Download Practice Exam I FR Answers and Explanations

ANSWERS AND EXPLANATIONS TO PRACTICE EXAM I
FREE RESPONSE QUESTIONS
1.
Mandatory Calculation - Equilibrium – 10 points total
2NOCl (g) R 2 NO (g) + Cl2 (g)
(a)
(i)
(1pt)
Initial pressure (atm) before decomposition:
Given grams, temperature and volume—think Ideal Gas Law
PV= nRT
First, convert grams to moles and Celsius to Kelvin.
95.3 g NOCl ×
1 mol NOCl
= 1.46 moles
65.45 g NOCl
35ºC + 273 = 308K
Substitute and solve.
(1pt)
P=
nRT
=
(1.46 mol) (.0821 L ⋅ atm) (308K)
(12.0 L) (mol ⋅ K)
V
(1pt)
(ii)
(b)
(1pt)
(1pt)
Find concentration of NOCl.
mol 1.46 mol
=
= .122 M
liter
12.0 L
Calculate Kc if NOCl is 34.7% decomposed at equilibrium.
Calculate the amount of NOCl present at equilibrium.
(34.7%) (1.46 mol) = .507 moles decomposed (this = 2x)
NOCl left = 1.46 - .507 = .95 moles
Calculation of new molarities.
Reaction
2NOCl (g)
Initial Concentration
Change
Equilibrium
Concentration
1.46 mol
-2x
1.46 – 2x
.95
0
+2x
2x
.507
New Molarities
.95 mol
.507 mol
= .079 M
12.0 L
(1pt)
(1pt)
= 3.08 atm
Set up the Kc expression.
2
[ NO] [Cl 2 ]
Kc =
2
[ NOCl]
Substitute and solve.
⎯⎯
→
←⎯
⎯
+ Cl2 (g)
2 NO (g)
12.0 L
0
+x
x
.254
= .0423M
.254 mol
12.0 L
= .0212 M
Kc =
2
[.0423] [.0212]
[.079]
2
= 6.1×10
−3
(c)
(1pt)
(2pt)
Calculate Kp for the reaction.
Set up the expression, substitute and solve.
Kp = Kc (RT)∆n
∆n = number of moles of gaseous product – number of moles of gaseous reactant
=3–2=1
Kp = (6.1 x 10-3) [(.0821 L·atm/mol ·K) (308K)]1
= .15
(d)
Is reaction endothermic or exothermic? Justify.
Must state and explain for the two points.
The reaction must be endothermic. When the temperature increases from 35ºC to
250ºC, the number of moles of product greatly increases. With energy as a reactant,
as in an endothermic reaction, equilibrium would shift to products in response to an
increase in temperature.
Energy + 2NOCl (g) W 2 NO (g) + Cl2 (g) <explanation>
2. Kinetics –9 points total
2A+Bÿ2C
Experiment
1
2
3
(1pt)
(1pt)
Initial [A]
(mol.LB1)
0.020
0.040
0.020
Initial [B]
(mol.LB1)
0.20
0.80
0.40
Initial Rate of
Increase of [C],
(mol.LB1.secB1)
6.0 x 10B5
9.6 x 10B4
2.4 x 10B4
(a) Write the rate law for the reaction.
Rate = k [A]m [B]n
First calculate the order with respect to each reactant.
Observe the data and notice that in experiment #1 and #3 the concentration of [A] is
held constant. While [A] is constant, [B] doubles and the rate is quadrupled. When
the concentration is doubled and the initial rate is quadrupled, the reaction is second
order with respect to that component. This reaction is second order with respect to
[B].
Notice that [B] is never held constant. Algebra will always work to solve problems of
this type. Compare two rates, plug in the given values for concentrations and initial
rates and solve. Showing the math always justifies the answer. Any two sets of data
can be used.
m
2
k 2 [A]2 [B]2
=
m
2
rate1 k1 [A]1 [B]1
m
2
-4
k 2 (.040) 2 (.80) 2
9.6 × 10
-5 =
m
2
6.0 × 10
k1 (.020)1 (.20)1
m
*the k’s cancel since they are constant
16 = 2 ⋅16
rate 2
1= 2
m
m=0
In other words, altering the concentration of A has no effect on the initial rate of the
reaction.
The rate law expression: rate = k [B]2
(1pt)
(1pt)
(b) Calculate the value of the rate constant and specify units.
Substitute into the equation and solve. Any set of data may be used.
rate = k[B]2
k=
(1pt) for units
(2.4 × 10
6.0 × 10
-5
mol/L ⋅ s
−3
=
1.5
×
10
L/mol ⋅ s
2
(.20 mol/L)
(c) Calculate the initial rate of decrease of [A] in experiment 3.
