Chemistry Answers Chapter 14

Answers Chapter 14 Exercises (a) Solar heating, solar electricity, hydroelectricity, wind power, biomass. Moles of CH4 = 8.00 × 107/891 = 8.98 × 104 mol (b) Fos sil fuels. Mass = 8.98 × 104 mol × 16.05 g mol−1 (d) Renewable sources are generally derived from recent solar energy. 2 (a) useful output energy = total input energy useful heat energy = 0.85 total input heat energy Heat energy produced by combustion = 4.00 × 107 kJ/0.85 = 4.71 × 107 kJ Moles of CH4 = 4.71 × 107/891 = 5.28 × 104 mol Mass = 5.28 × 104 mol × 16.05 g mol−1 = 8.48 × 105 g = 8.48 × 102 kg (b) Energy produced by combustion = 4.00 × 107 kJ/0.5 = 8.00 × 107 kJ = 1.44 × 106 g = 1.44 × 103 kg 3 (a) DHc / kJ mol−1 (c) Tidal is due to the presence of the moon, nuclear fission due to presence of radioactive elements found in the earth. Specific energy / kJ g−1 H2 = 2 × 1.01 =2.02 −286 = 286/2.02 =142 CH4 = 12.01 + (4 × 1.01) −891 = 891/16.05 Formula 1 M / g mol−1 = 55.5 = 16.05 (b) Assuming ideal behaviour: PV = nRT PV = (m/M)RT ρ = m/V = PM/RT With everything in SI units, the units of density are kg m−3 With the molar mass in g mol−1, the units of density are g m−3 STP conditions: T = 273 K and P = 100 kPa 76 10:02 Formula H2 Density / g m−3 Specific energy / kJ g−1 Energy / kJ cm−3 = 286/2.02 = 1.00 × 105 × 2.02/8.13 × 273 = 142 = 91.0 = 891/16.05 = 1.00 × 105 × 16.05/8.13 × 273 = 55.5 = 723 = (286/2.02) × 1.00 × 105 × 2.02/ (8.13 × 273) = 12 900 CH4 = (891/16.05) × 1.00 × 105 × 16.05/ (8.13 × 273) = 40 100 Note: energy density of a gas is not determined by the molar mass. 3 (b) Increases the yield of the more useful lower fractions used as fuels for cars. (c) Hydrogen is the best fuel. It is dangerous to store and burn. 4 (b) S + O2 → SO2 and 2SO2 + O2 → 2SO3 and H2O + SO2 → H2SO3 and H2O + SO3 → H2SO4 2N + O2 → 2NO and 2NO + O2 → 2NO2 and 2NO2 + H2O → HNO3 + HNO2 5 useful output energy = (a) total input energy useful heat energy = 0.38 total input heat energy Heat energy produced by combustion = 5.00 × 105 kJ s−1/0.38 Mass = 5.00 × 105 kJ s−1/(0.38 × 33.0 kJ g−1) = 39900 g s−1 = 39.9 kg s−1 (b) We have the unbalanced equation: CH + (5/4)O2 → CO2 + 21H2O no. of moles of CO2 = no. of moles of CH no. of moles of C H = 39900/(12.01 + 1.01) = 3062 mol mass of CO2 = 3062 × 44.01 = 135 000 g = 135 kg 6 CH3 CH3 C CH3 CH2 CH CH3 CH3 7 Produces the more reactive alkenes, which can be used as chemical feedstock, to make useful products such as plastics. (a) Empirical formula: C135H96O9NS. (It typically also contains trace elements of silicon, sodium, calcium, aluminium, nickel, copper, zinc, arsenic, lead and mercury.) (a) C11H24 → C2H6 + 3C3H6 or C11H24 → C5H12 +3C2H4 (c) Lower temperatures needed. Catalysts act selectively, increasing the yield of the desired product. 8 (a) Octane is one of the main components of petroleum: C8H18 Any compound between pentane and decane would be acceptable as an answer. (b) They are isolated from crude oil by fractional distillation. The mixture of hydrocarbons is heated, causing them to vaporize. As the vapour travels up the fractionating column the hydrocarbons condense at different heights, resulting in their separation. The different compounds have different boiling points: the lowest boiling point compounds condense at the top and the highest boiling point compounds condense at the bottom. As the relative molar mass increases, the attractive London dispersion forces between the molecules increase, leading to an increase in the boiling point. (c) The components of gasoline have boiling points above normal temperatures. They are volatile liquids. They can be easily vaporized in the car cylinder for reaction with oxygen. 77 10:02 Answers Lower molar mass compounds are gases which occupy too much volume, while higher molar mass compounds do not vaporize or burn easily. (d) Higher fractions can be cracked into smaller molecules; the larger molecules are heated with a catalyst and broken into smaller molecules. Alkenes formed in the cracking process can undergo alkylation reactions with lower molecular mass alkanes to further increase the yield of gasoline. Coal liquefaction Direct: 5C(s) + 6H2(g) → C5H12(l) 11H2(g) + 5CO(g) → C5H12(l) + 5H2O(l) 13 They are needed as chemical feedstocks. Their combustion adds carbon dioxide to the atmosphere, which contributes to global warming. 14 C6H10 + C10H24 CO(g) + H2(g) + O2(g) → CO2(g) + H2O(g): ∆Hc0 = −283 + (−286) kJ mol−1 = −569 kJ mol−1 Alkylation: (CH3)3CH + CH3CH=CHCH2CH3 → CH3CH–CHCH2CH3 | | (b) CH4(g) + 2O2(g) → CO2(g) +2H2O(g): ∆Hc0 = −891 kJ mol−1 The general pattern is: straight-chain alkanes < cycloalkanes < alkenes < aromatics (c) One mole of synthesis gas has the same volume as two moles of methane. One mole of synthesis gas produces 569 kJ and two moles of methane produce 2 × 891 kJ. pentane < cyclopentane < pentene < benzene 10 The general pattern is that the octane number of straight-chain alkanes decreases with an increase in chain length. Alcohols have very high octane numbers. heptane < hexane < pentane < ethanol 11 (a) High specific energy / energy density. As a liquid it is convenient to handle and deliver. Easy to vaporize, which to assists combustion. (b) It is formed by the partial decomposition of marine plants millions of years ago. (c) Compounds are separated by fractional distillation. Increase of petrol fraction by cracking. Further refining: reforming, alkylation, isomerization to increase octane number. 