl 2 (g) + 2 H 2 O (g) b) The decomposition of NO 2 at 300 C 2 NO (g) N 2 (g) + O 2 (g) 9

10 Examples: 4. A cylinder equipped with a piston expands against an external pressure of 1.58 atm. If the initial volume is L and the final volume is L, how much work (in J) is done? 5. When fuel is burned in a cylinder equipped with a piston, the volume expands from L to 1.45 L against an external pressure of 1.02 atm. In addition, 875 J is emitted as heat. What is E for the burning of the fuel? 6. Which statement is true of the internal energy of the system and surroundings following a process in which E sys = + 65 kj? a) The system and surroundings both lose 65 kj of heat. b) The system and surroundings both gain 65 kj of heat. c) The system loses 65 kj of energy and the surroundings gain 65 kj of energy. d) The system gains 65 kj of energy and the surroundings lose 65 kj of energy. 10

11 ENTHALPY AND ENTHALPY CHANGE Recall: Enthalpy = The Heat of Reaction, H, at constant pressure (q p ) Actually: Enthalpy = H = E + PV Since internal energy (E), pressure (P), and volume (V) are state functions, it follows that enthalpy (H) is also a state function. Therefore, the change in enthalpy (H) for any process occurring at constant pressure is given by the expression: H = E + P V H = (q p + w) + P V = q p + w w H = q p The enthalpy or reaction equals the Heat of Reaction at constant pressure. In practice, certain heats of reaction are measured and tabulated as heats of formation (H f ). H f = Standard Enthalpy of Formation of a Substance Enthalpy change for the formation of one mole of substance in its standard state from its elements in their reference form ( most stable form at 25C and 1.00 atm) and in their standard states (available in Appendix II in your textbook). The heat absorbed (endothermic) or released (exothermic) during a chemical reaction is defined as standard enthalpy change (H). H for a reaction can be calculated from known Hf values for reactants and products. H = n H f (products) n H f (reactants) 11

12 Examples: 1. Given the H f values below, calculate the heat absorbed or evolved in the reaction between NH 3 gas and CO 2 gas that yields aqueous urea (NH 2 CONH 2 ) and liquid water. 2 NH 3 (g) + CO 2 (g) CO(NH 2 ) 2 (aq) + H 2 O (l) H =??? H f (kj/mol) H = n H f (products) n H f (reactants) H = [( 319.2) + ( kj)] [2 x ( 45.9 ) + ( 393.5)] = kj Since H is negative, we conclude that the reaction is exothermic. 2. Use data in appendix II in your text to find H for the reactions shown below: a) NH 4 NO 3 (s) NO 2 (g) + 2 H 2 O (l) b) SiO 2 (s) + 3 C (graphite) SiC (s) + 2 CO (g) 12

13 SPONTANEOUS & NONSPONTANEOUS PRECESSES Spontaneous Processes: Physical or chemical changes that occurs by themselves, without outside assistance Spontaneous Physical Change: (ball rolling from the top of a hill) high potential energy Spontaneous Chemical Change: ( reaction of solid potassium with liquid water) low potential energy spontaneous 2 K(s) + 2 H 2 O (l) 2 KOH(aq) + H 2 (g) In order for spontaneous changes to go in the opposite direction work would have to be expended, and the changes would be called non-spontaneous. Non-spontaneous Physical Change: (Rolling a ball uphill) Non-spontaneous Chemical Change Change: (Obtaining K(s) from KOH(aq)) non-spontaneous 2 KOH (aq) + H 2 (g) 2 K(s) + 2 H 2 O(l) The Second Law of Thermodynamics determines if a reaction is spontaneous or not. Exothermic Reactions (H 0) are not necessarily spontaneous and Endothermic Reactions (H 0) are not necessarily non-spontaneous 13

