MASTER CLASS PROGRAM UNIT 4 — CHEMISTRY SEMESTER TWO 2014

Transcription

MASTER CLASS PROGRAM
UNIT 4 — CHEMISTRY
SEMESTER TWO 2014
WRITTEN EXAMINATION — SOLUTIONS
SECTION A — MULTIPLE CHOICE QUESTION
Question 1
B
Question 6
A
Question 11
C
Question 16
D
Question 2
C
Question 7
B
Question 12
B
Question 17
A
Question 3
C
Question 8
C
Question 13
D
Question 18
C
Question 4
B
Question 9
A
Question 14
C
Question 19
A
Question 5
D
Question 10
D
Question 15
D
Question 20
B
QUESTION 3
Answer is C
−1
Rate of decomposition = 2/3 rate of oxygen formation = 2/3 × 0.25 = 0.167 mol.s
The term ‘decomposition’ means that the rate will be a positive number.
QUESTION 5
Answer is D
The overall reaction is: H 2O( g )  H 2( g ) +
1
O2( g )
2
Equation 1: SO2(g) + 2H 2 O(g) + Br2(g) → H 2 SO4(l) + 2HBr(g)
ΔH = −137.3kJ/mol
Equation 2: 2H 2 O(g) + 2 SO2(g) + O2(g) → 2 H 2 SO4(l)
ΔH = -550.8 kJ/mol
Equation 3: 2HBr(g) → H 2(g) + Br2(g)
ΔH = + 103.7 kJ/mol
Equation 1 has water as a reactant so leave it as is.
Equation 2 has oxygen as a reactant so reverse this equation. The amount of oxygen also
needs to be halved (reverse the sign and halve the ∆H value).
Equation 3 has hydrogen as a product so leave it as is.
Cancel chemicals that appear as reactants and products then add the equations (and ∆H
values).
ΔH = −137.3 + 275.4 + 103.7 = 241.8 kJ/mol
 The School For Excellence 2014
Unit 4 Master Classes – Chemistry – Trial Exam
Page 1
QUESTION 6
Answer is A
The specific heat capacity of an object is the amount of energy required to increase the
temperature of 1 g of the substance by 1⁰C.
In this case, the heat capacity can be determined by the amount of heat released per gram
and per ⁰C change in temperature as the metal is immersed in water.
The amount of energy released into the water can be determined using the heat capacity of
water.
Δ T ( H 2 O ) = 25.7 -18.5 = 7.20°C
Ereleased
into water
= 50.0 × 7.20 × 4.18 = 1504.8 J
The heat capacity of the metal is then found by determining how much energy was released
per gram and per ⁰C change in temperature.
Note: the final temperature of the metal will be the same as the final temperature of the
water.
Specific heat capacity of metal =
QUESTION 7
1504.8
= 0.450 J/g.°C -1
45.0 × (100 - 25.7)
Answer is B
Initial reaction mixture 2X + Y2  2 XY
Diagram 1 is not a possible equilibrium state.
2 particles of X have reacted which means that 1 particle of Y2 should also react to give 2
particles of XY. In diagram I, there are not 2 particles of XY.
Diagram 2 is a possible equilibrium state. 4 particles of X have reacted with 2 particles of Y2
to give 4 particles of XY which is predicted by the mole ratio of the equation.
Diagram 3 is not a possible equilibrium state.
For a reaction at equilibrium, all reactants and products are present to some extent.
QUESTION 8
Answer is C
If the pressure of a gaseous system is increased due to a decrease in volume at constant
temperature, then the system will move to decrease the pressure by favouring the side of
fewer particles.
For the reaction O3(g) + NO(g)  NO2(g) + O2(g) the number of particles on the reactant and
product side is equal so there will be no shift in equilibrium position.
For the reaction 2SO2(g) + O2(g)  2 SO3(g) there are fewer particles on the product side so
the equilibrium position would shift to the right.
