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CHEMICAL EQUILIBRIUM Chapter 16

CHEMICAL EQUILIBRIUM Chapter 16. equilibrium vs. completed reactions equilibrium constant expressions Reaction quotient computing positions of equilibria: examples Le Chatelier’s principle - effect on equilibria of: addition of reactant or product pressure temperature.

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CHEMICAL EQUILIBRIUM Chapter 16

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  1. CHEMICAL EQUILIBRIUMChapter 16 • equilibrium vs. completed reactions • equilibrium constant expressions • Reaction quotient • computing positions of equilibria: examples • Le Chatelier’s principle - effect on equilibria of: • addition of reactant or product • pressure • temperature YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics) Entropy & Free Energy (Ch 20)

  2. a A + b B c C + d D THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type the following is a CONSTANT (at a given T) : If K is known, then we can predict concentrations of products or reactants. Entropy & Free Energy (Ch 20)

  3. [iso] 0.35 n-Butane iso-Butane 0.25 0.35 Q = = = 1.40 [n] 0.25 Q - the reaction quotient Q has the same form as K, . . . but uses existing concentrations All reacting chemical systems can be characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. Since K =2.5, system NOT AT EQUIL. To reach EQUILIBRIUM [Iso] must INCREASE and [n] must DECREASE. Entropy & Free Energy (Ch 20)

  4. Q/K 1 Q  Q = K Typical EQUILIBRIUM Calculations 2 general types: a. Given set of concentrations, is system at equilibrium ? Calculate Q compare to K IF: Q > K or Q/K > 1  REACTANTS Q < K or Q/K < 1  PRODUCTS Q=K at EQUILIBRIUM Entropy & Free Energy (Ch 20)

  5. H2(g) + I2(g) 2 HI(g) Kc = 55.3 Step 1Define equilibrium condition in terms of initial condition and a change variable [H2] [I2] [HI] At equilibrium 1.00-x 1.00-x 2x Step 2 Put equilibrium [ ] into Kc . Step 3. Solve for x. 55.3 = (2x)2/(1-x)2 Square root of both sides & solve gives: x = 0.79 [H2] = [I2] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M At equilibrium Entropy & Free Energy (Ch 20)

  6. EQUILIBRIUM AND EXTERNAL EFFECTS • The position of equilibrium is changed when there is a change in: • pressure • changes in concentration • temperature • The outcome is governed by LE CHATELIER’S PRINCIPLE Henri Le Chatelier 1850-1936 - Studied mining engineering - specialized in glass and ceramics. “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.” Entropy & Free Energy (Ch 20)

  7. Shifts in EQUILIBRIUM : Concentration • If concentration of one species changes, concentrations of other species CHANGES to keep the value of K the same (at constant T) • no change in K - only position of equilibrium changes. ADDING PRODUCTS - equilibrium shifts to REACTANTS ADDING REACTANTS - equilibrium shifts to PRODUCTS - GAS-FORMING; PRECIPITATION REMOVING PRODUCTS - often used to DRIVE REACTION TO COMPLETION Entropy & Free Energy (Ch 20)

  8. n-Butane Isobutane [iso] = 2.5 K = [n] Effect of changed [ ] on an equilibrium Solution A. Calculate Q with extra 1.50 M n-butane. INITIALLY: [n] = 0.50 M [iso] = 1.25 M CHANGE: ADD +1.50 M n-butane What happens ? 16_butane.mov (16m13an1.mov) Q = [iso] / [n] = 1.25 / (0.50 + 1.50) = 0.63 Q < K . Therefore, reaction shifts to PRODUCT Entropy & Free Energy (Ch 20)

  9. B A Butane/Isobutane B. Solve for NEW EQUILIBRIUM - set up concentration table [n-butane] [isobutane] Initial 0.50 + 1.50 1.25 Change - x + x Equilibrium 2.00 - x 1.25 + x Solution x = 1.07 M. At new equilibrium position, [n-butane] = 0.93 M [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane. Entropy & Free Energy (Ch 20)

  10. N2O4(g) 2 NO2(g) Effect of Pressure (gas equilibrium) Increase P in the system by reducing the volume. Increasing P shifts equilibrium to side with fewer molecules (to try to reduce P). Here, reaction shifts LEFT PN2O4 increases 16_NO2.mov (16m14an1.mov) PNO2 decreases See Ass#2 - question #6 Entropy & Free Energy (Ch 20)

  11. CO2(g) + H2O(liq) CO2(aq) + heat LOWER T HIGHER T EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature change  change in K • Consider the fizz in a soft drink Kc = [CO2(aq)]/[CO2(g)] • Change T: New equilib. position? New value of K? • Increase T • Equilibrium shifts left: [CO2(g)]  [CO2 (aq)] • K decreases as T goes up. • Decrease T • [CO2 (aq)] increases and [CO2(g)] decreases. • K increases as T goes down Entropy & Free Energy (Ch 20)

  12. N2O4 + heat 2 NO2 (colorless) (brown) Temperature Effects on Chemical Equilibrium Kc = 0.00077 at 273 K Kc = 0.00590 at 298 K Horxn = + 57.2 kJ Increasing T changes K so as to shift equilibrium in ENDOTHERMIC direction 16_NO2RX.mov (16m14an1.mov) Entropy & Free Energy (Ch 20)