14 ENTROPY (S) Entropy is the thermodynamic quantity that is a measure of the randomness or disorder in a system. Entropy is a state function: - the quantity of entropy in a substance depends only on variables that determine the state of the substance (temperature and pressure) Entropy is measured in J/K (SI unit) Entropy increases when the disorder in a sample of the substance increases: Example: Consider the melting of ice at 0C and 1 atm. MELTING ICE H 2 O molecules occupy regular fixed positions (crystal lattice) Ordered crystalline structure LIQUID WATER H 2 O molecules move about freely, giving a disordered structure Less-ordered structure Lower Entropy Higher Entropy S i = Initial Entropy = Entropy of Ice (41 J/K) S f = Final Entropy = Entropy of liquid water (63 J/K) melting at 0C and 1 atm H 2 O(s) H 2 O(l) S = 22 J/K S i = 41 J/K S f = 63 J/K In any natural process the degree of disorder (Entropy) increases. Natural processes are changes that occur by themselves (spontaneous changes). Second Law of Thermodynamics in reference to the system and its surroundings The Total Entropy of a System and its Surroundings always increases for a Spontaneous Process. ENTROPY ENERGY is created during a spontaneous change can be neither created or destroyed during a spontaneous change 14

15 ENTROPY & 2 ND LAW OF THERMODYNAMICS The 2 nd Law of Thermodynamics states that for any spontaneous process, the entropy of the universe increases (S univ > 0). Therefore, the criterion for spontaneity is the entropy of the universe. Processes that increase the entropy of the universe those that result in greater dispersal or randomization of energy occur spontaneously. Consider the example below: the expansion of a gas into a vacuum (a spontaneous process with no change in enthalpy). When the valve between the flasks opens, the gas in the left flask spontaneously expands into a vacuum. Since the pressure against the expansion is zero, no work is done by the gas. However, even though the total energy of the gas does not change, the entropy does change. This can easily be seen by the various ways the gas can distribute itself in the flasks (3 of which are shown below). Probability analysis indicates that it would be more probable to find the atoms in state C compared to either states A or B. And it can also be seen that state C has greater entropy than either state A or B. Therefore, the change of entropy in transitioning from state A to state C is positive because state C has greater entropy than state A. S = S final S initial S > 0 15

16 ENTROPY CHANGE FOR A PHASE TRANSITION The entropy of a sample of matter increases as it changes state from a solid to a liquid or from a liquid to a gas. This is due to the increase in disorder of molecules in liquid compared to solid and gas compared to liquid. We can therefore predict the sign of S for processes involving changes of state (or phase change). In general, entropy increases ( S > 0) for each of the following: The phase transition from a solid to a liquid the phase transition from a solid to a gas the phase transition from a liquid to a gas an increase in the number of moles of a gas during a chemical reaction Examples: 1. Predict the sign of S for each of the following processes: a) the boiling of water S = b) I 2 (g) I 2 (s) S = c) CaCO 3 (s) CaO (s) + CO 2 (g) S = 16

17 HEAT TRANSFER & CHANGES IN ENTROPY OF SURROUNDINGS The criterion for spontaneity is an increase in the entropy of the universe. However, there are several spontaneous processes that include a decrease in entropy. For example, when water freezes below 0C, the entropy of the water decreases, yet the process is spontaneous. The 2 nd Law of Thermodynamics states that for a spontaneous process, the entropy of universe increases ( S univ > 0). In the example above, even though the entropy of water decreases, the entropy of the universe must increase in order for the process to be spontaneous. Similar to the distinction of a system and surroundings in thermodynamics, we can distinguish between the entropy of the system and the surroundings. In the example of the freezing water, S sys is the entropy change of water, and the S surr is the entropy change of the surroundings. The entropy change of the universe is then the sum of the entropy changes of the system and the surrounding. S univ = S sys + S surr If a spontaneous process includes a decrease in S sys, then it would follow that there must be a greater increase in S surr in order for the S univ to increase. Since freezing of water is an exothermic process, it would follow that the heat given off by the process increases the entropy of the surroundings by a greater amount than the decrease of entropy of the system, making it a spontaneous process. To summarize: An exothermic process increases the entropy of the surroundings. An endothermic process decreases the entropy of the surroundings. 17

