AJC H2 CHEM P2 Solutions

©2014AndersonJC/CHEM 1 H2 Chemistry 9647 2014 JC2 Prelim P2 Suggested Solutions 1 (a) (i) The boiling point increases with concentration because the temperature has to be raised in order to increase the lowered vapour pressure to that of the ambient value. [1] (ii)  [1m] straight–line or curve showing a gradual increase.  [1m] line to begin on the y–axis with the value 100C (or 373 K) labelled with units. [there is no ecf here from (i) to (ii)] [2] (b)  [1m] for a diagram which shows some form of heating (not a water bath, but allow oil–bath or heating mantle) of an apparatus (50 cm 3 round–bottomed flask / boiling tube) containing the KCl solution.  [1m] for showing the position of the thermometer bulb is half–immersed in the solution – the full line in the diagram indicates the water level (to allow accurate determination of the temperature of the equilibrium). [2] 100 C boiling point 0 concentration of potassium chloride (not required) saturated

©2014AndersonJC/CHEM 2 (c) Experimental Procedure 1. Weigh accurately about 2 g of solid KC l in a 50 cm 3 round–bottomed flask and record your reading. 2. Using a 50 cm3 burette, add 20 cm 3 of deionised water into the round–bottomed flask. 3. Stopper and shake the flask to dissolve the solid. 4. Add anti–bumping granules (boiling chips) to the solution. 5. Set up the apparatus as shown in (b). 6. Using a heating mantle (or oil–bath), gently heat the flask containing KC l solution, until the temperature remains constant. 7. Record this temperature (i.e. the boiling point). 8. Repeat steps 1 to 7, using 3 g, 4 g, 5 g and 6 g KC l, respectively.  [1m] for general outline of procedure with coherent sequence of steps leading to determination of boiling point (take reading when temperature remains constant).  [2m] for a method, which gives details / a list of the masses / volumes of water to be used along with masses of solid KC l to produce a solution, provided that the total mass / volume of water does not exceed 100 g / 100 cm 3 AND 0.357(same) waterof volume KCl solid of mass max.  or 0.357 waterof volume smallest (same) KCl solid of mass   [1m] if at least five different concentrations of KCl are prepared. (A method based on adding varying volumes of the water to one mass of solute is allowed, e.g. 10 cm3, 15 cm3, 20 cm3, 25 cm3, 30 cm3)  [1m] for safety & reliability considerations: steps 3 and 4 (or in diagram) Calculation of Molality Taking density of water to be 1 g cm–3 Molality = 1.00 x used waterdeionised of volume 1000x74.6 used KCl solid of mass  [1m] for correct general expression to calculate molality. [6] (d) The experiment is limited by the solution becoming saturated. [1]

©2014AndersonJC/CHEM 3 2 (a) Breath alcohol concentration = 2100 1 x 80 mg / 100 cm3 = 0.03809 mg / 100 cm3 breath alcohol concentration (in mol cm–3) = 310 0.03809 x 46.0 1 x 100 1 = 8.28 x 10–9 mol cm–3 [1] [1] (b) (i) Eo(Cr2O7 2–/Cr3+) = +1.33 V Eo cell = +1.33 – 0.058 = +1.27 V 3C2H5OH + 2Cr2O7 2– + 16H+ 3CH3CO2H + 4Cr3+ + 11H2O [1] [1] (ii) Q = It = 0.1 x 5 = 0.5 C Q = nF  n(e–) = 0.5 / 96500 = 5.181 x 10–6 mol n(ethanol) = 5.181 x 10 –6 x 1/4 = 1.30 x 10–6 mol [1] [1] (iii) n(ethanol) per cm3 of exhaled air (breath) = 1.30 x 10–6 / 60.0 = 2.166 x 10–8 mol cm–3 n(ethanol) per cm 3 of blood = 2.166 x 10–8 x 2100 = 4.548 x 10–5 mol cm–3 mass of ethanol per cm3 of blood = 4.548 x 10–5 x 46.0 = 2.09 x 10-3 g cm–3 Since 2.17 x 10 –8 mol cm–3 >> 8.28 x 10–9 mol cm–3 (legal limit for breath alcohol concentration; answer to (a)), the driver is drink–driving. or Legal limit to drive is 80 mg of ethanol per 100 cm 3 of blood  80 x 10–3 x (1 / 100) = 8.00 x 10–4 g cm–3 (mass of ethanol per cm3 of blood) Since 2.09 x 10-3 >> 8.00 x 10–4 g cm–3, the driver is drink–driving. [1] [1] (iv) By measuring the change in intensity of the colour change of orange Cr 2O7 2– to green Cr3+. [1]