12 Coal gasification C(s) + H2O(g) → CO(g) + H2(g) (a) CO(g) + 12O2(g) → CO2(g): ∆Hc0 = −283 kJ mol−1 H2(g) + 12O2(g) → H2O(g): ∆Hc 0 = −286 kJ mol−1 (e) Cracking: C16H34 → C6H12 + C10H22 9 Carbon-containing fuels are non-renewable. 15 (a) Methane is the major component of natural gas. It has the formula CH4. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) (b) Natural gas is the cleanest of the fossil fuels to burn as it has a high H : C ratio. The combustion of natural gas produces minimal amounts of carbon monoxide, hydrocarbons and particulates. It does contribute to global warming but does not contribute to acid rain, unlike coal and oil. (c) Natural gas is the fossil fuel in the shortest supply and it is unevenly distributed around the world. Oil is expected to last a little longer and coal, which is distributed more evenly around the world, longer still. (d) Supplies of methane can be increased as a result of the cracking of larger hydrocarbons from oil or by coal gasification. CO(g) + 3H2(g) → CH4 + H2O 78 10:02 Cracking C4H10 → CH4 (g) + C3H6(g) 18 (c) They produce less carbon dioxide / have a smaller carbon footprint. Coal gasification C(s) + H2O(g) → CO(g) + H2(g) 19 (a) Predicted mass of deuterium nucleus = 2 (1.008 665 + 1.007 265) amu CO(g) + 3H2(g) → CH4 + H2O = 2.01593 amu ∆m = 2.01593 − 2.01355 amu = 0.00238 amu = 0.00238 × 1.66 × 10−27 kg (a) wide availability relatively cheap compared to other sources ∆E = 0.00238 × 1.66 × 10−27 × (3.00 × 108)2 J ease of transportation ∆E = 0.00238 × 1.66 × 10−27 × (3.00 × 108)2 × 6.02 × 1023 J mol−1 power stations can be built close to the source = 2.14 × 1011 J mol−1 high energy density concern over nuclear (b) limited productivity of other sources not possible to generate sufficient electrical energy without it 4 2 No. of nucleons = 2.14 × 108 kJ mol−1 can be used with existing technology Nuclide 16 Binding energy / kJ mol−1 He 4 2.73 × 109 many transport systems rely on fossil fuels 2.73 × 109/4 8 = 6.825 × 10 (b) Oil used to power internal combustion engines 17 Binding energy / kJ mol−1 per nucleon 2 1 H 2 2.14 × 108 2.14 × 108/2 = 1.07 × 108 It is more efficient It produces more thermal energy per unit of mass / has a higher specific energy / energy density It produces less CO2 per unit of output energy 18 (a) Fuel Methylbenzene Ethanol (b) Fuel Methylbenzene ∆Hc0 / kJ mol–1 −3910 −1367 Moles of CO2 produced / mol 7 × 236 Moles needed to produce 10 000 kJ Molar mass / g mol–1 Mass needed to produce 10 000kJ 10 000/3910 (12.01 × 7) + (8 × 1.01) 2.56 × 92.15 = 2.56 = 92.15 = 236 10 000/1367 (12.01 × 2) + (6 × 1.01) + 16.00 7.31 × 46.08 = 7.31 = 46.08 = 337 Mass of CO2 produced / g 7 × 236 × 44.01 = 72700 Ethanol 2 × 337 2 × 337 × 44.01 = 29700 79 10:02 Answers (c) We have the following energy changes as the nuclei of the two elements are formed: 22 2p + 2n → 2 12H: ∆E = 2 × 2.14 × 108 kJ mol−1 = 6.05 × 1011 J 2p + 2n → 42He: ∆E = 2.73 × 109 kJ mol−1 = 6.05 × 108 kJ Using Hess’ law we have for the following: (b) Mass of coal needed = 6.05 × 108 kJ/ 30 kJ g−1 = 20 × 106 g 2 2 1H → 2p + 2n: ∆E = −2 × 2.14 × 108 kJ mol−1 2p + 2n → 42He: ∆E = 2.73 × 109 kJ mol−1 Adding the equations: = 20 × 103 kg 23 E = 0.018884 × 10−3 kg × (3.00 × 108)2 ∆E = 2.30 × 109 kJ mol−1 = 1.70 × 1012 J mol−1 = 1.70 × 109 kJ mol−1 (d) 42He is more stable than 21H (a) 10n (b) 21 17 8 (a) E = Δmc2 Δm = 0.018884 amu = 0.018884 g mol−1 4 2 2 1H → 2He: ∆E = (−2 × 2.14 × 108) + (2.73 × 109) kJ mol−1 20 (a) Energy produced from 1 g = 6.72 × 10−6 × (3.00 × 108)2 J g−1 (b) mass of coal which produces this energy = 6.04 × 108 kJ/30 kJ g−1 = 2.02 × 107 g O = 2.02 × 104 kg (a) ΔE = hc/λ = 20.2 ×103 kg = 6.63 × 10−34 × 3.00 × 108/λ J = 6.02 × 1023 × 6.63 × 10−34 × 3.00 × 108/λ J mol−1 24 (a) W: atomic number = 90, mass number = 233 − 4 = 229: 229 90Th = 6.02 × 1020 × 6.63 × 10−34 × 3.00 × 108/λ kJ mol−1 X: atomic number = 0, mass number = 236 − 233 = 3: 310n = 0.00012/λ kJ mol−1 Y: atomic number = 93, mass number = 239: 239 93Np λ/m ΔE / kJ mol−1 656 × 10−9 0.00012/656 × 10−9 = 183 486 × 10−9 0.00012/486 × 10−9 = 247 434 × 10−9 0.00012/434 × 10−9 = 276 (b) By inspection we can see that transitions from n = 2 fall into this range ΔE / kJ mol−1 Transition 183 n=2→n=3 247 n=2→n=4 276 n = 2 → n =5 (c) At higher energy, energy levels become closer together; the energy differences between higher energy levels and the lower level (n = 2) become closer together and the difference in wavelength decreases. Z: atomic number = 34, mass number = 92: 92 34Se (b) Process II (fission) is used to produce electricity in nuclear power plants. (Process IV is also a fission reaction and is another potential reaction.) This process can be initiated as required by controlling the input of neutrons, whereas the other processes are natural ones and occur randomly. Process II is self-sustaining if the critical mass is available. It produces more neutrons than are needed initially and so a chain reaction occurs which can lead to the fission of more nuclei. (c) The mass of the products is less than the mass of the reactants. The difference is converted to energy according to the equation ΔE = Δmc2. 