18 TEMPERATURE DEPENDENCE OF S surr Because of the large increase in S surr, the freezing of water is spontaneous at low temperatures. However, freezing of water is NOT spontaneous at high temperatures. Why? Units of entropy are J/K. Therefore, entropy is a measure of energy dispersal (joules) per unit temperature (K). The higher the temperature, the lower the amount of entropy change for a given amount of energy dispersed. Water does not freeze spontaneously at high temperatures, because the increase in S surr is small compared to decrease in S sys, making S univ negative. S univ = S sys + S surr (for water freezing) 18

19 QUANTIFYING ENTROPY CHANGE OF SURROUNDINGS & SYSTEM We know that when a system exchanges heat with surroundings, it changes the entropy of the surroundings. At constant pressure, q sys can be used to quantify the entropy change of the surroundings. In general, A process that emits heat into the surroundings (q sys = negative) increases the entropy of the surroundings (positive S surr ) A process that absorbs heat from the surroundings (q sys = positive) decreases the entropy of the surroundings (negative S surr ) The magnitude of the change in entropy of the surroundings is proportional to the magnitude of the q sys. S surr α q sys It can also be shown that for a given amount of heat exchanged with the surroundings, the magnitude of S surr is inversely proportional to the temperature. S surr α 1/T Combining the above relationships: S = surr -q T sys At constant pressure, q sys = H sys. Therefore, S = surr - H T sys (constant P, T) 19

20 Examples: 1. Calculate S surr for the melting of ice (Heat absorbed = Heat of Fusion = H fus = 6.0 kj/mol) 3 - Hfus - 6.0x10 J S surr = = = -22 J/K T 273 K 2. Liquid ethanol, C 2 H 6 O (l), at 25 C has an entropy of 161 J/(mol x K). If the heat of vaporization (H vap ) at 25 C is 42.3 kj/mol, what is the entropy of the vapor in equilibrium with the liquid at 25 C? C 2 H 6 O (l) C 2 H 6 O (g) H vap = kj/mol S sys = (S gas S liq ) > 0 3 qsys H vap kj/mol 10 J S sys = = = x = 142 J/K mol T T 298 K 1 kj S gas = S liq + S sys = 161 J/K J/K = 303 J/K 3. Consider the combustion of propane gas: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) H sys = 2044 kj a) Calculate the S surr associated with this reaction occurring at 25C. b) Determine the sign of S sys c) Determine the sign of S univ. Will this reaction be spontaneous? 20

21 GIBBS FREE ENERGY The relationships between enthalpy change in a system and the entropy change in the surroundings is shown below: S = surr - H T sys Also, for any process, the entropy change of the universe is the sum of the entropy change of the system and the entropy change of the surroundings. S univ = S sys + S surr Combining the two relationships above leads to the following relationship: S = S - univ sys H T sys Multiplying equation above with T, we arrive at the expression: T S univ = H sys T S sys Willard Gibbs (American physicist) combined together the concepts of enthalpy (H) and entropy (S) into a single thermodynamic function called Gibbs Free Energy (G), defined as shown below: G = H TS The change in free energy ( G) is expressed as: G = H T S Since S univ is a criterion for spontaneity, G is also a criterion for spontaneity (but opposite in sign), and is commonly called chemical potential because it is analogous to mechanical potential energy. Just as mechanical systems tend toward lower potential energy, chemical systems tend toward lower Gibbs free energy. 21

22 GIBBS FREE ENERGY Summarizing Gibbs Free Energy (at constant T & P): G is proportional to the negative of S univ. A decrease in Gibbs free energy ( G < 0) corresponds to a spontaneous process. An increase in Gibbs free energy ( G > 0) corresponds to a nonspontaneous process. 0 + G < 0 G > 0 Reaction is spontaneous Product-favored reaction Forward reaction is favored Reaction is non-spontaneous Reactant-favored reaction Reverse reaction is favored Conclusion: G gives a composite of the two factors that contribute to spontaneity (H and S) 22

