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Equilibrium Constant (Keq)
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Keq: aA + bB cC + dD Keq = [C]c[D]d [A]a[B]b Please write
# expressing the [reactants] & [products] at equilibrium constant for every reversible rxn at equilibrium constant T & P) aA + bB cC + dD Keq = [C]c[D]d [A]a[B]b
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Keq = Products Reactants Magnitude of Keq
Please write Keq = Products Reactants Magnitude of Keq Keq = 1 Keq > 1 Keq < 1 means products = reactants Forward rxn is favoured, products > reactants reverse rxn is favoured, products < reactants
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Keq = [N2O4] [NO2]2 = [0.0014] [0.0172]2 Keq = 4.73
Given: 2NO2 N2O4 + E To calculate Keq, the concentrations must be given at equilibrium! What is the Keq if the [N2O4] is M & [NO2] is M at equilibrium? Keq = [N2O4] [NO2]2 = [0.0014] [0.0172]2 Keq = 4.73
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Keq = [N2O4] [NO2]2 = [0.00455] [0.031]2 Keq = 4.73 Same Keq!!! Given:
2NO2 N2O E Calculate Keq, if the [N2O4 ] is M & [NO2] is M at equilibrium. Keq = [N2O4] [NO2]2 = [ ] [0.031]2 Keq = 4.73 Same Keq!!!
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Important! Please write Keq is the same for a given reaction that is:
at equilibrium at the same temperature no matter what the initial concentrations were. Keq changes with temperature!
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H2 (g) + I2 (g) ⇄ 2HI (g) Keq = [HI]2__ [H2][I2] = __[3.0]2__
Please write Ex.1 H2 (g) + I2 (g) ⇄ 2HI (g) Find the Keq if the [H2] is 0.46M, [I2] is 0.39M & [HI] is 3.0M at equilibrium . Keq = [HI]2__ [H2][I2] = __[3.0]2__ [0.46][0.39] Keq = 50
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PCl5 (g) PCl3 (g) + Cl2 (g)
Please write PCl5 (g) PCl3 (g) + Cl2 (g) Ex.2 Find the Keq, if the initial [PCl5]is 0.70M & the equilibrium [Cl2] is 0.15M. Keq expression for equilibrium concentrations only!!! Use an ICE table
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PCl5 (g) ⇄ PCl3 (g) + Cl2 (g) Keq = [PCl3][Cl2]__ [PCl5]
Please write PCl5 (g) ⇄ PCl3 (g) + Cl2 (g) Ex.2 [PCl5] [PCl3] [Cl2] I Initial C Change E Equilibrium Based on ratio from balanced rxn! 0.70 - 0.15 + 0.15 + 0.15 0.55 0.15 0.15 Keq = [PCl3][Cl2]__ [PCl5] = [0.15][0.15]__ [0.55] =
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P318 #1 to #7
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NH3 (g) H2 (g) + N2 (g) Warm up question Ex.3 Initially a 5.0L flask contains 0.2M NH3 & 0.08M N2. At equilibrium [NH3] is 0.156M. Find Keq NH3 H2 N2 I C E 0.2 0.08 MOLE RATIO 0.156 0.066 0.102
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Keq = [H2]3[N2]__ [NH3]2 = [0.066]3[0.102]__ [0.156]2 Keq = 0.0012
solution = [0.066]3[0.102]__ [0.156]2 Keq =
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H2 (g) + I2 (g) 2HI (g) Keq = [HI]2 [H2] [I2] 50 = [x]2__
Please write Ex.4 Find the [HI] at equilibrium, if [H2] is 0.50M & [I2] is 0.50M at equilibrium (Keq = 50). Keq = [HI]2 [H2] [I2] 50 = [x]2__ [0.50][0.50] [HI] = 3.5 M at equi
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H2 (g) + I2 (g) 2HI (g) Please write Ex.5
What are the equilibrium concentrations of each substance if a flask initially contains only 0.5M H2 & 0.5M I2? Keq = 50. H2 I2 HI I C E 0.5 0.5 - x - x + 2x 0.5 - x 0.5 - x 2x
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√ / / Keq = [HI]2 [H2] [I2] At Equilibrium: 50 = [2x]2__
Please write Keq = [HI]2 [H2] [I2] At Equilibrium: [H2] = 2x = = 0.11M [I2] = 0.5 – x = = 0.11M [HI] = 2x = 2 x 0.39 = 0.78M 50 = [2x]2__ [0.5-x][0.5-x] √ / / 50 = [2x] 2_ [0.5-x] 2 7.07 = [2x]_ [0.5-x] 7.07(0.5 – x) = 2x x = 2x x = 0.39
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H2 (g) + I2 (g) 2HI (g) Ex.6 What are the equilibrium concentrations of each substance if a 0.5L flask initially contains 2 moles H2 and 2 moles of I2? Keq = 50 H2 I2 HI I C E 4 4 - x - x + 2x 4 - x 4 - x 2x
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At Equi: [H2] = 0.88M [I2] = 0.88M [HI] = 6.24M
If not same, multiply out then use quadratic formula Keq = [HI]2 [H2] [I2] 50 = [2x]2__ [4-x][4-x] 7.07 = [2x]_ [4-x] At Equi: [H2] = 0.88M [I2] = 0.88M [HI] = 6.24M 7.07(4 – x) = 2x x = 2x x = 3.12
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H2(g) + I2(g) 2HI(g) At 1100K the Keq 25.
Please write Ex.7 At 1100K the Keq 25. H2(g) + I2(g) 2HI(g) 2 moles of H2 and 3 moles of I2 aree placed in a 1 L container. Find the concentration of each substance at equilibrium?
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I Please write C E 2 M 3 M 0 M -x -x +2x 2-x M 3-x M 2x M
H2(g) I2(g) HI(g) H2 I2(g) HI I C E 2 M 3 M 0 M -x -x +2x 2-x M 3-x M 2x M
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Continuing the Problem
Please write Continuing the Problem So.. Carried over to next slide
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Using the Quadratic formula
Please write Using the Quadratic formula Quadratic formula, where: a=21, b=-125, c=150 Two solutions to formula But only one of the solutions will give a real answer!
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Choosing the Best Answers
Two solutions to formula Try finding the correct concentrations using the solutions Solve using 4.29 Solve using 1.67 Concentration cannot be negative! At equilibrium: [H2] = 0.33 mol/L, [I2] = 1.33 mol/L and [HI] = 3.34 mol/L Please write
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Simplifying I.C.E. tables(the 5% rule)
When doing an ICE table you may have to subtract a very small value from a relatively large value… for example 2.0 mol/L – 1.0x10-4 mol/L. In this case, don’t bother doing the subtraction, since by the time you change it to show significant digits, the result will be the same: 2.0 mol/L – mol/L = mol/L ≈ 2.0 mol/L. If the number you subtract is less than 5% of the original number, you can usually skip the subtraction.
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