Chemical energetics - student copy

YISHUN INNOVA JUNIOR COLLEGE 2021 JC2 H2 CHEMISTRY CHEMICAL ENERGETICS (FOUNDATION) ASRJC/2020/II/1c 1 (c) (i) Define the term standard enthalpy change of formation of a substance . [1] (ii) Table 1.1 enthalpy change value / kJ mol – 1 standard enthalpy change for P(s) + 2O 2 (g) + 3e –  PO 4 3 – (aq) – 1284 standard enthalpy change for K(s)  K + (aq) + e – – 251 standard enthalpy change for K 3 PO 4 (s)  3K + (aq) + PO 4 3 – (aq) – 2 Using a labelled energy cycle, and the enthalpy values given in Table 1.1, determine the standard enthalpy change of formation of solid potassium phosphate, K 3 PO 4 . [3] (iii) The value of  G O at 298 K for K(s)  K + (aq) + e – is – 284 kJ mol – 1 . Calculate  S O for the reaction, and explain its sign. [2] HCI/2020/II/5a 2 Titanium dioxide, TiO 2 , is a brilliant white pigment which imparts whiteness and opacity for paints, plastics, sunscreens, toothpaste, and food colouring. It is also coated on the paper you are reading now. (a) In the industrial manufacture of TiO 2 , titanium ( IV ) chloride, TiC l 4 , is oxidised by oxygen according to the equation: TiC l 4 (g) + O 2 (g)  TiO 2 (s) + 2C l 2 (g)  H o = – 175 kJ mol - 1 Use the following data, together with appropriate data from the Data Booklet , to calculate the average bond energy of the Ti − C l bond in TiC l 4 . standard enthalpy change of formation of TiO 2 (s) = – 939 kJ mol - 1 standard enthalpy change of atomisation of Ti(s) = +473 kJ mol−1 [3]

NJC/2020/II/4f 3 (f) Alternatively, 3-methylbutyl ethanoate can also be obtained from a reaction between ethanoyl chloride and 3-methylbutan-1-ol. compound  H f ꝋ / kJ mol  1 Boiling point / o C ethanoic acid  484 118 3-methylbutan-1-ol  353 132 3-methylbutyl ethanoate  478 142 ethanoyl chloride  273 51 water  286 100 hydrogen chloride  166  85 (i) Write an equation, with state symbols, for this reaction. [1] (ii) Using the data provided, determine the standard enthalpy change of this reaction. [1] (iii) Determine the standard enthalpy change of the reaction between CH 3 COC l and water. Hence explain why it is important to exclude water from the reaction between ethanoyl chloride and 3-methylbutanol by comparing the spontaneity of these two reactions. [3] NYJC/2020/II/1d 4 (d) Lead( II ) carbonate thermally decomposes according to the equation. PbCO 3 (s)  PbO(s) + CO 2 (g)  H decomp < 0 Some enthalpy changes are listed below. Enthalpy change of formation of PbCO 3 (s) = – 532 kJ mol – 1 Enthalpy change of atomisation of lead = +195 kJ mol – 1 Enthalpy change of combustion of carbon = – 394 kJ mol – 1 Lattice energy of lead( II ) oxide = – 3520 kJ mol – 1 First electron affinity of oxygen = – 141 kJ mol – 1 Second electron affinity of oxygen = +798 kJ mol – 1 By using data in this question and relevant data in the Data Booklet , construct a fully- labelled energy level diagram to determine the enthalpy change of decomposition of lead( II ) carbonate. [4] SAJC/2020/II/5e

Use the data in Table 2.1 to calculate the energy released per dm 3 of methane at 25 o C and 25.0 MPa. [1] (c) In an experiment, some butane was burned underneath a copper can containing 600 cm 3 of water. It was found that the temperature of water increased from 10°C to 70°C. This process was known to be only 80% efficient. Calculate the mass of butane used in the experiment. [2] (d) The global supplies of methane are depleting. Methods are being developed to produce methane from the fermentation of waste organic matter. (i) Aqueous propanoic acid disproportionate to produce methane and carbon dioxide as shown in the following equation. 4CH 3 CH 2 COOH (aq) + 2H 2 O ( l )  5CO 2 (g) + 7CH 4 (g) Construct the relevant half-equations for the reaction. [2] (ii) Suggest a method for removing the CO 2 from the gaseous product mixture in (d)(i) . [1] (iii) Some relevant standard enthalpy change of formation values,  H f o , are given in Table 2.2. Calculate the enthalpy change,  H r o , for the reaction represented by the given equation in (d)(i) , using the data in Table 2.2. Table 2.2 compound  H f o (kJ mol – 1 ) CH 3 CH 2 COOH – 510 H 2 O – 286 CH 4 – 75 CO 2 – 394 [2] ACJC/2020/III/2d 7 (d) An experiment was conducted to determine the enthalpy change of combustion of compound B and the following data was obtained. amount of compound B used = 0.0250 mol mass of water heated = 200 g temperature rise = 19 o C specific heat capacity of water = 4.2 J K  1 g  1