23 SPONTANEITY & FREE ENERGY Consider the following situations: Case 1: H is negative and S is positive (exothermic) (increase in entropy) Reaction is spontaneous at all temperatures Case 2: 2 N 2 O (g) 2 N 2 (g) + O 2 (g) H = kj (2 mol gas) (3 mol gas) H is positive and S is negative (endothermic) (decrease in entropy) Reaction is nonspontaneous at all temperatures 3 O 2 (g) 2 O 3 (g) H = kj (3 mol gas) (2 mol gas) Case 3: H is negative and S is negative (exothermic) (decrease in entropy) Reaction is spontaneous only at low temperatures Case 4: H 2 O (l) H 2 O (s) H = 6.01 kj H is positive and S is positive (endothermic) (increase in entropy) Reaction is spontaneous only at high temperatures H 2 O (l) H 2 O (g) H = kj (at 100C) 23

24 SPONTANEITY & FREE ENERGY Examples: 1. Predict the conditions (high temperature, low temperature, all temperatures or no temperatures) under which each reaction below is spontaneous: a) H 2 O (g) H 2 O (l) b) CO 2 (s) CO 2 (g) c) 2 NO 2 (g) 2 NO (g) + O 2 (g) (endothermic) 2. For the reaction shown below, calculate G at 25C and determine whether the reaction is spontaneous. If the reaction is not spontaneous at 25C, determine at what temperature (if any) the reaction becomes spontaneous. CCl 4 (g) C (s, graphite) + 2 Cl 2 (g) H = kj S = J/K 24

25 ENTROPY CHANGES IN CHEMICAL REACTIONS Similar to the H for a reaction, we can define the standard entropy change for a reaction (S) as the entropy for a process in which all reactants and products are in their standard states. The standard states are defined as: Gas: pure gas at pressure exactly 1 atm. Liquid or Solid: pure substance at most stable form and pressure of 1 atm and temperature of 25C. Solution: concentration of 1 M. Since entropy is a state function, the standard change in entropy of a reaction is: S = S products S reactants The standard molar entropy (S) of a substance is needed to calculate the S for a reaction. The standard molar entropy of a substance is measured against the absolute value of zero as defined by the 3 rd law of thermodynamics. Third Law of Thermodynamics: The third law of thermodynamics states that: The entropy of a perfect crystal at absolute zero (0 K) is zero A perfect crystal at 0 K has only one possible way to arrange its components. Standard molar entropy values for substances are tabulated in textbooks and available in Appendix II. 25

26 STANDARD MOLAR ENTROPY The standard entropy of any substance is affected by the state of the substance, molar mass, molecular complexity and the extent of dissolution. As the molar mass of a substance increases, the value of standard molar entropy also increases. For example, the standard molar entropy of noble gases increases progressively from He to Xe. Similarly, the standard molar entropy of a substance depends on its state, as shown in the example below: In general, S gas > S liq > S solid For a given state of matter, standard molar entropy also increases with molecular complexity. For example, consider the following: Examples: 1. Rank each set of substances below in order of increasing standard molar entropy (S): a) I 2 (g), F 2 (g), Br 2 (g), Cl 2 (g) b) NH 3 (g), Ne (g), SO 2 (g), CH 3 CH 2 OH (g), He (g) 26

27 CALCULATING S FOR A REACTION Recall that the entropy usually increases in the following situations: (a) A reaction in which a molecule is broken into two or more smaller molecules. Example: fermentation of glucose (grape sugar) to alcohol: C 6 H 12 O 6 (s) 2 C 2 H 5 OH (l) + 2 CO 2 (g) 1 large molecule 4 smaller molecules (b) A reaction in which there is an increase in moles of gas. This may result from a molecules breaking up, in which case Rule (a) and Rule (b) are related. Example: the burning of acetylene in oxygen: 2 C 2 H 2 (g) + 3 O 2 (g) 4 CO 2 (g) + 2 H 2 O (g) 5 moles of gas 6 moles of gas n gas = 6 5 = +1 The standard entropy change (S) for a reaction can also be calculated from the standard entropies of reactants and products as shown below: S = ns (products) ns (reactants) Examples: 1. For each pair of substances, choose the one with the higher expected value of standard molar entropy. Explain the reasons for your choices. a) CO (g) CO 2 (g) b) CH 3 OH (l) CH 3 OH (g) c) Ar (g) CO 2 (g) d) NO 2 (g) CH 3 CH 2 CH 3 (g) 27

