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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