Equilibrium Practice Problems Worksheet
1. In the reaction 2 NO2(g) ↔ N2O4(g) the initial concentration of NO2 was 0.250 M
and N2O4 was 0.000 M. At equilibrium, the concentration of N2O4 was measured
as 0.0133M.
a. Calculate the equilibrium concentration of NO2
b. Calculate Kc.
2. 0.05 mol H2(g) and 0.05 mol Br2(g) are placed together in a 5.0 L flask and
heated to 700 K. What is the concentration of each substance in the flask at
equilibrium if Kc = 64 at 700 K?
H2(g) + Br2(g) ↔ 2 HBr(g)
*Using the hundred rule says that we cannot ignore the x in the [H2] and [Br2]
3. Consider the following equilibrium 2NO(g) + O2(g)  2NO2(g)
Keq = 1.5
0.800 mol NO, 0.600 mol O2, and 0.400 mol NO2 are placed in a 2.0L vessel. Is
the system at equilibrium? What will happen to [O2] as equilibrium is
approached?
Q
= (0.200)2
=
0.833 < Keq = 1.5
(0.400)2(0.300)
Not at equilibrium
Shifts Right
[O2] will decrease.
4. Consider the following equilibrium:
SO3(g) + NO(g)  NO2(g) + SO2(g)
Keq = 0.500
Exactly 0.100 mol SO3 and 0.100 mol NO were placed in a 1.00 L flask and
allowed to go to equilibrium. Calculate the equilibrium concentration of SO 2.
SO3(g)
+
NO(g)

NO2(g)
+
SO2(g)
I
0.100
0.100
0
0
C
x
x
x
x
E
0.100 – x
0.100 – x
x
x
x2
=
0.500
=
0.7071
(0.100 – x)2
x
0.100 – x
1.7071x = 0.07071
x = 0.0414 M
[SO2] = 0.0414 M
5.
Given: At 25 ºC: [Ca2+(aq)] = 4.53x10−7 mol/L; [PO43−(aq)] = 3.02x10−7 mol/L
Required: Ksp of calcium phosphate at 25 ºC
6. Given: Ksp of Zn(OH)2(s) = 7.7x10−17 at 25 ºC
Required: molar solubility of Zn(OH)2(s) at 25 ºC
7. What is the solubility of AgI in a 0.274-molar solution of NaI. (Ksp of AgI =
8.52 x 10¯17)
Dissociation equation:
AgI (s) <===> Ag+ (aq) + I¯ (aq)
Ksp expression: Ksp = [Ag+] [I¯]
Let us substitute into the Ksp expression: 8.52 x 10¯17 = (x) (0.274)
The answer:[Ag+] = 3.11 x 10¯16 M
8. 2.1x10-3mol/L KCl(aq) is mixed with 6.3x10-3 mol/L Pb(NO3)2. Will a
precipitate form?
KCl(s)  K+(aq) + Cl-(aq)
Pb(NO3)2(s)  Pb2+(aq) + 2NO3-(aq)
2 possible products: KNO3 or PbCl2
PbCl2 (s)  Pb2+(aq) + 2Cl-(aq)
Q = [Pb2+][Cl-]2
Q = (6.3x10-3)(2.1x10-3)2
Q = 2.8x10-8
*Compare this value to the Ksp value found on the chart
Ksp = 1.2 x10-5
Therefore, since Q is less than K, the system shifts right and no precipitate will form