This is just stoichiometry from the reaction given at the beginning of the problem. A
decreases at the same rate that C increases since they have a 1:1 mole ratio.
k=
(1pt)
rate
2
[B]
-4
mol/L ⋅ s of C) ×
2 mol A
= − 2.4 × 10
-4
mol/L ⋅ s
2 mol C
Note: The negative sign is given to indicate that the rate is decreasing.
(2pt)
(d) Which mechanism is consistent? Justify.
In order to earn the points here, prove two things:
The mechanism adds up to the balanced equation and the rate law written from the
slow step matches the rate law calculated.
I.
A+BÆX
(fast)
X Æ C +Y
(slow)
A+YÆC
(fast)
II. 2B Æ X
A+XÆY+B
Y + AÆ2C
(slow)
(fast)
(fast)
Both mechanisms add up to give the overall balanced equation. Mechanism I would
give the following rate law: rate = k[X] (from slow step); [X] is an intermediate and
its formation depends upon the fast step before. Using substitution, the rate law
becomes: rate = k’[A] [B] This does not match our calculated rate law.
Mechanism II gives the following rate law: rate = k[B]2 This does match our
calculated rate law, so this is the best choice.
(1pt)
3.
(e)
Which substances are intermediates?
Intermediates are those substances that are produced in one elementary reaction and
used in a later reaction. The intermediates in the above reaction mechanism are X
and Y.
Stoichiometry and Colligative Properties—9 points total
(a)
Determine the empirical formula from the data given.
The empirical formula is the smallest whole number ratio of the elements in a
compound. The first step in solving this problem is to assume that you have
100.grams of your compound and convert everything into moles.
(1pt)
1 mol C
59.90 g C ×
= 4.9875 mol C
12.01 g C
20.15 g H ×
1 mol H
= 19.950 mol H
1.01 g H
1 mol O
19.95 g O ×
= 1.2461mol O
16.00 g O
The second step is to calculate the mole ratio between elements. This will give
subscripts. The moles of oxygen are the smallest so divide everything by the
moles of oxygen.
4.9875 mol C
=4
1.2461 mol O
(1pt)
19.950 mol H
=16
1.2461 mol O
1.2461mol O
1.2461 mol O
=1
The empirical formula is: C4 H16 O
(b)
Calculate the molar mass from the freezing point data given.
(1pt) for calculating molality
Molar mass = grams/ mole
The problem gives the number of grams but moles must be found. Begin with the
freezing point formula.
i = 1 since the compound is a nonelectrolyte
∆Tf = Kf · m · i
Calculate the temperature change:
Normal freezing temperature: 179.8 ºC
Freezing point with solute added: 89.2 ºC
∆Tf = 179.8 – 89.2 = 90.6 ºC
Solve the equation for molality:
o
ΔTf
90.6 C
m=
m=
= 2.27 m
o
40.0 C/m
Kf
(1pt) for calculating molar mass
Find the number of moles from the molality equation.
m=
mol solute
kg solvent
mol solute = m × kg solvent
= 2.27 mol/kg × .045 kg = .102 moles solute
Solve for molar mass: molar mass =
(c)
(1pt)
.102 mol solute
Calculate mole fraction for camphor.
Calculate the moles of each component.
8.175 g C 4 H16 O ×
45.00 g C10 H16 O ×
(1pt)
8.175 g solute
1 mol C 4 H16 O
= .102 mol
80.1g
1 mol C10 H16 O
= .2961mol
152.00 g
Calculate the mole fraction for camphor.
Χ camphor =
=
moles camphor
.2961
.2961 + .102
total moles
= .744
= 80.1g/mol
(d)
(1pt)
Calculate the boiling point.
Calculate the boiling point change. Substitute and solve.
ΔTb = K b ⋅ m ⋅ i
o
= (5.62 C/m) (2.27 m)
o
= 12.7 C
(1pt)
Add the boiling point change to the normal boiling point.
New boiling point: 207.42 º C + 12.7 º C = 220.1 º C
(e)
(1pt)
Calculate the vapor pressure of camphor above the solution.
Vapor pressure is lowered for a nonvolatile solute according to Raoult’s law.
∆Psolvent = (Xsolute) (Psolvent)
Calculate the mole fraction of the solute.
1.0 - .744 = .256
Calculate the change in pressure by substituting in to Raoult’s Law.
∆Psolvent = (.256) (745 mm Hg) = 191 mm Hg
Calculate the new vapor pressure.
