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**Equilibrium Law Calculations**

(with RICE charts)

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**Ratio, Initial, Change, Equilibrium**

Read 566 (from “Calculating Kc…”) to Follow the sample calculation carefully. Example pg 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

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**Ratio, Initial, Change, Equilibrium**

PE 9 - pg 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, pg. 589

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RE pg HBr 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

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RE pg 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] =

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**14.9 - pg. 570 CO + H2O CO2 + H2 R I C E CO H2O CO2 H2 1 1 1 1 0.100**

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

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**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 = x, x=0.156 H2 (I2 also): = M HI: 2(0.156) = M

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**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 x x x x Kc = [NO2][SO2] [SO3][NO] = [0.15 -x]2 [x]2 = 0.50 .707=x/[0.15-x], x=x, x=0.062 SO3, NO: = M NO2, SO2: = M

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**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 x x Kc = [CO2][H2] [CO][H2O] = [0.01+x]2 [ x]2 = 0.40 .6325=(0.01-x)/(0.01+x), x= CO, H2O: = M CO2, H2: = = M

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**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

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**Equilibrium calculations when Kc is small**

Looking at the equilibrium law for 14.10: 4x3 = small Kc [ x]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 x 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].

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**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 x 2x [N2][O2] Kc = [NO]2 = [0.033-x][ x] [2x]2 = 4.8 x 10-31

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**This is the equilibrium [NO2]**

PE 12 - pg N2 + O2 2NO [0.033-x][ 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]

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**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

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RE pg HCl 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

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**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

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