Page 2
For the reaction CH 4( g ) + 2 H 2 S( g )  CS2( s ) + 4 H 2( g ) there are fewer particles on the reactant
side so the equilibrium position would shift to the left (correct answer).
2+
2+
For the reaction Cu( NH3 )4( aq )  Cu( aq ) + 4 NH3( aq ) the change in pressure would have no
effect
since this is not a gaseous reaction.
QUESTION 9
Answer is A
The graph shows that for the reaction aA + bB ↔ cC + dD, as the pressure increases, so
does the % yield. This indicates that there has been a net forward reaction to partly oppose
the pressure increase. Therefore there must be fewer particles on the product side than on
the reactant side. So, c+d < a+b.
As the temperature increases, so does the % yield. This indicates that a net forward reaction
has occurred to partially oppose the increase in temperature. Therefore, the forward
reaction must be endothermic. Since the number of products has increased and the number
of reactants has decreased, the value of K will have increased (since K eq =
QUESTION 10
[ P]
).
[ R]
Answer is D
HCN( aq ) + H 2O( aq )  H3O(+aq ) + CN(−aq )
i. and iv are correct
i.
Dilution with water
If the solution is diluted, then the concentration of all reactants and products will
decrease and will not return to their initial concentrations once equilibrium is re+
established. Since the [ H 3 O ] has decreased, the pH will increase. Water is a
reactant so the dilution would push the equilibrium to the right and there would be an
increase in the % ionisation of the acid.
ii.
Addition of KCN
The addition of KCN would add CN − ions to the system. This would push the
+
equilibrium to the left and reduce the % ionisation. The [ H 3 O ] would go down so the
pH would increase.
iii.
Addition of 3 drops of water.
The addition of 3 drops of H 2 O would push the reaction to the right so that the %
+
ionisation increases and the [ H 3 O ] would increase which would decrease the pH.
Page 3
iv.
Addition of NaOH.
The NaOH would react with the H 3 O
+
pushing the equilibrium to the right to partially
+
oppose the removal of the H 3 O . Therefore the % ionisation would increase. The
+
amount of H 3 O will also increase while equilibrium is being re-established, but
+
+
because the net forward reaction only partially opposes the loss of H 3 O , the [ H 3 O ]
will not return to its original concentration. Therefore, at the new equilibrium the
[ H 3O + ] will be less than before and the pH will have increased.
QUESTION 11
Answer is C
+
Adding an acid raises the [ H3O ]. To partially oppose this increase, some of the H 3O
−
+
−14
reacts with OH − to form water. This lowers [OH ] until K w = 1 x 10 .
QUESTION 12
Answer is B
+
An increase in pH value corresponds to a decrease in [ H 3O ].
The amount that it decreases by is:
10difference in pH = 10−8.3−−10 = 101.7 = 50.12
Alternatively:
pH = 8.3
+
[H 3O ] = 10
pH = 10
−8.3
[H 3O + ] = 10 -10 M
M
Difference in [H 3O + ] =
QUESTION 13
10 -8.3
= 50.12
10−10
Answer is D
2−
3+
The balance half equation for the conversion of Cr2O7( aq) to Cr(aq ) is
Cr2 O72(−aq) + 14H (+aq) + 6e − → 2Cr(3aq+ ) + 7 H 2 O(l )
Therefore, 6 mole of electrons are needed.
Page 4
QUESTION 14
Answer is C
The relevant equations are shown below.
