Thermodynamics- Chapter 19 Schedule and Notes

Thermodynamics- Chapter 19 Schedule and Notes Date Topics Video cast DUE Assignment during class time One Review of thermodynamics 1_thermo_review AND Review of thermo Wksheet 2.1ch19_intro Optional: 1sc_thermo Two 19.1-4; state function demo 2.2ch19_qualitative_two AND 1, 2, 7, 11, 21, 29a, 30a, 31 3.1ch19_qualitative_three Optional: 3_sc_thermo Three 19.4-5 3.2video_thermo AND 5video_thermo 39, 40, 47, 51, 52 Optional: 4sc_thermo Four 19.5-6 6sc_thermo 53, 55, 57, 60, 64 Five 19.7 7video_thermo 71, 74, 75, 79 Six Enthalpy, Entropy and Gibbs Lab Seven Review Practice Tests Eight Nine Practice Test Review and in-class quiz Test

Sample Problems and Notes for Review of Thermodynamics VideoCast 2008 Energy and physical or chemical processes. Measured in 1 st law of Thermodynamics: Endo vs Exo: Heat at constant pressure is called: Diagrams: Ways to find Enthalpy of a Process: 1) Using info from an equation 2) Calorimetry 3) Hess Law 4) Enthalpy of Formation 5) Bond Enthalpies 1) Information from an Equation 2H 2 O 2 (l) 2H 2 O(l) + O 2 (g) H= -196 kj What is the heat (q) [or H, since it is at constant pressure)] when 5.0g of H2O2 decomposes? Think of the reverse of this question too.

2) Calorimetry At constant pressure, two equations: How much heat is needed to warm 250g of H 2 O from 22C to 98C. c H2O = 4.184 J/gK What is the molar heat capacity (Cp) of H 2 O? More calorimetry Examples: When a student mixes 50mL of 1.0M HCl and 50mL of 1.0M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 C to 27.5 C. Calculate the enthalpy change for the reaction, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100.0mL, that its density is 1.0g/mL, and that its specific heat is 4.184J/g C.

Constant Volume Calorimetry (Bomb Calorimetry): Methylhydrazine (CH 6 N 2 ) is commonly used as a liquid rocket fuel. The combustion of methylhydrazine with oxygen produces nitrogen, carbon dioxide and water. 2CH 6 N 2 + 5O 2 2N 2 + 2CO 2 + 6H 2 O When 4.00g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 C to 39.50 C. In a separate experiment the heat capacity of the calorimeter is measured to be 7.794 kj/ C. What is the heat of reaction for the combustion of a mole of CH 6 N 2? 3) Hess Law Calculate H for the reaction: 2C (s) + H 2(g) C 2 H 2(g) Given the following reactions and their respective enthalpy changes: C 2 H 2(g) + 5/2O 2(g) 2CO 2 + H 2 O H = -1299.6kJ C (s) + O 2(g) CO 2(g) H = -393.5kJ H 2(g) + 1/2O 2(g) H 2 O H = -285.8kJ 4) Standard Enthalpies of Formation, p. 191ff Use values from appendix C, pg 1123, to determine the enthalpy of reaction for the combustion of pentane gas to form carbon dioxide gas and liquid water. (p193)

5) Using Bond Enthalpies, p.332 Bond enthalpy is the enthalpy change, H, for the breaking of a particular bond in one mole of a gaseous substance. Use bond enthalpies from p.330 in your text to estimate the enthalpy of reaction for the combustion of gaseous ethane, producing carbon dioxide gas and water vapor. p. 332.

Thermochemistry Review Worksheet 1) The SI unit for energy is the Joule. Show how the units of meter, second and kilogram can be used to define Joule. 2) What is the first law of thermodynamics? 3) Thermodynamic quantities are always identified by a number and a sign (+ or -). What does each tell us about a system? 4) How much heat is released when 4.50g of methane gas is burned in a constant pressure system? (the H for the combustion of methane is 890kJ/mol) 5) Large beds of rocks are used in some solar-heated homes to store heat. a) calculate the quantity of heat absorbed by 50.0kg of rocks if their temperature increases by 12.0 C. (Assume the specific heat of the rocks is 0.82 J/gK) b) What temperature change would these rocks undergo if they absorbed 450kJ of heat?

