Chemistry 12
Unit 2
Tutorial #2 - Unit 1 – Equilibrium
Multiple Choice Explanations
The answer is C
The Fe3O4 and the Fe are left out of the Keq expression because they are solids and their
concentrations are constant. Remember the Keq expressions only contain gases and
aqueous species. Also remember that coefficients in the equation become exponents in the
Keq expression.
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Chemistry 12
Unit 2
The answer is B
This is a Trial Keq type of problem. The Keq expression is:
Keq = [O2]3
[O3]2
And the volume of the container is 1.0 L, so substituting
the initial concentrations of O2 and O3 into the Keq expression, we get:
3
Trial Keq = (0.10)2
(0.10)
= 0.10 The Trial Keq(0.10) < Actual Keq (65) so the
reaction will shift to the Right side as equilibrium is approached. In this process of
shifting to the right [O3] will decrease and [O2] will increase.
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Chemistry 12
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The Answer is B
If the equilibrium sample is turning from orange to yellow, it would mean that the solution that is added
causes this equilibrium to shift to the left (there would be lot’s of yellow CrO42- ) Answers “A”, “B” and
“C” are all soluble salts which dissociate completely in aqueous solution. Answer “D” is a weak acid
which partially ionizes. Here are equations describing what all of these do in aqueous solution:
A.
B.
C.
D.
KNO3(s)  K+(aq) + NO3-(aq)
NaOH (s)  Na+(aq) + OH-(aq)
NH4NO3 (s)  NH4+(aq) + NO3-(aq)
CH3COOH(aq) + H2O (l)  H3O+(aq) + CH3COO- (aq)
What you need to do is find out which of all the ions produced would cause the equilibrium: to shift to
the left.
K+, NO3- and Na+ will have no effect on this equilibrium since they are not contained in it and do not
react with anything in it.
Looking up NH4+ on the acid table, it is on the lower left side which means it is a weak acid, producing
some H3O+. This H3O+ will shift the equilibrium to the right, keeping it orange.
The CH3COOH in answer “D” also produces some H3O+ which would shift it right and keep it orange.
The OH- in answer “B”, will react with (neutralize) the H3O+ in the equilibrium, thus decreasing the
[H3O+] and causing the equilibrium to shift to the left and turn from orange to yellow.
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Chemistry 12
Unit 2
The Answer is B
To get the highest concentration of NH3 at equilibrium, we want the reaction to be shifted as far as
possible to the right.
Since the energy (heat) term is on the right side, decreasing the temperature (taking away heat) would
cause a shift to the right according to LCP (LeChatelier’s Principle), so we can eliminate “A” and “C” as
answers.
There are 4 moles of gas (mog) on the left side and 3 mog on the right side. Increasing the pressure
would cause the reaction to shift to the right (least mog) according to LCP. So answer “B” would be the
correct one.
The Answer is B
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Chemistry 12
Unit 2
The SnO2(s) and the Sn (s) are left out of the Keq expression because they are solids and have a constant
concentration. Remember the products are on top and the reactants are on the bottom of the expression.
Also, don’t forget that a coefficient in the balanced equation becomes an exponent in the Keq expression!
The Answer is D
Be careful with this one! The changes in “A” and “B” will cause the equilibrium to shift to the right but
this does NOT alter the value of Keq! When [H2] is increased, the equilibrium shifts to the right just
enough to compensate and keep the value of Keq constant. If we decrease the volume (increase the
pressure) as in answer “B”, again the equilibrium shifts to counteract our
change and keep the Keq value constant. Powdering the solid C as in answer
[H2]
“C” will increase the surface area and hence the rate of the reaction. But it
increases both the forward and the reverse rate, giving no difference in the
equilibrium mixture and hence does not affect Keq.
The reaction is exothermic and can be written with the heat term on the right:
C (s) + 2 H2(g)  CH4(g) + 74.8 KJ
Taking away heat (decreasing the temperature) will cause this reaction to shift to the right. This will
cause [CH4] to increase and [H2] to decrease. The Keq expression is:
So this will cause the ratio (and the value of Keq )to increase.
[CH4]
Keq =
[H2]2
Remember the temperature is the only thing that can change Keq!
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Chemistry 12
Unit 2
The Answer is C
Notice that all concentrations given are at equilibrium. Therefore we just write the Keq expression and
insert the equilibrium concentration values. (This is not an “ICE” problem!)
Keq =
[HI]2
[H2][I2]
=
[0.0156]2
= 5.03 x 101
[0.00220][0.00220]
The Answer is C
“A”, “B” and “D” are all ionic because they have a metal in the formula. “B” and “D” are bases because
they are ionic and form the OH- ion in aqueous solution. When –OH is bonded directly to a carbon (C)
atom, the bond is a strong covalent bond and does not break to release OH- ions in water.
Remember that compounds with “C’s and “H’s are molecular except for ones that end with the carboxyl
group (COOH) or (COO-). CH3OH is and example of an alcohol, which is a molecular compound.
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Chemistry 12
Unit 2
The Answer is D
In an endothermic reaction the products must have a HIGHER PE. So the answer must be
C or D. A catalyzed reaction will have a lower PE curve than an uncatalyzed reaction.
So the answer must be D.
I – is an uncatalyzed exothermic
II - is an catalyzed exothermic
III – is an uncatalyzed endothermic
The Answer is D
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Chemistry 12
Unit 2
All equilibrium systems MUST BE CLOSED, so nothing can escape. The macroscopic
properties are constant because the forward and reverse rates are equal. It is the
microscopic properties that are dynamic. (changing) The answer is D, because who cares
about the concentrations of the products and reactant as long as they are constant. Is is
possible to have more moles of products than reactants.
The Answer is D
If there is MORE moles of gas present at equilibrium then the reaction must have shifted right. Because
“initially” there was only 5.0 moles . So this means that the reaction proceeded right. For a reaction to
proceed right the ACTUAL Keq must be more than the Trial Keq. So Trial Keq< Keq.
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Chemistry 12
Unit 2
The Answer is B
There is 7 mog on the products side and 6 mog on the reactant side so the entropy INCREASED The
reaction is EXOTHERMIC so the enthalpy will DECREASE. Because the products have lower energy.
The Answer is A
To increase the moles of NO2 you need to shift the equilibrium to the LEFT. Adding N2O4 will
increase the [N2O4] which will shift the eq to the left to decrease what you added. Adding a
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Chemistry 12
Unit 2
catalyst does not shift the eq because it will increase the rate of the forward and reverse
reaction equally. Decreasing the T will shift it right and so will decreasing the volume.
The Answer is A
DO NOT include solids! DO NOT include liquids if they are not changing. So that only leaves
[H+]2 as the product which goes on top.
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Chemistry 12
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The Answer is B
Keq is Products over reactants if you switch it around it is still products over reactants but the Keq will
be 1/50 = 0.0200
The Answer is B
The Answer is C
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Chemistry 12
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The Answer is B
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Chemistry 12
Unit 2
The Answer is D
The Answer is C
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Chemistry 12
Unit 2
The Answer is C
The Answer is A
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Chemistry 12
Unit 2
Tutorial #2 - Unit 1 – Equilibrium
Multiple Choice Explanations
(student handout)
The answer is
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Chemistry 12
Unit 2
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