Practice Test problems: equilibrium I
1) The water gas reaction is exothermic:
1CO (g) + 1 H2O (g) < 1 CO2 (g) + 1H2 (g)

The effect of allowing the reaction to proceed with a suitable catalyst
No change in equilibrium position because it has already reached equilbrium

The law of mass action as it applies to this reaction

The speed of the forward reaction compared to the speed of the reverse reaction at equilibrium
Equal rates in both directions

The effect of adding more CO (g) if the reaction is 0th order with respect to that reactant
Again, it is at equilibrium so this is not a kinetic issue. Lowering concentration of CO will cause
the equilibrium to shift to the left. Concentration of CO2 and H2 will drop.

The effect at equilibrium of cooling the mixture
1CO (g) + 1 H2O (g) < 1 CO2 (g) + 1H2 (g)
Since the reaction is exothermic, lowering the temperature will allow it to give off heat easier. It
will give off heat when it runs forward so cooling it will cause the equilibrium to shift forward.
The equilibrium constant will increase.

The Effect of condensing CO2 (g) from the products at equilibrium
1CO (g) + 1 H2O (g) < 1 CO2 (g) + 1H2 (g)
Pulling out CO2 (g) will cause the equilibrium to shift to the right to make more of the CO2 (g) which was
reduced in concentration.

The effect of adding pressure by adding a non – reactive gas such as nitrogen

The effect of adding pressure by lowering the volume and keeping the concentration the same.
The increase in pressure will have no effect on this equilibrium because there are equal number of moles
of gas on both sides of the reaction.
AP chemistry
Ch14 – commonly occurring acids and bases
Monday, February 08, 2016
Practice Test problems: equilibrium II
2) The decomposition of calcium carbonate is endothermic:
CaCO3 (s) > CaO (s) + CO2 (g)

The effect of allowing the reaction to proceed with a suitable catalyst
The catalyst allow the reaction to reach equilibrium faster but will not shift equilibrium of the value of
the equilibrium constant

The law of mass action as it applies to this reaction
Keq = [CO2]1 remember that the pure solids are not included in equilibrium constants

The effect of adding more CaCO3 (s)
No change in equilibrium position, The concentration of a solid is constant. You can have more of it but
the concentration won’t change and neither will the equilibrium constant.

The effect of adding more CO2 (g)
The equilibrium will shift left until the pressure of CO2 (g) is back to the same level. The AMOUNT of CaO
(s) will reduce as it is changed into CaCO3 (s)

The effect at equilibrium of cooling the mixture
CaCO3 (s) > CaO (s) + CO2 (g)
This reaction is endothermic, it needs heat to run forward. Cooling it will prevent it from running
forward. It will be able to run in reverse better and that is why the equilibrium constant is reduced.

The Effect of condensing CO2 (g) from the products at equilibrium
CaCO3 (s) > CaO (s) + CO2 (g)
The equilibrium will shift left until the pressure of CO2 (g) is back to the same level. The AMOUNT of
CaCO3 (s) will reduce as it is changed into CaO (s)

The effect of adding pressure by adding a non – reactive gas such as nitrogen

The effect of adding pressure by lowering the volume and keeping the concentration the same.
This equilibrium has 1 mole of gas product and 0 moles of gas reactant. An increase in
concentration of gas will cause the equilibrium to shift towards less gas. Likewise a decrease will
cause the equilibrium to shift toward more gas.
AP chemistry
Ch14 – commonly occurring acids and bases
Monday, February 08, 2016
(In class group former AP probs on Acid/Base equilibriums)
The pKa value of butanoic acid is 4.83 . Find the pH of a 0.20 M solution of butanoic acid.
The pKa is the negative log of the Ka value. The Ka = 10 -4.83 = 1.48 x 10 -4
Butanoic acid is a 4 carbon carboxylic acid - C4O2H7
O
C4O2H7 == C4O2H6 + H+1
||
0.2 -x M
x M
x M
1.48 x 10 -4 = x2 / 0.2 -x assume x<<0.2
C - C - C - C - OH
1.48 x 10 -4 = x2 / 0.2
x = 0.00172 M [H+1]
5% of 0.2M = 0.01M (5% rule applies)
pH = - log (0.00172 ) = 2.76
pOH = 11.23
Dimethylamine (CH3)2NH has a Kb 5.4 x 10-4 A 1.5 M concentration of Dimethylamine has a pH of
(CH3)2NH
+
H2O ==
(CH3)2NH2+1 +
-1
OH remember to write equilibriums for bases with water
1.5 –x M
x M
5.4 x 10-4 = x2 /1.5 –x
assume x<<1.5 x =9x10-3
[-1OH] = 9x10-3
AP chemistry
x M
M pOH = - log (9x10-3 ) = 2.04
(5 % rule applies)
pH = 11.95
Ch14 – commonly occurring acids and bases
Monday, February 08, 2016