6) When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates. a) write the thermochemical equation for this process assuming H = 65.5kJ/mol b) the energy for a reaction such as this is sometimes called the energy of precipitation or heat of precipitation or H of formation. Calculate the H of formation of 0.200 mol of AgCl. c) Calculate the H when 0.350 mmol (yes, milli-moles!) of AgCl dissolves in water.

7) A coffee-cup calorimeter of the type shown in figure 6.5 on page 185 of your text contains 150.0 g of water at 25.1 C. A 121.0g block of Cu metal is heated to 100.4 C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/gK. The Cu is added to the calorimeter and after a time the contents of the cup reach a constant temperature of 30.1 C. a) Determine the amount of heat, in Joules, lost by the copper block. b) Determine the amount of heat gained by the water (the specific heat of water is 4.184 J/gK). c) The difference between your answers for (a) and (b) is due to the heat loss though the styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the cover) by 1K. Calculate the heat capacity of the calorimeter in J/K. d) If only one Styrofoam cup were used in the apparatus, rather than two, would you expect the heat capacity of the calorimeter to increase, decrease or stay the same? Explain. e) What would be the final temperature of the system if all the heat lost by the Cu block were absorbed by the water in the calorimeter?

8) Calculate H for the reaction NO (g) + O (g) NO 2(g) Given the following: NO (g) + O 3(g) NO 2(g) + O 2(g) H=-198.9kJ O 2(g) 2O (g) H=495.0kJ O 3(g) 3/2O 2(g) H=-142.3kJ 9) What is the standard enthalpy of formation? What does the little mean in the term H? How do you use standard enthalpies of formation to find H rxn? 10) Use values from Appendix 4 (page A22) in your text to calculate a) the enthalpy change for the combustion of 1 mol of ethanol b) the combustion of 4.5g of ethanol. c) Now compare the energy in (b) with the combustion of 4.5g of methane gas (#3 above) which is greater? Why? 11) Use bond enthalpies from p.330 in your text to estimate the bond enthalpy of a nitrogen-nitrogen triple bond, when it is produced in the following reaction: N 2 H 4 (g) N 2 (g) + 2H 2 (g) H = -86kJ

Videos 2.1 and 2.2 and 3.1 qualitative aspects of thermodynamics video notes on this page. Write neat and large and use lots of white space on this and other pages provided by you.

Sample Problems, Chapter 19 These note pages start with video 3.2video_thermo Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: a) when a piece of metal heated to 150 C is added to water at 40 C, the water gets hotter. b) Water at room temperature decomposes into H 2(g) and O 2(g). c) Benzene vapor, C 6 H 6(g), at a pressure of 1atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 C. If a process is nonspontaneous, does that mean that it cannot occur under any circumstances? The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is 38.9 C, and its molar enthalpy of fusion is H fusion = 2.29kJ/mol. What is the entropy change of the system when 50.0g of Hg (l) freezes at the normal freezing point? Extra The normal boiling point of ethanol is 78.3 C and its molar enthalpy of vaporization is 38.56 kj/mol. What is the change in entropy in the system when 68.3g of C 2 H 5 OH (g) at 1atm condenses to liquid at the normal boiling point? The second law of thermodynamics says that entropy..

19.3 Microstates not super important, but helps in understanding S = k lnw What is W? Predict whether S is positive or negative for each of the following processes, assuming each occurs at a constant temperature: a) H 2 O (l) H 2 O (g) b) Ag + (aq) + Cl - (aq) AgCl (s) c) 4Fe (s) + 3O 2(g) 2Fe 2 O 3(s) d) N 2(g) + O 2(g) 2NO (g) Choose the sample of matter that has greater entropy in each pair and explain your choice: a) 1 mol of NaCl (s) or 1 mol of HCl (g) at 25 C b) 2 mol of HCl (g) or 1 mol of HCl (g) at 25 C c) 1 mol of HCl (g) or 1 mol of Ar (g) at 298K If you are told that the entropy of a certain system is zero, what do you know about the system? 19.4 Calculate the S for the synthesis of ammonia from N 2(g) and H 2(g) at 298K: N 2(g) + 3H 2(g) 2NH 3(g)

