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Understanding Equilibrium in Chemical Reactions

Learn about the concept of equilibrium in chemical reactions, including its definition, factors affecting equilibrium position, equilibrium expressions, and calculations of equilibrium constant.

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Understanding Equilibrium in Chemical Reactions

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  1. EquilibriumEquilibrium is the state where the RATE of the forward reaction is equal to theRATE of the reverse reaction

  2. Equilibrium Alternate Definition of Equilibrium: The state where the concentration of all reactants and all products remain constant with time

  3. Reactions are reversible • A + B  C + D ( forward) • A + B C + D ( reverse) • Preferred way of writing: • Initially there is only A and B so only the forward reaction is possible in beginning • As C and D build up, the reverse reaction speeds up while the forward reaction slows down. • Eventually the rates are equal

  4. Reaction Rate vs Time ForwardReaction Equilibrium Position Reaction Rate  Reverse reaction Time

  5. Equilibrium Position • Whether the reaction lies far to the left or right depends on 3 factors: • Initial concentrations – higher concentrations means more collisions – faster reaction • Relative energies of reactants and products – nature goes to minimum energy • Degree of organization of reactants and products – nature likes maximum disorder called entropy

  6. What is equal at Equilibrium? • Rates are equal • Concentrations are NOT. • Rates are determined by concentrations • Dynamic:  constant change, activity, or progress

  7. equilibrium equilibrium equilibrium Start with NO2 Start with N2O4 Start with NO2& N2O4

  8. Law of Mass Action • For any reaction: • This is called the EQUILIBRIUM EXPRESSION • K is called the Equilibrium Constant. • Sometimes written as Kc or Keq MEMORIZE THIS EQUILIBRIUM EXPRESSION Carnegie Hall Time #17 p614

  9. Playing with K • If we write the reaction in reverse. Then the new equilibrium constant is K’ • Notice Product always in numerator

  10. Playing with K • If we multiply the equation by a constant (n) • Then the equilibrium constant is

  11. K is CONSTANT • No units • Temperature affects rate. • Each set of equilibrium concentrations is called an Equilibrium Position • There are an unlimited number of Equilibrium Positions. • Pure Solidsdo NOT appear in expressions • Pure Liquids do NOT appear in expressions • H2O(l) is pure so leave it out of expression, NOT H2O(g)

  12. 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 • Answer: • 6.02 x 10-2

  13. 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] = 0.203M • Answer: • 6.02 x 10-2

  14. Calculate K • N2 + 3H2 2NH3 • Initial At Equilibrium • [N2]0 = 2.00 M [N2] = 2.59 M • [H2]0 = 1.00 M [H2] = 2.77 M • [NH3]0 = 3.00 M [NH3] = 1.82 M • Answer: • 6.02 x 10-2 • Expression has same value regardless of amount of gases mixed together initially CH Time #19 p614

  15. Equilibrium and Pressure • Equilibria involving gases can be described using pressures. • PV = nRT (solve for P) • P = (n/V)RTmol / vol = M • P = MRT or P=CRT (solve for C) • C = P/RT C represents the molar concentration of the gas

  16. Equilibrium and Pressure • 2SO2(g) + O2(g)  2SO3(g) Write the Equilibrium Expression • Kp is equilibrium constant in terms of partial pressures of the gases. • Use ( ) instead of [ ] in expression

  17. Calculate Kp • The reaction for the formation of nitrosyl chloride • 2NO(g) + Cl2(g) 2NOCl(g) was studied at 25oC. The pressures at equilibrium were found to be; • PNOCl = 1.2 atm • PNO=5.0 X 10-2 atm • PCl2=3.0 X 10-1 atm • Write the Equilibrium Expression • Calculate the value of Kp for this reaction at 25oC Answer: 1.9 X 103 CH Time #25 p615

  18. Derivations of General Equations jA + kB lC + mD • Dn=(l+m)-(j+k)=Change, in moles of gas, SOOOOooo

  19. Heterogeneous Equilibria • So far every example dealt with reactants and products where all were in the same phase called Homogeneous Equilibria • Equilibria with more than 1 phase are called Heterogeneous Equilibria • The position of a Heterogeneous Equilibriadoes not depend on amounts of pure solids or liquids present.

  20. 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. • Don’t put pure solid or liquid in Equilibrium Expression

  21. For Example But the concentration of I2 does not change, So this is the correct expression: What you think expression will look like:

  22. Solving Equilibrium Problems Write the expressions for Kc and Kp for the following process: Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chlorine gas. Answer: Kc=[Cl2] and Kp=(PCl2) Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II) sulfate Answer: Kp = (PH2O)5CHT #29 p615

  23. Application of Equilibrium Constant New Condition CO=? H2O=? H2=? CO2=? Initial Condition CO=? H2O=? H2=? CO2=? Fill in the values for Initial and New Condition Keq = 6.25 Is the New Condition at Equilibrium?

  24. Application of Equilibrium Constant Initial Condition CO = 7 H2O = 7 H2 = 0 CO2 = 0 @ Equilibrium CO = 7-x H2O = 7-x H2 = +x CO2 = +x x CO Disappear x H2O Disappear x H2 Form x CO2 Form For the system to be at Equilibrium we must have the following ratio This is what Equilibrium looks like.

