Equilibrium Law Calculations (with RICE charts)

Ratio, Initial, Change, Equilibrium Read 566 (from “Calculating Kc…”) to 568. Follow the sample calculation carefully. Example 14.7 - pg. 567 H2 + I2  2HI R I C E H2 I2 HI 1 1 2 0.100 0.100 -0.08 -0.08 +0.16 0.02 0.02 0.16 Ratio, Initial, Change, Equilibrium Q - Try PE 9 on pg. 568 [H2] [I2] Kc= [HI]2 = [.020] [.020] [.160]2 = 64

Ratio, Initial, Change, Equilibrium PE 9 - pg. 568 PCl3 + Cl2  PCl5 R I C E PCl3 Cl2 PCl5 1 1 1 0.2 0.1 -0.08 -0.08 +0.08 0.12 0.02 0.08 Ratio, Initial, Change, Equilibrium [PCl3] [Cl2] Kc = [PCl5] = [.12] [.02] [.08] = 33.3 Q - Try 14.38, 14.39 pg. 589

RE 14.38 - pg. 589 2HBr  H2 + Br2 R I C E HBr H2 Br2 2 1 1 0.500 -0.260 +0.130 +0.130 0.240 0.130 0.130 [HBr]2 Kc = [H2] [Br2] = [.24]2 [.13] [.13] = 0.293

RE 14.39 - pg. 589 CH2O  H2 + CO R I C E CH2O H2 CO 1 1 1 0.100 -0.020 +0.020 +0.020 0.080 0.020 0.020 [CH2O] Kc = [H2][CO] = [.08] [.02][.02] = 0.0050

14.9 - pg. 570 CO + H2O  CO2 + H2 R I C E CO H2O CO2 H2 1 1 1 1 0.100 Read 570-1. Follow sample calculation carefully. 14.9 - pg. 570 CO + H2O  CO2 + H2 R I C E CO H2O CO2 H2 1 1 1 1 0.100 0.100 -x -x +x +x 0.10 - x 0.10 - x x x Kc = [CO2][H2] [CO][H2O] = [0.10 -x]2 [x]2 = 4.06 x/[0.10-x] = 2.01, x = 0.201-2.01x, 3.01x = 0.201 x=0.0668 PE 11, 14.40, 14.41 pg. 589 (notice [ ] for 14.41)

PE 11 - pg. 571 H2 + I2  2HI R I C E H2 I2 HI 1 1 2 0.200 0.200 -x -x -x -x +2x 0.2 - x 0.2 - x 2x [H2][I2] Kc = [HI]2 = [0.2 -x]2 [2x]2 = 49.5 2x/[0.2-x] = 7.04, 2x = 1.408-7.04x, x=0.156 H2 (I2 also): 0.2 - 0.156 = 0.044 M HI: 2(0.156) = 0.312 M

14.40 - SO3 + NO  NO2 + SO2 R I C E SO3 NO NO2 SO2 1 1 1 1 0.150 0.150 -x -x +x +x 0.15 - x 0.15 - x x x Kc = [NO2][SO2] [SO3][NO] = [0.15 -x]2 [x]2 = 0.50 .707=x/[0.15-x], 0.106-0.71x=x, x=0.062 SO3, NO: 0.15 - 0.062 = 0.088 M NO2, SO2: = 0.062 M

14.41 - CO + H2O  CO2 + H2 R I C E CO H2O CO2 H2 1 1 1 1 0.010 0.010 +x +x -x -x 0.01+x 0.01+x 0.01 - x 0.01 - x Kc = [CO2][H2] [CO][H2O] = [0.01+x]2 [0.01 - x]2 = 0.40 .6325=(0.01-x)/(0.01+x), x=0.00225 CO, H2O: 0.010 + 0.00225 = 0.0123 M CO2, H2: = 0.010 - 0.00225 = 0.0078 M

Equilibrium calculations when Kc is very small Thus far, problems have been designed so that the solution for x is straightforward If the problems were not so carefully designed we might have to use quadratic equation (or calculus) to solve the problem. If Kc is very large or very small we can use a simplification to make calculating x simple Setting up the RICE chart is the same, but the calculation of Kc is now slightly different Read pg. 572, 573

Equilibrium calculations when Kc is small Looking at the equilibrium law for 14.10: 4x3 = small Kc [0.100 - 2x]2 For Kc to be small, top must be small, bottom must be large (relative to top) For top to be small, x must be small If x is small, then 0.100 - 2x  0.100 Notice that we can only ignore x when it is in a term that is added or subtracted. Can we ignore x in: 4x, 3+x, 0.1-3x, 3x-x, x2+1? We can for these:    Try PE 12 (573). Concentrations are [initial].

PE 12 - pg. 573 N2 + O2  2NO R I C E N2 O2 NO 1 1 2 0.033 0.00810 -x -x -x +2x 0.033-x 0.00810-x 2x [N2][O2] Kc = [NO]2 = [0.033-x][0.0081-x] [2x]2 = 4.8 x 10-31

This is the equilibrium [NO2] PE 12 - pg. 573 N2 + O2  2NO [0.033-x][0.0081-x] [2x]2 = 4.8 x 10-31 Small Kc: Thus numerator is small and x must be small: x is negligible when adding or subtracting [0.033][0.0081] [2x]2 = 4.8 x 10-31 [2x]2 = 1.28 x 10-34 2x = 1.13 x 10-17 This is the equilibrium [NO2]

R I C E HCl H2 Cl2 2 1 1 2 1 -2x +x +x 2-2x 1+x x 2HCl  H2 + Cl2 Kc= 3.2 x 10–34 determine [equil], if [initial] are 2.0 M, 1.0 M, 0 M R I C E HCl H2 Cl2 2 1 1 2 1 -2x +x +x 2-2x 1+x x [HCl]2 Kc = [H2] [Cl2] = [2-2x]2 [1+x] [x] = [2]2 [1] [x] = 3.2 x 10–34 x = (3.2 x 10–34)(4) = 1.3 x 10–33 [equil] are 2, 1 and 1.3 x 10–33

RE 14.42 - pg. 590 2HCl  H2 + Cl2 R I C E HCl H2 Cl2 2 1 1 2 -2x +x +x 2-2x x x [2-2x]2 Kc = [x] [x] = [.24]2 [x]2 = 0.293

For more lessons, visit www.chalkbored.com 2Na + 2H2O  2NaOH + H2 R I C E Na H2O NaOH H2 2 2 2 1 0.100 0.100 -2x -2x +2x +x 0.10-2x 0.10-2x 2x x Kc = [NaOH]2[H2] [Na]2[H2O]2 = [2x]2[x] [0.10-2x]2 [0.10-2x]2 For more lessons, visit www.chalkbored.com