8. Reaction kinetics
A section of Chemistry, 9701
Listing 10 of 220 questions
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
9701_s17_qp_41
THEORY
2017
Paper 4, Variant 1
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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
9701_s17_qp_43
THEORY
2017
Paper 4, Variant 3
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Nitrogenmonoxide, NO, reacts with hydrogen, H2, under certain conditions. 2NO+ 2H2N2+ 2H2ODefine the term rate of reaction. Identify a change in the reaction mixture that would enable the rate of this reaction to be studied. The rate equation for this reaction is given. rate = k 2 The result of an experiment in which NO reacted with H2 is shown in the table. initial / mol dm–3 initial / mol dm–3 initial rate of reaction / mol dm–3 s–1 2.50 × 10–3 2.50 × 10–3 1.27 × 10–3 Use the data and the rate equation to calculate a value for the rate constant k. Give the units of k. k = units = A second experiment is performed at the same temperature. The initial concentration of H2is 4.60×10–3 mol dm–3. The initial rate of the reaction is 2.31×10–3 mol dm–3 s–1. Calculate the initial concentration of NO. initial concentration of NO= mol dm–3 State the order of the reaction with respect to NOand with respect to H2, and the overall order of the reaction. overall order The reaction is believed to proceed in three steps. 2NO N2O2 N2O2 + H2 N2O + H2O N2O + H2 N2 + H2O Deduce which of the three steps is the rate-determining step. Explain your answer to . A third experiment is performed under different conditions. A small amount of H2of concentration 0.0200 mol dm–3 is mixed with a large excess of NO. The concentration of H2is found to have a constant half-life of 2.00seconds under the conditions used. Define the term half-life. Use the axes below to construct a graph of the variation in the concentration of H2during the first 6seconds under the conditions used. 0.02 0.01 time / s / mol dm–3 NOacts as a catalyst in the oxidation of atmospheric sulfurdioxide. Give two equations to describe how NOacts as a catalyst in this process. equation 1 equation 2 Explain why NOcan be described as a catalyst in this reaction. Describe, with the aid of an equation, an environmental consequence of the oxidation of atmospheric sulfurdioxide.
9701_s18_qp_41
THEORY
2018
Paper 4, Variant 1
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Iodine monochloride, ICl, is a yellow-brown gas. It reacts with hydrogen gas under certain conditions as shown. 2ICl + H22HCl + I2Experiments are performed using different starting concentrations of ICl and H2. The initial rate of each reaction is measured. The following results are obtained. experiment / mol dm–3 / mol dm–3 relative rate of reaction 4.00 × 10–3 4.00 × 10–3 1.00 4.00 × 10–3 7.00 × 10–3 1.75 4.00 × 10–3 1.00 × 10–2 2.50 5.00 × 10–3 8.00 × 10–3 2.50 7.00 × 10–3 8.00 × 10–3 3.50 Identify a change, taking place in the reaction mixture, that would enable measurements of the rate of this reaction to be made. Use the data in the table to show that the reaction is first order with respect to H2. Use the data in the table to show that the reaction is first order with respect to ICl . Complete the rate equation for the reaction between ICl and H2. rate = Use experiment3 to calculate a numerical value for the rate constant, k. k = The reaction 2ICl + H22HCl + I2is first order with respect to ICl and first order with respect to H2. Suggest a mechanism for this reaction. You should assume ● the mechanism has two steps, ● the first step is much slower than the second step. first step second step An alternative method is used to show that the reaction is first order with respect to H2. This method uses a large excess of ICl and measures how the concentration of H2varies with time. Describe two ways of using these results to show the reaction is first order with respect to H2concentration. Explain the reason for using a large excess of ICl . A chemical reaction may be speeded up by the presence of a catalyst. Explain why a catalyst increases the rate of a chemical reaction.
9701_s18_qp_42
THEORY
2018
Paper 4, Variant 2
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Nitrogenmonoxide, NO, reacts with hydrogen, H2, under certain conditions. 2NO+ 2H2N2+ 2H2ODefine the term rate of reaction. Identify a change in the reaction mixture that would enable the rate of this reaction to be studied. The rate equation for this reaction is given. rate = k 2 The result of an experiment in which NO reacted with H2 is shown in the table. initial / mol dm–3 initial / mol dm–3 initial rate of reaction / mol dm–3 s–1 2.50 × 10–3 2.50 × 10–3 1.27 × 10–3 Use the data and the rate equation to calculate a value for the rate constant k. Give the units of k. k = units = A second experiment is performed at the same temperature. The initial concentration of H2is 4.60×10–3 mol dm–3. The initial rate of the reaction is 2.31×10–3 mol dm–3 s–1. Calculate the initial concentration of NO. initial concentration of NO= mol dm–3 State the order of the reaction with respect to NOand with respect to H2, and the overall order of the reaction. overall order The reaction is believed to proceed in three steps. 2NO N2O2 N2O2 + H2 N2O + H2O N2O + H2 N2 + H2O Deduce which of the three steps is the rate-determining step. Explain your answer to . A third experiment is performed under different conditions. A small amount of H2of concentration 0.0200 mol dm–3 is mixed with a large excess of NO. The concentration of H2is found to have a constant half-life of 2.00seconds under the conditions used. Define the term half-life. Use the axes below to construct a graph of the variation in the concentration of H2during the first 6seconds under the conditions used. 0.02 0.01 time / s / mol dm–3 NOacts as a catalyst in the oxidation of atmospheric sulfurdioxide. Give two equations to describe how NOacts as a catalyst in this process. equation 1 equation 2 Explain why NOcan be described as a catalyst in this reaction. Describe, with the aid of an equation, an environmental consequence of the oxidation of atmospheric sulfurdioxide.
9701_s18_qp_43
THEORY
2018
Paper 4, Variant 3
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Chlorate(ions undergo the following reaction under aqueous conditions. 2NH3 + Cl O– N2H4 + Cl – + H2O A series of experiments was carried out at different concentrations of Cl O– and NH3. The table shows the results obtained. experiment [Cl O–] / mol dm–3 / mol dm–3 initial rate / mol dm–3 s–1 0.200 0.100 0.256 0.400 0.200 2.05 0.400 0.400 8.20 Use the data in the table to determine the order with respect to each reactant, Cl O– and NH3. Show your reasoning. Write the rate equation for this reaction. rate = Use the results of experiment1 to calculate the rate constant, k, for this reaction. Include the units of k.  k =  units =  On the axes sketch a graph to show how the value of k changes as temperature is increased. rate constant, k temperature  In another experiment, the reaction between chlorate(ions and iodideions in aqueous alkali was investigated. A solution of iodide ions in aqueous alkali was added to a large excess of chlorate(ions and [I–] was measured at regular intervals. Describe how the results of this experiment can be used to confirm that the reaction is first-order with respect to [I–]. A three-step mechanism for this reaction is shown. step 1 Cl O– + H2O HCl O + OH– step 2 I– + HCl O HIO + Cl – step 3 HIO + OH– H2O + IO– Use this mechanism to deduce the overall equation for this reaction. Identify a step that involves a redox reaction. Explain your answer.  
9701_s19_qp_41
THEORY
2019
Paper 4, Variant 1
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Questions Discovered
220