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Equilibrium

Equilibrium. Reactions are reversible. A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down.

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Equilibrium

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  1. Equilibrium

  2. Reactions are reversible • A + B C + D ( forward) • C + D A + B (reverse) • Initially there is only A and B so only the forward reaction is possible • As C and D build up, the reverse reaction speeds up while the forward reaction slows down. • Eventually the rates are equal

  3. Forward Reaction Reaction Rate Equilibrium Reverse reaction Time

  4. What is equal at Equilibrium? • Rates are equal • Concentrations are not. • Rates are determined by concentrations and activation energy. • The concentrations do not change at equilibrium. or if the reaction is very slow.

  5. Law of Mass Action • For any reaction • jA + kB lC + mD • K = [C]l[D]m PRODUCTSpower [A]j[B]k REACTANTSpower • K is called the equilibrium constant. • is how we indicate a reversible reaction

  6. Playing with K • If we write the reaction in reverse. • lC + mD jA + kB • Then the new equilibrium constant is • K’ = [A]j[B]k = 1/K [C]l[D]m

  7. Playing with K • If we multiply the equation by a constant • njA + nkB nlC + nmD • Then the equilibrium constant is • K’ = [A]nj[B]nk= ([A] j[B]k)n= Kn[C]nl[D]nm ([C]l[D]m)n

  8. The units for K • Are determined by the various powers and units of concentrations. • They depend on the reaction. • Are usually dropped, unlike rate constant units which are specifically asked for.

  9. K is CONSTANT • At any temperature. • Temperature affects rate. • The equilibrium concentrations don’t have to be the same, only K. • Equilibrium position is a set of concentrations at equilibrium. • There are an unlimited number.

  10. Equilibrium Constant UNIQUE FOR EACH TEMPERATURE

  11. Calculate K • N2 + 3H2 2NH3 Initial At Equilibrium [N2]0 =1.000 M [N2] = 0.921M [H2]0 =1.000 M [H2] = 0.763M [NH3]0 =0 M [NH3] = 0.157M

  12. Calculate K • N2 + 3H2 2NH3 Initial At Equilibrium [N2]0 = 0 M [N2] = 0.399 M [H2]0 = 0 M [H2] = 1.197 M [NH3]0 = 1.000 M [NH3] = 0.157M • K is the same no matter what the amount of starting materials

  13. Equilibrium and Pressure • Some reactions are gaseous • PV = nRT • P = (n/V)RT • P = CRT • C is a concentration in moles/Liter • C = P/RT

  14. Equilibrium and Pressure • 2SO2(g) + O2(g) 2SO3(g) • Kp = (PSO3)2 (PSO2)2 (PO2) • K = [SO3]2 [SO2]2 [O2]

  15. Equilibrium and Pressure • K =(PSO3/RT)2 (PSO2/RT)2(PO2/RT) • K =(PSO3)2 (1/RT)2 (PSO2)2(PO2) (1/RT)3 • K = Kp (1/RT)2= Kp RT (1/RT)3

  16. General Equation • jA + kBlC + mD • Kp= (PC)l (PD)m= (CCxRT)l (CDxRT)m (PA)j (PB)k (CAxRT)j(CBxRT)k • Kp=(CC)l (CD)mx(RT)l+m(CA)j(CB)kx(RT)j+k Kp=K (RT)(l+m)-(j+k) = Kc(RT)Dn Dn=(l+m)-(j+k)=change in moles of gas

  17. Homogeneous Equilibria • So far every example dealt with reactants and products where all were in the same phase. • We can use K in terms of either concentration or pressure. • Units depend on reaction.

  18. Heterogeneous Equilibria • If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. • As long as they are not used up we can leave them out of the equilibrium expression.

  19. For Example • H2(g) + I2(s) 2HI(g) • K = [HI]2 [H2][I2] • But the concentration of I2 does not change. • K[I2]= [HI]2 = K’ [H2]

  20. Solving Equilibrium Problems Type 1

  21. The Reaction Quotient • Tells you the directing the reaction will go to reach equilibrium • Calculated the same as the equilibrium constant, but for a system not at equilibrium • Q = [Products]coefficient [Reactants] coefficient • Compare Q value to the equilibrium constant

  22. What Q tells us • If Q<K K > Q • Not enough products • Shifts right • If Q>K K < Q • Too many products • Shifts left • If Q=K system is at equilibrium

  23. Example For the reaction 2NOCl(g) 2NO(g) + Cl2(g) K = 1.55 x 10-5 M at 35ºC In an experiment 10.0 mol NOCl, 1.0 mol NO(g) and .10 mol Cl2 are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium?

  24. Solving Equilibrium Problems • Given the starting concentrations and one equilibrium concentration. • Use stoichiometry to figure out other concentrations and K. • Learn to create a table of initial and final conditions.

