9701_s17_qp_42
A paper of Chemistry, 9701
Questions:
8
Year:
2017
Paper:
4
Variant:
2
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1
Describe and explain the variation in the thermal stabilities of the carbonates of the Group2 elements. Suggest and explain a reason why sodium carbonate is more stable to heat than magnesiumcarbonate. Sodium hydrogencarbonate, NaHCO3, and potassium hydrogencarbonate, KHCO3, decompose on heating to produce gases and the solid metal carbonate. Write an equation for the decomposition of KHCO3. Predict which of NaHCO3 or KHCO3 will decompose at the lower temperature. Explain your answer. Use the data in the table below, and relevant data from the Data Booklet, to calculate the lattice energy, , of potassium oxide, K2O. energy change value / kJ mol–1 enthalpy change of atomisation of potassium, K+89 electron affinity of O–141 electron affinity of O–+798 enthalpy change of formation of potassium oxide, K2O–361 = kJ mol–1 State whether the lattice energy of Na2O would be more negative, less negative or the same as that of K2O. Give reasons for your answer.
10. Group 2  →  10.1. Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
11. Group 17  →  11.2. The chemical properties of the halogen elements and the hydrogen halides
4. States of matter  →  4.2. Bonding and structure
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Complete the table to show how both AgNO3and NH3could be used to distinguish between solutions of NaCl and NaI. test performed observation with NaCl observation with NaI Important information for this question ● In this question means ‘a solution in propanone’. ● Sodium iodide is soluble in propanone giving Na+and I–. ● Sodium chloride is insoluble in propanone. The reaction between 2-chlorobutane and sodium iodide in propanone is shown. CH3CH2CHCl CH3+ Na++ I–CH3CH2CHICH3+ NaCl The rate of this reaction can be investigated by measuring the electrical conductivity of the reaction mixture. The electrical conductivity changes as the reaction progresses due to the precipitation of the NaCl produced. Suggest how the electrical conductivity will change as the reaction proceeds. Explain your answer. Describe a suitable method for studying the rate of this reaction at a temperature of 40 °C, using the following. ● an electrical conductance meter which measures the electrical conductivity of solutions ● solutions of known concentrations of 2-chlorobutane in propanone and sodium iodide in propanone ● stopclock ● access to standard laboratory equipment The rate of this reaction was measured at different initial concentrations of the two reagents. The table shows the results obtained. experiment [CH3CH2CHCl CH3] / mol dm–3 [I–] / mol dm–3 relative rate 0.06 0.03 0.10 0.03 0.06 0.05 0.08 0.04 to be calculated Deduce the order of reaction with respect to each of [CH3CH2CHCl CH3] and [I–]. Explain your reasoning. order with respect to [CH3CH2CHCl CH3] order with respect to [I–]  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 = Suggest the mechanism for the reaction of 2-chlorobutane with iodide ions. Draw out the steps involved, including the following. ● all relevant lone pairs and dipoles ● curly arrows to show the movement of electron pairs ● the structure of any transition state or intermediate This reaction was carried out using a single optical isomer of 2-chlorobutane. 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. State the number of peaks that would be seen in the carbon-13 NMR spectrum of CH3CH2CHCl CH3. There are two isomers of CH3CH2CHCl CH3 that have fewer peaks in their carbon-13 NMR spectra than CH3CH2CHCl CH3. Draw the structures of the isomers and state the number of peaks for each isomer. isomer 1 isomer 2 number of peaks = number of peaks =
11. Group 17  →  11.4. The reactions of chlorine
8. Reaction kinetics  →  8.1. Rate of reaction