80 10:02 25 Less than 1% of uranium is the fissionable isotope: 235 92U There is less than the critical mass present. There is insufficient 235 92U to sustain the chain reaction. 26 As the ratio is close to one, the molecules have very similar physical properties and so are hard to separate. 28 (a) UO2 is a giant ionic; UF6 is molecular covalent. The +4 oxidation state forms ionic compounds but the +6 state forms covalent compounds as the U6+ ion would be too polarizing. (b) Enriched fuel can be used in the manufacture of nuclear weapons. 29 (b) Both compounds would have similar radioactivity as its depends on the nucleus of the atoms and not on the electron arrangements. (c) For fission reactors, the fuel (typically based on uranium) is usually based on the metal oxide; oxides are used rather than the metals themselves because the oxide melting point is much higher than that of the metal and because it cannot burn, being already in the oxidized state. 27 (a) The elements all have non-polar molecular covalent structures. The intermolecular forces are due to the interaction between an instantaneous dipole and induced dipole and so depend on the number of electrons. The boiling points increase as the number of electrons increase. (a) Fuel enrichment means that the amount of 235 U in the fuel is increased, which means more than the critical mass of 235U is available for fission so the reaction can be sustained. (a) mass which is converted to energy = 1/1000 × 1.0 kg = 1.0 × 10−3 kg ΔE = Δmc2 = 1.0 × 10−3 kg × (3.00 × 108 m s−1)2 = 9.0 × 1013 J = 9.0 × 1010 kJ (b) mass of coal which produces this energy = 9.0 × 1010 kJ/30 kJ g−1 = 3.00 × 109 g 30 Δm = 5.2 × 10−3 amu ΔE = Δmc2 = 5.2 × 10−3 × 1.661 × 10−27 × (3.00 × 108)2 = 7.77 × 10−13 J = 4.7 × 108 kJ mol−1 31 (b) UF6 has a lower boiling point than expected based on the trend in (a). Although it is a non-polar molecule it does contain polar bonds. The fluorine atoms all have a partial negative charge so repel the fluorine atoms in other molecules, which reduces the intermolecular forces. (a) The first neutron is needed to initiate a chain reaction and the reaction would not happen without it. (b) mass of reactants = 235.0439 + 1.0087 = 236.0526 amu mass of products = 95.9342 + 137.9112 + (2 × 1.0087) = 235.8628 amu Δm = 0.1898 amu ΔE = 0.1898 × 1.661 × 10−27 × (3.00 × 108)2J (c) It is volatile and the fluorine atoms only have one mass number so the mass difference is due to mass differences in the uranium atoms. (d) The separation depends on the ratio of the masses of the compounds containing the two isotopes = 352/349. = 0.1898 × 1.661 × 10−27 × (3.00 × 108)2 × 6.02 × 1023 J mol−1 = 1.71 × 1013 J mol−1 = 1.71 × 1010 kJ mol−1 32 (a) Th → 22488Rn + 24He 228 90 (b) mass of products = 228.022800 amu 81 10:02 Answers Δm = 0.005926 amu 35 235 92 ΔE = Δmc2 36 high level: contains fission products 144 1 U + 01n →………→ 90 38Sr + 54Xe + 2 0n Δm = 0.005926 × 1.66 × 10−27 kg low level: clothing / fuel cans / other ΔE = 0.005926 × 1.66 × 10−27 × (3.0 × 108)2J stored under water buried underground = 8.860 × 10−13 J encased in steel/concrete = 8.860 × 10−13 × 6.02 × 1023 J mol−1 vitrified / made into glass = 5.33 × 1011 J mol−1 = 5.33 × 108 kJ mol−1 37 (d) mass of coal burned = 5.33 × 108 kJ mol−1/28 kJ g−1 = 1.9 × 107 g Time / years = 1.9 × 104 kg 33 34 (a) Use the information in the question to compile a table: Time / half-lives Fraction remaining Activity / hr−1 g−1 0 60.0 5730 30.0 0 1 5730 × 2 15.0 1 0.5 5730 × 3 7.5 2 0.25 5730 × 4 3.75 3 0.125 4 0.0625 5 0.03125 6 0.015625 7 0.0078125 8 0.00390625 (b) One year is short compared to the half-life and so the number of plutonium nuclei does not change significantly. 9 0.001953125 (c) λ = ln2/t½ 10 0.000976563 The shell is approximately 5730 × 4 years = 22 920 years old. 38 (a) 238 94 4 Pu → 234 92U + 2He = ln2/(88 yr) = 0.007877 yr−1 (b) 99.9% ln([A]t)/([A]0) = −0.007877 yr−1 × 20 yr = −0.157533 First notice that 96.0 s = 96.0/19.2 = 5t½ ([A]t)/([A]0) = e−0.157 = 0.854 Use the information in the question to compile a table: = 85.4% Time / half-lives Time / s Activity / s−1 0 0 1200 1 19.2 600 2 3 38.4 57.6 300 150 4 76.8 75 5 96.0 37.5 The count rate would be 37 or 38 disintegrations per second. 