28 2. Predict the sign of S 0 for each reaction shown below. If you cannot predict the sign for any reaction, state why. a) C 2 H 2 (g) + 2 H 2 (g) C 2 H 6 (g) S = b) N 2 (g) + O 2 (g) 2 NO (g) S = c) 2 C (s) + O 2 (g) 2 CO (g) S = 3. Using tabulated S values in your text, calculate S for the reactions shown below: 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) 28

29 CALCULATING STANDARD FREE ENERGY CHANGES The standard free energy change (G) for a reaction is the criterion for its spontaneity, and can be calculated by 3 different methods. Method 1: Earlier in this chapter we learned how to use tabulated values of H f and S to calculate H and S for a reaction. We can now use the H and S values to calculate the G value for a reaction, using the equation: G = H TS Since H f and S values are tabulated at 25C, the equation above should only be valid at 25C. However, since H f and S values are affected very little by temperature, we can estimate the G values for temperatures other than 25C. Examples: 1. Given H f and S values shown below, calculate G for the reaction shown below, and determine if the reaction is spontaneous. SO 2 (g) + ½ O 2 (g) SO 3 (g) 2. Use the data in Appendix IIA of your text to calculate H, S and G at 25C for the reaction shown below. Is the reaction spontaneous? If not, would a change in temperature make it spontaneous? If so, should the temperature be raised or lowered from 25C? NH 4 Cl (s) HCl (g) + NH 3 (g) 29

30 Method 2: CALCULATING STANDARD FREE ENERGY CHANGES Because free energy is a state function, change in free energy or reactions (G) can also be calculated from the free energy of the reactants and products. In order to do this, we define the free energy of formation (G f ) for a substance as follows: The free energy of formation (G f ) is the change in free energy when 1 mole of a compound in its standard state forms from its constituent elements in their standard states. The free energy of formation of pure elements in their standard states is zero. The free energy change of a reaction can then be calculated using the equation below: G = ng f (products) mg f (reactants) Examples: 1. Calculate the standard free energy change (G) for the following reaction at 25 C, using standard free energies of formation given below: Na 2 CO 3 (s) + H + (aq) 2 Na + (aq) + HCO 3 (aq) G f (kj/mol) : G = [(2) x ( 261.9) + ( 587.1)] [( ) + 0] kj = 62.8 kj 2. Calculate the standard free-energy change (G) for the reaction shown below, and determine whether it is spontaneous or not. (Use G f values in your textbook). 2 NOCl (g) 2 NO (g) + Cl 2 (g) 3. Calculate the standard free-energy change (G) for the reaction shown below, and compare with the value obtained by method 1 on the previous page. (Use G f values in your textbook). NH 4 Cl (s) HCl (g) + NH 3 (g) 30

31 Method 3: CALCULATING STANDARD FREE ENERGY CHANGES Since free energy is a state function, the following relationships can be used to calculate G for a reaction: If a chemical equation is multiplied by some factor, then G for the reaction is also multiplied by the same factor. If a chemical equation is reversed, then G changes sign. If a chemical equation can be expressed as the sum of a series of steps, then G for the overall equation is the sum of the free energies of reactions for each step. Examples: 1. Find G for the reaction shown below: 3 C (s) + 4 H 2 (g) C 3 H 8 (g) Use the following reactions with known G values: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O (g) G = 2074 kj C (s) + O 2 (g) CO 2 (g) G = kj 2 H 2 (g) + O 2 (g) 2 H 2 O (g) G = kj 2. Determine G for the reaction shown below: Fe 2 O 3 (s) + 3 CO (g) 2 Fe (s) + 3 CO 2 (g) Use the following reactions with known G values: 2 Fe (s) + 3/2 O 2 (g) Fe 2 O 3 (s) G = kj CO (g) + ½ O 2 (g) CO 2 (g) G = kj 31

