IB Chemistry R1.3: Energy from fuels worked solutions

B. Balance oxygen in pairs and threes: \ce{4Al + 3O2 -> 2Al2O3}. There are four aluminium atoms and six oxygen atoms on each side. The other sequences fail this count.

D. The products are \ce{2CO + 3H2O} per ethanol, needing five O atoms. The alcohol already contains one; four more need two oxygen molecules. Thus one mole ethanol needs 2.00\,\mathrm{mol} oxygen.

D. Compare equal masses by dividing each molar release by its own molar mass. Ethane gives 1561/30.08=51.9\,\mathrm{kJ\,g^{-1}} and methanol gives 726/32.05=22.7\,\mathrm{kJ\,g^{-1}}. The unrounded ethane-to-methanol ratio is 2.29. Equal masses contain different amounts of molecules.

B. The same field previously produced food. Diverting it to fuel reduces the available food-producing land unless production is replaced elsewhere. Replanting and carbon uptake are possible benefits, but do not remove this competing use.

D. Find the smallest shared electron total: 2(6)=3(4)=12. Two methanol molecules supply what three oxygen molecules accept, giving the amount ratio 2:3.

D. The useful output is 27.4/1000=0.0274\,\mathrm{MJ} per gram of fuel. Divide the corresponding carbon dioxide mass by that output: 2.68/0.0274=97.8\,\mathrm{g\,MJ^{-1}}. This is already a useful-heat basis, so no further efficiency factor is needed.

B. Carbon in CO can still be oxidised to carbon dioxide, so some fuel carbon has not reached its fully oxidised product. Water, oxygen supply and heat transfer can occur during complete combustion too.

One litre contains 0.750\,\mathrm{kg} [1]. Therefore e_V=\rho e_m=0.750(28.0)=21.0\,\mathrm{MJ\,L^{-1}} [1]. This value applies at the stated storage density.

(a) Sustainable annual harvests can replace the residue feedstock on human timescales [1]. Coal formation requires geological times, so ongoing consumption is not replaced comparably [1]. (b) Existing residues can be used without dedicating further food-producing land to fuel [1]. (c) Collection, transport or processing can emit carbon dioxide; alternatively, residue removal can export soil nutrients [1]. One specific cost with a causal link is sufficient.

(a) \ce{2Mg(s) + O2(g) -> 2MgO(s)} [1]. (b) \ce{C(s) + O2(g) -> CO2(g)} [1]. (c) Three carbons and eight hydrogens require \ce{3CO2(g) + 4H2O(l)} [1]. These need ten O atoms; the alcohol supplies one. Hence \ce{C3H7OH(l) + 9/2O2(g) -> 3CO2(g) + 4H2O(l)} [1]. (d) One amine needs one carbon dioxide, five halves water and half a nitrogen molecule [1]. Scaling to integers gives \ce{4CH3NH2 + 9O2 -> 4CO2 + 10H2O + 2N2} [1]. This is the stipulated nitrogen-product model, not a universal claim about amine flames.

(a) During operation both devices convert chemical energy into electricity through redox reactions [1]. (b) Supply electrical energy to reverse the rechargeable battery discharge [1]. Replenish the fuel cell reactants to continue its supplied reaction [1]. (c) A dependable hydrogen delivery and storage system is required; exhaustion of the external supply stops the cell [1].

(a) Oxygen requires two waters. Add four hydrogen ions to supply their H atoms, then four electrons to balance charge: \ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}. Award one point for atom balance and one for charge balance [2]. (b) Extra hydrogen supplies no oxygen atoms for the stipulated water formation [1]. It also cannot replace the cathode oxygen-reduction step that accepts the circuit electrons [1]. (c) The useful output is electrical energy [1].

(a) Divide by molar mass: 891/16.04=55.5486\,\mathrm{kJ\,g^{-1}} [1]. Since one kJ per gram equals one MJ per kilogram, the result is 55.5\,\mathrm{MJ\,kg^{-1}} [1]. (b) \ce{CH4 + 2O2 -> CO2 + 2H2O} gives two oxygen molecules per methane [1]. Thus n(\ce{O2})=2(4.80/16.04)=0.599\,\mathrm{mol} [1].

(a) e_{V,L}=32.0(0.800)=25.6\,\mathrm{MJ\,L^{-1}} [1]. e_{V,G}=48.0(0.240)=11.52=11.5\,\mathrm{MJ\,L^{-1}} [1]. (b) Choose L for a volume limit because each litre stores more chemical energy [1]. G needs less fuel mass because its energy per kilogram is greater [1]. These rankings omit container mass and compare the given storage conditions.

(a) Multiply the anode by two and the cathode by three so both transfer twelve electrons [1]. Cancel electrons, hydrogen ions and two waters, leaving \ce{2CH3OH + 3O2 -> 2CO2 + 4H2O} [1]. (b) n(\ce{O2})=(3/2)(0.0800)=0.120\,\mathrm{mol} [1]. Electrons travel from the anode through the external circuit to the cathode, rather than through the electrolyte [1].

(a) The chemical-input account is 42.0+8.00-36.0=14.0\,\mathrm{g\,MJ^{-1}} [1]. Divide by efficiency: 14.0/0.400=35.0\,\mathrm{g\,MJ^{-1}} useful output [1]. (b) Regrowth offsets only 36.0 of the 50.0\,\mathrm{g} released [1]. There is a positive net addition even before any unlisted land or transport effects, so renewability alone does not imply neutrality [1].

