1. Equilibrium Law Calculations
(with RICE charts)

2. 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

3. 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

4. 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

5. 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] =

6. 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)

7. 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

8. 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

9. 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

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
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].

12. 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

13. 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]

14. 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

15. 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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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
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