39 (a) λ = ln2/(28.4 yr) = 0.0244 yr−1 A = A0e−0.0244 × 80 = 5.1 × 1016 × e−0.0244 × 80 = 7.24 × 1015 s−1 (b) Initial activity is very high and still highly radioactive after 80 years. This poses severe health risks and problems of disposal. 40 Eneeded = 1216 kJ mol−1 = 1216 × 1000 J mol−1/(6.02 × 1023 mol−1) 82 10:02 E = hν 44 ν = 1216 × 1000 J mol−1/(6.02 × 1023 mol−1 × 6.63 × 10−34 J s) Other sources: wind and hydroelectricity. (b) Advantage: renewable and has little environmental impact. = 3.05 × 1015 s−1 λ = 3.00 × 108 m s−1/3.05 × 1015 s−1 Disadvantages: photosynthesis is not very efficient so relatively little of the available solar energy is trapped. = 9.85 × 10−8 m This is in the UV region of the electromagnetic spectrum, as displayed in section 4 of the IB data booklet. 41 45 (c) Conjugated system of double and single bonds (d) Process: fermentation (b) The double bonds disappear as the molecules undergo an addition reaction with bromine. Equation: C6H12O6 → 2C2H5OH + 2CO2 Conditions: acidity / absence of oxygen / below 40 °C (c) Benzene absorbs radiation in the UV region. 42 (a) Increased conjugation (increased n) moves the absorption band λmax towards longer wavelength. Yeast provides enzyme 46 (c) Particulates (soot), hydrocarbons, carbon monoxide (d) Fossil fuels are running out. Biomass is a renewable source. 47 Some radiation is reflected or heats the surface of the earth Plants do not cover all the earth (c) Photosynthesis: 6CO2 + 6H2O → C6H12O6 + 6O2 C6H5−(CH=CH)6−C6H5 absorbs in purple/ blue region and is probably orange. The conjugated system includes eleven C= C bonds and so absorbs in the visible region. The molecule absorbs blue light and so appears orange. (a) 1% (b) Wavelength of radiation not absorbed by chlorophyll (c) C6H5−(CH=CH)5−C6H5 absorbs in the purple region and is probably yellow. 43 (a) Methane (b) Carbon monoxide and hydrogen (b) The first members of the series are colourless as they absorb in the UV region, but the later members (n > 2) are coloured as they absorb in the visible region. (d) Only λmax and not the full spectrum is given so it not possible to give a precise answer. (a) 6CO2 + 6H2O → C6H12O6 + 6O2 (b) Chlorophyll (a) CH2=CH– CH=CH2 is a conjugated system with two double bonds separated by a single bond which extends along the full length of the molecule. In nitrobenzene, conjugation between the benzene ring and the nitro group allows radiation of longer wavelength to be absorbed. This radiation occurs in the visible region of the spectrum and so the compound is coloured. (a) Fossil fuels and biomass are derived from the sun through photosynthesis. (d) Production of biogas Production of ethanol / fermentation 48 Biodiesel is renewable. Biodiesel is carbon neutral. Plants use the same amount of CO2 to make the oil that is released when the fuel is burned. Biodiesel is rapidly biodegradable and completely non-toxic, meaning spillages 83 10:02 Answers represent far less risk than petroleum diesel spillages. dioxide. More carbon dioxide is then released, which amplifies the initial change. Biodiesel has a higher flash point than petroleum diesel, making it safer in the event of a crash. Blends of 20% biodiesel with 80% petroleum diesel can be used in unmodified diesel engines. Biodiesel can be used in its pure form but engines may require certain modifications to avoid maintenance and performance problems. (d) Increased carbon dioxide increases the rate of photosynthesis, producing more phytoplankton, which further reduce levels of carbon dioxide. 51 The removal of CO32−(aq) will cause the equilibrium HCO3−(aq) H+(aq) + CO32−(aq) to shift to the right. The resulting decreased levels of HCO3−(aq) will cause H2CO3(aq) H+(aq) + HCO3−(aq) to shift to the right and so on until CO2(g) CO2(aq) shifts to the right with the absorption of CO2(g). 52 pH = 8.2: [H+] = 10−8.2 Biodiesel can be made from recycled vegetable and animal oils or fats. 49 (a) Distant from localized areas of pollution; data present an accurate measure of global levels of CO2. (b) % increase = (increase/initial value) × 100% pH = 8.1: [H+] = 10−8.1 = [(384 − 316)/316] × 100% = 21.5% % increase = (10−8.1 − 10−8.2)/10−8.2 × 100% (c) Combustion of fossil fuels. (d) The annual variation is due to CO2 uptake by growing plants. The uptake is highest in the northern hemisphere springtime. = (100.1 − 1) × 100% = 26% 53 (e) Photosynthesis: 6CO2 + 6H2O → C6H12O6 + 6O2 (b) methane produced from anaerobic decomposition CO2 dissolves in water: CO2 + H2O H2CO3(aq) (c) smoke particulates: block out sunlight (d) high-energy short-wavelength radiation passes through the atmosphere; (f) Decreased level of photosynthesis: less CO2 taken in by plants. lower energy / longer wavelength radiation from the earth’s surface is absorbed by vibrating bonds in CO2 molecules (g) CO2 absorbs infrared radiation, which leads to increased vibrations and bending and stretching of the bonds. 