(a) Four carbons form four carbon monoxides and ten hydrogens form five waters [1]. Their nine O atoms require \ce{9/2O2}: \ce{C4H10 + 9/2O2 -> 4CO + 5H2O} [1]. (b) A quarter of four carbon atoms gives one soot carbon, leaving three CO [1]. Eight product O atoms require four oxygen molecules: \ce{C4H10 + 4O2 -> C + 3CO + 5H2O} [1]. (c) Both CO and soot can still undergo oxidation to carbon dioxide [1]. That further step releases energy, so either partial route releases less than complete oxidation when the initial fuel amount and final physical states are matched [1].

(a) Equate accumulated savings with the initial release: 1.85t=27.0 [1]. Thus t=27.0/1.85=14.6\,\mathrm{years} [1]. (b) After eight years, 1.85(8.00)=14.8\,\mathrm{tonnes} have been avoided [1]. The remaining excess is 27.0-14.8=12.2\,\mathrm{tonnes}, so cumulative emissions have not yet fallen [1]. (c) The annual crop uptake/release cycle concerns recently fixed carbon [1]; clearing releases a separate woodland stock, which stays in the cumulative boundary [1]. This model assumes the supplied annual saving remains constant and the listed boundary is complete.

(a) The allowance is 160\,\mathrm{g} carbon dioxide. With M=44.01\,\mathrm{g\,mol^{-1}}, n=160/44.01=3.64\,\mathrm{mol}; this permits the same methanol amount [1]. The combustion-energy basis gives (160/44.01)(726)/1000=2.64\,\mathrm{MJ} [1]. Multiply by the capture fraction using unrounded values: E_{\rm electrical}=(160/44.01)(726/1000)(0.430)=1.13\,\mathrm{MJ} [1]. (b) Six electrons quantify the carbon oxidation-state change and charge transfer [1]. Each methanol has one carbon atom, so one carbon dioxide forms, not six [1]. (c) Sustainable replacement of a biological feedstock with low processing emissions could reduce net atmospheric addition [1]. The direct allowance remains unchanged. The specified efficiency uses combustion release as an energy basis; this is not a calculation of maximum thermodynamic electrical work.

(a) Component masses are 0.350(0.800)=0.280\,\mathrm{kg} and 0.650(0.600)=0.390\,\mathrm{kg} [1]. Their releases are 8.40 and 15.6\,\mathrm{MJ}, totalling 24.0\,\mathrm{MJ} [1]. Divide by 0.670\,\mathrm{kg} to obtain 35.8\,\mathrm{MJ\,kg^{-1}}; the additive volume is 1.00\,\mathrm L, giving 24.0\,\mathrm{MJ\,L^{-1}} [1]. (b) The required chemical input is 18.0/0.450=40.0\,\mathrm{MJ} [1]. Required volume is 40.0/24.0=1.67\,\mathrm L [1]. (c) The densities differ, so a volume fraction is not a mass fraction. Energy per kilogram must weight by component mass [1].

(a) q=mc\Delta T [1], so q=400(4.18)(24.0)=40128\,\mathrm J=40.1\,\mathrm{kJ} [1]. (b) n=2.30/46.07=0.0499240\,\mathrm{mol} [1]; complete-combustion release is (2.30/46.07)(1367)=68.2461\,\mathrm{kJ}=68.2\,\mathrm{kJ} [1]. The fraction is 40.128/68.2461=0.588 [1]. (c) Heat absorbed by the burner, container or surrounding air does not enter the water; alternatively, incomplete oxidation leaves chemical energy in CO or soot [1]. The available-energy calculation assumes complete combustion with the tabulated product states.

(a) The ethanol formed before separation is 0.240/0.800=0.300\,\mathrm{mol} [1]. From \ce{C6H12O6 -> 2C2H5OH + 2CO2}, glucose consumed is 0.150\,\mathrm{mol} and fermentation carbon dioxide is 0.300\,\mathrm{mol} [1]. (b) Both routes reach the same carbon dioxide/water products, so \Delta H_{\rm ferm}+2\Delta_{\rm c}H_{\rm ethanol}=\Delta_{\rm c}H_{\rm glucose} [1]. Thus \Delta H_{\rm ferm}=-2803-2(-1367)=-69\,\mathrm{kJ\,mol^{-1}} glucose [1]. (c) The fermentation releases 0.150(69)=10.35\,\mathrm{kJ}, reported as 10.4\,\mathrm{kJ} [1]. Recovered ethanol releases 0.240(1367)=328.08\,\mathrm{kJ}, reported as 328\,\mathrm{kJ} [1]. The unrecovered ethanol still carries chemical energy; recovery yield is distinct from fermentation conversion.

(a) p=100000\,\mathrm{Pa} and V=0.00248\,\mathrm{m^3} [1]. n=pV/(RT)=100000(0.00248)/(8.31(298))=0.100146\,\mathrm{mol}=0.100\,\mathrm{mol} [1]. (b) \ce{H2 -> 2H+ + 2e-} [1]. Twice the unrounded hydrogen amount gives 0.200\,\mathrm{mol} electrons [1]. (c) \ce{2H2 + O2 -> 2H2O} gives 0.100\,\mathrm{mol} water [1]. Mobile ions carry the internal current; hydrogen ions travel towards the cathode in this acidic model [1]. Electrons pass through the external circuit.