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Equilibrium Law Calculations
(with RICE charts)
Example 14.7 - pg. 567 H2 + I2 2HI
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H2 I2 HI
1 1 2
0.100 0.100 0
-0.08 -0.08 +0.16
0.02 0.02 0.16
Ratio, Initial, Change, Equilibrium
[H2] [I2]Kc=
[HI]2=
[.020] [.020]
[.160]2= 64
Q - Try PE 9 on pg. 568
• Read 566 (from “Calculating Kc…”) to 568. Follow the sample calculation carefully.
PE 9 - pg. 568 PCl3 + Cl2 PCl5
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PCl3 Cl2 PCl51 1 1
0.2 0.1 0
-0.08 -0.08 +0.08
0.12 0.02 0.08
Q - Try 14.38, 14.39 pg. 589
Ratio, Initial, Change, Equilibrium
[PCl3] [Cl2]Kc =
[PCl5]=
[.12] [.02]
[.08]= 33.3
HBr H2 Br2
2 1 1
0.500 0 0
+0.130+0.130-0.260
0.240 0.130 0.130
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RE 14.38 - pg. 589 2HBr H2 + Br2
[HBr]2Kc =
[H2] [Br2]=
[.24]2
[.13] [.13]= 0.293
CH2O H2 CO1 1 1
0.100 0 0
-0.020 +0.020 +0.020
0.0200.0200.080
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RE 14.39 - pg. 589 CH2O H2 + CO
[CH2O] Kc =
[H2][CO]=
[.08]
[.02][.02]= 0.0050
14.9 - pg. 570 CO + H2O CO2 + H2
CO H2O CO2
1 1 1
0.100 0.100 0
-x -x +x
0.10 - x 0.10 - x x
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H2
1
0
+x
x
= 4.06=[0.10 -x]2
[x]2 Kc = [CO2][H2]
[CO][H2O]x/[0.10-x] = 2.01, x = 0.201-2.01x, 3.01x =
0.201x=0.0668
Read 570-1. Follow sample calculation carefully.
PE 11, 14.40, 14.41 pg. 589 (notice [ ] for 14.41)
PE 11 - pg. 571 H2 + I2 2HI
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H2 I2 HI1 1 2
0.200 0.200 0
-x -x +2x
0.2 - x 0.2 - x 2x
2x/[0.2-x] = 7.04, 2x = 1.408-7.04x, x=0.156[H2][I2]
Kc = [HI]2=
[0.2 -x]2[2x]2
= 49.5
H2 (I2 also): 0.2 - 0.156 = 0.044 M
HI: 2(0.156) = 0.312 M
14.40 - SO3 + NO NO2 + SO2
SO3 NO NO2
1 1 1
0.150 0.150 0
-x -x +x
0.15 - x 0.15 - x x
.707=x/[0.15-x], 0.106-0.71x=x, x=0.062
=[0.15 -x]2
[x]2= 0.50
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SO2
1
0
+x
x
Kc = [NO2][SO2][SO3][NO]
SO3, NO: 0.15 - 0.062 = 0.088 MNO2, SO2: = 0.062 M
14.41 - CO + H2O CO2 + H2
CO H2O CO2
1 1 1
0.010 0.010 0.010
+x +x -x
0.01+x 0.01+x 0.01 - x
.6325=(0.01-x)/(0.01+x), x=0.00225
=[0.01+x]2
[0.01 - x]2= 0.40
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H2
1
0.010
-x
0.01 - x
Kc = [CO2][H2][CO][H2O]
CO, H2O: 0.010 + 0.00225 = 0.0123 MCO2, 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 smallLooking at the equilibrium law for 14.10:
[0.100 - 2x]2
4x3
= 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 smallIf x is small, then 0.100 - 2x 0.100Notice 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
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N2 O2 NO1 1 2
0.033 0.00810 0
-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
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]
HCl H2 Cl22 1 1
2 1 0
+x+x-2x
2-2x 1+x x
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2HCl H2 + Cl2 Kc= 3.2 x 10–34 determine [equil], if [initial] are 2.0 M, 1.0 M, 0 M
[HCl]2Kc =
[H2] [Cl2] =[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
HCl H2 Cl22 1 1
2 0 0
+x+x-2x
2-2x x x
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RE 14.42 - pg. 590 2HCl H2 + Cl2
[2-2x]2Kc =
[x] [x]=
[.24]2
[x]2
= 0.293
2Na + 2H2O 2NaOH + H2
Na H2O NaOH2 2 2
0.100 0.100 0
-2x -2x +2x
0.10-2x 0.10-2x 2x
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H2
1
0
+x
x
[0.10-2x]2 [0.10-2x]2=
[2x]2[x] Kc =
[NaOH]2[H2]
[Na]2[H2O]2
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