Scruggs chap 19

Scruggs chap 19 - Chapter 19 Thermodynamics q Study of...

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Unformatted text preview: Chapter 19 Thermodynamics q Study of energy changes in chemical reactions useful to understand the nature of chemical changes q Invented to try to improve steam engines q Recall Thermochemistry (study of heat changes) from Chapter 5 Types of processes q Spontaneous - occurs without external intervention q Nonspontaneous - does not occur unless energy is added from an external source q At Equilibrium - not all reactions go to completion; reversible reactions (can move back and forth along the same path). 19.1 Spontaneous Processes q Spontaneous processes qhot object cools qgases expand qiron rusts qCs reacts with H2O qWhat do all these processes have in common? qThey happen with no external intervention. qA process that is spontaneous in one direction is not spontaneous in the reverse direction. What makes a process spontaneous? q Tendency to go to a state of lower energy. qEnthalpy: H < 0 (exothermic) q Tendency to become more disordered or random qEntropy: S > 0 (increase in randomness) Review: Enthalpy and the First Law of Thermodynamics q 1st Law = Law of Conservation of Energy: energy is neither created nor destroyed in a process q Energy is converted to a different form during a process q Types of energy of interest: internal energy (E), heat (q), work (w) q Internal energy (E) = energy stored within a system In open systems, w is zero or so small that it is Review negligible q So E q q q = H at constant pressure q So E H 19.2 Entropy and the Second Law of Thermodynamics q Matter tends to change spontaneously to a state of lower energy and greater disorder q Use H to measure energy changes. q Consider the expansion of a gas: qThis is at constant temperature, so no heat is transferred. Thus H = 0. There is no opposing pressure, so no work is done. Why does the gas expand? q Which is more likely - a random distribution of molecules throughout the volume of the gas, or an ordered distribution with all the gas molecules on one side? q A disordered or random system is more probable. Entropy q How to measure disorder? Use entropy (S). q The more disordered the system, the larger the value of the entropy. q Entropy is a state function: S = Sfinal - Sinitial q S > 0 corresponds to an increase in randomness or freedom of motion. 19.3 The Molecular Interpretation of Entropy q The entropy of a system indicates its degree of disorder. q Which gas has more disorder? q A decrease in number of molecules leads to a decrease in entropy. 1 Which has the Greater Entropy? q Separate gases or mixture of gases? N2 O2 air (N2 + O2) 1 Which of these states of water has the highest entropy? Solid Liquid 3 3 % 3 3 % Gas 3 3 % 1. 2. 3. Solid Liquid Gas 1 2 3 1 Which has the greater entropy? q Solute and solvent or solution? q Solution has more arrangements, more motion 1 q Entropy change positive or negative? 1 Which has the greater entropy? Reactants or products? q 2SO2 (g) + O2 (g) 2SO3 (g) q 2O3 (g) 3O2 (g) q 2H2(g) + O2(g) 2H2O(l) q More motion, more arrangements for the substances that have more particles 1 What is the sign of S for the following reactions? NaCl(s) Na+(aq) + Cl-(aq) 20% 20% 20% 20% 20% CO(g) + 2H2(g) CH3OH(g) 2C(s) + O2(g) 2CO(g) 1. 2. 3. 4. 