Equilibrium Constant (Keq)
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
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
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 0.00140 M & [NO2] is 0.0172M at equilibrium? Keq = [N2O4] [NO2]2 = [0.0014] [0.0172]2 Keq = 4.73
Keq = [N2O4] [NO2]2 = [0.00455] [0.031]2 Keq = 4.73 Same Keq!!! Given: 2NO2 N2O4 + E Calculate Keq, if the [N2O4 ] is 0.00455 M & [NO2] is 0.031M at equilibrium. Keq = [N2O4] [NO2]2 = [0.00455] [0.031]2 Keq = 4.73 Same Keq!!!
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!
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
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
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] = 0.041
P318 #1 to #7
NH3 (g) H2 (g) + N2 (g) Warm up question Ex.3 2 3 1 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 - 0.044 + 0.066 + 0.022 0.156 0.066 0.102
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 = 0.0012
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
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
√ / / Keq = [HI]2 [H2] [I2] At Equilibrium: 50 = [2x]2__ Please write Keq = [HI]2 [H2] [I2] At Equilibrium: [H2] = 2x = 0.5 - 0.39 = 0.11M [I2] = 0.5 – x = 0.5 - 0.39 = 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 3.54 - 7.07x = 2x x = 0.39
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
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 28. 28 - 7.07x = 2x x = 3.12
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?
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) 2HI(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
Continuing the Problem Please write Continuing the Problem So.. Carried over to next slide
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!
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
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 – 0.0001 mol/L = 1.9999 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.