9701_w24_qp_42
A paper of Chemistry, 9701
Questions:
8
Year:
2024
Paper:
4
Variant:
2

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The equation for reaction 1 is shown. reaction 1 X 2Y Reaction 1 is first order with respect to the concentration of X. The half-life of the reaction, t1 2 , is 900 s at 20 °C. A solution of X with a concentration of 0.180 mol dm–3 is prepared at 20 °C. Calculate the average rate of reaction 1 over the first 1800 s. average rate of reaction 1 = Complete the rate equation for reaction 1. rate = Show that the rate constant, k, is 7.70 × 10–4 s–1 at 20 °C. Calculate the initial rate of reaction 1 when the concentration of X is 0.150 mol dm–3. Include units. rate = units Catalysts may be homogeneous or heterogeneous. Platinum is a transition element. Explain why transition elements behave as catalysts. Name the metal catalyst in the Haber process and explain why it is a heterogeneous catalyst. metal Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, NO2, from the exhaust gases of car engines. Describe the mode of action of a platinum catalyst in this process. NO2 acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, SO2. Write equations for the two reactions that occur. equation 1 equation 2 SO2 dissolves in water, forming H2SO3. H2SO3 can be oxidised under acidic conditions. The relevant electrode reaction and its E o– value are shown. SO4 2– + 4H+ + 2e– H2SO3 + H2O E o– = +0.17 V Four more half-equations for reactions occurring under acidic conditions, and their E o– values, are shown. H3BO3 + 3H+ + 3e– B + 3H2O E o– = –0.73 V BiO+ + 2H+ + 3e– Bi + H2O E o– = +0.28 V S + 2H+ + 2e– H2S E o– = +0.14 V Sb + 3H+ + 3e– SbH3 E o– = –0.51 V Select the oxidising agent that could oxidise H2SO3 to SO4 2– ions under acidic conditions. Write an equation, and give the E o– cell value, for the reaction that occurs. oxidising agent equation E o– cell = V
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Predict and explain the variation in enthalpy change of hydration for the ions Na+, Mg2+ and Al 3+. shows an incomplete energy cycle. change 1: lattice energy of magnesium chloride, ΔHlatt MgCl 2change 3: enthalpy change of solution of magnesium chloride, ΔHsol MgCl 2change 2: line A: MgCl 2line C: line B: Mg2++ 2Cl –Complete line C on . Include state symbols. Use both words and symbols to identify change 2 on . Use changes 1 and 3 as examples of how this should be done. Calculate a value for the lattice energy of magnesium chloride, ΔHlatt MgCl 2, by selecting and using appropriate data from Table 2.1. Table 2.1 energy change value / kJ mol–1 enthalpy change of solution of magnesium chloride –155 enthalpy change of formation of magnesium chloride –642 first ionisation energy of magnesium +736 second ionisation energy of magnesium +1450 electron affinity of chlorine –349 enthalpy change of hydration of Mg2+ –1920 enthalpy change of hydration of Cl – –364 ΔHlatt MgCl 2= kJ mol–1 Define entropy. At 25 °C the enthalpy change of solution of compound Z is +26 kJ mol–1. The entropy change of solution of Z at the same temperature is +52 J K–1 mol–1. Calculate the value of the Gibbs free energy change, ΔG, for the solution of Z at 25 °C. ΔG = kJ mol–1 Use your answer to to predict whether or not Z is soluble in water at 25 °C. Explain your answer. Predict whether Z becomes more or less soluble as the water is heated from 25 °C to 95 °C. Explain your answer.
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The pH of a saturated solution of calcium hydroxide is 12.35 at 298 K. Show that the concentration of hydroxide ions in a saturated solution of calcium hydroxide is 0.0224 mol dm–3 at 298 K. Use data given in to calculate the solubility product, Ksp, of calcium hydroxide at 298 K. Include the units of Ksp in your answer. Ksp = units A spatula measure of solid calcium chloride is stirred into a sample of saturated calcium hydroxide solution. All of the calcium chloride dissolves. Describe one other observation that would be made and give an estimated value of the pH of the solution obtained. Explain both your answers. observation pH of solution explanation Calcium hydroxide reacts with dilute sulfuric acid to form calcium sulfate. Barium hydroxide behaves in a similar way, forming barium sulfate. Explain why calcium sulfate is more soluble in water than barium sulfate. Some solid calcium is added to an excess of aqueous ethanoic acid, CH3COOH, and left until all the calcium has reacted. The resulting mixture, mixture D, contains no undissolved solids. Write an equation for the reaction of calcium with CH3COOH. Use formulae of molecules and ions to identify two conjugate acid–base pairs present in mixture D. Pair 1 should consist of organic species. pair 1: conjugate acid conjugate base pair 2: conjugate acid conjugate base Write the expression for the Ka of CH3COOH. Ka = The concentration of calcium ethanoate, (CH3COO)2Ca, in mixture D is 0.394 mol dm–3. The concentration of CH3COOH in mixture D is 0.270 mol dm–3. The Ka of CH3COOH is 1.74 × 10–5 mol dm–3 at 298 K. Calculate the pH of mixture D. pH = Write two equations to show how mixture D can act as a buffer solution. equation 1 equation 2