  25. The Reaction Quotient • Used to determine the direction system must shift to reach equilibrium • Calculated the same as the equilibrium constant, but for a systemNOT at equilibrium • Compare Q to K

  26. Compare Q : K 3 POSSIBLE CASES • If Q<K • Not enough products • Shift to right more product is made • If Q>K • Too many products • Shift to left more reactant is formed • If Q=K, system is at equilibrium

  27. Example For the synthesis of ammonia at 500oC, the equilibrium constant is 6.0 X 10-2. Predict the direction in which the system will shift to reach equilibrium in the following case: [NH3]0=1.0X10-3M; [N2]0=1.0X10-5M; [H2]0=2.00X10-3M Steps: • Write the balanced equation • Calculate Q • Compare Q to K: 1.3X107 > 6.0x10-2 • Because Q > K the concentration of product must be decreased and the system will shift to the left • N2 + 3H2 2NH3 CHT #33

  28. 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. • ICE is nICE – coming soon to your paper!

  29. Calculating Equilibrium Concentrations Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentration of all species if 1.000 mol of each component is mixed in a 1.000L flask. Calculate Initial Concentrations [CO]0=[H2O]0=[CO2]0=[H2]0 = 1.000mol/1.000L = 1.000M

  30. Calculate Q Compare Q to K: Because Q < K the system is not at equilibrium yet and needs to shift to the right. “But how much?” We define all the reactants and products in terms of x (change from initial conc.) Example This is nICE: Set up a table to enter information

  31. Sign of X determined by the shift direction. Plug into the equilibrium expression. Solve for x = 0.387mol/L (It’s algebra folks) Plug the value for x back into the original equation and calculate the concentrations at equilibrium. [CO]= [H2O] = 0.613M and [CO2] = [H2] = 1.387M

  32. ICE is nICE • Putting all the concentration values into a table is a convenient way of organizing the data. It’s called an ICE table. Be COOL - Use it on all problems! Carnegie Time # 46 p.616

  33. Steps in Solving Equilibrium Problems • Write the balancedequation • Write the equilibrium expression • List the initial concentrations (Ice Table) • Calculate Q & determine direction • Define change iCe(ICe Table) • Sub. equilibrium conc. into equilibrium expression and solve icE(IcE Table) • Check your answers Study Sample Exercise 13.11

  34. 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.

  35. Example • H2(g) + F2(g) 2HF(g) • K = 1.15 x 102 at 25ºC • Calculate the equilibrium concentrations if a 3.00 L container initially contains 6.06 g of H2 and 228 g F2 • [H2]0 = (6.06g/2.02)/3.00 L = 1.000M • [I2]0 = (228g/38.)/3.00L = 2.000M • [HI]0 = 0

  36. Q=0<K so more product will be formed because the reaction will shift to the right • Assumption since K is large, reaction will go to completion. • Stoichiometry tells us I2 is LR, it will be smallest at equilibrium What is your mathematical proof? • Set up ICE table.

  37. Use x to represent the number of moles changed • Change will be –x for the reactants because they are getting smaller by some amount called x • Change in productis 2x because the product coefficient for HF is 2 and x represents the moles of HF produced

  38. Concentration at Equilibrium is the sumof Initial and Change, plug it into Equilibrium Expression

  39. We can NOT take the square root of both sides and arrive at a simple solution. • Must use the quadratic equation

  40. Now plug these values into the equilibrium expression and calculate K The value is very close to the original K. So the calculations are correct.

  41. Problems with small KAKA Approximation Process / 5% Rule K< .01

  42. Approximation Process / 5% Rule • Set up table of initial, change, and final concentrations. • Choose X to represent a very small number. • For a small K the product concentration is small • This will greatly reduce the math needed

  43. Example For the reaction: 2NOCl 2NO +Cl2 K= 1.6 x 10-5 (This is a small number) If 1.00 mol NOCl is placed in a 2.00 L container What are the equilibrium concentrations? Set up your ICE table

  44. The equilibrium must satisfy the following equation: But when you cross multiply and collect all your terms on the same side of the equation you get an x3 term which really complicates things.

  45. The Logic Steps • K = 1.6 x 10-5 is a small number • The system will NOT go very far to the right to reach equilibrium • The original reactant concentration will NOT get much smaller • SO the value for NOCl will not get smaller • Which means the value of x is relatively small • (0.50-x) can be approximated to (0.50) eliminating the x in the denominator

  46. Making the approximation allows us to simplify the expression Solve for x = 1.0 x 10-2 How valid is the approximation? Plug the value of x into original (0.50-2(1.0 x 10-2) = 0.48 The difference between 0.50 and 0.48 is 0.02 which is 4% of the original concentration of NOCl

  47. 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 • Our assumption was valid.

  48. Practice Problem • For the reaction 2ClO(g) Cl2 (g) + O2 (g) • K = 6.4 x 10-3 • In an experiment • 0.100 molClO(g) • 1.00 mol O2 • 1.00 x 10-2mol Cl2 are mixed in a 4.00 L container. • What are the equilibrium concentrations.

  49. Mid-range K’s .01<K<10

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