CH12 Unit03 PEA01 VIC - worksheet PDF

Preview

Summary

worksheet...

Description

Heinemann Chemistry 2 5e

Chemistry Unit 3 Practice Exam 1 Answers Section A—Multiple-choice questions Question 1 C

A thermochemcial equation is balanced with respect to the number of atoms of each element and in the case of ionic equations, electric charge. The enthalpy change, ΔH, for the reaction and symbols of state are also included. Since heat is generated the reaction is exothermic, which is indicated by the negative sign in front of the value of the enthalpy change for the reaction.

Question 2 D

1 MJ = 1000 kJ 1000

n(CH3OH) = 726 mol 1000

m(CH3CH2OH) = 726 × 32.0 = 44.1 g (3 sig. figs) Question 3 B

0.74

n(C4H10O) = 74.0 = 0.010 mol m(water) = 500.0 g energy released by 0.010 mole C4H10O = mcΔT = 500.0 × 4.18 × 11.0 J = 22 990 J 22 990

energy released by 1.0 mole C4H10O = 0.01 = 2 299 000 J =

2 299 000 1000 = 2299 kJ

= 2.3 × 103 kJ enthalpy of combustion = 2.3 × 103 kJ mol−1 (2 sig. figs) Question 4 C

The combustion of both biodiesel and petrodiesel produces carbon dioxide. The carbon dioxide from the combustion of biodiesel is offset by the consumption of carbon dioxide during photosynthesis, which occurs in the chlorophyll of plants used to manufacture biodiesel.

Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4886 1125 4 Page 1

Question 5 C

There is no proton transfer so the reaction is not an acid–base one. There is a transfer of electrons from Zn to H+ to form Zn2+ and H2. The Zn loses electrons so is oxidised by the H+, which is therefore the oxidising agent.

Question 6 C

Consider the following standard reduction potentials. Cu2+(aq) + 2e−

Cu(s) + 0.34 V

Pb2+(aq) + 2e−

Pb(s) – 0.13 V

Zn2+(aq) + 2e−

Zn(s) – 0.76 V

If metal M reacts with two of the solutions but not three, the standard reduction potential of M2+(aq) + 2e− M(s) must be between –0.76 V and −0.13 V. Question 7 C

From the reduction potentials in Question 6, Zn is the strongest reducing agent.

Question 8 B

Oxidation occurs at the anode. The anode is negative since oxidation reactions produce electrons. Hydrogen gas is oxidised at the anode. H2(g) + 2OH−(aq) → 2H2O(l) + 2e−

Question 9 D

The standard electrode potentials can be obtained from the VCAA Data Book. O2(g) + 2H2O(l) + 4e− 2H2O(l) + 2e−

4OH−(aq)

H2(g) + 2OH−(aq)

E0 = 0.40 V E0 = −0.83 V

cell voltage = E0cathode reaction − E0anode reaction = 0.40 − (−0.83) = 1.23 V Question 10 B

The half-equation at the ring is Au3+(aq) + 3e− → Au(s). As this is a reduction, it will take place at the cathode. In order to attract the positively charged Au3+(aq) ions, the ring needs to be attached to the negative terminal of the power supply.

Question 11 B

The oxidation reactions that occur at the anode in a galvanic cell generate electrons, which gives the anode a negative polarity. The external power source in an electrolytic cell withdraws electrons from the anode, which has a positive charge. The external power source provides electrons to the cathode, which has a negative charge.

Question 12 C

Since PbO2 has the greater standard electrode potential it is reduced to PbSO4 during discharge, while Pb is oxidised to PbSO4. By definition, reduction occurs at the cathode in both galvanic and electrolytic cells. The equation for the cathode reaction as the cell discharges is: PbO2(s) + SO4 2−(aq) + 4H+(aq) + 2e− → PbSO4(s) + 2H2O(l) As the cell is recharged, the cathode reaction is: PbSO4(s) + 2e− → Pb(s) + SO4 2−(aq) Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4886 1125 4 Page 2

Question 13 A

The rate of reaction increases, and continues to increase, as both pressure and temperature increase.

Question 14 D

The reaction is exothermic (ΔH < 0) so an increase in temperature will favour the reverse reaction. Therefore, the value of the equilibrium constant will decrease and so will the equilibrium yield of NO(g).

Question 15 A

The reaction is endothermic, so an increase in temperature will shift the equilibrium to the right, thus producing more products. There are 2 moles of reactants in the equation but 4 moles of products. Therefore, an increase in pressure will favour the reverse reaction and the amount of H2(g) will decrease. A catalyst only affects the reaction rate, not the product yield.

Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4886 1125 4 Page 3

Section B * equals 1 mark (4 marks)

Question 1 a. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)* ΔH = −899 kJ mol−1*

(1 mark for correctly balanced equation including symbols of state, 1 mark for correct sign and units for heat of combustion) b. 1 MJ = 1000 kJ 1000

n(CH4) = 899 = 1.12 mol* pV = nRT nRT

V= P =

1.12 × 8.31 × 298 100

= 27.74 L* (Alternatively, multiply n(CH4)* by molar volume at SLC* (24.76 mol L−1).) (8 marks)

Question 2 Property

Biodiesel

Petrodiesel

Source

Plants*

Crude oil*

Chemical composition

Fatty acid ester molecules that have between 12 and 22 carbon atoms.* The formula of a typical molecule is C16H34O2.