50 (e) melting of polar ice caps; thermal expansion of oceans will lead to rise in sea levels, which can cause coastal flooding; crop yields reduced; changes in flora and fauna distribution; drought; increased rainfall; desertification (a) Carbon dioxide has polar C= O bonds and the oxygen atoms have lone pairs. It can form hydrogen bonds with water molecules. (b) Relatively strong hydrogen bonds are formed: ΔH is negative. The solubility decreases with increasing temperature, as the equilibrium shifts to the endothermic (reverse) direction as the temperature increases. (c) Increased temperatures due to increased atmospheric carbon dioxide concentrations could result in reduced solubility of carbon (a) respiration, volcanic eruption, complete aerobic decomposition of organic matter, forest fires 54 (a) incoming radiation from the sun is of short wavelength long-wavelength infrared radiation leaves earth’s surface and some is absorbed by gases in the atmosphere results in increased vibration of bonds in molecules which then re-radiate heat back to the earth 84 10:02 (b) Natural: (evaporation from) oceans −0.03 = −8.31 × 298/(2 × 9.65 × 104) × lnQ Artificial: burning (any specified) fossil fuel 0.03 × 2 × 9.65 × 104/(8.31 × 298) = lnQ (c) CO2 is more abundant but CH4 absorbs the radiation more effectively / has a larger greenhouse factor 55 2.34 = lnQ Q = e2.34 = 10.1 10.1 = [Zn2+(aq)]/0.100 (a) coal / diesel (fuel) / wood Zn2+(aq)] = 1.01 mol dm−3 (b) CH4 + O2 → C + 2H2O 56 (b) The equilibrium needs to be shifted to the right. (a) E = E0 − (RT/nF)lnQ Cell reaction: Zn(s) + Ni2+(aq) Ni(s); Q = [Zn2+(aq)]/[Ni2+(aq)] [Zn2+(aq)] could be decreased or [Pb2+(aq)] could be increased Zn2+(aq) + E0 = −0.26 − (−0.76) = +0.50 V In detail: Q = 0.100/0.00100 = 100 0.65 = 0.63 − (8.31 × 298/(2 × 9.65 × 104)) × lnQ E = 0.50 − (8.31 × 298/(2 × 9.65 × 104)) × ln100 V −0.02 = 8.31 × 298/(2 × 9.65 × 104) × lnQ = 0.50 − 0.059 −0.02 × 2 × 9.65 × 104/(8.31 × 298) = lnQ = 0.441 V −1.55873 = lnQ Q = e−1.55873 (b) E = E − (RT/nF)lnQ 0 Cell reaction: Mn(s) + Pb (aq) Pb(s); Q = [Mn2+(aq)]/[Pb2+(aq)] 2+ Q = 0.210 = [Zn2+(aq)]/0.100 Mn (aq) + 2+ Zn2+(aq)] = 0.021 mol dm−3 E0 = −0.13 − (−1.18) = 1.05 V (c) Increase the temperature. This would increase the size of the second negative term in the Nernst equation. Q = 0.100/0.00010 = 1.0 × 103 E = 1.05 − (8.31 × 298/(2 × 9.65 × 104)) × ln(1.0 × 103) V 58 = 1.05 − 0.088633 = 0.96 V (a) The half reaction at the cathode is: Zn2+(aq) Zn(s) + 2e− The equilibrium will be to the right for high [Zn2+(aq)] (c) E = E0 − (RT/nF)lnQ Cell reaction: Zn(s) + Fe2+(aq) Fe(s); Q = [Zn2+(aq)]/[Fe2+(aq)] Solution B should be used at the cathode. Zn2+(aq) + (b) E = E0 − (RT/nF)lnQ E0 = −0.45 − (−0.76) = +0.31 V E = −(RT/nF)(ln2) Q = 1.50/0.100 = 15.0 = 0.0089 V E = 0.31 − (8.31 × 298/(2 × 9.65 × 104)) × ln15.0 V 59 (a) CH3OH(l) + (3/2)O2(g) → CO2(g) + 2H2O(l) −167 = 0.31 − 0.035 = 0.28 V 57 0 −394.4 2(−237.1) ∆Gf0 / kJ mol−1 ∆Greaction = ∑∆Gf0 (products) − ∑∆Gf0 (reactants) (a) E = E0 − (RT/nF)lnQ 2+ Cell reaction: Zn(s) + Pb (aq) Pb(s); Q = [Zn2+(aq)]/[Pb2+(aq)] Zn (aq) + 2+ E0 = −0.13 − (−0.76) = +0.63 V Q = [Zn2+(aq)]/0.100 0.60 = 0.63 − (8.31 × 298/(2 × 9.65 × 104)) × lnQ = (−394.4 + 2(−237.1) − (−167) kJ mol−1 = −701.6 kJ mol−1 −∆Gsys −701.6 × (b) Efficiency = × 100% = −∆Hsys −726 100% = 97% 85 10:02 Answers 60 (a) Oxidation number increases from 0 to +2. Pb(s) is oxidized. 64 (b) Pb(s) + SO4 2−(aq) → PbSO4(s) + 2e− (c) PbSO4 is insoluble: the Pb2+ ions do not disperse into solution. (b) Boron has only three outer electrons so it produces a positive hole in the lattice to give a p-type semiconductor. (d) Advantage: delivers large amounts of energy over short periods. Disadvantage: heavy mass; lead and sulfuric acid could cause pollution. 61 (a) Bacteria oxidize the substrate: they live at the negative electrode or cathode. 65 Arsenic, as it has five outer electrons and a similar atomic radius. 66 Si is doped with As to produce an n-type semiconductor and with Ga to produce a p-type semiconductor. Light stimulates electron flow from the n-type to the p-type semiconductor through an external circuit. (b) Sucrose is oxidized by the bacteria. 12 Cs are oxidized from 0 to + 4 so 48 electrons are released 24 Os are needed so 13 H2O are needed Disadvantages no pollution low power output Check that the charges are balanced: zero on both sides no moving parts so no maintenance C12H22O11(aq) + 13H2O(l) → 12CO2(g) + 48H+(aq) + 48e− no need for refuelling as sunlight is unlimited needs a large surface area battery / storage facilities needed in absence of light produces less noise conserves petroleum for other uses 48 Hs are needed on both sides: 48H+(aq) + 12O2(g) + 48e− → 24H2O(l) 68 (d) C12H22O11(aq) + 13H2O(l) → 12CO2(g) + 48H+(aq) + 48e− C12H22O11(aq) + 12O2(g) → 12CO2(g) + 11H2O(l) (a) H2 + 2OH− → 2H2O + 2e− O2 + 2H2O + 4e− → 4OH− (b) Less waste heat produced and more chemical energy converted to useful energy. They use a renewable energy source which is more efficient. Metals conduct electricity well, insulators do not; semiconductors have intermediate conductivities. The conductivity of