SRJC H2 CHEM P3 ANS

1 Turn Over] SRJC 2014 Prelim H2 Chemistry Paper 3 Answers 1 Singapore has hosted the Formula One Grand Prix since 2008. This car racing event sees race car drivers from around the world compete in a circuit to win the top honour of a race car champion. These race cars are fuelled by petrol, which is a mixture of hydrocarbons. One such hydrocarbon is pure liquid octane, C8H18. (a) (i) Define the term standard enthalpy change of combustion. Standard enthalpy change of combustion (Hc ) of a substance is the energy evolved when one mole of the substance is completely burnt in oxygen under standard conditions. OR C8H18 (l) + 2 25 O2 (g)  8CO2 (g) + 9H2O(l) Hc  < 0, standard conditions (ii) Use the data in the table below to calculate the standard enthalpy change of combustion of octane. Hence calculate the heat evolved per gram of octane. species ΔHf / kJ mol-1 C8H18(l) -269 CO2(g) -394 H2O(l) -286 C8H18(l) + 2 25 O2(g)  8CO2(g) + 9H2O(l) 8C(s) + 9H2(g) + 2 25 O2(g) By Hess’ Law, ΔHc = 269 + 8(-394) + 9(-286) = - 5460 kJ mol-1 OR ΔHc = ΔHf (products) – ΔHf (reactants) = 269 + 8(-394) + 9(-286) = - 5460 kJ mol-1 Heat evolved per gram of octane = 114 5460 = 47.9 kJ g-1 -269 8(-394) + 9(-286)

2 Turn Over] (iii) Methanol, CH 3OH, is also commonly used as a fuel for high -performance engines in race cars. Due to the high temperature in car engines, methanol is converted to the vapour state. Use the bond energies given in the Data Booklet to calculate the standard enthalpy change of combustion of methanol. Hence calculate the heat evolved per gram of methanol. CH3OH(l) + 2 3 O2(g)  CO2(g) + 2H2O(g) Bonds broken in the reactants : 3C-H, C-O, O-H, 2 3 O=O Bonds formed in the products: 2C=O, 4 O-H ΔHc = ∑BE(reactants) – ∑BE(products) = 3(410) + 360 + 460 + 2 3 (496) – 2(740) – 4(460) = -526 kJ mol-1 Heat evolved per gram of methanol = 32 526 = 16.4 kJ g-1 (iv) Using your answers in (a)(ii) and (a)(iii), suggest an advantage of using liquid octane over methanol as the fuel in race cars. It gives out more heat energy per gram of fuel. [6] (b) Methylbenzene is a component of lead -free petrol. It is synthesised by cyclising and dehydrogenating heptane. Methylbenzene can be made by catalytic reforming of heptane. This process causes straight chain hydrocarbons between 6 to 8 carbon atoms to rearrange into compounds containing benzene rings. This process requires a platinum catalyst, high temperature of 500 oC and pressure of about 20 atm. C7H16(g) C7H8(g) + 4H2(g) ∆H > 0 (i) State Le Chatelier’s Principle. Le Chatelier’s Principle states that when a system in equilibrium is disturbed , the system will react to counteract the effect of the change until a new equilibrium is reached.

3 Turn Over] (ii) Suggest reasons why these operating conditions are used.  At low pressure, by Le Chatelier’s Principle, the position of equilibrium shifts to the right to increase total amount (in mol) of gases. Hence, high yield of methylbenzene is obtained.  In addition, the reaction reaches equilibrium faster.  A catalyst is used to increase the rate of reaction / to achieve equilibrium at a shorter time.  At high temperature, by Le Chatelier’s Principle, the position of equilibrium shifts to the right to absorb heat , favouring the endothermic reaction. Hence, high yield of methylbenzene is obtained. [4] (c) Compound P has the molecular formula C5H9ON.  P exhibits optical isomerism, is neutral, and reacts with sodium metal.  On reacting P with lithium aluminium hydride, compound Q, C 5H13ON, is formed.  When Q is refluxed with aqueous acidified potassium manganate( VII), R, C5H11O2N is obtained.  The addition of thionyl chloride, SOCl2 to R produces S.  Under suitable conditions, S forms T, C 5H9ON, which is neutral, along with white fumes of HCl. Explain and identify the structures of P, Q, R, S and T. [10] Deductive statements  P exhibits optical isomerism  P has a chiral carbon that is bonded to 4 different substituents.  P is neutral and undergoes redox with sodium metal  P contains an alcohol functional group  P undergoes reduction with lithium aluminium hydride to form compound Q.  P contains a nitrile functional group; Q contains an amine functional group.  Q undergoes oxidation with aqueous potassium manganate(VII), to form R.  R contains a carboxylic acid functional group.  R undergoes nucleophilic substitution with SOCl2 to form S.  S contains an acid chloride/ acyl chloride functional group.  S forms a cyclic amide under suitable conditions via nucleophilic substitution.

4 Turn Over] Total 20 marks Structures P HO CN OR HO CN Q HO CH2NH2 OR HO CH2NH2 R HO CH2NH2 O OR HO CH2NH2 O S Cl CH2NH2 O OR Cl CH2NH2 O T O N H OR O N H

5 Turn Over] 2 Iron forms compounds mainly in the +2 and +3 oxidation states. Iron compounds are coloured and were widely used in paintings during the 19th century. (a) Explain why iron exhibits variable oxidation states but not calcium. [2]  Iron possesses variable oxidation states due to the small energy level difference between the 3d and 4s electrons. This results in different numbers of 3d and 4s electrons lost to form stable ions and compounds of different oxidation states.  Calcium is restricted to oxidation numbers of +2 (Group II) because once the s electrons are removed, further removal of inner shell electrons will require too much energy. (b) Explain why most iron compounds are coloured. [3]  The presence of ligands splits the 3d orbitals of the iron atom or ion into two energy levels. This is called d-orbital splitting.  A 3d electron absorbs certain wavelengths of light energy from the visible region of the electromagnetic spectrum.  The 3d electron undergoes d-d transition and is promoted to a higher energy d orbital.  The complementary wavelengths are transmitted/reflected and appear as the colour observed. (c) When Vincent van Gogh painted the “The Starry Night” in 1889, he used a lot of the pigment known as Prussian Blue, which is an iron complex, Fe4(Fe(CN)6)3. Draw the structure of the complex ion in Prussian Blue, stating the oxidation number of iron in the complex ion. [2] Fe CN CN CN CN NC NC Oxidation number of Fe in complex ion = +2 4-

6 Turn Over] (d) When water ligands in a hydrated metal ion are substituted by other ligands, the equilibrium constant for the reaction is referred to as the stability constant, Kstab of the new complex. [Fe(H2O)6]3+ + nL [Fe(H2O)6–nLn](3–n)+ + nH2O The following table lists two stability constants for the above reaction. (where n is a whole number) L n Kstab SCN– 1 9 x 102 CN– 6 1 x 1034 (i) Write the expression for the stability constant, Kstab, for L = SCN–. Kstab = ]SCN][)OH(Fe[ ])OH)(SCN(Fe[ 3 62 2 52   (ii) Use the data given in the table to predict and explain which is the predominant complex formed when a solution containing equal concentrations of both SCN– and CN– ions was added to a solution containing Fe3+(aq) ions. [3] Since Kstab for the complex formed by CN – and Fe 3+ ions (1 x 10 34) is much larger than that by SCN – and Fe3+ ions (9 x 10 2), the position of equilibrium for [Fe(CN)6]3– lies more to the right. [Fe(CN)6]3– is more stable than [Fe(SCN)(H 2O