(i) Draw a labelled experimental set-up that can be used to obtain the data above. [2] (ii) Use the given data to calculate the enthalpy change of combustion of compound B . [2] (iii) The actual enthalpy change of combustion of compound B is  715 kJ mol  1 . Explain why your answer in (d)(ii) differs from the actual enthalpy change of combustion. [1] SAJC/2020/III/3b 8 (b) In an experiment, 50.0 cm 3 of 1.00 mol dm − 3 hydrochloric acid is added to 50.0 cm 3 of 2.00 mol dm − 3 ethylamine. The temperature increased from 26.0°C to 31.5°C. The heat transfer for this reaction is 85% efficient. TMJC/2020/III/1b 9 (b) (i) Construct a Born-Haber cycle for MgO. Use relevant data from the Data Booklet and the following data to calculate the lattice energy of MgO. standard enthalpy change of atomisation of Mg = +150 kJ mol − 1 1 st electron affinity of oxygen = −142 kJ mol − 1 2 nd electron affinity of oxygen = +844 kJ mol − 1 standard enthalpy change of formation of MgO = −602 kJ mol − 1 [5] (ii) Suggest why the value in (b)(i) differs from the theoretical value for the lattice energy of MgO. [1] VJC/2020/III/4c (i) Calculate ∆ H r for the reaction. Assume that the specific heat capacity of the solution is 4.2 J g − 1 K − 1 and that the density of the solution is 1.0 g cm − 3 . [2] (ii) Suggest and explain how the enthalpy change of reaction in (i) would differ when hexylamine is used in place of ethylamine. [1]

(a) The equation that links the melting point of a naphthalene – diphenylamine mixture with its enthalpy change of fusion,  H fusion , is shown. fusion m log Y = A 2.30 RT H    Y = mole fraction of naphthalene R = molar gas constant T m = melting point of naphthalene in K A is a constant Mole fraction of naphthalene, Y is calculated as shown. N N D n Y = n + n n N = amount in moles of naphthalene n D = amount in moles of diphenylamine The melting point and freezing point of a substance are the same. The melting point, T m , of a substance can be found by recording the temperature at which the substance freezes, measured when crystals first start to appear on cooling. Table 6.1 shows the results of a series of experiments using 0.100 mol of naphthalene and different amount of diphenylamine. Table 6.1 amount of diphenylamine, n D / mol temperature at which crystals appear, T m / K 0.00 353 0.0088 349 0.0178 345 0.0266 341 0.0355 338 0.0444 334 0.0533 331 0.0621 329 0.0769 325 (i) Suggest why the melting point decreases as the amount of diphenylamine in the mixture increases. [1]

Using the results in Table 6.1, log Y and m 1 T are calculated. A graph is then plotted to show the relationship between log Y and m 1 T . Fig. 6.1 shows a graph of log Y against m 1 T . m 1 T / 10 − 3 K − 1 Fig. 6.1 (ii) Calculate the gradient of the graph in Fig. 6.1 and hence, determine the value of the enthalpy change of fusion of naphthalene,  H fusion , in kJ mol – 1 . [2] (b) Fig. 6.2 shows  H 1 ,  H fusion and  H mixing relationship in an energy cycle. C 10 H 8 (s)  H fusion C 10 H 8 (l) C 10 H 8 (l) mixed with diphenylamine  H mixing  H 1 Fig. 6.2 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 0.00 2.80 2.85 2.90 2.95 3.00 3.05 3.10 log Y

[2] EJC/2020/II/5a-f 3 Table 5.1 and Table 5.2 show some physical properties of the Group 1 and Group 13 elements, respectively. Table 5.1 element radius / nm / kJ mol – 1 / kJ mol – 1 / kJ mol – 1 metallic ionic (M + ) Li 0.152 0.076 +162 – 278.5 – 293.3 Na 0.186 0.102 +108 – 240.3 – 261.9 K 0.227 0.138 +89.6 – 252.1 – 283.3 Rb 0.248 0.152 +82.0 – 251.1 – 284.0 Cs 0.265 0.167 +78.2 – 258.0 – 292.0 Table 5.2 element ionisation energy / kJ mol – 1 / kJ mol – 1 / kJ mol – 1 first second third fourth A l 577 1820 2740 11600 – 528.4 – 485.3 f H      M g f H      M aq  f G      M aq  f H      3 M aq  f G      3 M aq 

