9701_s17_qp_41
A paper of Chemistry, 9701
Questions:
6
Year:
2017
Paper:
4
Variant:
1
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Describe and explain the variation in the solubilities of the hydroxides of the Group 2 elements. The table lists the standard enthalpy changes of formation, , for some compounds and aqueous ions. species / kJ mol–1 Ba2+–538 OH––230 CO2–394 BaCO3–1216 H2O–286 Reaction 1 occurs when CO2is bubbled through an aqueous solution of Ba(OH)2. Use the data in the table to calculate the standard enthalpy change for reaction 1, 1. Ba(OH)2+ CO2BaCO3+ H2Oreaction 1 1 = kJ mol–1 If CO2is bubbled through an aqueous solution of Ba(OH)2 for a long time, the precipitated BaCO3dissolves, as shown in reaction 2. BaCO3+ CO2+ H2OBa(HCO3)2reaction 2 The standard enthalpy change for reaction 2, 2, = –26 kJ mol–1. Use this information and the data in the table to calculate the standard enthalpy change of formation of the HCO3 –ion. HCO3 –= kJ mol–1 The overall process is shown by reaction 3. Use your answer to , and the data given in the table, to calculate the standard enthalpy change for reaction3, 3. Ba(OH)2+ 2CO2Ba(HCO3)2reaction 3 3 = kJ mol–1 How would the value of 3 compare with the value of 4 for the similar reaction with Ca(OH)2as shown in reaction4? Explain your answer. Ca(OH)2+ 2CO2Ca(HCO3)2reaction 4 The standard entropy change for reaction 1 is 1. Suggest, with a reason, how the standard entropy change for reaction3 might compare with 1.
10. Group 2  →  10.1. Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
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One atom of each of the four elements H, C, N and O can bond together in different ways. Two examples are molecules of cyanic acid, HOCN, and isocyanic acid, HNCO. The atoms are bonded in the order they are written. Draw ‘dot-and-cross’ diagrams of these two acids, showing outer shell electrons only. HOCN, cyanic acid HNCO, isocyanic acid Suggest the values of the bond angles HNC and NCO in isocyanic acid. HNC NCO Suggest which acid, cyanic or isocyanic, will have the shorter C–N bond length. Explain your answer. Isocyanic acid is a weak acid. HNCO H+ + NCO– Ka = 1.2 × 10–4 mol dm–3 Calculate the pH of a 0.10 mol dm–3 solution of isocyanic acid. pH = Sodium cyanate, NaNCO, is used in the production of isocyanic acid. Sodium cyanate is prepared commercially by reacting urea, (NH2)2CO, with sodium carbonate. Other products in this reaction are carbon dioxide, ammonia and steam. Write an equation for the production of NaNCO by this method. Barium hydroxide, Ba(OH)2, is completely ionised in aqueous solutions. During the addition of 30.0 cm3 of 0.100 mol dm–3 Ba(OH)2 to 20.0 cm3 of 0.100 mol dm–3 isocyanic acid, the pH was measured. Calculate the [OH–] at the end of the addition. [OH–] = mol dm–3 Use your value in to calculate [H+] and the pH of the solution at the end of the addition. final [H+] = mol dm–3 final pH = On the following axes, sketch how the pH changes during the addition of a total of 30.0 cm3 of 0.100 mol dm–3 Ba(OH)2 to 20.0 cm3 of 0.100 mol dm–3 isocyanic acid. volume of Ba(OH)2 added / cm3 pH The cyanate ion, NCO–, can act as a monodentate ligand. State what is meant by the terms monodentate, ligand. Silver ions, Ag+, react with cyanate ions to form a linear complex. Suggest the formula of this complex, including its charge. When heated with HCl , organic isocyanates, RNCO, are hydrolysed to the amine salt, RNH3Cl, and CO2. RNCO + H2O + HCl RNH3Cl + CO2 A 1.00 g sample of an organic isocyanate, RNCO, was treated in this way, and the CO2 produced was absorbed in an excess of aqueous Ba(OH)2 according to the equation shown. The solid BaCO3 precipitated weighed 1.66 g. Ba(OH)2+ CO2BaCO3+ H2OCalculate the number of moles of BaCO3 produced. moles of BaCO3 = Hence calculate the Mr of the organic isocyanate RNCO. Mr of RNCO = The R group in RNCO and RNH3Cl contains carbon and hydrogen only. Use your Mr value calculated in to suggest the molecular formula of the organic isocyanate RNCO. molecular formula of RNCO Suggest a possible structure of the amine RNH2, which forms the amine salt, RNH3Cl.
