1. When a system is at equilibrium
a) the forward and reverse reaction rates are equal.
Observable macroscopic properties do not change
2a) This reaction is exothermic, so heat is on the product side. Reducing temperature will make it shift to the product side maximizing the production of NO.
2b) Decreasing pressure by increasing volume will favour the product side since there are 10 molecules on the right and 9 on the left.

2. 2c A platinum catalyst would have no effect since it speeds up both forward and reverse reactions equally.
2d Rate equations depend on concentration and adding He would increase pressure but not change the concentration so there would be no effect.

3. H2(g) I2(g)  HI(g)
mol/L
Initial mol/2.00L
2.00
Shift
@E mol/2.00L
0.20
+0.20
+0.20
-0.40
1.60
0.20
Ke = / (0.20)2 = 64

4. NO(g) O2(g)  2NO2(g)
mol/L
Initial
Shift @E x
0.606
0.190
6.45 x 105= (0.190)2/ (0.606)x2
(6.45 x 105)(0.606)x2 = (0.190)2
x2 = (0.190)2 / [(6.45 x 105)(0.606)]
x = 3.04 x 10-4

5. 5. H2(g) + CO2(g)  H2O(g) + CO(g)
mol/L
Initial
Shift @E
12.0/20.0
0.600
8.0/20.0
0.400
10.0/20.0
0.500 x
4.40 = (0.500)x/ (0.400)(0.600)
x = (4.40) (0.400)(0.600) / 0.500
x = 2.11 mol/L x 20.0 L = 42.2 mol

6. 6. 2A(s) + 3B(g)  2C(g) + D(s)
Remember in a heterogeneous system solids are not included in the equilibrium constant
Ke = [C]2 / [B]3
1.00 x 10-2 = [C]2 / 0.203
[C]2 = (1.00 x 10-2) (0.203)
[C] = 8.90 x 10-3 mol/L

7. 7. HCHO(g) <==> H2(g) + CO(g)
mol/L
Initial mol/4.00L
1.00
Shift @E
-0.20
+0.20
+0.20
0.80
0.20
0.20
Ke =(0.20)2 / 0.80
Ke = mol/L

8. H2(g) I2(g)  HI(g)
mol/L
Initial
Shift @E -x -x
+2x x x 2x
0.22
0.22
1.6
49.0 = (2x)2 / (1.00 – x)2
7.0 = 2x / 1.00 – x
7.00 – 7x = 2x
9x = 7
x = 7/9 = 0.78

9. 9. SO2(g) + NO2(g)  SO3(g) + NO(g)
mol/L
Initial
Upset x
Shift @E
-0.20
-0.20
+0.20
+0.20
0.40 x
1.00
(0.80)(0.30)/(0.20)(0.60) = (0.50)(1.00)/(x)(0.40)
x = (0.20)(0.60)(0.50) / (0.40)(0.80)(0.30)
x =0.625 mol/L x 2.0 L = 1.25 mol

10. 10. H2(g) I2(g)  HI(g)
mol/L
Initial
Upset
Shift @E
2.00 -x -x
+2x
2.1 - x x x
Ke =(0.60)2 / (0.20)(0.10) = 18
18 = ( x)2 / (2.1-x)(0.20-x)
18 = x + 4x2 / x-0.20x+x2
x+18x2 = x + 4x2
14x2-43.8x+7.2=0 Use x = -b+or-(b2-4ac)1/2/2a

11. x = ( ((43.8)2-4(14)(7.2))1/2) / 2(14)
= ( ( – 403.2)1/2) / 28
= ( ( )1/2) / 28
= (43.8 – 38.93) / 28
= 0.174
[HI] = x = (0.174) = 0.95 mol/L
Ke =(0.60)2 / (0.20)(0.10) = 18
18 = ( x)2 / (2.1-x)(0.20-x)
18 = x + 4x2 / x-0.20x+x2
x+18x2 = x + 4x2
14x2-43.8x+7.2=0 Use x = -b+or-(b2-4ac)1/2/2a

12. 11. CO(g) + Cl2(g)  COCl2(g)
mol/L
Initial
Shift @E -x -x x x x x
=1.15
1-0.85=0.15
0.85 mol/L
5.00 = x / (1.00 – x)(2.00-x)
5.00 = x / 2.00 –3.00x + x2
10 – 15x + 5x2 = x
5x2 – 16 x + 10 = 0
x = (16+-(256-4(5)(10))1/2 / 10
x = / 10 = 0.85

13. 12. A(g) + B(g)  C(g) + D(g)
mol/L
0.80
0.80 c
-0.60
-0.60
0.60
0.60 e
0.20
0.20
0.60
0.60
(0.60)
(0.60)
Ke = =
9.0
(0.20)
(0.20)

14. 13. CH4(g) + 2 H2S(g)  CS2(g)+ 4 H2(g)x y z -z
-2z 4z
0.100
0.030 z 4z
(z) (4z)4
100 =
(0.100)(0.030)2
256 z5 = (100)(0.100)(0.030)2
z = mol/L
x= = mol/L
y = (0.129) = mol/L

15. 14. 2 NH3(g)  N2(g)+ 3 H2(g)
1.00
mol/L
0.100
0.300
0.800
0.100
0.300
Ke = (0.300)3(.100)
= 4.22 x 10-3 mol/L
0.82
b) If V increases the equilibrium shifts to side with more molecules so NH3 decreases
c) Ke increases