5. +,+,+ -,-,+,-,+,-,+ -,-,+ 1 2 3 4 5 1 Which has the greater entropy? q Low temperature or high temperature? q Consider a gas at different temperatures: q High T q more motion, more arrangements 1 Effect of Temperature q Entropy decreases with decreasing temperature because motion decreases. Can we reach a point of no motion? What is the reference point for measurement of entropy? q We can measure absolute entropies (S). The 3rd Law of Thermodynamics: S of perfect crystal = 0 at 0 K 1 Third Law of Thermodynamics q Entropy = 0 (at 0 K) Entropy > 0 1 Entropy and Temp 2 Absolute Entropy (J/mol K) q Calculate S from the amount of heat required to raise the temperature from 0 K q Slow increase in S with T (difficult to calculate); but large increase in S at phase change (no T change, and S = H/T) P4O10 2 Entropy Changes for Phase Changes q S = H/T q Example For vaporization of water (at 100C), H = +44.01 kJ/mol. q S = (+44.01 kJ/mol) / 373 K = 0.118 kJ/mol K = 118 J/mol K 2 q Values of S can be obtained from 19.4 Calculation of Entropy Changes measurements of the variation in heat capacity with temperature. q Results of calculations and measurements have been collected into tables q Table 19.2, Appendix C in your text q Tables usually list standard molar entropies, S. q Standard state defined as pure solids or 2 Entropy Changes q For phase changes or chemical reactions, can calculate S because entropy is a state function (does not depend on the path) q So = nproductsSoproducts - nreactantsSoreactants where n is the appropriate coefficient from the balanced equation for the process 2 Entropy Changes q So = nproductsSoproducts - nreactantsSoreactants where n is the appropriate coefficient from the balanced equation for the process q Calculate So for the vaporization of water H2O(l) H2O(g) q So = SoH2O(g) - SoH2O(l) 2 Calculate the entropy change for the reaction below: 2H2(g) + O2(g) 2H2O(l) 1. 2. 3. 4. -88.75 J/K -146.89 J/K -275.64 J/K -326.35 J/K So in J/mol K H2 130.57 O2 205.03 H2O(g) 188.71 H2O(l) 69.91 2% 5 2% 5 2% 5 2% 5 1 2 3 4 2 Second Law of Thermodynamics q 2nd Law: In any spontaneous process, the entropy of the universe increases. q Suniv = Ssys + Ssurr: The change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings. q Entropy is not conserved: Suniv is positive. 2 Second Law of Thermodynamics q Suniv = Ssys + Ssurr> 0 for any spontaneous process q Examples of Spontaneous Processes: 2H2 + O2(g) 2H2O(g) Exothermic NH4NO3(s) NH4+(aq) + NO3-(aq) Endothermic 2 Second Law of Thermodynamics q For a process at Equilibrium: Suniv = 0 q For a spontaneous process (i.e. irreversible): Suniv > 0 q For a nonspontaneous process: Suniv < 0 2 G from 2nd Law of Thermodynamics q Suniv = Ssurr + Ssys ( Suniv > 0 for Sp. Proc.) q Ssurr = Hsurr/T q Ssurr = - Hsys/T q Substitute: Suniv = - Hsys/T + Ssys q Mult. By (-T): -T Suniv = Hsys -T Ssys q Given that: -T Suniv = G q Substitute: G = Hsys -T Ssys q (See box on p. 825 in text) 3 19.5 Gibbs Free Energy ( G) q How do we combine H and S to decide about spontaneity? q G = Hsys -T Ssys q Or just G = H -T S q Gibbs free energy change or free energy change = G = maximum amount of energy available to do work on the surroundings 3 Free Energy Changes ( G) q G = H - T S q G < 0 q G > 0 q G = 0 A spontaneous process Nonspontaneous State of equilibrium 3 Review H H Enthalpy Change ( H) 3 Freezing H2O q H2O (l) H2O (s); endothermic or exothermic? S > 0 or S < 0? q Heat is gained by surroundings (lost by H2O), therefore is exothermic q Products are more ordered, so S <0 3 Freezing H2O q At T > 273K q-T S> H, so H T S >0, net freezing does not occur q At T = 273K q H-T S=0, system is in equilibrium q At T < 273K q-T S< H, so H T S < 0, freezing happens 3 Standard Free Energy Changes q Can get Go from Go = Ho - T So qUse Go to predict spontaneity in the standard state q Can also get values of Go from free energies of formation: Gof (formation from the elements) q Gof = 0 for an element in its stable form q Hess's Law: Go = nprod Gfoprod - nreact Gforeact 3 Standard States G means the standard free energy change. q o q Reactants and products must be in standard state (table 19.3, p. 824): q Solid: pure solid q Liquid: pure liquid 3 of q Hess's Law: Go = nprod Gfoprod - nreact Gforeact 3 q Go = nproducts Gof products - nreactants Gof reactants q Similarly, Ho = nproducts Hof products - nreactants Hof S = nproductsSoproducts - nreactantsSoreactants q Values of Gof, Hof, So are listed for standard