  25. Consider the following reaction at 600ºC • 2SO2(g) + O2(g) 2SO3(g) In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol SO3 were present. Calculate the equilibrium concentrations of O2 and SO2, K and KP

  26. LARGE K K >100

  27. What if you’re not given equilibrium concentration? • The size of K will determine what approach to take. • First let’s look at the case of a LARGE value of K ( >100). • Allows us to make simplifying assumptions.

  28. Example • H2(g) + I2(g) 2HI(g) • K = 7.1 x 102 at 25ºC Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.9 g of H2 294 g I2. • [H2]0 = (15.7g/2.02)/5.00 L = 1.56 M • [I2]0 = (294g/253.8)/5.00L = 0.232 M • [HI]0 = 0

  29. Q= 0<K so more product will be formed. • Assumption since K is large reaction will go to completion. • Stoichiometry tells us I2 is LR, it will be smallest at equilibrium let it be x • Set up table of initial, change in and equilibrium concentrations.

  30. H2(g) I2(g) 2HI(g) initial 1.56 M 0.232 M 0 M change eqX • Choose X so it is small. • For I2 the change in X must be X-.232 M • Final must = initial + change

  31. H2(g) I2(g) 2HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 eq X • Initial + change = final, • Change = final – initial • Using to stoichiometry we can find: • Change in H2 = = x -.232 M • Change in HI = (2(.232-x) = (.464-2x)

  32. H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 X-0.232 0.464-2Xeq X • Now we can determine the final concentrations by adding.

  33. H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 X-0.232 0.464-2X eq 1.328+X X 0.464-2X • Now plug these values into the equilibrium expression • K = (0.464-2X)2 = 7.1 x 102 (1.328+X)(X) • Now you can see why we made sure X was small.

  34. Why we chose X • K = (0.464-2X)2 = 7.1 x 102 (1.328+X)(X) • Since X is going to be small, we can ignore it in relation to 0.464 and 1.328 • So we can rewrite the equation • 7.1 x 102 = (0.464)2 (1.328)(X) • Makes the algebra easy

  35. H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 X-0.232 0.464-2X eq 1.328+X X 0.464-2X • When we solve for X we get 2.8 x 10-4 • So we can find the other concentrations • I2 = 2.8 x 10-4 M • H2 = 1.328 M • HI = 0.464 M

  36. Checking the assumption • The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid. • If not we would have had to use the quadratic equation. • More on this later………..anyone have it programmed already?????? • Our assumption was valid.

  37. Practice For the reaction Cl2 + O2 2ClO(g) K = 156 In an experiment 0.100 mol ClO, 1.00 mol O2 and 0.0100 mol Cl2 are mixed in a 4.00 L flask. • If the reaction is not at equilibrium, which way will it shift? • Calculate the equilibrium concentrations of all species.

  38. Problems with small K K< .01

  39. Process is the same • Set up table of initial, change, and final concentrations. • For a small K the product concentration is small.

  40. For example nitrogen + oxygen yields nitrogen monoxide K= 4.10 x 10-4 If .50 moles of nitrogen and .86 moles of oxygen are mixed in a 2.0 L container, what are the equilibrium concentrations of all 3 substance? Q =

  41. N2 O 2NO Initial Change eq K = Because K is small, assume x is negligible compared to .25 and .43

  42. Simplified K expression: • K = • Always check the assumption. • X must be less than 5% of the original concentration.

  43. Mid-range K’s .01<K<100

  44. QUADRATIC EQUATION TI-82 PROGRAM Hit PRGM key Using right arrow, highlight NEW and hit ENTER Give the program a name (i.e. QUAD) and hit ENTER Hit PRGM key Using right arrow, highlight I/O Highlight Input and hit ENTER Hit ALPHA key and then hit the letter A, then hit ENTER Repeat for B and C Hit PRGM key Using right arrow, highlight I/O Using down arrow, highlight Disp and hit ENTER Type in the following formula: (-B+√(B^2-4*A*C))/(2*A) and hit ENTER Repeat for the following: (-B-√(B^2-4*A*C))/(2*A) and hit ENTER To execute the program: Hit 2nd and Quit to exit the editing of the program Hit PRGM key Highlight EXEC and use down arrow to highlight QUAD and hit ENTER PRGMQUAD should show up at the top of the screen –hit ENTER ENTER first coefficient after ? (i.e. 1) ENTER second coefficient after ? (i.e. -5) ENTER constant term after ? (i.e. 6) The answers of 3 and 2 DONE should show up in the lower right corner of the screen.

  45. No Simplification • Choose X to be small. • Can’t simplify so we will have to solve the quadratic • H2(g) + I2 (g) HI(g) K=38.6 What are the equilibrium concentrations if 1.800 mol H2, 1.600 mol I2 and 2.600 mol HI are mixed in a 2.000 L container?

  46. Problems Involving Pressure • Solved exactly the same, with same rules for choosing X depending on KP • For the reaction N2O4(g) 2NO2(g) KP = .131 atm What are the equilibrium pressures if a flask initially contains 1.000 atm N2O4?

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