CHE1102, Chapter 18 Learn, 1 Chapter 18 Thermodynamics

CHE1102, Chapter 18 Learn, 2 Spontaneous process – any process, once started, that proceeds without the external input of energy Nonspontaneous process – any process which requires the continual, external input of energy to keep the process going Any process that is spontaneous in one direction will be nonspontaneous in the reverse direction ! Spontaneous vs Nonspontaneous Change

CHE1102, Chapter 18 Learn, 3 Nonspontaneous Change Occurs only with outside assistance Never occurs by itself: –Room gets straightened up –Pile of bricks turns into a brick wall –Decomposition of H 2 O by electrolysis Continues only as long as outside assistance occurs: –Person does work to clean up room –Bricklayer layers mortar and bricks –Electric current passed through H 2 O

CHE1102, Chapter 18 Learn, 4 Spontaneous Change What factors influence spontaneity? Spontaneous Change –Occurs by itself –Without outside assistance until finished e.g. –Water flowing over waterfall –Melting of ice cubes in glass on warm day

CHE1102, Chapter 18 Learn, 5

CHE1102, Chapter 18 Learn, 6 Spontaneous Processes Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. Above 0  C, it is spontaneous for ice to melt. Below 0  C, the reverse process is spontaneous.

CHE1102, Chapter 18 Learn, 7 Spontaneous processes tend to favor: 1. Decrease in Energy 2. Increase in Disorder

CHE1102, Chapter 18 Learn, 8

CHE1102, Chapter 18 Learn, 9 Direction of Spontaneous Change Many reactions which occur spontaneously are exothermic: –Iron rusting –Fuel burning  H and  E system are negative –Heat given off –Energy leaving system Thus,  H is one factor that influences spontaneity

CHE1102, Chapter 18 Learn, 10 Direction of Spontaneous Change Some endothermic reactions occur spontaneously: –Ice melting –Evaporation of water –Expansion of CO 2 gas into vacuum –The operation of a chemical “cold pack”  H and  E are positive –Heat absorbed –Energy entering system Clearly other factors influence spontaneity

CHE1102, Chapter 18 Learn, 11 Entropy (Symbol S ) Thermodynamic quantity Describes number of equivalent ways that energy can be distributed Can be thought of as a measure of the randomness or disorder of a system The greater the statistical probability of a particular state the greater the entropy! –Larger S, means more possible ways to distribute energy and that it is a more probable result

CHE1102, Chapter 18 Learn, 12 Disorder Driven by Statistical Probability Statistical probability – refers to the number of possible arrangements of a system A disordered state is more probable because it can be achieved in more statistical ways the larger the number of different possible combinations, the greater the probability of getting a disordered state

CHE1102, Chapter 18 Learn, 13 Examples of Spontaneity Spontaneous reactions –Things get rusty spontaneously Don't get shiny again –Sugar dissolves in coffee Stir more—it doesn't undissolve –Ice  liquid water at RT Opposite does NOT occur –Fire burns wood, smoke goes up chimney Can't regenerate wood Common factor in all of these: –Increase in randomness and disorder of system –Something that brings about randomness more likely to occur than something that brings order

CHE1102, Chapter 18 Learn, 14 There are 2,598,960 possible five-card poker hands

CHE1102, Chapter 18 Learn, 15 Entropy, S less disorder S = smaller # more disorder S = larger # Ludwig Boltzmann 1844 – 1906

CHE1102, Chapter 18 Learn, 16

CHE1102, Chapter 18 Learn, 17 order disorder is favored to occur spontaneously order disorder ΔS = + # 2 nd Law of Thermodynamics – the entropy of the universe is increasing over time disorder order ΔS = - #

CHE1102, Chapter 18 Learn, 18 Entropy, S State function Independent of path  S = Change in entropy For chemical reactions or physical processes Factors that affect entropy – volume, temperature, physical state, number of particles

CHE1102, Chapter 18 Learn, 19 Effect of Volume on Entropy For gases, entropy increases as volume increases a)Gas separated from vacuum by partition b)Partition removed, more ways to distribute energy c)Gas expands to achieve more probable particle distribution More random, higher probability, more positive S

CHE1102, Chapter 18 Learn, 20 3 rd Law of Thermodynamics – entropy decreases as temperature decreases (and vice-versa) Effect of Temperature on Entropy

CHE1102, Chapter 18 Learn, 21 Effect of Physical State on Entropy Crystalline solid very low entropy Liquid higher entropy, molecules can move freely –More ways to distribute KE among them Gas highest entropy, particles randomly distributed throughout container –Many, many ways to distribute KE

CHE1102, Chapter 18 Learn, 22 Entropy Affected by Number of Particles Adding particles to system Increase number of ways energy can be distributed in system So all other things being equal Reaction that produces more particles will have positive  S

CHE1102, Chapter 18 Learn, 23 Learning Check Which represents an increase in entropy? A. Water vapor condensing to liquid B. Carbon dioxide subliming C. Liquefying helium gas D. Proteins forming from amino acids

