RI 2015_H1_CHEM_P2_ANS Prelims

Suggested Solutions to RI Y6 2015 H1 Prelims Paper 2 Section A (40 marks) 1 (a) (i) (ii) The order of reaction is the sum of all the powers of all the reactants’ concentrations in the rate equation. Or If rate = k[A]m[B]n where k is a rate constant and [A], [B] are concentrations of A and B respectively, Order of reaction = m + n (iii) From the graph, time taken, t1, for [product] to increase from 0 to ½Co = 1.45 min time taken, t2, for [product] to increase from ½Co to ¾ Co = 1.35 min If t1 ≈ t2≈1.40 min, then t1/2 is constant. Since t1/2 is constant, the reaction is first-order with respect to the reactant. Rate = k[N2O5] (iv) t1/2 = ln 2/ k k = ln 2/ 1.40 mins = 0.495 min–1

3 (b) (i) Pressure used is high as a high pressure favours the forward reaction. By Le Chatelier’s Principle, when pressure increases, the system will favour the side with fewer number of gaseous molecules. Hence, position of equilibrium shifts to the right and this increases yield. Since the forward reaction is exothermic, a lower temperature would result in a higher yield of ammonia. However, the rate of production is too slow at low temperature. On the other hand, a high temperature increases the rate of production but results in lower yield and higher production cost. Thus, a compromise is needed and a moderately high temperature (of 450 oC) is used to ensure a reasonable rate of production and yield. (b) (ii) This reduces the surface area of the catalyst, hence reducing the activity of the catalyst. (b) (iii) There is a general increasing trend in the ionisation energies, from left to right, as we remove one additional electron. This is because once the first electron is removed from the neutral atom, each successive electron is removed from an ion of increasing positive charge which attracts the electrons more strongly. There is a significant difference between the third and the fourth ionisation energies because the fourth electron is removed from an inner principal quantum shell. Therefore there are 3 electrons in the outermost shell. (iv) 1s2 2s2 2p6 2 (a) In graphite, the carbons are arranged in six -membered hexagonal rings with 3 σ bonds for each carbon, the single electron not used for bonding forms delocalised pi bonds /electron cloud above and below the layers of graphite rings, allowing for the movement of electrons for electr ical conductivity in a direction parallel to the layers. In diamond, the carbons are arranged in a network of tetrahedral carbons with four σ bonds each, this network of σ bonds do not allow for electrical conductivity. (b) (i) K2Cr2O7, H2SO4 heat with immediate distillation NaOH(aq) heat (ii) NaOH, I2(aq), heat o