745 mm Hg – 191 mm Hg = 554 mm Hg
Alternatively, the problem could be solved using the following:
Psolution = (Χsolvent) (Pº solvent)
Psolution = (0.744) (745 mm Hg) = 554 mm Hg
4.
Net Ionic Equations –choose 5 of 8 (3 points for each equation for a total of 15
points)
For each reaction:
1 point is awarded for correct reactants and 2 points for correct products.
All substances that ionize (strong acids, strong bases and soluble salts) must be written in
ionic form. All spectator ions must be cancelled.
1 point is awarded for balancing the reaction
1 point is awarded for a correct answer to the descriptive chemistry question
You can practice these to your heart’s content at : http://dwb4.unl.edu/AP2/
(a) Dilute hydrochloric acid is added to a solution of potassium carbonate.
(i) 2 H+ + CO32− → H2O + CO2
(ii) Describe a test to positively identify the gaseous product.
A glowing splint will be extinguished in the presence of the carbon dioxide gas.
(b) Water is added to a sample of solid magnesium nitride.
(i) 6 H2O + Mg3N2 → 3 Mg(OH)2 + 2 NH3
(ii) Describe any change in color or odor.
The smell of ammonia will be noticeable as the reaction proceeds.
(c)
Ethanol is burned in oxygen.
(i) C2H5OH + 3 O2 → 2 CO2 + 3 H2O
(ii) Identify a possible source of the activation energy for this reaction.
A lit match, spark or other ignition source could ignite this mixture.
5. Atomic Structure and Bonding---9 points total
In an essay of this type, be sure to read the prompt and address whatever is asked. In this essay it
is important to discuss the bonding and/or draw Lewis structures to demonstrate understanding.
When two substances are referred to in the question, be sure and refer to both substances in the
explanation.
(a) Perchloric acid is stronger than chloric acid. Explain.
(1pt) Draw each of the substances.
O
O
Cl
O
H
O
(1pt)
O
Cl
O
H
O
Explain.
Perchloric acid has four oxygen atoms compared to only three found in chloric acid. Since
oxygen is so highly electronegative it causes the net pull away from hydrogen to be greater
with each additional oxygen, making it easier for water molecules to surround and remove
the hydrogen of perchloric acid compared to that of chloric acid. An acid is considered to be
strong if it ionizes close to 100% in solution. The less energy needed to remove the
hydrogen, the stronger the acid.
(b) C4H10 is a liquid and CH4 is a gas.
(1pt) Draw each of the substances.
(1pt)
(1pt)
Both butane and methane are covalently bonded and are nonpolar molecules. The
predominant type of intermolecular forces holding the molecules together are London
Dispersion forces. (LDFs)
Butane, C4H10, is a larger molecule with many more electrons than methane, CH4. The more
electrons in a substance, the greater chance for polarizability. Thus, butane is more
polarizable and is held by LDF more tightly than methane. The greater the intermolecular
force, the more condensed the state of matter.
(c) PF3 is polar and PF5 is not.
(1pt) Draw each of the substances.
F
F
P
F
F
F
F
PF
P
F
F
5
PF3
PF5
Central Atom
Central Atom
Total Electron Pair:
4
Total Electron Pair:
5
Bonding Pair:
3
Bonding Pair:
5
Non-Bonding Pair:
1
Non-Bonding Pair:
0
P
F
F
F
F
P
F
F
F
F
Molecular Geometry:
Trigonal pyramidal
Molecular Geometry:
Trigonal bipyramidal
(1pt) PF3 has a trigonal pyramidal shape. The one unshared electron pair on the central atom pulls
with greater force on phosphorous than do the shared pairs of electrons with fluorine. Because of
this unequal pull on the central atom, the molecule has a net dipole moment and is polar. PF5 is a
trigonal bipyramidal structure. The five shared pairs of electrons pull equally on the central atom
causing the molecule to have no net dipole moment and thus nonpolar.
(d) Bond lengths in the carbonate ion are all identical but are shorter than a single bond.
(1pt) Draw all of the resonance structures.
↔
(1pt)
↔
The carbonate ion appears to have one double bond and two single bonds. Single bonds are
longer than double bonds. None of the resonance structures above actually exist. The true
structure is a composite of the three diagrams. Therefore, all of the bonds are about 1 and 1/3
of a bond ---not single and not double.
6. Laboratory Question—9 points total
(a) Steps necessary to carry out the experiment.
When writing steps in an experiment, a brief list of the essential steps are a must.
Steps for this experiment might include:
1.