You try: Using the standard entropies in Appendix C, calculate the standard entropy change, S, for the following reaction at 298K: Al 2 O 3(s) + 3H 2(g) 2Al (s) + 3H 2 O (g) 19.5 What are the criteria for spontaneity in terms of: a) entropy b) free energy Calculate the standard free-energy change for the following reaction at 298K: P 4(g) + 6Cl 2(g) 4PCl 3(g)

What is G for the reverse of this reaction? C 3 H 8(g) + 5O 2(g) 3CO 2(g) + 4H 2 O (l) H = -2220 kj Without using data from appendix C, predict whether the Gibbs Free Energy for this reaction is more negative or less negative than H. Use data from appendix C to calculate the standard free energy change for the reaction at 298K. Is your prediction correct? 19.6 The Haber Process: N 2(g) + 3H 2(g) 2NH 3(g) Assume that H and S for this reaction do not change with temperature. a) predict the direction in which G for this reaction changes with increasing temperature. b) Calculate the values of G for the reaction at 25 C and 500 C

Using standard enthalpies of formation and standard entropies in appendix C, calculate H and S at 298K for the following reaction: 2SO 2(g) + O 2(g) 2SO 3(g) Using the values obtained, estimate G at 400K. 19.7 The Haber Process: N 2(g) + 3H 2(g) 2NH 3(g) Calculate G at 298K for a reaction mixture that consists of 1.0atm N 2, 3.0 atm H 2, and 0.50 atm NH 3.

Use standard free energies of formation to calculate the equilibrium constant, K, at 25 C for the reaction involved in the Haber process: The Haber Process: N 2(g) + 3H 2(g) 2NH 3(g) The standard free-energy change for this reaction was calculated in an earlier practice problem it was 33.3kJ/mol = -33300 J/mol. Use data from appendix C to calculate the standard free-energy change, G, and the equilibrium constant, K, at 298K for the reaction: H 2(g) + Br 2(l) 2HBr (g)

1991 D (Required) BCl 3 (g) + NH 3 (g) Cl 3 BNH 3 (s) The reaction represented above is a reversible reaction. (a) Predict the sign of the entropy change, S, as the reaction proceeds to the right. Explain your prediction. (b) If the reaction spontaneously proceeds to the right, predict the sign of the enthalpy change, H. Explain your prediction. (c) The direction in which the reaction spontaneously proceeds changes as the temperature is increased above a specific temperature. Explain. (d) What is the value of the equilibrium constant at the temperature referred to in (c); that is, the specific temperature at which the direction of the spontaneous reaction changes? Explain. Answer: (a) Because a mixture of 2 gases produces a single pure solid, there is an extremely large decrease in entropy, S < 0, i.e. the sign of S is negative. (b) In order for a spontaneous change to occur in the right direction, the enthalpy change must overcome the entropy change which favors the reactants (left), since nature favors a lower enthalpy, then the reaction must be exothermic to the right, H < 0. (c) G = H T S, the reaction will change direction when the sign of G changes, since H < 0 and S < 0, then at low temperatures the sign of G is negative and spontaneous to the right. At some higher T, H = T S and G = 0, thereafter, any higher temperature will see G as positive and spontaneous in the left direction. (d) At equilibrium, K = e G/RT, where G = 0, K = e o = 1 1993 D 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) The reaction represented above is spontaneous at 25 C. Assume that all reactants and products are in their standard state. (a) Predict the sign of S for the reaction and justify your prediction. (b) What is the sign of G for the reaction? How would the sign and magnitude of G be affected by an increase in temperature to 50 C? Explain your answer. (c) What must be the sign of H for the reaction at 25 C? How does the total bond energy of the reactants compare to that of the products? (d) When the reactants are place together in a container, no change is observed even though the reaction is known to be spontaneous. Explain this observation. Answer: (a) S<0. The number of moles of gaseous products is less than the number of moles of gaseous reactants. OR A liquid is formed from gaseous reactants. (b) G<0. G becomes less negative as the temperature is increased since S < 0 and G = H T S. The term T S adds a positive number to H. (c) H<0. The bond energy of the reactants is less than the bond energy of the products. (d) The reaction has a high activation energy; OR is kinetically slow; OR a specific mention of the need for a catalyst or spark.

