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K eq Calculations (p.63m)

K eq Calculations (p.63m). Last class we talked about Keq and the equilibrium expression. To calculate Keq you need the concentrations of all of the reactants and products. Therefore, the actual value of Keq can only be found by doing an experiment.

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K eq Calculations (p.63m)

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  1. Keq Calculations (p.63m) • Last class we talked about Keq and the equilibrium expression. • To calculate Keq you need the concentrations of all of the reactants and products. • Therefore, the actual value of Keq can only be found by doing an experiment. • We will do some Keq calculations based on provided numbers. • In a couple of classes we will do a lab where we do an experiment to calculate Keq.

  2. Example A (p.63b) • A 2.0 L bulb (flask) contains 6.00 mol NO2, 3.0 mol NO, and 0.20 mol O2 at eqm. What is Keq? 2 NO(g) + O2(g) 2 NO2(g) • What is the Keq expression? • What type of values do we plug into Keq? [NO2] = [NO] = [O2] = • Keq = = 40 (you can ignore units for Keq)

  3. Example B (p.64t) • 4.00 mol NO2 put in 2.00L bulb. When eqm is reached 0.500 mol NO is present. What is Keq? • What is the Keq expression? (always write this out) • Notice we are told there is an initial situation and then a shift to eqm. Whenever this is the case, we will use an ICE table. I = initial concentrations C = E =

  4. I C E Example B (cont.) • 4.00 mol NO2 put in 2.00L bulb. When eqm is reached 0.500 mol NO is present. What is Keq? 2 NO(g) + O2(g) 2 NO2(g) • Keq = 392 (why different from Example A?) • Always do C and E in terms of “x” (not like in the book) • The C values will always be +ve on one side and –ve on the other.

  5. I C E Example C (p.65b) • Some NO2 put in 5.00L bulb. At eqm [NO]=0.800M. If Keq=24.0, how many moles NO2 were added originally? • Use “Y” as original [NO2] 2 NO(g) + O2(g) 2 NO2(g) • Keq = 16.4 mol • Note that you will never have to use the quadratic eqn in Chem 12.

  6. Example D (p.67t) • Keq=49. If 2.0 mol NO, 0.20 mol O2 and 0.40 mol NO2 are put in 2.0L bulb, which way will the rxn shift to reach eqm? • Whenever you start with reactants and products you should calculate Ktrial to see which way eqm shifts (same as Keq calc). • Ktrial = (0.20)2 / (1.0)2 (0.10) = 0.40 • Since Ktrial (0.40) < Keq (49) there needs to be more products, therefore rxn shifts right.

  7. I C E Example E (p.68t) • Keq=3.5. If 4.0 mol SO2 and 4.0 mol NO2 are placed in a 5.0L bulb, what are all conc’s at eqm? SO2 + NO2  SO3 + NO • x = 0.52 M (again, look for a square root) • [NO] = [SO3] = 0.52 M • [SO2] = [NO2] = 0.28 M

  8. I C E Example F (p.69t) • A 1.0L flask contained 1.0 mol SO2, 4.0 mol NO2, 4.0 mol SO3, 4.0 mol NO at eqm. If 3.0 mol SO2 is added, what is [NO] at eqm? • NOTE: we start at eqm (calc Keq). • NOTE: the “I” for [SO2] is after 3.0 mol is added. SO2 + NO2  SO3 + NO • x = 1.33 M • [NO] = 5.3 M

  9. Homework • Hebden #47, 48, 51, 52, 54, 56-60, 62, 66*. • Next class we’ll have a quiz and a work period (finish #47, 48, 51, 52 by next class) • Lab 12B will be next, next class.

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