©2014AndersonJC/CHEM 4 3 (a) The energy absorbed or evolved when one mo le of a compound is formed from its constituent elements at standard states, under standard conditions of 298 K and 1 atm. [1] (b) (i) ∆Hr o = ∑∆Hf o(products) – ∑∆Hf o(reactants) = 2(–393.5) + 4(–241.8) – [83.3 + 2(9.10)] = –1855.7 kJ mol–1 [1m] for correct substitution [1m] for final answer given to 1 d.p. [1] [1] (ii) Entropy change is highly positive because there is a large increase in the number of moles of gaseous particles (from 0 to 9). Hence, there are more ways in which particles and energies of the particles can be distributed. [1] (iii) ∆G = ∆H – T∆S Decreasing the temperature decreases the magnitude of T ∆S (–T∆S becomes less negative). ∆G becomes less negative and hence the reaction becomes less spontaneous. [1] (c) (i) 2N2H4 + N2O4 4H2O + 3N2 [1] (ii) NN H H H H 2 + NN OO OO 4 O HH + 3N N ∆Hr o = ∑BEreactants – ∑BEproducts = [8(BEN–H) + 2(BEN–N) + 2(BEN=O) + 2(BEN–O) + BEN–N] – [8(BEO–H) + 3(BEN≡N)] = [8(390) + 2(160) + 2(607) + 2(201) + 160] – [8(460) + 3(994)] = –1446 kJ mol –1 [1] (iii) The bond energies in the Data Booklet are average values. or In their standard states at 298 K, N 2H4 and N 2O4 are liquids whereas the ∆Hr o calculated in (c)(ii) is based on reactions in the gas phase. [1] (d) (i) Mr UDMH = 60.0; Mr N2O4 = 92.0 Since UDMH : N 2O4 = 1 : 2, Mass of UDMH = 244 x 2(92.0) 60 60  = 60.0 kg [1] (ii) Since UDMH : product gases = 1 : 9, No. of moles of UDMH = 60 x 10 3 / 60.0 = 1000 mol No. of moles of product gases = 1000 x 9 = 9000 mol [1] (iii) pV = nRT V = nRT / p = 9000 x 8.31 x (–10 + 273) / 600 = 3.28 x 10 4 m3 [1]

©2014AndersonJC/CHEM 5 (e) (i) Gradient calculated from graph = 0.00152) - 0.0010 1.1 - 8.5 = –14230.7 Gradient = –Ea / R –14230.7 = –Ea / 8.31 Ea = 14230.7 x 8.31 = 118257 = +118 kJ mol–1 [1] [1] (ii) Yes, the mechanism is consistent with the given rate equation. Two molecules / moles of NO 2 are involved in slow step / rate–determining step. The sum of the 2 steps gives the overall balanced equation of the reaction. [1] [1] (f) (i) Rate = k (pNO)a (pH2)b Comparing Expt. 1 & 2, when pNO decreases by 2 times (from 120 to 60 torr), the initial rate decreases by 4 times (from 8.66 x 10–2 to 2.17 x 10–2 torr s–1).  order of reaction w.r.t. pNO = 2; i.e. a = 2 Comparing Expt. 2 & 3, when pH2 increases by 3 times (from 60 to 180 torr), the initial rate also increases by 3 times (from 2.17 x 10–2 to 6.62 x 10–2 torr s–1).  order of reaction w.r.t. pH2 = 1; i.e. b = 1 (Mathematical method is accepted here) rate = k (pNO)2 (pH2) Using results from Expt. 1, k = (8.66 x 10–2) / (1202 x 60) = 1.00 x 10–7 torr–2 s–1 [1m] for final answer [1m] for units [1] [1] [1] [2] (ii) rate = ∆pN2O / ∆t = –½ (∆pNO / ∆t) Initial rate of formation of N 2O = 1.00 x 10–7 x 2002 x 100 = 0.400 torr s–1 = ∆pN2O / ∆t Initial rate of disappearance of NO = ∆pNO / ∆t = 2 x 0.400 = 0.800 torr s–1 [1] [1] (iii) rate = k (pNO)2 (pH2) As pNO >> pH2, rate = k’ pH2 where k’ = k (pNO)2 k’ = 1.00 x 10–7 x (800)2 = 0.064 s–1 t1/2 = ln 2 / k’ = In 2 / 0.064 = 10.8 s [1] [1]

©2014AndersonJC/CHEM 6 4 (a) NH3 +