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Transition metal atoms and transition metal ions form complexes by combining with species called ligands. When NaOHis added to an aqueous solution containing [Co(H2O)6]2+ a precipitation reaction occurs accompanied by a colour change. In this reaction two of the water ligands each lose one H+ ion. The H+ ions are gained by OH– ions from the NaOH. State the colour change seen in this precipitation reaction. from to Complete the ionic equation for this precipitation reaction. [Co(H2O)6]2+ + + This precipitation reaction can also be described as a different type of reaction. Name this type of reaction. L is an uncharged tridentate ligand. L donates three lone pairs to a metal atom or ion. Cobalt forms an octahedral complex ion, E, with L. Complex ion E has a 2+ charge. Give the formula of E. Identify the oxidation state of cobalt in E. The d-orbitals of the cobalt atom or ion present in E are split in energy. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level. number of d-orbitals at a higher energy level number of d-orbitals at a lower energy level Define the term non-degenerate d-orbitals. The mineral chromite contains a compound which has the formula FeCrnO4. The oxidation state of iron in FeCrnO4 is +2. A sample of 4.18 g of FeCrnO4 is dissolved in an excess of sulfuric acid. The resulting solution is made up to 250 cm3. This is solution F. All the Fe2+ ions in 25.0 cm3 of solution F are oxidised to Fe3+ ions by exactly 18.7 cm3 of 0.0200 mol dm–3 KMnO4. One MnO4 – ion reacts with five Fe2+ ions. Assume no other oxidation reaction occurs. Write an equation for the reaction of Fe2+ ions with MnO4 – ions in acid solution. Calculate the number of moles of Fe2+ ions in 25.0 cm3 of solution F. number of moles of Fe2+ ions = Calculate the Mr of FeCrnO4 and use your answer to deduce the value of n. Mr of FeCrnO4 = value of n =
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Benzene reacts with chlorine gas to form chlorobenzene. This reaction can be described as the reaction between benzene molecules and Cl + ions. The Cl + ions are formed by adding a suitable catalyst to the chlorine gas. Give the name or formula of a catalyst that can be used for this reaction. The mechanism for this reaction is shown. Cl + Cl Cl x y step 1 step 2 diagram 1 diagram 2 diagram 3 H + The movement of a pair of electrons is represented by x in diagram 1. • State where this pair of electrons is before step 1 takes place. • State where this pair of electrons is after step 1 has taken place. The movement of another pair of electrons is represented by y in diagram 2. • State where this pair of electrons is before step 2 takes place. • State where this pair of electrons is after step 2 has taken place. There are six carbon atoms in diagram 2. State how many of these carbon atoms are sp hybridised, sp2 hybridised, and sp3 hybridised. sp hybridised sp2 hybridised sp3 hybridised Complete the equation for this reaction between benzene and chlorine. C6H6 + + The mechanism for this reaction is electrophilic substitution. Complete the following sentence. Write formulae in the gaps provided. During this reaction, the electrophile is and a atom in benzene is substituted by a atom. Chloroethane reacts with NaOH. Chlorobenzene does not. Name the mechanism of the reaction that chloroethane undergoes with NaOH, and identify the major organic product that is formed. mechanism major organic product Explain the difference in reactivity of chloroethane and chlorobenzene when treated with NaOH.
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The amino acid serine, HOCH2CH(NH2)COOH, exists in two optically active forms. These optical isomers, isomer P and isomer Q, are shown in . isomer P isomer Q HOOC COOH H H NH2 H2N CH2OH CH2OH C C Isomer P and isomer Q have identical physical and chemical properties, with the exception of two specific properties. One of these two properties is their differing effect on plane polarised light. State the other property by which they differ. A solution of pure isomer P of a particular concentration rotates plane polarised light by 5.0° in a clockwise direction. Describe how a solution of pure isomer Q of the same concentration affects plane polarised light. State another term, in addition to stereoisomers, optical isomers and non-superimposable mirror images, which can be used to describe this pair of chiral compounds, isomer P and isomer Q. Give the term used to describe a mixture containing equal amounts of isomer P and isomer Q. Describe one way in which a single pure optical isomer of serine can be produced, instead of making a mixture of isomer P and isomer Q. Complete Table 8.1 to describe the peaks seen in the proton (1H) NMR spectrum of HOCH2CH(NH2)COOH dissolved in D2O. Use as many rows in Table 8.1 as you need to, leaving the other rows blank. Table 8.1 group responsible for peak name of splitting pattern shown by peak explanation for splitting pattern Proline is a naturally occurring amino acid. The skeletal formula of proline is shown. O proline OH NH State the number of peaks in the carbon-13 (13C) NMR spectrum of proline. Glutamic acid is a naturally occurring amino acid. The skeletal formula of glutamic acid is shown. HO O NH2 O OH glutamic acid The isoelectric point of glutamic acid is pH 3. A sample of glutamic acid is dissolved in a solution of pH 1. A strong alkali is then added until the pH of the mixture reaches pH 14. During this process all possible ionised forms of glutamic acid are present at different times, depending on the pH of the solution. Complete the boxes below to show four different ionised forms of glutamic acid that are present at the stated pH values. at pH 1 at pH 3 at pH 9 at pH 14