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
12. Nitrogen and sulfur  →  12.1. Nitrogen and sulfur
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Bubbling air through different aqueous mixtures of CoCl 2, NH4Cl and NH3 produces various complex ions with the general formula [Co(NH3)6–nCl n]3–n. Determine the oxidation state of the cobalt in these complex ions. Name the two types of reaction undergone by the cobalt ions during the formation of these complex ions. The complex [Co(NH3)4Cl 2]+ shows isomerism. Draw three-dimensional structures of the two isomers, and suggest the type of isomerism shown here. isomer 1 isomer 2 type of isomerism What is meant by the term co-ordination number? Complete the table by predicting appropriate co-ordination numbers, formulae and charges for the complexes C, D, E and F. complex metal ion ligand co-ordination number formula of complex charge on complex C Cr3+ CN– 3 – D Ni2+ H2NCH2CH2NH2 E Pt2+ Cl – 2– F Fe3+ –O2C–CO2 – [Fe(O2CCO2)3] Iron(forms complexes in separate reactions with both SCN– ions and Cl – ions. Fe3++ SCN–[FeSCN]2+equilibrium 1 Fe3++ 4Cl –[FeCl 4]–equilibrium 2 Write the expressions for the stability constants, Kstab, for these two equilibria. Include units in your answers. Kstab1 = units = Kstab2 = units = An equilibrium can be set up between these two complexes as shown in equilibrium 3. [FeCl 4]–+ SCN–[FeSCN]2++ 4Cl –equilibrium 3 Write an expression for Keq3 in terms of Kstab1 and Kstab2. Keq3 = The numerical values for these stability constants are shown. Kstab1 = 1.4 × 102 Kstab2 = 8.0 × 10–2 Calculate the value of Keq3 stating its units. Keq3 = units =
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
11. Group 17  →  11.3. Some reactions of the halide ions
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Carvone occurs in spearmint and a stereoisomer of carvone occurs in caraway seeds. Treating either isomer with hydrogen over a nickel catalyst produces a mixture of isomers with the structural formula X. O OH carvone X H2 + Ni State the type of stereoisomerism carvone can show. Explain your answer. Write an equation, using molecular formulae, for this conversion of carvone to X. X can be synthesised from methylbenzene by the following route. X NO2 step 1 methylbenzene step 2 OH NH2 N2 +Cl – step 3 step 5 OH step 6 step 4 Name the mechanism in step1. What type of reaction is occuring in the following steps? step 3 step 5  Suggest reagents and conditions for each of the following steps. step 1 step 2 step 3 step 4 During step 6, hydrogen is added to the benzene ring to produce the cyclohexane ring in X. The six hydrogen atoms are all added to the same side of the benzene ring. State the reagents and conditions needed for this reaction. Complete the part structure to show the structure of the isomer of X that would most likely be obtained during this reaction. X
21. Organic synthesis  →  21.1. Organic synthesis
14. Hydrocarbons  →  14.2. Alkenes
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Compounds J, K, L and M are isomers of each other with the molecular formula C9H11NO. All four isomers contain a benzene ring. Two of the isomers contain a chiral centre. The results of six tests carried out on J, K, L and M are shown in the table. test observations with each isomer J K L M add cold HCl soluble soluble soluble insoluble add 2,4-DNPH reagent orange ppt. orange ppt. orange ppt. no reaction add NaOH+ I2pale yellow ppt. no reaction pale yellow ppt. no reaction warm with Fehling’s solution no reaction red ppt. no reaction no reaction heat with NaOHno reaction no reaction no reaction P(C6H7N) and Q(C3H5O2Na) produced diazotization and addition of alkaline phenol no dye produced orange dye produced no dye