H 2 O2( aq) + 2H (+aq) + 2e − → 2H 2 O(l )
2+
−
+
MnO 4(aq)
+ H (aq)
+ 5e → Mn (aq)
+ 4H 2 O (l)
(purple)
MnO 2(s) + 4H
(pink)
+
(aq)
2+
+ 2e → Mn (aq)
+ 2H 2 O (l)
(black/brown)
(pink)
E 0 = +1.49V
E 0 = +1.23V
Cu(2aq+ ) + 2e − → Cu( s )
E 0 = +0.34V
Fe(2aq+ ) + 2e − → Fe( s )
E 0 = −0.44V
Mn (2aq+ ) + 2e − → Mn( s )
E 0 = −1.03V
(pink)
When KMnO(aq) / Cu( s ) are mixed (shown above), no reaction will occur. This is because,
even though KMnO(aq) is a strong oxidant and Cu(s ) a strong reductant, there is no acid
present which is required for the reduction of KMnO(aq) .The combinations in all the other
2+
options would result in the pink Mn(aq ) ion being formed.
QUESTION 15
Answer is D
Non standard conditions will result in voltages other than expected. In option (b) and (d), the
conditions are not standard. Temperatures that are higher than the standard (option b)
would result in higher reaction rates and are more likely to also result in an increase in
voltage rather than a decrease. Using a lower concentration than the standard (option c) is
more likely to produce a voltage smaller than the expected voltage.
The use of a porous salt bridge would not affect the voltage since it is still acting to neutralise
the charge in both sides of the beaker. The anions used would also have no affect on the
voltage since they are spectator ions.
QUESTION 16
Answer is D
In the galvanic cell shown, lead ions would be reduced and magnesium metal would be
oxidised.
Pb(2aq+ ) + 2e − → Pb( s )
Mg ( s ) → Mg (2aq+ ) + 2e −
2+
2+
Electrons move from the Mg ( aq ) / Mg ( s ) half cell to the Pb( aq) / Pb( s ) half cell.
Mg(s ) is being oxidised.
Mg(s ) is the negative electrode.
Page 5
QUESTION 17
Answer is A
2+
2+
When Fe( aq) / Fe( s ) was connected to X ( aq ) / X ( s ) , a positive voltage was recorded with
Fe(2aq+ ) / Fe( s ) being attached to the negative end of the voltmeter. This indicates that the
Fe(2aq+ ) / Fe( s ) half cell was acting as the anode and Fe(s ) was being oxidised.
2+
2+
When Fe( aq) / Fe( s ) was connected to Y( aq) / Y( s ) , a positive voltage was recorded with
Fe(2aq+ ) / Fe( s ) being attached to the positive end of the voltmeter. This indicates that the
Fe(2aq+ ) / Fe( s ) half cell was acting as the cathode and that Fe(s ) was being reduced.
These results can be summarised into an electrochemical series as follows.
2+
2+
The voltage when X ( aq ) / X ( s ) and Y( aq) / Y( s ) are combined would be 0.35 + 1.77 = 2.12V .
QUESTION 18
2+
Answer is C
2+
If X ( aq ) / X ( s ) and Y( aq) / Y( s ) are connected, the following species would react.
2+
X (aq
) would be reduced.
Y( s ) would be oxidised.
2+
2+
Electrons would flow from Y( aq) / Y( s ) to X ( aq ) / X ( s ) .
Y( s ) would form the anode.
Page 6
QUESTION 19
Answer is A
The information from the question tells us that hydrogen is the reductant when the cell is
discharging. From that we can determine the following:
Discharging
Recharging
Hydrogen is the reductant.
Nickel ions are the reductant.
Nickel ions are the oxidant.
Hydrogen is the oxidant.
−
−
The equation NiOOH( s ) + H 2 O(l ) + e → Ni(OH ) 2( s ) + OH ( aq) shows a nickel compound
acting as an oxidant and being reduced in the process. Therefore, this reaction is occurring
as the cell is discharging and would be occurring at the cathode which is +ve in a galvanic
system.
