Equilibrium Law Calculations
(with RICE charts)
• Read 566 (from “Calculating Kc…”) to 568.
Follow the sample calculation carefully.
Example 14.7 - pg. 567 H2 + I2  2HI
H2
I2
HI
R
I
1
0.100
1
0.100
2
0
C
E
-0.08
0.02
-0.08
0.02
+0.16
0.16
Ratio, Initial, Change, Equilibrium Q - Try
[HI]2
[.160]2
PE
9
on
Kc=
= 64
=
pg. 568
[H2] [I2]
[.020] [.020]
PE 9 - pg. 568
PCl3
R
1
0.2
I
-0.08
C
E
0.12
PCl3 + Cl2  PCl5
Cl2
PCl5
1
1
0.1
0
-0.08
+0.08
0.02
0.08
Ratio, Initial, Change, Equilibrium
[PCl5]
[.08]
Kc =
=
= 33.3
[PCl3] [Cl2]
[.12] [.02]
Q - Try 14.38, 14.39 pg. 589
RE 14.38 - pg. 589 2HBr  H2 + Br2
R
I
C
E
Kc =
HBr
2
H2
1
Br2
1
0.500
-0.260
0
+0.130
0
+0.130
0.240
0.130
0.130
[H2] [Br2]
[HBr]2
[.13] [.13]
=
[.24]2
= 0.293
RE 14.39 - pg. 589 CH2O  H2 + CO
R
I
C
E
Kc =
CH2O
1
H2
1
CO
1
0.100
-0.020
0
+0.020
0
+0.020
0.080
0.020
0.020
[H2][CO]
[CH2O]
[.02][.02]
=
[.08]
= 0.0050
Read 570-1. Follow sample calculation carefully.
14.9 - pg. 570 CO + H2O  CO2 + H2
CO
H2O
CO2
H2
R
1
1
1
1
0.100
I 0.100
-x
-x
C
E 0.10 - x 0.10 - x
0
+x
0
+x
x
x
2
[CO
][H
]
[x]
2
2
Kc =
=
= 4.06
2
[0.10 -x]
[CO][H2O]
x/[0.10-x] = 2.01, x = 0.201-2.01x, 3.01x = 0.201
x=0.0668
PE 11, 14.40, 14.41 pg. 589 (notice [ ] for 14.41)
R
I
C
E
PE 11 - pg. 571 H2 + I2  2HI
H2
I2
HI
1
1
2
0.200
0.200
0
-x
-x
+2x
0.2 - x
0.2 - x
2x
2
2
[HI]
[2x]
Kc =
=
= 49.5
2
[H2][I2]
[0.2 -x]
2x/[0.2-x] = 7.04, 2x = 1.408-7.04x, x=0.156
H2 (I2 also): 0.2 - 0.156 = 0.044 M
HI: 2(0.156) = 0.312 M
SO3 + NO  NO2 + SO2
NO
NO2
SO2
1
1
1
0.150
0
0
I 0.150
-x
-x
+x
+x
C
E 0.15 - x 0.15 - x
x
x
14.40 SO3
R
1
2
[NO
][SO
]
[x]
2
2
Kc =
=
= 0.50
2
[SO3][NO]
[0.15 -x]
.707=x/[0.15-x], 0.106-0.71x=x, x=0.062
SO3, NO: 0.15 - 0.062 = 0.088 M
NO2, SO2: = 0.062 M
R
I
C
E
14.41 CO + H2O  CO2 + H2
CO
H2O
CO2
H2
1
1
1
1
0.010
0.010
0.010
0.010
+x
+x
-x
-x
0.01+x 0.01+x 0.01 - x 0.01 - x
2
[CO
][H
]
[0.01
x]
2
2
Kc =
=
= 0.40
2
[CO][H2O]
[0.01+x]
.6325=(0.01-x)/(0.01+x), x=0.00225
CO, H2O: 0.010 + 0.00225 = 0.0123 M
CO2, 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 small
Looking at the equilibrium law for 14.10:
4x3
= small Kc
2
[0.100 - 2x]
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 0.100 - 2x  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].
PE 12 - pg. 573
N2
R
1
I
C
E
N2 + O2  2NO
O2
NO
1
2
0.033
-x
0.00810
-x
0
+2x
0.033-x
0.00810-x
2x
[NO]2
[2x]2
Kc =
=
= 4.8 x 10-31
[N2][O2] [0.033-x][0.0081-x]
PE 12 - pg. 573
N2 + O2  2NO
[2x]2
= 4.8 x 10-31
[0.033-x][0.0081-x]
Small Kc: Thus numerator is small and x must be
small: x is negligible when adding or subtracting
[2x]2
= 4.8 x 10-31
[0.033][0.0081]
[2x]2 = 1.28 x 10-34
2x = 1.13 x 10-17
This is the equilibrium [NO2]
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
2
H2
1
Cl2
1
2
-2x
1
+x
0
+x
2-2x
1+x
x
[1] [x]
[H2] [Cl2] [1+x] [x]
–34
Kc =
=
3.2
x
10
=
=
[HCl]2
[2]2
[2-2x]2
x = (3.2 x 10–34)(4) = 1.3 x 10–33
[equil] are 2, 1 and 1.3 x 10–33
RE 14.42 - pg. 590 2HCl  H2 + Cl2
R
I
C
E
Kc =
HCl
2
H2
1
Cl2
1
2
-2x
0
+x
0
+x
2-2x
x
x
[x]2
[x] [x]
[2-2x]2
=
[.24]2
= 0.293
R
2Na + 2H2O  2NaOH + H2
Na
H2O
NaOH
H2
2
2
2
1
0.100
I 0.100
-2x
C -2x
E 0.10-2x 0.10-2x
Kc =
[NaOH]2[H2]
[Na]2[H2O]2
0
+2x
0
+x
2x
x
[2x]2[x]
=
[0.10-2x]2 [0.10-2x]2
For more lessons, visit
www.chalkbored.com