VJC H2 Chem 2013 Prelim P3 Soln

 VJC 2013 9647/03/PRELIM/13 [Turn over 1 Victoria Junior College 2013 H2 Chemistry Prelim Exam 9647/3 Suggested Answers 1 During strenuous exercise, pyruvic acid, CH 3COCOOH, is converted to lactic acid, CH3CH(OH)COOH, which builds up in our muscles due to limited oxygen availability. (a) Such anaerobic respiration is catalysed by the enzyme lactate dehydrogenase, with concomitant conversion of NADH to NAD +. NADH plays a key role in the production of energy through redox reactions. NADH and NAD+ are cofactors, substances that act with and are essential to the activity of an enzyme. (i) Write a balanced equation for anaerobic respiration. Using the data given in the table below, calculate the value of E cell for this reaction. C C H 3COCOOH + NADH + H+  CH3CH(OH)COOH + NAD+ Eo = +0.89 – (-0.32) = +1.21V (ii) It is not possible to use Eo values reliably to decide whether a chemical reaction will occur. Suggest a reason for this. The reaction may be kinetically unfeasible due to high activation energy, even though it is thermodynamically feasible indicated by a positive Eo cell value. (iii) The activity of lactate dehydrogenase hinges on the active site Fe 2+ contained in a heme group which can be oxidized to Fe 3+. Suggest a mechanism for the catalysis of anaerobic respiration by lactate dehydrogenase. Support your answers with relevant E values given above and from the Data Booklet. You may represent the reduced and oxidised forms of the enzyme as LDH Fe2+ and LDHFe3+. Step 1: 2LDHFe2+ + CH3COCOOH + 2H+  2LDHFe3+ + CH3CH(OH)COOH E cell = +0.89 – (+0.77) = +0.12V (> 0, hence feasible) Step 2: 2LDHFe3+ + NADH  2LDHFe2+ + NAD+ + H+ E cell = +0.77 – (-0.32) = +1.09V (> 0, hence feasible) (iv) After exercising, we do not stop panting immediately because we need to breathe in additional oxygen so that lactic acid can be reconverted back to pyruvic acid. Both breathing and heart rates increase until the excess lactic acid level normalizes. Write a balanced equation for the conversion of lactic acid back to pyruvic acid. Calculate the value of E  cell for this reaction, with the help of the Data Booklet. O2 + 2CH3CH(OH)COOH  2H2O + 2CH3COCOOH Eo = +1.23 – (+0.89) = +0.34V [8] Electrode reaction E /V 2e- + H+ + NAD+ NADH –0.32 2e- + 2H+ + CH3COCOOH CH 3CH(OH)COOH +0.89 1528

 VJC 2013 9647/03/PRELIM/13 [Turn over 2 (b) (i) Using bond energy data obtained from the Data Booklet, calculate the enthalpy change of reaction, ∆H, for the conversion of pyruvic acid into lactic acid. CH3COCOOH(l) + H2(g)  CH3CH(OH)COOH(l) ∆H = 740 + 436 – 410 – 360 – 460 = –54 kJ mol-1 (ii) Given that the actual enthalpy change of this reaction is 110 kJ mol1, suggest a reason for the discrepancy between your calculated ∆H value in (b)(i) and the actual ∆H v a l u e . Bond energy is the heat required to break 1 mole of covalent bonds in the gaseous state but pyruvic acid and lactic acid exist as liquid. OR Bond energy for a polyatomic molecule is an average value. (iii) Acid-base neutralisation takes place when pyruvic acid reacts with barium hydroxide Rank barium hydroxide, pyruvic acid, and its salt in increasing melting point. Explain your answer. Pyruvic acid, barium pyruvate, barium hydroxide. Pyruvic acid has a simple molecular structure with weak hydrogen bonding between its molecules, hence it has the lowest melting point. Barium pyruvate and sodium hydroxide, on the other hand, have giant ionic structure with strong ionic bonds be tween the oppositely charged barium and pyruvate or hydroxide ions. Hence, they have higher melting points as compared to pyruvic acid. Strength of ionic bond is dependent on lattice energy, which is in turn proportional to       r r q q . Since the cation Ba 2+ is the same and charges of the OH- and pyruvate anions are the same at -1 for both barium hydroxide and barium pyruvate, lattice energy is only dependent on the size of the anions. And since pyruvate ion is much larger than OH -, lattice energy for barium pyruvate is less exothermic, hence less energy is required to break the ionic bonds and barium pyruvate has a lower melting point compared to barium hydroxide. [6] 1529