32 FREE ENERGY and USEFUL WORK In a spontaneous reaction: the free energy is lowered as reactants change to products, the change in free energy is released as free-energy change( G) this G can be harnessed to perform useful work A spontaneous reaction can be used to obtain useful work The change in free energy of a chemical reaction represents the maximum amount of energy available (or free) to do work (if G is negative). Maximum useful work = w max = G completely In an ideal situation: G Useful work converted into converted In real situations: G Some Useful Work + Some Entropy into Conclusions: 1. In theory, all of the free-energy decrease liberated during a spontaneous chemical change can be used to do work (this would be w max ) 2. In practice, less work is obtained and the difference appears as an increase in entropy. C (s, graphite) + 2 H 2 (g) CH 4 (g) H = 74.6 kj S = 80.8 J/K G = 50.5 kj Why is only 50.5 kj available as free energy, even though 74.6 kj is produced in the reaction? How is this reaction spontaneous ( G< 0) when entropy decreases? 32

33 FREE ENERGY CHANGES FOR NONSTANDARD STATES The standard free energy changes for a reaction ( G) are only useful for a very narrow set of conditions where reactants and products are in their standard states. For example, consider the standard free energy change for evaporation of liquid water: H 2 O (l) H 2 O (g) G = kj/mol Since G for this process is positive, it is nonspontaneous. Yet when water is spilled under ordinary conditions, it evaporates spontaneously. Why? This is because ordinary conditions are not standard states. For example, for water vapor, standard conditions are those were P H2O would be 1 atm. Based on the G value given above, under these conditions, the water vapor would condense spontaneously. Free energy changes under nonstandard conditions ( G) can be calculated from G according to the equation below: G = G + RT lnq Q = Thermodynamic Reaction Quotient for reactions involving gases Q is obtained from partial pressures. for reactions in solutions, Q is obtained from molar concentrations R = Gas Law Constant in units of energy = J/mol K G = Standard Free-Energy Change G = Free-Energy Change when Reactants in nonstandard states are changed to products in non-standard states 33

34 FREE ENERGY CHANGES FOR NONSTANDARD STATES The relationship of free energy change for evaporation of water and pressure of water can be seen from diagram below: Under standard condition, Q = P H2O = 1 atm and G = G, as expected. G = G + RT ln (1) = kj/mol At equilibrium, where water vapor has a pressure of atm, G = 0 as shown below: G = G + RT ln (0.0313) = (8.314 J/molK)(298) ln (0.0313) = kj/mol 8.59 kj/mol = 0 Under nonstandard condition, P H2O = atm and G can be calculated as shown below: G = G + RT ln (0.005) = (8.314 J/molK)(298) ln (0.005) = kj/mol 13.1 kj/mol = 4.5 kj/mol Under these conditions, since G< 0, evaporation would be spontaneous. 34

35 FREE ENERGY CHANGES FOR NONSTANDARD STATES Examples: 1. For the reaction shown below: 2 NO (g) + O 2 (g) 2 NO 2 (g) G = 71.2 kj Calculate G under the following conditions: P NO = atm P O2 = atm P NO2 = 2.00 atm 2. Calculate G for the reaction shown below if the reaction mixture consists of 1.0 atm N 2, 3.0 atm H 2 and 1.0 atm NH 3. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G = kj 35

36 FREE ENERGY AND EQUILIBRIUM G determines the spontaneity of a reaction when reactants and products are in their standard states. Previously we learned that equilibrium constant (K) determines how far a reaction proceeds, also a measure of spontaneity. Therefore, it would follow that free energy changes and equilibrium constant are related. This relationship is shown below: G = G + RT ln Q At equilibrium 0 = G + RT ln K G = RT ln K It follows that: 1. For reactions involving only gases: K = K p 2. For reactions involving only solutes in liquid solutions: K = K c 3. For net ionic equations: K = K sp 36