QUESTION 20
Answer is B
The relevant equations from the electrochemical series are
Reduction reaction
E 0 value (V)
H 2 O2( g ) + 2H (+aq) + 2e − → 2H 2 O(l )
+1.77
O2( g ) + 4H (+aq) + 4e − → 2H 2 O(l )
+1.23
2+
( aq )
−
+ 2e → Pb( s )
-0.13
Fe(2aq+ ) + 2e − → Fe( s )
-0.44
2H 2 O(l ) + 2e − → H 2( g ) + 2OH (−aq)
-0.83
Pb
Mg
2+
( aq)
−
+ 2e → Mg( s )
-2.34
In cell A, the reactions occurring are:
Cathode (- electrode):
2H 2 O(l ) + 2e − → H 2( g ) + 2OH (−aq)
Anode (+ve electrode):
2H 2 O(l ) → O2( g ) + 4H (+aq) + 4e −
Therefore, no mass would be deposited on the negative electrode.
In cell B, the reactions occurring are:
Pb(2aq+ ) + 2e − → Pb( s )
2H 2 O(l ) → O2( g ) + 4H (+aq) + 4e −
In cell C, the reactions occurring are:
Fe(2aq+ ) + 2e − → Fe( s )
2H 2 O(l ) → O2( g ) + 4H (+aq) + 4e −
Page 7
Lead and iron will be deposited in cells B and C respectively. The mass is determined by
the mole of electrons that passes through the cell.
Q=I × t
= 3.00 × 30 × 60
= 5400 C
n(e − ) =
Q 5400
=
= 0.0600 mol
F 96500
1
1
n(e− ) = × 0.0600 = 0.0300mol
2
2
m(Pb) = 0.0300 × 207.2 = 6.22g
n(Pb) =
1
1
n(e− ) = × 0.0600 = 0.0300mol
2
2
m(Fe) = 0.0300 × 55.8 = 1.67g
n(Fe) =
Total mass deposited at the negative electrodes = 6.22 + 1.67
= 7.89 g
Page 8
SECTION B — SHORT ANSWER QUESTIONS
QUESTION 1
a.
(i)
The rest of the heat is used to heat the calorimeter.
(ii)
E = VIt = 2.30 × 3.00 × 2.00 × 60 = 828 J
Only 70% of the energy is used to heat the water:
70
× 828 = 579.6J
100
Energy absorbed by water = specific heat capacity of water × ΔT × m(water being heated)
579.6 = 4.18 × ΔT × 70
579.6
ΔT =
= 1.98°C
4.18 × 70
(iii)
C.F . =
Energy released into the calorimeter
change in temperature
(Note: The change in temperature of the calorimeter and its contents will be the
same as the change in temperature of the water.)
C .F . =
b.
828
= 418 J / °C
1.98
ΔT( due to methane) = 86.4 - (20 + 1.98) = 64.4°C
(i)
Ereleased by methane = ΔT × C.F.
= 64.4 × 418
= 26919.2J
= 26.9 kJ
n(CH 4 ) =
Hc =
0.50
= 0.03125 mol
16
26.9
= 861kJ / mol
0.03125
∴ Molar heat of combustion of CH 4 = 861 kJ/mol
(ii)
% accuracy =
=
experimental value
× 100
theoretical value
861
× 100 = 96.9%
889
Page 9
c.
ΔHc (methanol) = −725kJ/mol
Energy released into calorimeter by methane = 26.9kJ
1 mol methanol
x mol methanol
x=
725 kJ
26.9 kJ
26.9
= 0.0371 mol
725
m(methanol) = 0.0371 × 32 = 1.19 g
d.
(i)
Use the data booklet.
The heats of combustion increase with molecular mass of the hydrocarbons. As the
hydrocarbons become larger, the number of C-H bonds increases. If the molecules
contain more bonds, then there is more potential chemical energy being stored and
which can be released upon combustion.
(ii) The corresponding alcohol has a lower heat of combustion.
(iii) It suggests that the bonding in the alcohol is (overall) stronger than the bonding in
the corresponding alkane (due to the C-O and O-H bond).
QUESTION 2
a.
C6 H5COOH( aq ) + H 2O(l )  C6 H5COO−( aq ) + H3O+ ( aq )
b.