 VJC 2013 9647/03/PRELIM/13 [Turn over 3 (c) impulse. Calcium causes muscles to contract while magnesium relaxes them. Y is another element in Group II. Element Mg Ca Y E /V –2.37 –2.87 –2.95 Decomposition temperature of carbonates /K 470 1170 1770 (i) Predict the relative melting points of magnesium and calcium. Magnesium has a higher melting point than calcium, since Mg 2+ is smaller than Ca2+. (ii) With the help of the Data Booklet , calculate the enthalpy change, ∆ H when Mg and Ca ionise to Mg2+ and Ca2+ respectively. M g M g 2+ + 2e- Mg+ + e- By Hess’s law, ∆H = 1st IE + 2nd IE ∆H (Mg  Mg2+) = +736 + 1450 = +2186 kJ mol1 ∆H (Ca  Ca2+) = +590 + 1150 = +1740 kJ mol1 (iii) How does the trend in ∆H calculated in (c)(ii) relate to the Eo values? Hence, predict the relative reducing powers of magnesium and calcium. The less endothermic the ∆H, the more negative the E o values, the stronger is the reducing agent. Hence, calcium has a greater reducing power than magnesium. (iv) Account for the difference between the decomposition temperature of CaCO 3 and YCO3. C a 2+ has a higher charge density than Y 2+, hence it is more polarising and causes greater distortion to the electron cloud and CaCO 3 is thus less stable to heat. [6] [Total: 20] ∆H 1st 2nd IE 1530

 VJC 2013 9647/03/PRELIM/13 [Turn over 4 2 (a) Compound X, CH 3CH=CHCHClCH3, reacts with alcoholic KCN according to the following equation: CH3CH=CHCHClCH3 + CN–  CH3CH=CHCH(CN)CH3 + Cl– To investigate the above reaction, two separate experiments were performed using different initial concentrations of CN . In each experiment, the concentration of X in the reaction mixture was determined at different times as the reaction progresses. The same graph was obtained in both experiments, as shown below: The following results were obtained: (i) Explain why the reaction requires an alcoholic instead of an aqueous medium to take place. Halogenoalkane X is not soluble in water as it cannot form hydrogen bond with water. [1] (ii) Referring to the graphs above, deduce the order of reaction with respect to each of the reactants, CN – and X. For [X]-time graph, it shows two constant half-lives of 115 min. Hence, reaction is first order w.r.t. X. Gradient of the tangent at time=0 remains the same for two different initial [CN –], the reaction rate is independent of [CN –]. Hence, reaction is zero order w.r.t. CN–. [2] [X] /mol dm-3 time /min [CN]1 and [CN]2 1531

 VJC 2013 9647/03/PRELIM/13 [Turn over 5 + (iii) State, with a reason, whether the reaction proceeds by the S N1 or S N2 mechanism. Since the reaction is 1 st order w.r.t. X and zero order w.r.t. CN –, the rate- determining / slow step involves 1 molecule of X and no CN–. Mechanism is therefore SN1. [1] (iv) Hence, sketch a well labelled reaction pathway diagram for the reaction, given that the reaction is exothermic. [2] (v) The reaction of the compound, CH 3CH2CH2CHClCH3 with alcoholic KCN occurs via a mixture of SN1 and SN2 mechanisms. Propose a rate equation for the reaction. Rate = k1[CH3CH2CH2CHClCH3] + k2[CH3CH2CH2CHClCH3