metals decreases with temperature; the conductivity of semiconductors increases with temperature. Conductors have low ionization energies; insulators have high ionization energies. high capital cost easily damaged Conductivity increases as the gallium has one less electron than silicon. Electron holes are introduced, which make it a p-type conductor so electrons can move into these holes. Arsenic has one more electron than silicon. An extra electron is introduced so it is a n-type conductor. The extra electrons are free to move. 48H+(aq) + 12O2(g) + 48e− → 24H2O(l) 63 Advantages 67 48 Hs are needed on both sides (c) 48 electrons will reduce 12 O2 molecules 62 (a) Ionization energies decrease down a group so the electrons are easier to remove in germanium compared to diamond, resulting in more free electrons. 69 (a) Process Photovoltaic DSSC Light absorption Silicon atom is ionized to create hole and electron (Si+ and e−) pair Electron excited in conjugated organic molecule Charge separation Electron and hole move in opposite directions due to electric field in p–n semiconductor Election is accepted by semiconductor. Positive ion loses its charge as the ion is reduced by electrolyte 86 10:02 (b) If the electron directly recombines with an ion the light energy is converted to useless heat energy. This corresponds to 22 = Z2, where Z is the atomic number of helium. This is a general result: the energy level is proportional to the square of the atomic number. Challenge yourself 1 680 kJ of heat is given out. Energy efficiency is 32%. 2 The decrease in entropy in your brain is accompanied by increases elsewhere in the universe. Coal and oil are fossilized decayed plants or animals that contain amino acids. The amino acid cysteine contains sulfur. Coal and oil with a higher percentage of sulfur are considered ‘dirty’ because of the sulfur dioxide pollution that they produce on combustion. Sulfur dioxide results in acid rain. 3 (a) Plotting a graph we see that the data follow an approximate straight-line relationship: 4 100 7 The 1% difference in molecular mass between the molecules formed from the two isotopes is due only to the difference in masses of the uranium isotopes. 8 As many stages are involved (up to 4000), a large area of ground is needed. 9 Other covalent fluorides with the non-metal in a high oxidation state. An example is SF6. 10 C40H56 The position of the double bond in the hexagon on the right is different in the two isomers. 11 (a) 2C16H23O11 + 19H2O + O2 → xH2 + yCO + zCO2 −1/16 +1 +2 +1 −2 0 0 −2 +4 C H −2 O (b) C is oxidized O and H are reduced octane number 80 (c) Balancing the H atoms: 60 2C16H23O11 + 19H2O + O2 → 42H2 + yCO + zCO2 octane number = −31.23n + 216.64 40 x = 42 20 (d) Total change of oxidation number of H = −84 0 0 −20 1 2 3 4 5 6 7 8 This suggests an octane number of −33. In practise the graph is a curve rather than a straight line and octane has an octane number of −19. (b) Assuming the same straight-line trend, for an octane number of 100, n = 3.73. So propane has an octane number above 100. 5 The isomers of octane have the essentially the same number of C–C and C–H bonds. This suggests that they are not a key factor in a molecule’s octane number. 6 Total change of oxidation number of O = −2 number of carbon atoms The ratio between the values is approximately 4 Total increase in oxidation number of C =+86 Balancing the change in oxidation numbers: 2y + 4z − 2 = +86 Balancing the C atoms: y + z = 32 Solving the equations: 2z = 88 − 64 = 22 z = 11 and y = 21 2C16H23O11 + 19H2O + O2 → 42H2 + 21CO + 11CO2 (e) 42H2 + 21CO → 21CH3OH 2 molecules of wood produce 21 molecules of methanol 1 molecule produces 10.5 molecules 87 10:02 Answers 12 The solubility of carbon dioxide increases at lower temperatures and higher pressures. high-level waste is vitrified / made into glass / buried underground / in granite / in deep mines / under water / in steel containers / in cooling ponds / OWTTE Increased levels of CO2(g) will lead to the formation of the more soluble Ca(HCO3)2(aq). low-level waste is stored under water / in steel containers / in cooling ponds /filtered / discharged directly into sea / OWTTE CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq) 13 − ∆Gsys −∆Hsys + T∆Ssys × Efficiency = × 100% = − ∆Hsys −∆Hsys 100 = 1 + Accept cooling ponds/steel containers/under water/concrete containers only once. [3] T∆Ssys × 100 −∆Hsys The entropy decrease is smaller for the reaction which produces one mole of gaseous water, which leads to a larger efficiency. 3 lower temperature / lower energy consumption / more control of product [2] Practice questions 4 1 (a) long wavelength / infrared / IR radiation from Earth’s surface Do not accept ‘trapped’ or ‘blocked’. 5 Do not award mark for ‘IR from sun’. 