state conditions in Appendix C q Can use tables to predict the value of Go and the spontaneity of chemical reactions, even ones that are not yet observed. reactants o of 3 Hess's Law applies to G just like H (See sect. 5.6) reaction 1 reaction 2 -------------reaction 3 Go3 = Go2 + Go1 q Add equations, add Go values q Reverse equation, change sign of Go q Multiply equation by a constant, also multiply the value of Go by that constant 4 Adding Equations q H2(g) + 1/2 O2(g) H2O(l) Go= -237.18 kJ CaO(s) + H2O(l) Ca(OH)2(s) Go= -55.38 kJ -------------------------------------------------------------CaO(s) + H2(g) + 1/2 O2(g) Ca(OH)2(s) Go= ? q Go= (-237.18 kJ) + (-55.38 kJ) = -292.56 kJ 4 For the reactions below, calculate Go to determine if each is spontaneous (S) or not (NS). 2H2(g) + O2(g) 2H2O(g) CaO(s) + H2O(l) Ca(OH)2(s) 1. 2. 3. 4. NS, NS NS, S S, NS S, S 25% 25% 25% Gfo in kJ/mol CaO(s) -604.2 Ca(OH)2(s) -896.76 H2 0.00 H O(g) -228.59 25%2 H2O(l) -237.18 O2 0.00 1 2 3 4 4 Given reaction 1 and 2 below, find Go for reaction 3. 1: N2(g) + O2(g) 2NO(g) 2: 2NOCl(g) 2NO(g) + Cl2(g) Go = 173.38 kJ Go = 41.18 kJ 3:1/2 N2(g) + 1/2 O2(g) + 1/2 Cl2(g) NOCl(g) 1. 2. 3. 4. 214.56 kJ 107.28 kJ 66.10 kJ -107.28 kJ 2% 5 2% 5 2% 5 2% 5 1 2 3 4 4 19.6 Free Energy and Temperature 4 19.6 Free Energy and Temperature H= -227 kJ/mol, S = -309 J/mol K, T = 1450 K. Is this process spontaneous? (Watch units!) q G = 221 kJ/mol, so it is not spontaneous. q At what temperature will it be spontaneous? qT= H/ S= (-227,000 J/mol)/(-309 J/mol K) q = 735 K (below this value) 4 What is the lowest temperature (in K) at which this process will be spontaneous? H2O(l) H2O(g) H = 44.01 kJ/mol S = 118.8 J/mol K 25% 25% 25% 25% 1. 2. 3. 4. 0.37 K 170 K 273 K 370 K 1 2 3 4 4 19.7 Free Energy and the Equilibrium Constant q How can we determine G under nonstandard conditions? q How can we relate Go to the position of equilibrium (K)? 4 o q Q can be used to relate G to Go G = G + RT ln Q o R = 8.314 J/mol K T = absolute temperature (K) 4 o 15 10 5 ln (Q) 0 1.00E-09 1.00E-07 1.00E-05 1.00E-03 1.00E-01 1.00E+01 1.00E+03 1.00E+05 -5 -10 -15 -20 -25 Q 4 q G = Go +oRT ln Q q no products: Q = 0 q no reactants: Q = q standard state: all P=1, all [X]=1, so Q = 1 and ln Q = ln 1 = 0, G = Go q Between the extremes, 0 < Q < and G changes as the reaction progresses (increases as the reaction goes forward) 5 CO(g) + H2O(g)oh CO2 (g) + H2(g) x q Q = PCO2PH2/PCOPH2O q Initially, Q = 0, but then it increases q Initially G is more negative than Go q As the reaction progresses, G increases until G = 0 and net reaction ceases and the system is at equilibrium (forward and reverse process occur at the same rate). 5 How can we relate Go to K? o q When G = 0 (at equilibrium) q 0 = Go + RT ln Q qQ=K 5 Thus, Go = -RT ln K o q Go > 0 when K << 1 (lies toward reactants) q Go < 0 when K >>1 (lies toward products) q Go = 0 when K = 1 5 q Predict whether the equilibrium constant o at room temperature will be greater than, less than, or equal to 1. Explain. q C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) q (Hint: This is a combustion reaction) 5 q Go = - RT ln K o q Calculate K for the following reaction at 25oC. H2(g) + Br2(g) h 2HBr(g) Go = - 53.2 x kJ/mol q -53,200 J/mol = - 8.314 J/mol K 298 K ln K q ln K = 21.47 q K = e21.47 q K = 2.12 x 109 (K has no units) q When K is this large, the reaction proceeds far toward products to reach equilibrium. 5 What is K for this reaction at 0oC? 100oC? H2(g) + Br2(g) o 2HBr(g) Go = - 53.2 kJ/mol x 25% 25% 25% 25% 1. 2. 3. 4. 1.50 x 1010, 2.80 x 107 1.50 x 107, 2.80 x 1010 1.023, 1.017 6.70 x 10-11, 3.60 x 10-8 1 2 3 4 5 q Go = - RT ln K o q G = Go + RT ln Q H2(g) + Br2(g) h 2HBr(g) Go = - 53.2 x kJ/mol q Suppose this system starts at equilibrium. What will happen to the value of G when Br2 is removed from the system? 5 ...
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