Ga 577 1980 2960 6190 – 211.7 – 159.0 I n 558 1821 2706 5350 – 105.0 – 98.0 T l 589 1971 2878 4934 +196.6 +214.6 The lighter Group 13 metals form compounds in the +3 oxidation state predominantly, although T l tends to form compounds in the +1 oxidation state. (a) What does the data in Table 5.2 suggest about the stability of the M 3+ (aq) ions formed? [1] (b) From Table 5.1, it can be seen that is more negative than the corresponding for the Group 1 metals. (i) What does this suggest about the sign of for the Group 1 metals? [1] (ii) Suggest a reason for the sign of in (b)(i) . [1] (iii) However, from Table 5.2, is significantly less negative than for the tripositive Group 13 cations, especially A l and Ga. Suggest a possible reason in terms of structure and bonding. [1] (c) An energy cycle relating the standard enthalpy change of formation of M + (aq) to three other energy terms, I , II and III , is shown in Fig. 5.1. Fig. 5.1 (i) Name the three energy terms. I : .....................................................................................................................  f G      3 M aq f G      M aq  f H      M aq  f S      M aq  f S      M aq  f G      3 M aq  f H      3 M aq  M(g) M + (g) + e – I M + (aq) + e – M(s) II III

This occurrence of an oxidation state which is 2 less than the group valency, as exemplified by T l , is sometimes referred to as the “inert - pair effect”. (f) However, the term “inert - pair effect” is somewhat inaccurate since it implies that the energy required to involve the valence n s 2 electron pair in bonding increases in the sequence A l < Ga < I n < T l . Using appropriate data from Table 5.2, justify why the term “inert - pair effect” is inaccurate. [2] TJC/2020/II/5a-c 4 Octane, C 8 H 18 , is a liquid hydrocarbon used as a fuel for motor vehicles. A scientist conducted a thermometry experiment as shown in Fig 5.1, to determine the enthalpy change of combustion of octane. The heat evolved from the combustion of octane is used to raise the temperature of the water and calorimeter. Fig. 5.1 For an accurate determination of the enthalpy change of combustion of octane from this experiment, the heat absorbed by the calorimeter has to be accounted for, which can be used to determine the heat capacity. (a) Define standard enthalpy change of combustion for octane . [1] (b) In one experiment, the scientist followed the procedure below to determine the heat capacity of the copper calorimeter. Step 1: Measure 50 cm 3 of cold water using a 50 cm 3 measuring cylinder into an insulated copper calorimeter. Step 2: Using the same measuring cylinder, measure 50 cm 3 of cold water into a 100 cm 3 beaker. Step 3: Heat up the water in the 100 cm 3 beaker to about 90 o C. Step 4: Stir the cold water in the copper calorimeter with a thermometer and note the initial temperature.

Step 5: When the temperature of the water in the 100 cm 3 beaker is about 90 o C, remove it from the heat source and measure accurately the temperature of the hot water. Step 6: Immediately pour the hot water from the 100 cm 3 beaker to the cold water in the copper calorimeter. Stir with a thermometer and record the highest temperature reached in the mixture. The data collected is shown in Table 5.1. Table 5.1 Initial temperature of cold water / o C 27.6 Initial temperature of hot water / o C 81.3 Highest temperature of mixture reached / o C 41.5 (i) With reference to the procedure above, suggest why it is not necessary for the temperature of water in Step 3 to be accurate? [1] (ii) Using the data in Table 5.1, calculate the temperature rise of the cold water in the calorimeter and temperature fall of hot water in the beaker. [1] (iii) With reference to your answer in (b)(ii) , calculate the heat absorbed by the calorimeter. Hence, determine the heat capacity of the calorimeter in J o C -1 . [2] (c) (i) After the heat capacity is determined, the same calorimeter was washed and cooled, before 250 cm 3 of water was added at room temperature. 1.35 g of octane was then burnt in excess oxygen and the maximum rise in temperature of the water was found to be 45 o C. Using your answer in (b)(iii) , calculate the enthalpy change of combustion of octane in kJ mol -1 . [2] (ii) The enthalpy change of combustion of octane can also be determined by considering the enthalpy change of vapourisation of octane and water. Enthalpy change of vapourisation / kJ mol -1 octane + 58.0 water + 44.0 Using the information above and relevant values from the Data Booklet , construct a suitable energy cycle to determine another value for the enthalpy change of combustion of octane in kJ mol -1 . [3]