1994 D 2 H 2 S(g) + SO 2 (g) 3 S(s) + 2 H 2 O(g) At 298 K, the standard enthalpy change, H for the reaction represented above is 145 kilojoules. (a) Predict the sign of the standard entropy change, S, for the reaction. Explain the basis for your prediction. (b) At 298 K, the forward reaction (i.e., toward the right) is spontaneous. What change, if any, would occur in the value of G for this reaction as the temperature is increased? Explain your reasoning using thermodynamic principles. (c) What change, if any, would occur in the value of the equilibrium constant, K eq, for the situation described in (b)? Explain your reasoning. (d) The absolute temperature at which the forward reaction becomes nonspontaneous can be predicted. Write the equation that is used to make the prediction. Why does this equation predict only an approximate value for the temperature? Answer: (a) S is negative ( ). A high entropy mixture of two kinds of gases forms into a low entropy solid and a pure gas; 3 molecules of gas makes 2 molecules of gas, fewer gas molecules is at a lower entropy. (b) G < 0 if spontaneous. G = H T S Since S is neg. ( ), as T gets larger, T S will become larger than +145 kj and the sign of G becomes pos. (+) and the reaction is non spontanseous. (c) When T S < +145 kj, K eq > 1, when T S = +145 kj, K eq = 1, when T S > +145 kj, K eq < 1, but > 0 (d) G = 0 at this point, the equation is T = H ; this assumes that H and/or S do not change with temperature; not a perfect assumption leading to errors in the S calculation. 1996 B C 2 H 2 (g) + 2 H 2 (g) C 2 H 6 (g) Information about the substances involved in the reaction represented above is summarized in the following tables. (a) Substance S (J/mol K) H f (kj/mol) C 2 H 2 (g) 200.9 226.7 H 2 (g) 130.7 0 C 2 H 6 (g) - - - - 84.7 Bond Bond Energy (kj/mol) C C 347 C=C 611 C H 414 H H 436 If the value of the standard entropy change, S, for the reaction is 232.7 joules per mole Kelvin, calculate the standard molar entropy, S, of C 2 H 6 gas. (b) Calculate the value of the standard free-energy change, G, for the reaction. What does the sign of G indicate about the reaction above? (c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K. (d) Calculate the value of the C C bond energy in C 2 H 2 in kilojoules per mole.

Answer: (a) 232.7 J/K = S (C2 H 6 ) [2(130.7) + 200.9] J/K (b) S (C2 H 6 ) = 229.6 J/K H = H ƒ (products) H ƒ (reactants) = 84.7 kj [226.7 + 2(0)] kj = 311.4 kj G = H T S = 311.4 kj (298K)( 0.2327 kj/k) = 242.1 kj A G < 0 (a negative G ) indicates a spontaneous forward reaction. (c) K eq = e G/RT = e ( 242100/(8.314)(298)) (d) = 2.74 10 42 H = bond energy of products bond energy of reactants 311.4 kj = [(2)(436) + E c c + (2)(414)] [347 + (6)(414)] kj E c c = 820 kj 1997 D For the gaseous equilibrium represented below, it is observed that greater amounts of PCl 3 and Cl 2 are produced as the temperature is increased. PCl 5 (g) PCl 3 (g) + Cl 2 (g) (a) (b) (c) (d) What is the sign of S for the reaction? Explain. What change, if any, will occur in G for the reaction as the temperature is increased? Explain your reasoning in terms of thermodynamic principles. If He gas is added to the original reaction mixture at constant volume and temperature, what will happen to the partial pressure of Cl 2? Explain. If the volume of the reaction mixture is decreased at constant temperature to half the original volume, what will happen to the number of moles of Cl 2 in the reaction vessel? Explain. Answer: (a) The sign of S is (+). There is an increase in the number of gas molecules as well as a change from a pure gas to a mixture of gases. (b) G = H T S. Both S and H are (+). As temperature increases, at some point the sign of G will change from (+) to ( ), when the system will become spontaneous. (c) There will be no change in the partial pressure of the chlorine. Without a volume or temperature change, the pressure is independent of the other gases that are present. (d) The number of moles of Cl 2 will decrease. The decrease in volume will result in an increase in pressure and, according to LeChatelier s Principle, the equilibrium system will shift to the left (the side with fewer gas molecules) to reduce this increase in pressure. This will cause a decrease in the number of moles of products and an increase in the number of moles of reactant.

