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Keq Calculations in Equilibrium Chemistry Class

Learn about calculating Keq with provided concentrations using examples and ICE tables. Practice problems and upcoming lab work included.

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Keq Calculations in Equilibrium Chemistry Class

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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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