CHE1102, Chapter 18 Learn, 24 Entropy Changes in Chemical Reactions Reactions without gases - Calculate number of mole molecules  n = n products – n reactants If  n is positive, entropy increases More molecules, means more disorder Reactions involving gases - Calculate change in # of moles of gas,  n gas If  n gas is positive,  S is positive  n gas is more important than  n molecules

CHE1102, Chapter 18 Learn, 25 Predict Sign of  S for Following Reactions CaCO 3 (s) + 2H + (aq)  Ca 2+ (aq) + H 2 O (l) + CO 2 (g) –  n gas = 1 mol – 0 mol = 1 mol –since  n gas is positive,  S is positive 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) –  n gas = 4 mol + 1 mol – 2 mol = 3 mol –since  n gas is positive,  S is positive OH – (aq) + H + (aq)  H 2 O (l) –  n gas = 0 mol –  n = 1 mol – 2 mol = –1 mol –since  n is negative,  S is negative

CHE1102, Chapter 18 Learn, 26 Learning Check Which of the following has the most entropy at standard conditions? A. H 2 O(l ) B. NaCl(aq) C. AlCl 3 (s) D. Can’t tell from the information

CHE1102, Chapter 18 Learn, 27 Standard entropy of reaction – the (entropy, or disorder) that accompanies ANY reaction under standard conditions (units are J/K) = [(sum S  products) – (sum S  reactants)] 1. write and balance the reaction 2. use table of thermodynamic data of S  to calculate Standard entropy, S  – disorder of a substance at standard conditions (units are J/mole·K)

CHE1102, Chapter 18 Learn, 28 N 2 O (g) + O 2 (g) NO 2 (g) 34 Determine the standard entropy of reaction, 2 = – 93.0 J/K = – #, suggests the reaction is nonspontaneous

CHE1102, Chapter 18 Learn, 29 Δ H rxn only slightly varies with temperature Δ S rxn is highly dependent on temperature Recap: 2 forces in nature which drive processes to occur spontaneously 1. Decrease in energy ΔH rxn = – # (exo) 2. Increase in disorder ΔS rxn = + # (disorder)

CHE1102, Chapter 18 Learn, 30 Willard Gibbs 1839 – 1903 ΔG rxn = ΔH rxn – T ΔS rxn ΔH rxn is the heat of reaction ΔS rxn is the entropy of reaction T is the temperature in Kelvin ΔG rxn is the Gibbs Free Energy

CHE1102, Chapter 18 Learn, 31 ΔG rxn or Gibbs Free Energy, is the ultimate, final deciding factor as to whether a reaction will occur spontaneously, anywhere in the universe Bottom Line: When ΔG rxn = – #, the reaction is spontaneous. When ΔG rxn = + #, the reaction is nonspontaneous. When ΔG rxn = 0, the reaction is at equilibrium (has no tendency to go one way or the other)

CHE1102, Chapter 18 Learn, 32 Is the following reaction spontaneous at 175 °C ? Δ G rxn = + #, the reaction is nonspontaneous Δ G rxn = kJ 42 N 2 O (g) + O 2 (g) NO 2 (g) 3 = – 93.0 J/K favors nonspontaneous favors spontaneous = – 28.0 kJ Who wins ?? ΔG rxn = ΔH rxn – T ΔS rxn

CHE1102, Chapter 18 Learn, 33

CHE1102, Chapter 18 Learn, 34

CHE1102, Chapter 18 Learn, 35 Learning Check For the reaction 2NO(g) + O 2 (g)  2NO 2 (g),  H° = kJ/mol and  S° = J/K mol. Which of these statements is true? A. The reaction is spontaneous at all temperatures. B. The reaction is only spontaneous at low temperatures. C. The reaction is only spontaneous at high temperatures. D. The reaction is at equilibrium at 25°C under standard conditions.

CHE1102, Chapter 18 Learn, 36 Learning Check At what temperature (K) will a reaction become nonspontaneous when  H = –50.20 kJ mol –1 and  S = J K –1 mol –1 ? A. 298 K B K C K D. The reaction cannot become non-spontaneous at any temperature We are looking for when ΔG > 0 So ΔH – TΔS > 0 or ΔH > TΔS ΔH/ΔS > T ( kJ mol -1 )/( kJ K -1 mol -1 ) > T K > T so not possible

CHE1102, Chapter 18 Learn, 37 Calculate  S° for reduction of aluminum oxide by hydrogen gas Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) SubstanceS ° (J/ K mol) Al(s)28.3 Al 2 O 3 (s)51.00 H2(g)H2(g)130.6 H 2 O(g)188.7

CHE1102, Chapter 18 Learn, 38 Calculate ΔS°  S ° = 56.5 J/K J/K – (51.00 J/K J/K )  S ° = J/K Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) = + #, suggests the reaction is spontaneous

CHE1102, Chapter 18 Learn, 39 Standard Free Energy Changes Standard Free Energy Change,  G ° –  G measured at 25 °C (298 K) and 1 atm Two ways to calculate, depending on what data is available Method 1:  G ° =  H ° – T  S ° Method 2:

CHE1102, Chapter 18 Learn, 40 Calculate  G ° Method 1 Al 2 O 3 (s) + 3H 2 (g)  2Al (s) + 3H 2 O (g) Step 1: Calculate  H° for reaction using heats of formation below Calculate  G ° for reduction of aluminum oxide by hydrogen gas

CHE1102, Chapter 18 Learn, 41 Calculate  G ° Method 1  H ° = 0.0 kJ – kJ – 0.00 kJ – (– kJ)  H ° = kJ

CHE1102, Chapter 18 Learn, 42 Calculate  G ° Method 1 Step 2: Calculate  S °  see Example 3  S ° = J/K Step 3: Calculate  G ° =  H ° – ( K)  S °  G ° = kJ – (298 K)(179.9 J/K)(1 kJ/1000 J)  G ° = kJ – 53.6 kJ = kJ –  G ° is positive –Indicates that the reaction is not spontaneous

CHE1102, Chapter 18 Learn, 43 Calculate  G ° Method 2 Use Standard Free Energies of Formation Energy to form 1 mole of substance from its elements in their standard states at 1 atm and 25 °C

CHE1102, Chapter 18 Learn, 44 Calculate  G ° Method 2 Calculate  G ° for reduction of aluminum oxide by hydrogen gas. Al 2 O 3 (s) + 3H 2 (g)  2Al(s) + 3H 2 O(g) Substance (kJ/mol) Al (s) 0.0 Al 2 O 3 (s) – H2(g)H2(g) 0.0 H2O(g)H2O(g) – 228.6

CHE1102, Chapter 18 Learn, 45 Calculate  G ° Method 2  G° = 0.0 kJ – kJ – 0.00 kJ – (– kJ)  G° = kJ Both methods same within experimental error

CHE1102, Chapter 18 Learn, 46  G ° = Maximum Possible Work  G ° is maximum amount of energy produced during a reaction that can theoretically be harnessed as work –Energy that need not be lost to surroundings as heat –Energy that is “free” or available to do work

CHE1102, Chapter 18 Learn, 47 How K is related to  G °  G ° = –RT lnK K = e –  G°/RT

CHE1102, Chapter 18 Learn, 48  G° and Position of Equilibrium When  G ° > 0 (positive) –Position of equilibrium lies close to reactants –Little reaction occurs by the time equilibrium is reached –Reaction appears nonspontaneous When  G ° < 0 (negative) –Position of equilibrium lies close to products –Mainly products exist by the time equilibrium is reached –Reaction appears spontaneous

CHE1102, Chapter 18 Learn, 49  G ° and Position of Equilibrium When  G ° = 0 –Position of equilibrium lies ~ halfway between products and reactants –Significant amount of both reactants and products present at time equilibrium is reached –Reaction appears spontaneous, whether start with reactants or products Can Use  G ° to Determine Reaction Outcome –  G ° large and positive No observable reaction occurs –  G ° large and negative Reaction goes to completion

CHE1102, Chapter 18 Learn, 50 Calculating  G ° from K K sp for AgCl(s) at 25 °C is 1.8  10 –10 Determine  G ° for the process Ag + (aq) + Cl – (aq)  AgCl(s) Reverse of K sp equation, so  G ° = –RT lnK = –( J/K mol)(298 K) × ln(5.6  10 9 ) × (1 kJ/1000 J)  G ° = –56 kJ/mol Negative  G ° indicates precipitation will occur

CHE1102, Chapter 18 Learn, 51 Determining Effect of Temp on Spontaneity Calculate  G ° at 25 °C and 500 °C for the Haber process N 2 (g) + 3H 2 (g)  2NH 3 (g) Assume that  H ° and  S ° do not change with temperature –  H ° = – kJ –  S ° = – J/K

CHE1102, Chapter 18 Learn, 52 Determining Effect of Temp on Spontaneity Step 1. Calculate  G ° for the reaction at 25 °C using  H ° and  S ° N 2 (g) + 3H 2 (g)  2NH 3 (g) –  H ° = –92.38 kJ –  S ° = –198.4 J/K  G ° =  H ° – T  S °  G ° = –92.38 kJ – (298 K)(–198.4 J/K)  G ° = –92.38 kJ kJ = –33.3 kJ So the reaction is spontaneous at 25 °C

CHE1102, Chapter 18 Learn, 53 Determining Effect of Temp on Spontaneity Step 2. Calculate  G ° for the reaction at 500 °C using  H ° and  S °. –T = 500 °C = 773 K –  H ° = – kJ –  S ° = – J/K  G ° =  H ° – T  S °  G ° = –92.38 kJ – (773 K)(–198.4 J/K)  G ° = –92.38 kJ kJ = 61 kJ So the reaction is NOT spontaneous at 500 °C

CHE1102, Chapter 18 Learn, 54 Your Turn! Calculate the equilibrium constant for the decomposition of water at 298 K given  G ° = kJ mol. A × B x 10 0 C.7 × D × 10 41