1. To measure enthalpy. 2. To calculate entropy. 3. To calculate Gibbs free energy. THERMODYNAMICS ENTHALPY, ENTROPY and GIBBS FREE ENERGY OBJECTIVES The Gibbs free energy is (delta G) is a measure of the spontaneity of a process and of the useful energy available from it. Negative delta G indicates a spontaneous process. Positive delta G indicates a nonspontaneous process or a process spontaneous in opposite direction Delta G is calculated from the equation: Delta G= delta H T delta S In today s experiments we will measure experimentally change in enthalpy, delta H, and obtain change in entropy, delta S, from the table data. After completing the calculations answer the following questions separately for each experiment: 1. Is the dissolving exothermic or endothermic? 2. Does the entropy decreases or increases during dissolving? 3. Is the Gibbs free energy positive or negative? Is the process of dissolving spontaneous?

Teacher Notes -- EXAMPLES OF CALCULATIONS: 1. Measurement and calculations of delta H. 22.0 g of magnesium sulfate was dissolved in 300. ml of water at 23.3 degree C in a coffee-cup calorimeter. After dissolving, the temperature increased to 36.6 degree C. Write the thermochemical equation for the reaction, that is, calculate the enthalpy for the reaction: MgSO4(s) --> Mg 2+ (aq) + SO4 2 (aq) Assume that the calorimeter loses only a negligible quantity of heat, density of the solutions is 1.00 g/ml, and the specific heat of the solutions is 4.184 J/g K. Since adding the crystals doesn t change the volume noticeably, assume the mass of solution to be 300. g. Step 1. Calculate the heat produced in the reaction: heat = sp. heat of solution x mass of solution x delta T or qsoln = csoln x masssoln x delta Tsoln qsoln =4.184 J/g K x 300. g x 13.3 K = 16694 J = 16.7 kj Step 2. Calculate how many moles are 22.0 g of magnesium sulfate. 22.0 g x 1 mol/120.4 g = 0.183 mol MgSO4 Step 3. Calculate the heat evolved per one mole of magnesium sulfate. This is the sought delta H value. Enthalpy will be negative, ( ), since it is an exothermic reaction (the temperature went up). 1 mol x 16.7 kj/0.183 mol = 91.3 kj Step 4. Rewrite the equation with state symbols and with the delta H value. Enthalpy is expressed per number of moles in the equation, in this case, per one mole. MgSO4(s) >Mg 2+ (aq) + SO4 2 (aq) delta H = - 91.3 kj 2. Delta S from standard tabulated data. MgSO4(s) 91.6 S (J/K mol) Mg 2+ (aq) - 138.1 SO4 2 (aq) 20.1 Delta S = S products - S reactants Delta S = [Mg 2+ (aq) + SO4 2 (aq)] - MgSO4(s) Delta S = [-138.1 J/K mol + 20.1 J/K mol] - 91.6 J/K mol = - 209.6 J/K mol This value indicates that entropy decreases during the dissolving. This is a rare case. It is related to formation of regular structures of water surrounding the ions. 3. Calculation of the Gibbs free energy. In all calculations we will use the standard temperature 298 K which is close to a room temperature. Enthalpy and entropy doesn t change significantly in our range of temperatures. Remember that H is usually expresses in kj and S in J. Perform appropriate conversions before calculating G. Delta G= delta H T delta S Delta G = - 91300 J/mol - [298 K x (- 209.6 J/K mol)] = - 28800 J = - 28.8 kj/mol Negative value of delta G indicates that dissolving magnesium sulfate is spontaneous.