  13. EQUILIBRIUM AND EXTERNAL EFFECTS Catalytic exhaust system • Add catalyst ---> no change in K • A catalyst only affects the RATE of approach to equilibrium. Entropy & Free Energy (Ch 20)

  14. CHEMICAL EQUILIBRIUMChapter 16 • equilibrium vs. completed reactions • equilibrium constant expressions • Reaction quotient • computing positions of equilibria: examples • Le Chatelier’s principle - effect on equilibria of: • addition of reactant or product • pressure • temperature YOU ARE NOT RESPONSIBLE for section 16.7 (relation to kinetics) Entropy & Free Energy (Ch 20)

  15. Entropy and Free Energy(Kotz Ch 20) • Spontaneous vs. non-spontaneous • thermodynamics vs. kinetics • entropy = randomness (So) • Gibbs free energy (Go) • Go for reactions - predicting spontaneous direction • Grxn versus Gorxn • predicting equilibrium constants from Gorxn Entropy & Free Energy (Ch 20)

  16. Entropy and Free Energy ( Kotz Ch 20 ) • some processes are spontaneous; others never occur. WHY ? • How can we predict if a reaction can occur, given enough time? THERMODYNAMICS 9-paper.mov 20m02vd1.mov • Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur. • To predict if a reaction can occur at a reasonable rate, one needs to consider: KINETICS Entropy & Free Energy (Ch 20)

  17. Thermodynamics • state of a chemical system (P, T, composition) • From a given state, would a chemical reaction decrease the energy of the system? • If yes, system is favored to react — a product-favored system which will have aspontaneousreaction. • Most product-favored reactions are exothermic. • Spontaneous does not imply anything about time for reaction to occur. (kinetics) Entropy & Free Energy (Ch 20)

  18. Thermodynamics versus Kinetics • Diamond to Graphite • spontaneous from thermodynamics • but not kinetically favored. • Paper burns. • - product - favored reaction. • - Also kinetically favored once reaction is begun. Entropy & Free Energy (Ch 20)

  19. Product-Favored Reactions In general, product-favored reactions are exothermic. E.g. thermite reaction Fe2O3(s) + 2 Al(s)  2 Fe(s) + Al2O3(s) DH = - 848 kJ Entropy & Free Energy (Ch 20)

  20. Non-exothermic spontaneous reactions But many spontaneous reactions or processes are endothermic . . . NH4NO3(s) + heat  NH4+ (aq) + NO3- (aq) Hsol = +25.7 kJ/mol or have H = 0 . . . Entropy & Free Energy (Ch 20)

  21. Entropy, S One property common to product-favored processes is that the final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. The thermodynamic property related to randomness is ENTROPY, S. Reaction of K with water Entropy & Free Energy (Ch 20)

  22. PROBABILITY - predictor of most stable state WHY DO PROCESSES with  H = 0 occur ? Consider expansion of gases to equal pressure: This is spontaneous because the final state, with equal # molecules in each flask, is much more probable than the initial state, with all molecules in flask 1, none in flask 2 SYSTEM CHANGES to state of HIGHER PROBABILITY THIS IS USUALLY the more RANDOM state. Entropy & Free Energy (Ch 20)

  23. Gas expansion - spontaneity from greater probability Consider distribution of 4 molecules in 2 flasks P1 < P2 P1 > P2 P1 = P2 With more molecules (>1020) P1=P2 is most probable by far Entropy & Free Energy (Ch 20)

  24. WHAT about EXOTHERMIC REACTIONS ? Consider 2 H2 (g) + O2 (g)  2 H2O (l) • HIGHLY EXOTHERMIC • final state (liquid) is MUCH MORE ORDERED • (less random arrangement) than initial state (2 gases) • BUT PROBABILITY of final state is higher when one • considers change in the surroundings. WHY ? • Heat evolved increases motion of molecules in the surroundings Increased disorder of surroundings decreased disorder of system > MUST consider change in disorder in BOTH SYSTEM and SURROUNDINGS to predict DIRECTION of SPONTANEITY Entropy & Free Energy (Ch 20)

  25. Directionality of Reactions How probable is it that reactant molecules will react? PROBABILITY suggests that a product-favored reaction will result in the dispersal of energyordispersal of matteror both. Entropy & Free Energy (Ch 20)

  26. Directionality of Reactions Probability suggests that a product-favored reaction will result in the dispersal of energy or of matter or both. Matter Dispersal Energy Dispersal 9_gasmix.mov 20m03an1.mov 9_exorxn.mov 20m03an2.mov Entropy & Free Energy (Ch 20)

  27. Directionality of ReactionsEnergy Dispersal • Exothermic reactions involve a release of stored chemical potential energy to the surroundings. • The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules. • The final state—with energy dispersed—is more probable and makes a reaction product-favored. Entropy & Free Energy (Ch 20)

  28. Standard Entropies, So • Every substance at a given temperature and in a specific phase has a well-defined Entropy • At 298o the entropy of a substance is called • So - with UNITS of J.K-1.mol-1 • The larger the value of So, the greater the degree of disorder or randomness • e.g. So (in J.K-1mol-1) : Br2 (liq) = 152.2 • Br2 (gas) = 245.5 • For any process: So =  So(final) -  So(initial) So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1 Entropy & Free Energy (Ch 20)

  29. Vapour Ice Water Entropy, S S (gases) > S (liquids) > S (solids) So (J/K•mol) H2O(gas) 188.8 H2O(liq) 69.9 H2O (s) 47.9 Entropy & Free Energy (Ch 20)

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