3  Chemical Equilibrium
Answers to NChO Practice Items
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
(D) Kc = 1 / [O2]5
Solids are not included in the Kc expression.
(A) 1 only
Note that H > 0, so the reaction is endothermic and “heat” is a reactant.
The words “increase the quantity of Fe(s)” means shift to the right.
1. will shift the equilibrium to the right.
2. will shift the equilibrium to the left.
3. will have no effect on the equilibrium.
(A) 1 only
1. adding a catalyst lowers the activation energy by providing an alternate reaction pathway.
2. the same equilibrium concentrations will be achieved… only more quickly.
3. the H will remain the same… it only depends on the energy of the reactants and products.
(C) 1 and 2 only
Temperature is the only factor that changes the value of the equilibrium constant, K. Because K
changes, the equilibrium concentrations will be affected.
(C) 2HI(g) H2(g) + I2(g)
When n = 0, the term (RT)n becomes 1 and Kp = Kc.
(C) K = [XeF4] / ( [Xe] [F2]2 )
Products over reactants raised to the powers of their coefficients.
(B) K = [NH3]2[CO2]
Solids do not appear in the equilibrium expression.
[NH3] is squared because the coefficient for NH3(g) is 2.
(D) lowering the temperature
Temperature is the only factor that will change the equilibrium constant.
(A) AlCl3
Three clues lead to choice (A):
(1) “colorless aqueous solution” eliminates blue CuSO4.
(2) ppt with NaOH eliminates Ba(NO3)2 because Ba(OH)2 is soluble.
(3) white ppt with AgC2H3O2 AgI is a yellow precipitate; AgCl is a white precipitate.
(C) 7.3 x 10-14
PbCO3 is a 1:1 compound so Ksp = s2
s = 2.7 x 10-7
Ksp = (2.7 x 10-7)2 = 7.3 x 10-14
(D) PbCl2
Solubility rules help eliminate (A), (B), and (C). (NO3-, Na+, and K+ are always soluble.)
(D) 8.8 x 10-9
PbI2 is a 1:2 compound so Ksp = 4s3
s = 0.0013 M Ksp = 4(0.0013)3 = 8.8 x 10-9
(C) MgSO4 and Pb(NO3)2
PbSO4 precipitate is formed. (solubility rules)
(A) 1 only
Ca(OH)2 + 2 NH4Cl  2 NH3(g) + 2 H2O + CaCl2
“1” is true
“2” CaCl2 is formed, but it is not a precipitate
“3” is a complex ion.
NH3 is a good ligand, but you would need concentrated NH3 to form the complex ion.
15.
16.
17.
18.
19.
20.
(B) BaCO3 only
Look for salt with anion from a weak acid.
BaCO3; CO32- from H2CO3 (weak)
BaSO4; SO42- from H2SO4 (strong)
(B) CaCl2 + K2CO3 
CaCO3 is a precipitate (solubility rules)
(C) 4.4 x 10-4
CuCl is a 1:1 compound, so Ksp = s2
Ksp = 1.9 x 10-7 = s2
s = 1.9 x10 7 = 4.4 x 10-4
(B) 1.5 x 10-3
PbI2 is a 1:2 compound so Ksp = 4s3
Ksp = 1.4 x 10-8
s = 3 K sp /4 = 1.5 x 10-3
(A) BaCO3
see explanation for 1996-40. above. HCl, HNO3, H2SO4 are all strong acids.
(D) CaCl2
Look for salt with anion NOT from a weak acid. HF, H2CO3, and H2C2O4 are all weak acids.