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Equilibrium Law Calculations (with RICE charts)

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Example pg. 567 H 2 + I 2 2HI R I C E H2H2 I2I2 HI Ratio, Initial, Change, Equilibrium [H 2 ] [I 2 ] Kc= [HI] 2 = [.020] [.160] 2 = 64 Q - Try PE 9 on pg. 568 Read 566 (from Calculating Kc…) to 568. Follow the sample calculation carefully.

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PE 9 - pg. 568 PCl 3 + Cl 2 PCl 5 R I C E PCl 3 Cl 2 PCl Q - Try 14.38, pg. 589 Ratio, Initial, Change, Equilibrium [PCl 3 ] [Cl 2 ] Kc = [PCl 5 ] = [.12] [.02] [.08] = 33.3

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HBrH2H2 Br R I C E RE pg HBr H 2 + Br 2 [HBr] 2 Kc = [H 2 ] [Br 2 ] = [.24] 2 [.13] = 0.293

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CH 2 OH2H2 CO R I C E RE pg. 589 CH 2 O H 2 + CO [CH 2 O] Kc = [H 2 ][CO] = [.08] [.02][.02] =

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pg. 570 CO + H 2 O CO 2 + H 2 COH2OH2OCO x +x x x R I C E H2H x x = 4.06 = [0.10 -x] 2 [x] 2 Kc = [CO 2 ][H 2 ] [CO][H 2 O] x/[0.10-x] = 2.01, x = x, 3.01x = x= Read Follow sample calculation carefully. PE 11, 14.40, pg. 589 (notice [ ] for 14.41)

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PE 11 - pg. 571 H 2 + I 2 2HI R I C E H2H2 I2I2 HI x +2x x 2x 2x/[0.2-x] = 7.04, 2x = x, x=0.156 [H 2 ][I 2 ] Kc = [HI] 2 = [0.2 -x] 2 [2x] 2 = 49.5 H 2 (I 2 also): = M HI: 2(0.156) = M

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SO 3 + NO NO 2 + SO 2 SO 3 NONO x +x x x.707=x/[0.15-x], x=x, x=0.062 = [0.15 -x] 2 [x] 2 = 0.50 R I C E SO x x Kc = [NO 2 ][SO 2 ] [SO 3 ][NO] SO 3, NO: = M NO 2, SO 2 : = M

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CO + H 2 O CO 2 + H 2 COH2OH2OCO x -x 0.01+x x.6325=(0.01-x)/(0.01+x), x= = [0.01+x] 2 [ x] 2 = 0.40 R I C E H2H x x Kc = [CO 2 ][H 2 ] [CO][H 2 O] CO, H 2 O: = M CO 2, H 2 : = = 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: [ x] 2 4x 3 = small Kc 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 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, x 2 +1? We can for these: Try PE 12 (573). Concentrations are [initial].

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PE 12 - pg. 573 N 2 + O 2 2NO R I C E N2N2 O2O2 NO x +2x x x2x [N 2 ][O 2 ] Kc = [NO] 2 = [0.033-x][ x] [2x] 2 = 4.8 x

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PE 12 - pg. 573 N 2 + O 2 2NO [0.033-x][ x] [2x] 2 = 4.8 x 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 [2x] 2 = 1.28 x x= 1.13 x This is the equilibrium [NO 2 ]

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HClH2H2 Cl x -2x 2-2x1+xx R I C E 2HCl H 2 + Cl 2 Kc= 3.2 x 10 –34 determine [equil], if [initial] are 2.0 M, 1.0 M, 0 M [HCl] 2 Kc = [H 2 ] [Cl 2 ] = [2-2x] 2 [1+x] [x] = [2] 2 [1] [x] x = (3.2 x 10 –34 )(4) = 1.3 x 10 –33 [equil] are 2, 1 and 1.3 x 10 –33 = 3.2 x 10 –34

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HClH2H2 Cl x -2x 2-2xxx R I C E RE pg HCl H 2 + Cl 2 [2-2x] 2 Kc = [x] = [.24] 2 [x] 2 = 0.293

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2Na + 2H 2 O 2NaOH + H 2 NaH2OH2ONaOH x +2x x 2x R I C E H2H x x [0.10-2x] 2 = [2x] 2 [x] Kc = [NaOH] 2 [H 2 ] [Na] 2 [H 2 O] 2 For more lessons, visit

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