Mass out an alloy sample on the analytical balance.
2.
Place the alloy into a beaker and add excess nitric acid. Be sure to perform
this step under the fume hood.
3.
Stir the beaker and its contents with a stirring rod to be sure that all metal is in
solution.
4.
Precipitate the silver ions by adding an excess of the sodium chloride solution.
Stir with the stirring rod as sodium chloride is added. Continue this process
until no more white precipitate forms.
5.
Mass a piece of filter paper and set up the vacuum filtration system.
6.
Empty the entire contents of the beaker into the Buchner funnel. Rinse the
beaker with small amounts of distilled water to remove any residue. Empty
this residue into the Buchner funnel.
7.
Collect the solid and the filter paper and place in a drying oven.
8.
Mass the solid and the filter paper.
9.
Clean up the lab area.
No two students will write the same procedure. The most essential steps that will earn points
would include:
(1pt) React the alloy with nitric acid.
(1pt) Precipitate the silver with sodium chloride.
(1pt) Dry the sample of silver chloride.
(2pt)
(b) List measurements:
Mass of the alloy
Mass of the filter paper
Mass of the filter paper with the solid AgCl
(one point for any two)
(c) Calculations necessary to find % silver.
mass of Ag from AgCl = mass of AgCl ×
(2pt)
% Ag =
mass Ag
×100
total mass of alloy
(one point for each equation)
(d) Would NaBr be appropriate? Explain.
MM of Ag
MM of AgCl
(2pt)
NaBr would be a good substitute. The purpose of NaCl was to precipitate the silver ions and
sodium bromide would be able to do this. Silver forms insoluble salts with chloride, bromide
and iodide.
Ag+ + Br- → AgBr ↓
7. Extra Practice Essay ---Electrochemistry—8 points total
(2pt)
(1pt)
(2pt)
(2pt)
(1pt)
(a) Predict sign of Eº and explain.
The sign of Eº must be positive. The prompt gives a K value of 1.5 × 1011 which means
that the products are favored at equilibrium. Since the reaction proceeds as written, the
voltage should be positive.
(b) Identify reducing agent.
Cd changes oxidation states from 0 to +2—thus, it is oxidized. Whatever species is
oxidized is known as the reducing agent.
(c) At a higher temperature, how would the cell potential change?
Explain questions such as this with mathematical formulas if at all possible. There are
two equations that allow this prediction.
∆Gº = - nFEº and ∆Gº = - RT lnK
Therefore, - nFEº = - RT lnK
Simplifying to solve for Eº:
RT lnK
Eo =
nF
The voltage should increase with an increase in temperature, all other factors held
constant since voltage and temperature are directly proportional to each other.
(d) Changing concentration of cadmium ions has what effect?
Explain mathematically if possible.
.0592
E = Eo −
ln Q
n
[Cd 2+ ] .20
Q=
=
[Cu 2+ ] 1.0
The voltage will increase with a decrease in the concentration of cadmium. The ln of
a number less than one will be negative. A negative multiplied by a negative will
yield a positive a number, which will add to the standard voltage.
(e) Cell potential at equilibrium.
By definition, cell potential at equilibrium is zero
8. Thermodynamics—8 points total
(2pt)
(1pt)
(2pt)
(1pt)
(2pt)
(a) Predict entropy change. Explain.
∆S will be negative. Two moles of gaseous reactants form one mole of solid so the
system becomes more ordered.
(b) Predict free energy change at equilibrium. Explain
By definition, ∆Gº = 0 at equilibrium.
(c) Temperature is increased ∆G changes sign. Explain.
The forward reaction is spontaneous at low temperatures, so ∆G is negative. The
Gibb’s equation is useful in explaining this.
∆Gº = ∆Hº - T∆Sº
Because ∆S is negative, ∆H must also be negative if the reaction is spontaneous.
Therefore, if the temperature increases so much that the T∆S term is greater than the
∆H term, the free energy value becomes positive and the reaction becomes nonspontaneous.
(d) Write an expression for calculating the specific temperature referred to in part c.
Begin with Gibb’s equation: ∆Gº = ∆Hº - T∆Sº
At equilibrium ∆Gº = 0 and the equation becomes:
0 = ∆Hº - T∆Sº
Rearrange to solve for T:
ΔHo
T=
ΔSo
(e) What effect does adding more NH4Cl have on the value of the equilibrium constant?
Explain.
If the system is at equilibrium when more ammonium chloride is added, the equilibrium
constant will remain the same. A solid is considered to have constant concentration and
therefore, does not appear in the equilibrium expression.