(i)
[H 3O +(aq) ] = 10 − pH = 10 −4.32 M
[C6 H 5COO − ( aq ) ] = [ H 3O + ( aq ) ] = 10 −4.32 M =4.79 × 10 −5 M
c.
[H 3O + ] 2
(i) 6.4 × 10 =
[C6 H 5COOH] − [H 3O + ]
−5
K a obtained from the Data Booklet
(10 −4.32 )2
6.4 × 10 =
[C6 H 5COOH] − 10 -4.32
−5
6.4 × 10 −5 × ([C6 H 5COOH] − 10 -4.32 ) = (10 -4.32 )2
[C6 H 5COOH] − 10 -4.32 =
[C6 H 5COOH] =
(10 -4.32 )2
6.4 × 10 −5
(10 -4.32 )2
+ 10 -4.32
−5
6.4 × 10
= 1.13 × 10-4 M
Page 10
(ii)
d.
% ionisation =
10 -4.32
× 10 0 = 42.4%
1.13 × 10 -4
Ionisation is not a good measure of the strength of an acid since even weak acids can
ionise to a significant extent if diluted as shown in this question (dilution pushes the
equilibrium to the right and increases ionisation). The K a value is a better measure of
acid strength.
QUESTION 3
a.
b.
The reactants are more stable than the products.
c.
ΔH = −220 − −390 = +170kJ / mol
d.
The reverse reaction has been halved, therefore, halve the Ea for the reverse reaction.
Ea =
50
= +25 kJ / mol
2
Page 11
QUESTION 4
a.
(i)
2 NOCl( g ) ⇔ 2 NO( g ) + Cl2( g )
Initial
Amount converted
Amount at
equilibrium
K eq =
=
[ NOCl( g ) ]
[ NO( g ) ]
[ Cl 2 ( g ) ]
2.50
2
= 1.25
28
x 1.25
100
= 0.350 ↓
1.25 − 0.350 = 0.900
0
0
0.350 ↑
0.350
2
= 0.175 ↑
0.175
0.350
[Cl2( g ) ][ NO( g ) ]2
[ NOCl( g ) ]2
0.175 × 0.350 2
= 0.0265 M
0.900 2
(ii) The equation has been halved so the initial equilibrium value needs to be square
rooted. It has also been reversed so the reciprocal of the initial equilibrium value
needs to be found.
K new =
b.
1
1
=
= 6.14M
0.0265
Kinitial
(i)
NOCl
NO
Cl2
*Observe mole ratios!
Page 12
(ii) The increase in volume has caused the pressure to decrease. Therefore the
reaction will favour a net forward reaction since the product side produces the
greatest number of particles and will partially oppose the pressure decrease.
c.
(i)
The circled area shows what happens to both the forward and reverse reaction
rates immediately after the change is made. The increase in volume means that
both rates will decrease sharply but the reverse reaction rate is affected to a
greater extent.
(ii) There is a net forward reaction as the equilibrium moves to partially oppose the
volume increase/pressure decrease. Therefore the forward reaction rate must be
greater than the reverse reaction rate.
d.
Since there is a net forward reaction, the mass of NO increases but since the volume
has also increased, the [ NO ] will have decreased.
Page 13
QUESTION 5
a.
(i)
Al(3l+) + 3e− → Al( s )
(ii) Balance the number of electrons in the half equations and then add.
4 Al(3l+) + 12e − → 4 Al( s ) (x 4)
3C( s ) + 6O(2l−) → 3CO2( g ) + 12e −
(x3)
4 Al(3l+) + 3C( s ) + 6O(2l−) → 4 Al( s ) + 3CO2( g )
Simplify: 2 Al2O3(l ) + 3C( s ) → 4 Al( s ) + 3CO2( g )
b.