(b) The binary compounds of lithium, LiX (X = H, F, C l , Br, and I ), all crystallise in the cubic lattice similar to NaC l lattice. One unit cell of this structure (small circles = Li + , large circles = X – ) is shown below. The length of the edge of the unit cell is abbreviated  . compound  / pm LiH 408.3 LiF 403.5 LiC l 513.6 LiBr 548.9 Li I 601.9 1 pm = 1  10 – 10 cm (i) Given that there are four Li + ions and four X – ions in one unit cell, calculate the mass of one unit cell and hence the density of LiC l in g cm – 3 . [Density,  = mass / volume] [2] The melting points of the lithium halides and of lithium hydride are plotted against the unit cell edge length,  , as shown below. (ii) Explain the relationship observed between  values and melting points of the lithium halides. [2] LiH LiF LiBr 700 800 900 1000 1100 1200 400 450 500 550 600 650 melting point / K  / pm LiC l Li I

(iii) Lithium hydride and lithium fluoride have similar  values . Suggest an explanation on why the melting point of lithium hydride is much lower than that of lithium fluoride. [2] ACJC/2020/III/5a 6 Calcium fluoride, CaF 2 is a Group 2 halide which occurs naturally in the mineral fluorspar . It is the major source of hydrogen fluoride, a chemical used to produce refrigerants and herbicides. The data in Table 5.1 will be useful in this question. Table 5.1 standard enthalpy change of formation of CaF 2 (s)  1220 kJ mol  1 standard enthalpy change of formation of Ca 2+ (aq)  543 kJ mol  1 standard enthalpy change of formation of F  (aq)  333 kJ mol  1 standard enthalpy change of hydration of Ca 2+ (g)  1579 kJ mol  1 standard enthalpy change of hydration of F  (g)  524 kJ mol  1 (a) (i) Define the term standard enthalpy change of solution of CaF 2 , Δ H o solution . [1] (ii) The standard Gibbs free energy change of solution of CaF 2 , Δ G o solution , is +64.4 kJ mol  1 . Use relevant data given in Table 5.1 to calculate the Δ H o solution , and the standard entropy change of solution, Δ S o solution , of CaF 2 . [3] (iii) Construct a fully labelled energy cycle relating the lattice energy of CaF 2 and the Δ H o solution of CaF 2 . Use your cycle to calculate the lattice energy of CaF 2 . [2] (iv) Suggest, with reasons, how the magnitude of the lattice energy of CaF 2 might compare to that of  CaC l 2  CaO [2]

NYJC/2020/III/4a-e 8 An Ellingham diagram is commonly used to show the variation in the free energy change of reaction with temperature. Since enthalpy change and entropy change are essentially constant with temperature unless a phase change occurs, the free energy versus temperature plot can be drawn as a series of straight lines. The graph showing the relationship between the Gibbs free energy change and the temperature of some oxides as well as the melting point of some elements are provided below. Table 1 Element Melting point / K Carbon 3823 Calcium 1115 Magnesium 923 (a) The equation that relates the three thermodynamics factors, the Gibbs free energy change , the enthalpy change of energy and the entropy change of energy is known to be ∆ G = ∆ H  T ∆ S . Define entropy . [1] (b) Given that the standard entropy change of energy is the negative gradient of each line in the Ellingham diagram, (i) Explain why the gradient for oxidation of carbon to carbon dioxide is almost zero. [1]

(ii) Calculate the entropy change of the reaction, 2C  O 2  2CO, in J mol  1 K  1 . [2] (c) Using the Ellingham diagram and the information from Table 1, suggest an explanation for the shape of MgO graph from 0 K to 1373 K. Your answer should include:  the general shape of the graph  the difference in the slope from 0 K to 923 K and 923 K to 1373 K. [3] (d) In an effort to decrease the amount of greenhouse gases, the decomposition of CO 2 to CO and O 2 is a potential remedy. 2CO 2  2CO  O 2 Using the equation provided and the data from the Ellingham diagram on Page 22 , construct a suitable energy cycle to calculate the enthalpy change of the decomposition reaction. [3] (e) By using the Ellingham diagram on Page 22 , deduce which metal, Mg or Ca is the stronger reducing agent. [3] TMJC/2020/III/5a 9 This question is about the chemistry of nitrogen and oxygen containing compounds. (a) Dinitrogen tetroxide, N 2 O 4 , is a powerful oxidiser that spontaneously reacts on contact with hydrazine, making the pair a popular bipropellant for rockets. Fig 5.1 shows how the entropy of N 2 O 4 varies with temperature at fixed pressure. N 2 O 4 exists as a solid at point A and gas at point D . Fig 5.1 - 0 Temperature/K A B C D Entropy/ kJ mol – 1