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Equilibrium Law Calculations

Equilibrium Law Calculations. (with RICE charts). R. I. C. E. [HI] 2. [.160] 2. Kc=. =. Q - Try PE 9 on pg. 568. [H 2 ] [I 2 ]. [.020] [.020]. Example 14.7 - pg. 567 H 2 + I 2  2HI. H 2. I 2. HI.

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Equilibrium Law Calculations

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  1. Equilibrium Law Calculations (with RICE charts)

  2. R I C E [HI]2 [.160]2 Kc= = Q - Try PE 9 on pg. 568 [H2][I2] [.020][.020] Example 14.7 - pg. 567 H2 + I2 2HI H2 I2 HI • Read 566 (from “Calculating Kc…”) to 568. Follow the sample calculation carefully. 1 1 2 0.100 0.100 0 -0.08 -0.08 +0.16 0.02 0.02 0.16 Ratio, Initial, Change, Equilibrium = 64

  3. R I C E [.08] [PCl5] Kc = = [.12][.02] [PCl3][Cl2] PE 9 - pg. 568 PCl3 + Cl2 PCl5 PCl3 Cl2 PCl5 1 1 1 0.2 0.1 0 -0.08 -0.08 +0.08 0.12 0.02 0.08 Ratio, Initial, Change, Equilibrium = 33.3 Q - Try 14.38, 14.39 pg. 589

  4. R I C E [H2][Br2] [.13][.13] Kc = = [HBr]2 [.24]2 RE 14.38 - pg. 589 2HBr  H2 + Br2 HBr H2 Br2 2 1 1 0.500 0 0 -0.260 +0.130 +0.130 0.240 0.130 0.130 = 0.293

  5. R I C E [H2][CO] [.02][.02] Kc = = [CH2O] [.08] RE 14.39 - pg. 589 CH2O  H2 + CO CH2O H2 CO 1 1 1 0.100 0 0 -0.020 +0.020 +0.020 0.080 0.020 0.020 = 0.0050

  6. R I C E [CO2][H2] [x]2 Kc = = [0.10 -x]2 [CO][H2O] 14.9 - pg. 570 CO + H2O  CO2 + H2 CO H2O CO2 H2 1 1 1 1 Read 570-1. Follow sample calculation carefully. 0.100 0 0 0.100 -x -x +x +x 0.10 - x 0.10 - x x x = 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)

  7. R I C [HI]2 [2x]2 E Kc = = [H2][I2] [0.2 -x]2 PE 11 - pg. 571 H2 + I2 2HI H2 I2 HI 1 1 2 0.200 0.200 0 -x -x +2x 0.2 - x 0.2 - x 2x = 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

  8. R I C E [NO2][SO2] [x]2 Kc = = [SO3][NO] [0.15 -x]2 14.40 - SO3 + NO  NO2 + SO2 SO3 NO NO2 SO2 1 1 1 1 0.150 0 0 0.150 -x -x +x +x 0.15 - x 0.15 - x x x = 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

  9. R I C E [CO2][H2] [0.01 - x]2 Kc = = [CO][H2O] [0.01+x]2 14.41 - CO + H2O  CO2 + H2 CO H2O CO2 H2 1 1 1 1 0.010 0.010 0.010 0.010 +x +x -x -x 0.01+x 0.01+x 0.01 - x 0.01 - x = 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

  10. 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

  11. Equilibrium calculations when Kc is small 4x3 = small Kc [0.100 - 2x]2 Looking at the equilibrium law for 14.10: 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].

  12. R I C E [NO]2 [2x]2 Kc = = [N2][O2] [0.033-x][0.0081-x] PE 12 - pg. 573 N2 + O2 2NO N2 O2 NO 1 1 2 0.033 0.00810 0 -x -x +2x 0.033-x 0.00810-x 2x = 4.8 x 10-31

  13. [2x]2 [2x]2 [0.033-x][0.0081-x] [0.033][0.0081] PE 12 - pg. 573 N2 + O2 2NO = 4.8 x 10-31 Small Kc: Thus numerator is small and x must be small: x is negligible when adding or subtracting = 4.8 x 10-31 [2x]2 = 1.28 x 10-34 2x = 1.13 x 10-17 This is the equilibrium [NO2]

  14. R I C E [1] [x] [H2][Cl2] [1+x] [x] Kc = = = [HCl]2 [2]2 [2-2x]2 2HCl  H2 + Cl2 Kc= 3.2 x 10–34determine [equil], if [initial] are 2.0 M, 1.0 M, 0 M HCl H2 Cl2 2 1 1 2 1 0 -2x +x +x 2-2x 1+x 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

  15. R I C E [x][x] [x]2 Kc = = [2-2x]2 [.24]2 RE 14.42 - pg. 590 2HCl  H2 + Cl2 HCl H2 Cl2 2 1 1 2 0 0 -2x +x +x 2-2x x x = 0.293

  16. R I C E [NaOH]2[H2] [2x]2[x] Kc = = [Na]2[H2O]2 2Na + 2H2O  2NaOH + H2 Na H2O NaOH H2 2 2 2 1 0.100 0 0 0.100 -2x -2x +2x +x 0.10-2x 0.10-2x 2x x [0.10-2x]2 [0.10-2x]2 For more lessons, visit www.chalkbored.com

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