6.25% remains 4 half-lives = 6400 years causes (increased) vibration in bonds 6 re-radiates heat back to the Earth Accept ‘re-transmits’ 235 92 U undergoes a fission reaction due to neutron capture thermal expansion of oceans / rise in sea levels / coastal flooding reaction produces neutrons so chain reaction occurs stated effect on agriculture (e.g. crop yields changed) mass of products less than reactants corresponding energy released according to [4] E = mc2 changes in flora / plant / fauna / animal / insect distribution / biodiversity (c) the mass needed that allows fission to be sustained [1] Accept specific example. stated effect on climate (e.g. drought / increased rainfall / desertification) 7 (a) wide availability produce energy at appropriate rate Do not accept ‘climate change’ alone. ease of transportation Do not allow ‘increased temperature / global warming’ (given in question). current technology is based on fossil fuels Award [1] each for any three. [3 max] high-level waste has longer half-life / low-level waste has shorter half-life [2] (a) energy transferred to surroundings / from system which is no longer available for use / cannot be used again [2] (b) [2 max] (b) melting of polar ice caps / glaciers melting 2 light nuclei for fusion and heavy nuclei for fission more massive nucleus produced in fusion as nuclei joined together and two lighter nuclei produced in fission as nuclei split apart [2] (some of this radiation) is absorbed (by gas) Do not accept ‘reflects/bounces’ Catalytic cracking: used to produce moderate length alkanes for fuels high energy density / specific energy [2 max] (b) energy produced per unit mass / kg / stored per unit mass / kg [1] (c) uranium / hydrogen [1] 88 10:02 (d) 6CO2 + 6H2O → C6H12O6 + 6O2 (d) Efficiency = 600 × 106/energyinput 8 C6H12O6 → 2C2H5OH + 2CO2 (ii) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) (a) coal about 90% and petroleum about 84% and natural gas 75% [1] ∆Gf0 /kJ mol−1 −175 (c) hydrogen has a very high specific energy / energy density Formula ∆Hc / kJ mol−1 C 3H 8 44.11 −2219 2219/44.11 = 50.31 C4H10 58.14 −2878 2878/58.14 = 49.50 C 3H 8 11 [2] (a) the octane number indicates the resistance of a motor fuel to knock / premature ignition octane numbers are based on a scale on which 2,2,4-trimethylpentane (isooctane) is 100 (minimal knock) and heptane is 0 (maximum knock) [2] Specific energy / kJ g−1 (b) octane number of straight-chain alkanes decrease with an increase in chain length straight chain alkanes < aromatics alcohols have very high octane numbers: [2] hexane (24.8) < pentane (61.7) < benzene (105.8) < ethanol (108) [4] The octane numbers are given for reference and are not expected to be recalled. Energy / kJ cm−3 (2219/44.11) × 1.00 × 105 × 44.11/(8.13 × 273) 12 5 = 2219 × 1.00 × 10 / (8.13 × 273) (a) 23 5 92 U → 23910Th + 42He atomic number = 90 and element = Th mass number = 231 =100 000 C4H10 [3] they oxidize the biomass (b) ρ = PM/RT Formula 3(−237.1) (f) bacteria live at the anode it is clean burning producing only H2O when it is burned [2] M /g mol−1 2(−394.4) thermodynamic efficiency = −1325.1/−1367 = 97% produce less pollution / smaller carbon foot print [2 max] (a) 0 ΔG = 2(−394.4) + 3(−237.1) − (−175) = −1325.1 kJ mol−1 liquid or gaseous state make them more convenient to use / easier to transport [2] (b) To answer this question you need to know that the half-life of 235 U is 704 million years. 92 (2878/58.14) × 1.00 × 105 × 58.14/(8.13 × 273) U will After one half-life the amount of 235 92 have decayed to 50% of its original amount. = (2878 × 1.00 × 105)/ (8.13 × 273) 10 [2] Mass of fuel = 2000 × 106/60 × 103 = 33.3 × [3] 103 g s−1 (b) they have higher specific energy 9 [1] (e) (i) fermentation Energyinput = 600 × 106/0.30 J s−1 = 2000 × 106 J s−1 = 130 000 [3] (c) propane as it the smallest molar mass [1] (a) photosynthesis [1] 2 × 704 million years = 1408 million years [2] It will take 1408 million years for 75% of the 235 U to decay. [2] 92 (b) chlorophyll conjugated structure U will After two half-lives the amount of 235 92 have decayed to 25% of its original amount, i.e. 75% will have decayed. (c) 6CO2 + 24H+ + 24e− → C6H12O6 + 6H2O [1] 89 10:02 Answers 13 (a) effects due to: increased carbon dioxide levels, global warming unburned hydrocarbons and carbon monoxide non-metal / sulfur / nitrogen oxides produce acid rain 13 (b) particulates cause global dimming Strategy [3 max] Action Increased energy efficiency when not in use of transport such as hybrid cars or public transport Award [1 max] for action Reduced dependence on carbon-based energy hydropower, wave, tidal or nuclear power of biomass depends on the processes by which it is converted to energy Award [1 max] for action Capture and storage of carbon from fossil fuels or within plants and soil supporting the plants. Alternatively, carbon dioxide can be captured either before or after fossil fuel is burned and then be stored (sequestered) within the earth [6] produce more CO2 and have higher carbon footprint [3] (a) octane C8H18 16.05 55.5 methanol CH3OH −726 32.05 22.7 [2] (b) CH4: Ox (C) = −4; CH3OH: Ox (C) = −2 [1] (c) the more oxidized the C the lower the specific heat [1] (b) fractional distillation mixture of hydrocarbons is heated causing them to vaporize different compounds have different boiling points; the lowest boiling point compounds condense higher up the column as the size of the molecule increases the attractive van der Waals forces increase [4] (c) larger molecules can be cracked into smaller molecules when heated with a catalyst and broken into smaller molecules equation for cracking: C14H30 → C7H14 + C7H16 (d) methanol C6H12O6 [2] −2803 180.18 Specific energy / kJ g−1 −891 Relative molecular mass methane CH4 [2] Formula any other compound with C5–C10 Fuel 14 Formula Fuel have less efficient vehicles Specific energy / kJ g−1 (a) Standard enthalpy of combustion ΔH / kJmol−1 15 use more gasoline Standard enthalpy of combustion ΔH / kJmol−1 (c) countries with cheaper gasoline on average: Relative molecular mass Award [1 max] for action 15.6 oxidation number of C = 0 the result supports the hypothesis [3] 90 10:02 16 (a) they are too viscous devices / fuel cells require constant supply of reactants and batteries have stored chemical energy / provide power until stored as there are strong intermolecular forces between the triglyceride molecules [2] chemicals are used up / batteries can be recharged and fuel cells do not need recharging (have a continuous supply of fuel) / fuel cells are more expensive than rechargeable batteries / the reactions in a rechargeable battery are reversible and in a fuel cell are not [2] (b) transesterification reaction with methanol or ethanol with strong acid or base catalyst produces a methyl / ethyl ester 17 [2] (a) Cathode: 2H2O + O2 + 4e− → 4OH– Anode: 2H2 + 4OH– → 4H2O + 4e− If both equations given but at wrong electrodes award [1]. 19 so electrons can flow through the material more easily [2] (p-type) has small amount of / is doped with a group 3 element / B / In / Ga (b) Cathode: nickel hydroxide / Ni(OH)2 which produces electron holes / positive holes Anode: cadmium hydroxide / Cd(OH)2 Cell equation: Cd + 2H2O + 2NiO(OH) → Cd(OH)2 + 2Ni(OH)2 18 sun / photons cause release of electrons electrons move from n-type to p-type material [3] (a) contains no lithium / metal / uses lithium salt in an organic solvent (as electrolyte) involves movement of lithium ions (between electrodes) [2] 20 (a) 56 [5 max] Fe [1] (b) the energy released when nuclides form from separate nuclei into separate nucleons (c) 23 5 92 [1] U + 01n → 145 16Ba + 39 26Kr +2 10n two neutrons produced may cause two further fissions (b) Anode: LiC6 → Li+ + 6C + e− / Li+ ions dissociate from anode (and migrate to cathode) Cathode: Li+ + e− + MnO2 → LiMnO2 / Li+ + e− + CoO2 → LiCoO2 / Li+ + e− + FePO4 → LiFePO4 / Li+ + e− + NiO2 → LiNiO2 / Li+ ions are inserted into metal oxide / phosphate (structure) Award [1] if electrodes are reversed. Si has a lower ionization energy (than P or S) [2] (c) Similarity: both convert chemical energy directly into electrical energy / both use spontaneous redox reactions (to produce energy) / both are electrochemical cells / voltaic cells / galvanic cells Difference: fuel cells are energy conversion devices and rechargeable batteries are energy storage producing four neutrons which may produce four further fissions [3] (d) one reaction: ∆E = 3.1 × l0−28 × [3 × 108]2/2.8 × l0−11 (J) = 2.8 × 10−11 × 6.02 × 1023 = 1.69 × 1013 (J mol−1) = 1.69 × 1010 (kJ mol−1) [3] (e) mass of 239 52U in 1 g = 3.00 × 10−2 g n = 3.00 × 10−2/235 = 0.000128 mol specific energy = 1.69 × 1010 × 0.000128 = 2.16 × 109 kJ g−1 [3] (f) enrichment process converted to gaseous UF6 position in centrifuge / diffusion rate depends on the masses [3] 91 10:02 Answers 21 (a) nuclear fusion [1] excited electron injected into TiO2 semiconductor ionized dye reduced by electrode at anode (b) Δm = 0.00535 amu = 1.66 × 10 10–30 kg −27 electrons pass through external circuit and reduce electrolyte at cathode × 0.00535 kg = 8.88 × ΔE = Δmc2 = 8.88 × 10−30 × (3.00 × 108)2 J = 7.99 × 10−13 J = 4.18 × 1011 J mol−1 = 4.18 × 108 kJ mol−1 [4] (c) 32He this is the source of the energy produced [2] (d) confining the hot plasma 22 [1] Photovoltaic cell: an impurity such as arsenic and gallium added to dope silicon layers to produce p–n junction electron emitted as photon ionizes Si atom when p-type and n-type silicon are put together, an electric field is produced which prevents recombination electrons need to pass through external circuit Dye sensitised cell: organic molecules with conjugated absorbs electrons 23 [8] (a) 42% efficient so 2.00 × 104/0.42 kJ = 4.76 × 104 kJ required every second 4.76 × 107/2.8 × 10−11 = 1.70 × 1018 fissions each second = 6.12 × 1021 fissions each hour moles needed = 6.12 × 1021/6.02 × 1023 = 1.02 × 10−2 so mass needed = 235 × 1.02 × 10−2 = 2.39 g [3] (b) U-238 is present U-238 captures a neutron (to produce plutonium) plutonium can be used as fuel for (fast breeder) reactors [3] (c) 0.693/(2.40 × 104 yr) = 2.89 × 10−5 yr−1 [1] (d) −ln(0.01)/2.89 × 10−5 = 1.59 × 105 yr [2] 10:02