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CIE Nov 2013 v3 Paper 4 Q3
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![Examiners For Use (a) Write equations, with state symbols, to de ne the following. (i) the CBr bond energy in CH3Br(ii) the Al Cl bond energy in Al Cl 3[3] (b) (i) Describe and explain the trend in bond energies of the bonds in Cl 2, Br2 and I2.(ii) Fluorine, F2, does not follow this trend. Suggest a possible reason why.[3] (c) (i) Use data from the Data Booklet to calculate the enthalpy change of the following reaction. H2(g) + X2(g)2HX(g) when X = Cl when X = I H =kJ mol1 H =kJ mol1 (ii) Use these results to describe and explain the trend in the thermal stabilities of the hydrides of Group VII.[5]UCLES 2013 9701/43/O/N/13<br />
Examiners For Use (d) Bromine reacts with hot NaOH(aq) to give a solution which on cooling produces white crystals of compound A. A has the following percentage composition by mass: Na, 15.2; O, 31.8; Br, 53.0. The remaining solution contains mostly NaBr, with a little of compound A. (i) Calculate the empirical formula of A. (ii) Construct an equation for the reaction between Br2 and hot NaOH(aq).[4] [Total: 15]UCLES 2013 9701/43/O/N/13 [Turn over<br />
BLANK PAGEUCLES 2013 9701/43/O/N/13<br />](https://learnah.org/wp-content/uploads/2018/08/November_2013_v3_QP_-_Paper_4_CIE_Chemistry_A-level_3-0.jpg)
![Page 4 Mark Scheme Syllabus Paper GCE A LEVELOctober/November 2013 9701 43 (a) (i) CH3Br(g)CH3(g) + Br(g) (ii) /3 AlCl 3(g)/3Al(g) + Cl(g) or AlCl3(g)AlCl 2(g) + Cl(g) (AlCl 3(g)Al(g) + 3Cl(g) for (1) mark) (b) (i) bond energies decrease from Cl 2 to I due to increasing bond length or increase in number of electron shells which causes less effective orbital overlap or less attraction for the shared pair (ii) either because fluorine is electronegative, (hence each F wants to keep its electrons to itself) or because the bond length is so short there is repulsion between the lone pairs (on F) or repulsion between the nuclei (of F) [4 max 3] [1] [2] [3] [1] [1] [1] [1] [2] [1] [1] [1] [5] (c) (i) for chlorine: H = E(HH) + E(ClCl)2E(HCl) = 436 + 242(2431) = 184 kJ mol for iodine: H = E(HH) + E(II)2E(HI) = 436 + 151(2299) = 11 kJ mol (ii) Hydrides become less thermally stable down the group from Cl to I as the HX bond energy decreases (more than does the XX bond energy) (d) (i) Na Br 15.2 / 23 31.8 / 16 53.0 / 79.9 [1]0.661 1.99 0.6630.661 1.0 3.0 1.0 thus NaBrO3 [1] (ii) 3Br2 + 6NaOH NaBrO3 + 5NaBr + 3H2O or 3Br2 + 6OH BrO3 + 5Br + 3H2O species [1] balancing [1] [4] [Total: 15]Cambridge International Examinations 2013<br />](https://learnah.org/wp-content/uploads/2018/08/November_2013_v3_MS_-_Paper_4_CIE_Chemistry_A-level_3.jpg)