37 RELATIONSHIP OF G, K, AND SPONTANEITY G = RT ln K When K > 1 When K 1 When K <1 Reactants Products Reactants Products Reactants Products log K > 0 log K 0 log K < 0 G < 0 G 0 G > 0 Reaction is spontaneous Reaction gives an equilibrium mixture Forward reaction is non-spontaneous Reverse reaction is spontaneous Examples: 1. Use the standard free energies of formation given to determine the equilibrium constant (K) for the following reaction at 25 C: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G f (kj/mol) Use the standard free energies in your text to calculate G and K for the following reaction at 25 C: 2 H 2 (g) + O 2 (g) 2 H 2 O (g) 37

38 CHANGE OF FREE ENERGY WITH TEMPERATURE How can G be found for temperatures other than standard temperatures (25 C)? An approximate method used to calculate G T is based on the assumption that both H and S are constant with respect to temperature (only approximately true) Then: G T = H - TS (a convenient approximation for G T ) NOTE: G T is strongly temperature dependent G T = Change in free Energy for a substance: at 1 atm of pressure (standard pressure) and at the specified temperature, T (nonstandard temperature) depends on the value of determined by SPONTANEITY G T TEMPERATURE Meaning: It follows that SPONTANEITY IS TEMPERATURE DEPENDENT! Some chemical changes may be non-spontaneous at one temperature but spontaneous at another temperature. Examples: 1. Sodium Carbonate, Na 2 CO 3, can be prepared by heating sodium hydrogen carbonate, NaHCO 3. Given the H f and S values below, estimate the temperature at which NaHCO 3 decomposes to products at 1 atm. 2 NaHCO 3 (s) Na 2 CO 3 + H 2 O (g) + CO 2 (g) H f (kj/mol) : S (J/K) :

39 CHANGE OF FREE ENERGY WITH TEMPERATURE Examples: 2. Given H and S values at 25C, estimate G at 400 K for the reaction shown below: 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) H = kj S = J/K 3. Methanol can be prepared from reaction of CO and H 2, as shown below: CO (g) + 2 H 2 (g) CH 3 OH (g) H = 90.7 kj S = J/K a) Is this reaction spontaneous under standard conditions at 25C? b) Calculate G at 500 K. Is the reaction spontaneous under standard conditions at this temperature? 39

40 Answers to In-Chapter Problems: Page 5 Example No. Answer 1 positive 2 negative 3 positive 7 1 E = +71 J 2 E = 727 kj kj 9 3a 3b kj 0 J J 10 5 E = 998 J 6 d a 2b 1a 1b 1c 3a H = kj H = kj positive negative positive S surr = +6.86x10 3 J/K 20 3b S sys = c 1a 1b 1c 2 S univ = + ; reaction is spontaneous spontaneous at low temperature; nonspontaneous at high temperature nonspontaneous at low temperature; spontaneous at high temperature nonspontaneous at low temperature; spontaneous at high temperature G = kj; reaction is not spontaneous The reaction becomes spontaneous at T = 673 K 40

41 Page Example No. 1a 1b 1a 1b 1c 1d 2a 2b 2c Answer CO 2 (g) ; greater molar mass and complexity CH 3 OH (g) ; gas phase CO 2 (g); greater molar mass and complexity CH 3 CH 2 CH 3 ; greater complexity S = negative S = inconclusive S = positive 3 S = J/K G = 70.9 kj ; spontaneous 2 H = kj S = J/K G = 91.2 kj nonspontaneous; becomes spontaneous at high temperature 2 G = kj ; nonspontaneous 3 G = 91.2 kj ; value is comparable to that calculated by method 1 1 G = 23 kj 2 G = 29.4 kj 1 G = 50.7 kj 2 G = 41.5 kj 1 K= 6.9x K= 1.39x T= K 2 G T = kj 39 3a 3b G = 24.7 kj ; reaction is spontaneous at 25C G T = kj ; reaction is nonspontaneous at 500 K 41

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