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In a molecule of SOCl 2 the sulfur atom has four bonds. Draw a 'dot-and-cross’ diagram of SOCl 2. Show the outer shell electrons only. When SOCl 2 is reacted with a carboxylic acid to produce an acyl chloride, two acidic gases are formed. SOCl 2+ RCO2HRCOCl + SO2+ HCl A 1.00 g sample of a carboxylic acid RCO2H was treated in this way, and the gases were absorbed in 60.0 cm3 of 0.500 mol dm–3 NaOH, an excess. Write equations for the reactions between NaOH and HCl , NaOH and SO2. The excess NaOH was titrated with 0.500 mol dm–3 H+. It required 10.8 cm3 of the H+solution to reach the end-point. Calculate the total number of moles of NaOH that reacted with the SO2 and HCl. moles of NaOH = Calculate the number of moles of RCO2H that produced the SO2 and HCl. moles of RCO2H = Hence calculate the Mr of the carboxylic acid, RCO2H. Mr RCO2H = The R group contains carbon and hydrogen only. Suggest the molecular formula of RCO2H. The following synthetic route shows how a carboxylic acid can be converted into an amine. SOCl 2 NH3 step 3 CH3CO2H CH3COCl CH3CONH2 CH3CH2NH2 Suggest a reagent for step 3. Angelic acid, C5H8O2, is a natural product isolated from the roots of the angelica plant. ● Angelic acid reacts with H2 + Ni to form T, C5H10O2. ● T undergoes the above synthetic route to form the amine U, C5H13N. ● U can also be made by reacting 1-bromo-2-methylbutane with ammonia. Both angelic acid and T exist as stereoisomers. Suggest structures for angelic acid, T and U. angelic acid T U State the type of stereoisomerism shown by angelic acid and T. angelic acid compound T
18. Carboxylic acids and derivatives  →  18.1. Carboxylic acids
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
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A number of isomers with the formula Cr(H2O)6Cl 3 exist. Their general formula is [Cr(H2O)6-nCl n]Cl 3-n.nH2O. Each isomer contains a six co-ordinated Cr(ion in an octahedral complex. Water molecules not directly bonded with the Cr atom are held in the crystal lattice as water of crystallisation. The Cr–Cl bond is not easily broken and so chloride bonded with the Cr(ion in the complex does not react. 1.00 g samples of three of the isomers, A, B and C, were dissolved in separate samples of water. An excess of AgNO3was added to each and the mass of AgCl formed was measured. Ag++ Cl –AgCl The number of moles of AgCl formed was calculated. The table shows the results. isomer moles of AgCl formed from 1.00 g of isomer A 3.75 × 10–3 B 7.50 × 10–3 C 1.13 × 10–2 Calculate the Mr of Cr(H2O)6Cl 3. Mr Cr(H2O)6Cl 3 = Use the data in the table above to calculate the value of n for each of the isomers, A, B and C. Complete the table below with the values of n and the molecular formula of each isomer, in the style of the general formula given above. Show your working for at least one calculation of n. isomer n molecular formula A B C Two isomers have the same shape and their formula is Ni(R3P)2(CN)2, where R = CH3. Only one of these isomers has a dipole moment. Name the type of isomerism shown by Ni(R3P)2(CN)2. Draw structures of these two isomers. isomer 1 isomer 2 State which isomer has a dipole moment. Explain your answer.
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
13. An introduction to AS Level organic chemistry  →  13.4. Isomerism: structural isomerism and stereoisomerism
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1,2-diaminoethane, en, H2NCH2CH2NH2, is a bidentate ligand. What is meant by the terms bidentate and ligand? bidentate ligand There are three isomeric complex ions with the formula [Cr2Cl 2]+. Complete the three-dimensional diagrams of the isomers in the boxes. You may use N N to represent en. Cr Cr Cr Copper forms complexes with NH3 and en according to equlibria 1 and 2. equilibrium 1 Cu2++ 4NH3[Cu(NH3)4]2+equilibrium 2 Cu2++ 2en[Cu2]2+Write the expressions for the stability constants, Kstab1 and Kstab2, for equilibria 1 and 2. Include units in your answers. Kstab1 = units = Kstab2 = units = An equilibrium is set up when both en and NH3 ligands are added to a solution containing Cu2+as shown in equilibrium3. equilibrium 3 [Cu(NH3)4]2++ 2en[Cu2]2++ 4NH3Write an expression for the equilibrium constant, Keq3, in terms of Kstab1 and Kstab2. Keq3 = The numerical values for these stability constants are shown. Kstab1 = 1.2 × 1013 Kstab2 = 5.3 × 1019 Calculate the value of Keq3 stating its units. Keq3 = unit = ΔS o values for equilibria 1 and 2 differ greatly, as can be seen in the table. All values are at a temperature of 298 K. equilibrium ΔH o / kJ mol–1 ΔS o / J K–1 mol–1 ΔG o / kJ mol–1 –92 –60 –74 –100 +40 Explain why is so different from . Calculate at 298 K. = kJ mol–1 What conclusion can be made about the relative feasibility of equilibria 1 and 2? Explain your answer. Using data from the table, suggest a value of ΔH o for equilibrium 3. State the type of reaction that is occurring in equilibrium 2.
