PJC H2 Chem 2012 Prelim P2 Soln

© PJC 2012 9647/02/JC2 Prelim/2012 2 Answer all the questions in the spaces provided. 1 Planning (P) You are provided with solutions FA 1, FA 2 and FA 3. FA 1 and FA 2 are either 1.0 mol dm –3 nitric acid or 1.0 mol dm–3 ethanoic acid whereas FA 3 is 2.0 mol dm–3 sodium hydroxide. You are to plan an experiment that will enable you to identify FA 1 and FA 2 , and hence, calculate the enthalpy change of neutrali sation for the reaction between ethanoic acid and sodium hydroxide. You are provided with the following apparatus: a thermometer, a polystyrene cup, other common apparatus in the laboratory (a) State the independent and dependent variable s when determining the identity of each solution, FA 1 and FA 2. [1] Independent variable – Strength of acid Dependent variable – Rise in temperature (b) Plan an experiment to identify which of the two solutions, FA 1 and FA 2 , is 1.0 mol dm–3 nitric acid and which is 1.0 mol dm–3 ethanoic acid. Your plan must identify the acids solely based on the change in temperature. Mathematical processing of the temperature change is thus not required. Your plan should give a step by step description of the method including: · the apparatus used for measurement · appropriate volumes of reagents · how you would measure the various variables [3] Step 1: Using a 100 cm 3 measuring cylinder, measure out 25 cm3 of FA 3 into the polystyrene cup and record its initial temperature using a thermometer. Rinse the thermometer. Step 2: Using ANOTHER 100 cm3 measuring cylinder, measure 50 cm3 of FA 1. Record its initial temperature. Step 3: The initial temperature is given by the average of the initial temperature of FA 1 and FA 3.

© PJC 2012 9647/02/JC2 Prelim/2012 3 Step 4: Carefully transfer the 50 cm3 of FA 1 into the polystyrene cup containing FA 3 , stir with the thermometer and record the highest temperature reached. Rinse the cup. Step 5: Repeat steps 1 to 4, this time replacing FA1 with FA2. (c) Either a burette with 0.1 cm3 interval or a measuring cylinder with 1 cm3 interval can be used to measure the volume of FA 3 required. It is known that the error (or uncertainty) that is associated with each reading when using a measuring cylinder with 1 cm3 interval is ±0.5 cm3, while that using a burette with 0.1 cm3 interval is ±0.05 cm3. Assuming that the volume of FA 3 measured is V cm 3, calculate the maximum total percentage error (or uncertainty) in the measurement of the volume of FA 3, in terms of V, when using: (i) a measuring cylinder with 1 cm3 interval, Percentage error in measuring V cm3 = ± (0.5/ V) x 100% = ± 50/V % (ii) a burette with 0.1 cm3 interval. Percentage error in measuring V cm3 = ± [(0.05/ V) x 100%] x 2 = ± 10/V % [2] (d) Explain clearly how you could determine the identity of the acids using the difference in temperature rise in the experiment proposed in (b). [2] Ethanoic acid is a weak acid and nitric acid is a strong acid. The temperature rise will be lower when 1.0 mol dm -3 ethanoic acid is used compared to 1.0 mol dm-3 nitric acid for the same number of moles of water formed. This is because some of the e nergy evolved from the neutralisation process is used to further dissociate the ethanoic acid completely. (e) (i) Define the term standard enthalpy change of neutralisation. The heat evolved when one mole of water is formed when an acid neutralises a base under standard conditions of 298K and 1 atm. (ii) Show the mathematical expression for the enthalpy change of neutralisation for the reaction between ethanoic acid and sodium hydroxide, using the volumes proposed in (b). The temperature change measured in (b) should be represented by ∆T. [You may assume that 4.2 J of heat energy raised the temperature of 1 cm 3 of any solution by 1oC.]

© PJC 2012 9647/02/JC2 Prelim/2012 4 Heat evolved by neutralisation reaction = mc∆T = (vol. of FA1 or FA2 + vol of FA3) x c x ∆T = (50 + 25) x 4.2 x ∆T = 315∆T J ∆Hneutralisation = – 315∆T / amount of water formed = – 315∆T / 0.050 = – 6300∆T J mol-1 [3] (f) A student suggests titrating FA 3 against FA 1 and FA 2 separately, using phenolphthalein as indicator, to identify th e acids . Explain why this proposed method cannot work. [1] Both FA 1 and FA 2 are of the same concentration and basicity (monobasic). Both acids would hence require the same amount of NaOH for a complete reaction, and hence same volume. The volume of FA 3 (NaOH) required at end point would hence be the same. [Total: 12] 2 Nitrogen dioxide, NO2 undergoes dimerisation to form dinitrogen tetraoxide, N2O4. 2NO2(g) ⇌ N2O4(g) (a) (i) Draw the dot-and-cross diagram of NO2. O N Oxx x x .. .. .. x .. .. .. NO2 molecule (ii) State the shape and bond angle of the O–N–O bond in NO2. Shape of NO2: Bent Bond angle: accept any value where 110o < angle < 120o (iii) Draw the dot -and-cross diagram of NO 2- ion. With reference to your answer in (a)(i) and (ii), suggest a bond angle of the O –N–O in NO2- ion, as compared to NO2. Explain your answer.

© PJC 2012 9647/02/JC2 Prelim/2012 5 O N Oxx x.... .. x .. .. . NO2 - ion x x . - Bond angle of N O2- ion: bond angle must be smaller than NO 2, but still within range of 110o < angle < 120o Lone pair of electrons present in NO 2- ion occupies a larger volume of space than the single lone electron present in NO 2 molecule. Hence, lone pair-bond pair repulsion in NO 2- ion is greater than lone electron-bond pair repulsion in NO 2, causing NO 2- ion to have a smaller bond angle than NO2. [5] The general rate equation for the dimerisation of nitrogen dioxide can be represented as: Rate = k(pNO2)n, where n represents the order of reaction with respect to nitrogen dioxide. To determine the value of n, an experiment was carried out to collect experiment data on the partial pressure of nitrogen dioxide (p NO2) measured at a constant temperature of 500 K. The results were given below. Time/ s p NO2/ atm Rate /atm s-1 (pNO2)2/ atm2 0 0.917 9.48 x 10-5 0.841 1000 0.827 7.75 x 10-5 0.683 2000 0.753 6.45 x 10-5 0.567 3000 0.691 5.45 x 10-5 0.477 4000 0.638 4.67 x 10-5 0.407 Table 1 (b) (i) Given that n is 2, process the results in the Table 1 to produce data that would enable you to plot a straight line graph.

© PJC 2012 9647/02/JC2 Prelim/2012 6 (ii) Hence, use the processed data to plot the graph on the grid below. (iii) A student repeate d the experiment at the temperature of 250 K. Sketch, on the same axes in (ii) , the graph obtained by this student. Label this graph as 250 K. Straight line graph still obtained, with y-intercept = 0. Gradient (k) decreases. [4] [Total: 9] 250 K 500 K (pNO2)2 /atm2 Rate / atm s-1 0

© PJC 2012 9647/02/JC2 Prelim/2012 7 3 The following reaction scheme shows the chemistry of some chromium -containing species in aqueous solution. Cr2O7 2- SO2 I A II Zn [Cr(H2O)6]2+ blue solution Na2CO3 (aq)III B grey-green ppt IV V NaOH(aq) excess NH3 (aq) + CO2 [Cr(OH)6]3- [Cr(NH3)6]3+ VIII excess H2NCH2CH2NH2 C VI H2O2 VII di