Hydrocarbon molecules that have between 10 and 15 carbon atoms.* The formula of a typical molecule is C12H23.

Relative viscosity and implications for use

Biodiesel is more viscous than petrodiesel and cannot be used at low temperatures as it blocks the flow of fuel in fuel lines.*

Petrodiesel is less viscous than biodiesel and can be used at lower temperatures.*

Relative amount of greenhouse gases generated in their extraction and use

The carbon dioxide generated by the combustion of biodiesel is offset by the carbon dioxide absorbed by plants during photosynthesis. A relatively small amount of carbon dioxide is produced during growing, extraction and transport of biodiesel.*

Large amounts of carbon dioxide are produced by the combustion of petrodiesel. One molecule of C12H23 produces 12 molecules of carbon dioxide. The extraction, processing and transport of petrodiesel also produce large amounts of carbon dioxide.*

(4 marks)

Question 3 Equation

Explanation +

2+

a. Mg(s) + 2H (aq) → Mg (aq) + H2(g)*

Mg has lost 2e− so it has been oxidised to Mg2+. Two H+(aq) have gained e− so they have been reduced to H2.*

Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4886 1125 4 Page 4

Equation

Explanation

b. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)*

The oxidation number of C in CH4(g) is −4 and in CO2(g) it is +4. As the oxidation number has increased, the carbon has been oxidised. The oxidation number of O in O2 is 0 and in H2O it is −2. As the oxidation number has decreased, the oxygen has been reduced.*

Question 4

(4 marks)

a. The blue becoming lighter indicates that the Cu2+(aq) ions are being consumed, so the cathode reaction is Cu2+(aq) + 2e− → Cu(s)*. At the anode, the reaction must be M(s) → M2+(aq) + 2e− So metal M causes the reduction of the Cu2+(aq) and therefore M is the stronger reducing agent*. b. M is the reducing agent and Cu2+(aq) must be the oxidising agent. E0cell = E0 oxidant − E0reductant 1.52 = 0.34 (from data book) – E0reductant* E0reductant = 0.34 − 1.52 = −1.18 V = standard reduction potential for the M2+/M half-cell* Question 5

(5 marks)

a. energy transformations in a power station that burns fossil fuels chemical energy → heat energy → mechanical (kinetic) → electrical energy* energy transformations in a fuel cell chemical energy → electrical energy* b. Fuels cells have a more efficient conversion of chemical energy to electrical energy.* OR Fuel cells that use hydrogen produce only water and a small amount of heat as waste products. They do not generate greenhouse gases such as carbon dioxide.* OR Fuels such as hydrogen, methane, methanol and ethanol can be used in fuel cells.* (1 mark each for any two of the above advantages) c.

Fuel cells generate a direct current (DC) which must be converted to an alternating current (AC) for domestic and industrial use.* OR Some of the fuels used in fuel cells (such as hydrogen) are explosive.*

Question 6 a.

(8 marks)

at the cathode: Cu2+(aq) + 2e− → Cu(s)* at the anode: Cu(s) → Cu2+(aq) + 2e− * Zn(s) → Zn2+(aq) + 2e− * Pb(s) → Pb2+(aq) + 2e− *

Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4886 1125 4 Page 5

b. As most of the metal in the anode oxidises, the silver and gold impurities remain unreacted* as they are weak reducing agents* and drop to the bottom of the tank. c.

Zn2+(aq) and Pb2+(aq), which are formed at the anode, are weaker oxidising agents than Cu2+(aq)* and so will not be reduced at the cathode but remain in solution*. (6 marks)

Question 7 a. Energy (kJ mol–1) without catalyst

186 kJ mol–1

with catalyst

100 kJ mol–1 26 kJ mol–1

HI

I2, H2

b. i.

Number of particles

(1 mark for correct graphs for uncatalysed and catalysed reactions, 1 mark for correct placement of ΔH value, 1 mark for correct placement of EA values for uncatalysed and catalysed reactions)

Catalyst

EA

Kinetic energy

ii. A catalyst enables the reaction to take place via an alternative reaction pathway that has lower activation energy.* Consequently, a greater number of particles have a kinetic energy equal to or greater than the activation energy, resulting in a greater number of successful collisions.* (8 marks)

Question 8 2+

a.

[FeSCN ]

Kc = Fe3+[SCN−]* [FeSCN2+] = Kc[Fe3+][SCN−] = 1.05 × 103 × 0.01 × 0.01 = 0.105 M*

b. (1 mark for each correct colour change in column 1, 2 marks for each correct explanation) Change The mixture is heated.

Intensity of red colour of mixture increased/decreased/the same

Explanation in terms of reaction rates and position of equilibrium

becomes lighter red*

The forward reaction is exothermic. Increasing the temperature increases the rate of both forward and reverse reactions but the rate of the endothermic reaction (reverse reaction here) is increased more.* So the equilibrium shifts to the left and less red FeSCN2+ is present.*

Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4886 1125 4 Page 6

Change A small amount of solid NaSCN is added.

Intensity of red colour of mixture increased/decreased/the same

Explanation in terms of reaction rates and position of equilibrium

becomes darker red*

At the instant the concentration of one of the reactants increases, the rate of the forward reaction increases and is therefore greater than the rate of the reverse reaction.* So at the new equilibrium position, more products have formed. More FeSCN2+ causes the darker colour.*

Copyright © Pearson Australia 2016 (a division of Pearson Australia Group Pty Ltd) ISBN 978 1 4886 1125 4 Page 7...