7. Equilibria  →  7.1. Chemical equilibria: reversible reactions, dynamic equilibrium
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The table lists some organic acids and their pKa values. acid formula pKa ethanoic acid CH3CO2H 4.76 chloroethanoic acid Cl CH2CO2H 2.86 aminoethanoic acid H2NCH2CO2H 9.87 State the relationship between pKa and the strength of an acid. State the mathematical relationship between pKa and the acidity constant Ka. Give reasons for why the pKa value for chloroethanoic acid is smaller than that for ethanoic acid. Use the zwitterionic structure for aminoethanoic acid in aqueous solution to write an equation for its dissociation giving H+ions. Calculate the pH of a 0.100 mol dm–3 solution of aminoethanoic acid. pH = A 10.0 cm3 sample of 0.100 mol dm–3 aminoethanoic acid was titrated with 0.100 mol dm–3 NaOH. After 20.0 cm3 of NaOH, an excess, had been added, the pH was found to be 12.5. Using the following axes, sketch a graph showing how the pH changes during this titration. volume of NaOH added / cm3 pH
7. Equilibria  →  7.2. Brønsted–Lowry theory of acids and bases
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Compounds W, X, Y and Z are isomers of each other with the molecular formula C8H7Cl O. All four isomers contain a benzene ring. Only one of the isomers contains a chiral centre. The results of six tests carried out on W, X, Y and Z are shown in the table. test observations with each isomer W X Y Z add cold AgNO3white ppt. forms immediately none white ppt. forms very slowly none heat with NaOH, then add dilute HNO3 + AgNO3white ppt. none white ppt. none add NaOH+ I2none pale yellow ppt. none none warm with Fehling’s solution none none red ppt. none add cold, dilute, acidified KMnO4no change no change no change decolourises add Br2no change no change no change decolourises and forms white ppt. Use the experimental results in the table above to determine the group, in addition to the benzene ring, present in the four isomers W, X, Y and Z. Complete the table below, identifying the grouppresent in each isomer. groupin compound W X Y Z Isomers W, X, Y and Z all have the molecular formula C8H7Cl O. Use the information in to suggest a structure for each of these isomers and draw these in the boxes. W X Y Z Draw a circle around the chiral centre in one of the above structures.
13. An introduction to AS Level organic chemistry  →  13.1. Formulas, functional groups and the naming of organic compounds
13. An introduction to AS Level organic chemistry  →  13.4. Isomerism: structural isomerism and stereoisomerism
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The amino acid tyrosine can be synthesised from phenol by the route shown. step 1 step 2 step 5 step 4 1. PCl 5 2. H2O step 3 HO CO2H NH2 HO CO2H Cl HO CO2H OH HO HO HO CHO CN OH tyrosine phenol Name the mechanism occurring in the following steps. step 1 step 2 What type of reaction is occurring in step 3? Suggest reagents and conditions for each of the following steps. step 1 step 2 step 3 step 5 Draw the structures of the products of the reactions of tyrosine with an excess of each of the following reagents. with NaOHwith HCl with Br2 Question 8 continues on the next page. The dipeptide phe-tyr has the following structure. CO2H H N O OH H2N A mixture of this dipeptide (phe-tyr) and its two constituent amino acids (phe and tyr) was subjected to electrophoresis in a buffer at pH 12. At the end of the experiment the following results were seen. Spots R and S remained very close together. + – mixture applied here P R S The three spots are due to the three species phe, tyr and phe-tyr. Which species is responsible for spot P? Explain your answer. Suggest why the other two species give spots R and S that are so close together.
21. Organic synthesis  →  21.1. Organic synthesis
14. Hydrocarbons  →  14.2. Alkenes
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