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Equilibrium

Equilibrium. Chapter 17. Equilibrium – occurs when the forward and reverse rxns happen at the same time Dynamic equilibrium – chemical rxn happens continuously with no net change

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Equilibrium

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  1. Equilibrium Chapter 17

  2. Equilibrium – occurs when the forward and reverse rxns happen at the same time • Dynamic equilibrium – chemical rxn happens continuously with no net change Ex. saturated sugar in iced tea ~ the sugar is undergoing molecular solvation at the same time that its re-crystalizing so you can’t see a net change of sugar on the bottom Ex. Ocean ~ water is evaporating and condensing at the same time

  3. A + B  C + D rate = k[A][B] After time, C & D react together to make A & B. (the reverse rxn) C + D  A + B rate = k’[C][D] So aA + bBcC + dDKc = [C]c[D]d [A]a[B]b Kc = conc. of products (or activities) conc. of reactants For aqueous activity = concentration gas activity = pressure solid activity = 1

  4. Equilibrium Constant, Kc • the equilibrium constant tells the tendency of the rxn to either form products or stay as reactants if Kc is large, greater than 1 = rxn likes to form products (product favored) if Kc is small, less than 1 = rxn likes to stay as reactants (reactant favored) if Kc = 1, then the rxn is at equilibrium (very rare)

  5. Finding Kc expression When finding K, don’t use liquids or solids, also K has no units Ex. 1) N2(l) + 3H2(g) 2NH3(g) Kc = Ex. 2) MgCO3(s) Mg2+(aq) + CO32-(aq) Kc =

  6. For a rxn at a specific temp, Kc is always the same value when the concentrations are at equil. Ex. 3) 2SO2(g) + O2(g)  2 SO3(g) Exp. 1 = 0.344 M 0.172 M 0.056 M Exp. 2 = 0.424 M 0.212 M 0.076 M For experiments 1 & 2 find Kc at 1500K. Kc1= Kc2 =

  7. Q, the reaction quotient • Q measures the progress of the reaction If… Q = Kc the rxn is at equilibrium Q < Kc the rxn proceeds to the right (produces products) Q > Kc the rxn proceeds to the left (reactants are produced)

  8. Ex. 4) At a high temp, Kc = 65.0 for this rxn: 2HI(g) H2(g) + I2(g) Using the following conc. determine if the system is at equilibrium. If not, which direction will the rxn proceed to reach equilibrium? [HI] = 0.500 M, [H2] = 2.80 M, [I2] = 3.40 M

  9. Le Chatelier’s Principle of shift • A rxn at equilibrium will tend to stay at equilibrium • If you disturb the rxn with a stress, the equil. will shift in a direction to minimize the stress. • 3 types of stressors • Concentration • Pressure • Temp/heat

  10. 1. Concentration – adding or removing reactants/products Ex. 5) H2(g) + CO(g) H2CO(aq) Add more CO - Remove H2CO - Remove H2 -

  11. 2. Pressure • for gases only. Why? s & l are only slightly compressible. • look at the number of moles to determine which way the rxn will shift • If P decreases, V increases; shift to the side with the larger # of gas moles • If P increases, V decreases; shift to the side with the smaller # of gas moles Ex. 6) 3H2(g) + N2(g) 2 NH3(g) Increase pressure Ex. 7) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Increase pressure

  12. 3. temp/heat • Reminder - heat is a part of the rxn. • Endothermic – ∆H is a reactant • Exothermic – ∆H is a product • Note: K of an exothermic rxn decreases with increasing temp, and K of an endothermic rxn increases with increasing temp.

  13. Magnitude • The magnitude of Kc is a measure of the extent to which reactions occur. For any balanced eqn. the value of Kc • is constant at a given temp • changes if temp. changes • does not depend on the initial concentrations.

  14. Notes and Example Problems Chemical Equilibrium There are basically six types of equilibrium problems in this chapter. All are approached in the same way except for the solving of x. All can be solved using RICE. R = reaction, I = Initial conc, C = Change in conc, E = Equilibrium or ending amount • The following are examples of each kind of problem.

  15. I. All beginning amounts and at least one equilibrium amount is given. Ex. 8) What is the value of Kc for N2 + 3 H2 ↔ 2 NH3, if you start with2.00M N2, 4.00M H2, and 5.00M NH3 and at equilibrium you have 2.00M H2. • R • I • C • E

  16. II. Only the beginning amounts of reactants are given, and x can be solved by taking the square root of each side of the equation. Ex. 9) K for A + B ↔ C + D is 49. If 0.400 moles of each A and B are placed in a 2.00 L flask, what are the concentrations of all species at equilibrium? (see page 669 for more help)

  17. III. Only beginning amounts of reactants are given and the quadratic equation must be used to solve for x. Ex. 10) For A + B ↔ C + D, the equilibrium constant is 49 at a certain temperature. If 0.600 moles of A and 0.200 moles of B are placed in a 2.00L flask, what concentration of all species are present at equilibrium?

  18. IV. Only beginning amounts are given, however, equation may be simplified so that the quadratic equation is not required. To be able to use this simplification, K must be small. (in the order of 1x10-4 or smaller)  rule Ex. 11) A .200 M HF solution has a Ka of 7.2x10-4. What is the equilibrium concentration of H+? HF(aq) ↔ H+(aq) + F-(aq)

  19. V. Gaseous equilibrium: worked exactly like other equilibrium problems, but the concentrations are given in atmospheres. The equilibrium expression takes on a different look since P is used in place of the brackets. *solids and pure liquids are not included in the equil expression Ex. 12) A(s) ↔ B(g) + C(g) Solid A is placed in a vessel. After equilibrium is established, the total pressure is 0.75 atm. What is the value for Kp?

  20. VI. Beginning amounts are given for all reactants and products, but no equilibrium amounts are given. Q, the reaction quotient, must be used. Ex. 13) H2 + I2 ↔ 2HI, K for this reaction is 50. The following concentrations are initial concentrations: H2 = 0.080 M, I2 = 0.060 M, HI =0 .79 M. What concentrations of H2, I2, and HI will be present at equilibrium?

  21. Relationship between Kp & Kc Kp = Kc(RT)∆n or Kc = Kp(RT) -∆n ∆n = (moles gas products) – (moles gas reactants) Ex. 14) For the rxn 2S(g) + 3O2(g) 2SO3(g), Kp = 2.2 x 108 at 301oC. What is Kc?

  22. Ex.15) Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25oC, in a vessel in which the total pressure is 0.25 atmosphere. 2NOBr(g) ↔ 2NO(g) + Br2(g) A) What is the value of Kp? B) What is Kc?

  23. Relationship between ∆Gorxn & KFree Energy and Equilibrium Thermodynamic View of Equilibrium • Equilibrium point occurs at the lowest value of free energy available to the reaction system ∆Gorxn = -RTlnK R = 8.314 J/molK b/c of ∆Go ∆Gorxn< 0 K > 1 products favored ∆Gorxn = 0 K = 1 at equilibrium (very rare) ∆Gorxn > 0 K < 1 reactants favored Ex. 16) Find the Kp for N2O4(g) 2NO2(g) for the temp at 25oC.

  24. K at different temperatures van’t Hoff eqn. allows us to find K at various temps lnKT2 = ∆Ho1 - 1 KT1 R T1 T2 Ex. 17) Estimate Kp at 250. oC for the rxn of 2NO2(g) 2NO(g) + O2(g)Kp = 4.3 x 10-13at 25oC and ∆Ho = 114kJ/mole.

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