Q=I × t
= 170000 × 24 × 60 × 60
= 1.47 × 1010 C
1.47 × 1010
n(e ) =
= 1.52 × 10 5 mol
96500
−
n(Al) =
1
1.52 × 10 5
n(e − ) =
= 5.07 × 10 4 mol
3
3
m(Al) = 5.07 × 10 4 × 27 = 1.37 × 106 g = 1.37 × 10 3 kg
c.
If the charge on the aluminium was +2 rather than +3, less electrons would be needed
per mole to convert aluminium ions into aluminium metal. Therefore more Al would be
produced.
d.
If the solution was aqueous, water would be reduced instead of aluminium since water
is a stronger oxidant than aluminium ions.
Page 14
QUESTION 6
a.
C2 H 5OH (l ) + 3H 2O(l ) → 2CO2( g ) + 12 H (+aq ) + 12e−
O2( g ) + 4 H + ( aq ) + 4e− → 2 H 2O(l )
b.
c.
m(ethanol) = 50 × 0.789 = 39.45g
n(ethanol) =
39.45
= 0.8576 mol
46
n(e- ) = 12 × n(ethanol) = 12 × 0.8576 = 10.2912 mol
Q = 10.2912 × 96500 = 993,110.87C
Q = I ×t
993,110.87 = 20.0 × t
t=
d.
993,110.87
= 49655.54 s = 827.59 min = 13.79 hrs
20
Any two of:
The lead-acid battery is rechargeable whereas the fuel cell is not
Reactants are continuously supplied to the fuel cell whereas there is a finite amount of
reactants in a lead-acid fuel cell.
Fuel cells require auxiliary systems in order to produce electricity, lead acid batteries do
not.
e.
Ethanol can be produced in a carbon neutral manner. It is produced from the
fermentation of biomass which is a renewable resource. Growing the biomass also
reduces the concentration of carbon dioxide in the atmosphere which helps reduce
global warming.
Page 15
QUESTION 7
a.
Under the given voltage, 1.6 V , the only metals that could be deposited are Cu , Sn
and Co .
b.
First layer will not be pure. Predominant reaction occurring will be the reduction of
2+
Cu (aq
) , however, small amounts of Sn and Co will also be deposited, as sufficient
voltage is passing through the cell to enable these reactions to occur.
c.
Zn(2aq+ ) + 2e − → Zn( s )
Sn(2aq+ ) + 2e − → Sn( s )
Co(2aq+ ) + 2e − → Co( s )
Assuming that all the metals ions were consumed (cathode mass stopped changing)
then
0.500 mol of Zn had been produced. This required 1.000 mol of electrons.
0.500 mol of Sn had been produced. This required 1.000 mol of electrons.
0.500 mol of Co had been produced. This required 1.000 mol of electrons.
Total number of mole of electrons required = 3.000 mol
Number of coulomb = 3 × 96,500 = 289,500 C = 2.90 × 10 5 C
Page 16
QUESTION 8
a.
CH 3 ( CH 2 )14 COOCH 2
|
CH 3 ( CH 2 )14 COOCH
+ 3CH 3OH → 3CH 3 ( CH 2 )14 COOCH 3 + C3 H 8O3
|
CH 3 ( CH 2 )14 COOCH 2
b.
Renewable
Carbon neutral
c.
Energy density describes how much energy can be derived from a fuel per unit mass.
The higher the energy density, the more energy that can be derived from the fuel per
unit mass and hence the better the fuel.
d.
39.55 MJkg −1 = 39.55 ×106 J /1000 g =
39.55 × 106
J / g = 35.55 kJ / g
103
Ethanol molar enthalpy ΔH = −1364 kJ / mol (Data Booklet)
In 1.00 g: n(C2 H 5OH ) =
m 1.00
=
= 0.02174 mol
M 46.0
0.02174 mol → x kJ
1 mol → −1364 kJ
∴ x = −29.65217 kJ
Energy density is 29.7 kJg −1
Biodiesel is the better fuel.
Page 17

MASTER CLASS PROGRAM UNIT 4 — CHEMISTRY SEMESTER TWO 2014

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