produced no dye produced Use the experimental results in the table above to determine the group, in addition to the benzene ring, present in each of the four isomers J, K, L and M. Complete the table below, identifying the grouppresent in each isomer. groupin compound J K L M Name the type of reaction occurring in test 5 that converts M into P + Q. Suggest structures for compounds P and Q. P (C6H7N) Q (C3H5O2Na) Isomers J, K, L and M all have the molecular formula C9H11NO. Use the information in to suggest a structure for each of these isomers and draw these in the boxes. Draw circles around all chiral centres in K and L. J K L M Compound N is another isomer which has the same molecular formula C9H11NO and also contains a benzene ring. N contains the same functional group as M. When heated with NaOH, N produces ethylamine and a sodium salt W. Suggest the structure of W. W
15. Halogen compounds  →  15.1. Halogenoalkanes
14. Hydrocarbons  →  14.2. Alkenes
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
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The reaction between 1-chloro-1-phenylethane and hydroxide ions to produce 1-phenylethanol is: C6H5CHCl CH3 + OH– C6H5CH(OH)CH3 + Cl – 1-chloro-1-phenylethane 1-phenylethanol The rate of this reaction can be studied by measuring the amount of hydroxide ions that remain in solution at a given time. The reaction can effectively be stopped if the solution is diluted with an ice-cold solvent. Describe a suitable method for studying the rate of this reaction at a temperature of 40 °C, given the following. ● a solution of 0.10 mol dm–3 1-chloro-1-phenylethane, labelled A ● a solution of 0.10 mol dm–3 sodium hydroxide, labelled B ● 0.10 mol dm–3 HCl ● volumetric glassware ● ice-cold solvent ● stopclock ● access to standard laboratory equipment and chemicals The rate of this reaction was measured at different initial concentrations of the two reagents. The table shows the results obtained. experiment [C6H5CHCl CH3] / mol dm–3 [OH–] / mol dm–3 relative rate 0.05 0.10 0.5 0.10 0.20 1.0 0.15 0.10 1.5 0.20 0.15 to be calculated Deduce the order of reaction with respect to each of [C6H5CHCl CH3] and [OH–]. Explain your reasoning. order with respect to [C6H5CHCl CH3] order with respect to [OH–]  Write the rate equation for this reaction, stating the units of the rate constant, k. rate = mol dm–3 s–1 units of k = Calculate the relative rate for experiment 4. relative rate for experiment 4 = Use your answers in to help you to draw the mechanism for the reaction of 1-chloro-1-phenylethane with hydroxide ions, including the following. ● all relevant lone pairs and dipoles ● curly arrows to show the movement of electron pairs ● the structures of any transition state or intermediate This reaction was carried out using a single optical isomer of 1-chloro-1-phenylethane. Use your mechanism in to predict whether the product will be a single optical isomer or a mixture of two optical isomers. Explain your answer. The proton NMR spectrum of a sample of 1-phenylethanol shows four peaks: a multiplet for the C6H5 protons and three other peaks as shown in the table. When the sample is shaken with D2O and the proton NMR spectrum recorded, fewer peaks are seen. Complete the table for the proton NMR spectrum of 1-phenylethanol, C6H5CH(OH)CH3. Use of the Data Booklet might be helpful. δ / ppm number of 1H atoms responsible for the peak group responsible for the peak splitting pattern result on shaking with D2O 1.4 2.7 4.0 7.2-7.4 C6H5 multiplet peak remains
8. Reaction kinetics  →  8.1. Rate of reaction
8. Reaction kinetics  →  8.2. Effect of temperature on reaction rates and the concept of activation energy
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