8.1. Rate of reaction
A subsection of Chemistry, 9701, through 8. Reaction kinetics
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The rate of the reaction H2+ I22HIis studied. A small amount of H2is mixed with a large excess of I2at a temperature of 400 K and the reaction is monitored. The graph obtained is shown. relative concentration of H2time / seconds Suggest why a large excess of I2is used in this experiment. The reaction is first order with respect to H2. Use data from the graph to confirm this statement. Three separate experiments were carried out at 400 K with different starting concentrations of H2and I2. The results are shown in the table. experiment / mol dm–3 / mol dm–3 rate of reaction / mol dm–3 s–1 1.0 × 10–2 1.0 × 10–2 2.0 × 10–17 1.0 × 10–1 1.0 × 10–1 2.0 × 10–15 5.0 × 10–1 5.0 × 10–1 5.0 × 10–14 Use the data, and the order of reaction with respect to H2given in , to deduce the order of reaction with respect to I2. Explain your answer, giving data in support of your explanation. Use information from and your answer to to write the rate equation for the forward reaction. rate = Use your rate equation and data from experiment1 to calculate the value of the rate constant, k, for the forward reaction at 400 K. Include units for k.  k = units =  At 400 K the rate constant for the forward reaction is approximately 1000times greater than the rate constant for the backward reaction. The overall orders of the forward and backward reactions are the same. forward reaction H2+ I2→ 2HIbackward reaction 2HI→ H2+ I2Use this information to explain what will happen if equal concentrations of HI, H2and I2are mixed at 400 K. You should comment on: ● the relative initial rates of the forward and backward reactions ● the position of the equilibrium reached. At 700 K the rate constant for the forward reaction is approximately 50times greater than the rate constant for the backward reaction. Use this information and the information in to deduce the signs of the ∆H values of the forward and backward reactions. Explain your answer. 
9701_w20_qp_42
THEORY
2020
Paper 4, Variant 2
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Explain what is meant by the following terms: homogeneous catalyst heterogeneous catalyst  Iodide ions react with peroxydisulfate ions. 2I– + S2O8 2– → I2 + 2SO4 2– This reaction is slow, but it is catalysed by Fe2+ ions. Write two equations to explain how this reaction is catalysed by Fe2+ ions.  Suggest why the alternative route in the presence of Fe2+ ions has a lower activation energy than the route in the absence of a catalyst. Nitrogenmonoxide reacts with oxygen. 2NO+ O2→ 2NO2This reaction is second order with respect to nitrogenmonoxide and first order with respect to oxygen. Under certain conditions the value of the rate constant, k, is 8.60×106 mol–2dm6 s–1. Construct the rate equation for this reaction. rate =  Calculate the initial rate of the reaction under these conditions when the initial concentration of nitrogen monoxide is 7.20×10–4 mol dm–3 and the initial concentration of oxygen is 1.90×10–3 mol dm–3.  rate of reaction = mol dm–3 s–1 The drug cisplatin, Pt(NH3)2Cl 2, hydrolyses in water. Pt(NH3)2Cl 2 + H2O → [Pt(NH3)2Cl (H2O)]+ + Cl – The rate equation is shown. rate = k [Pt(NH3)2Cl 2] The value of k is 2.50×10–5 s–1 under certain conditions. This reaction has a constant half-life. Explain why this is the case. Use the information in this question to show that the half-life of this reaction is 2.77×104 s.  8.00×10–6 moles of Pt(NH3)2Cl 2 are added to 100 cm3 of water. Calculate the time taken for the concentration of Pt(NH3)2Cl 2 to fall to 2.50×10–6 mol dm–3.  time taken = s 
9701_w22_qp_41
THEORY
2022
Paper 4, Variant 1
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Nitrogenmonoxide, NO, reacts with ozone, O3. NO+ O3→ NO2+ O2This reaction is first order with respect to both NO and O3. At 298 K, the rate constant k = 11 500 mol–1 dm3 s–1. Complete the rate equation for this reaction. rate =  A reaction is carried out in which the initial concentrations of NO and O3 are both 1.20×10–6 mol dm–3. Calculate the initial rate of the reaction. State its units.  rate of reaction = units = The reaction described in is monitored over a period of time. Predict whether or not the graph of against time, under these conditions, shows that the reaction has a constant half-life. Explain your answer. prediction explanation Nitrousoxide, N2O, decomposes into its elements. 2N2O→ 2N2+ O2At a high temperature, a small amount of platinum wire is added to a large amount of nitrousoxide. The reaction follows zero order kinetics. The platinum wire behaves as a catalyst. Sketch a graph, on the axes below, of reaction rate against time for the catalysed decomposition of N2O under these conditions. reaction rate time  Sketch a graph, on the axes below, of against time for this reaction. time  Platinum behaves as a heterogeneous catalyst in this reaction. Describe the mode of action of a heterogeneous catalyst. Suggest a reason why this reaction has zero order kinetics when the amount of nitrousoxide is large and the amount of platinum is small. 
9701_w22_qp_42
THEORY
2022
Paper 4, Variant 2
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Explain what is meant by the following terms: homogeneous catalyst heterogeneous catalyst  Iodide ions react with peroxydisulfate ions. 2I– + S2O8 2– → I2 + 2SO4 2– This reaction is slow, but it is catalysed by Fe2+ ions. Write two equations to explain how this reaction is catalysed by Fe2+ ions.  Suggest why the alternative route in the presence of Fe2+ ions has a lower activation energy than the route in the absence of a catalyst. Nitrogenmonoxide reacts with oxygen. 2NO+ O2→ 2NO2This reaction is second order with respect to nitrogenmonoxide and first order with respect to oxygen. Under certain conditions the value of the rate constant, k, is 8.60×106 mol–2dm6 s–1. Construct the rate equation for this reaction. rate =  Calculate the initial rate of the reaction under these conditions when the initial concentration of nitrogen monoxide is 7.20×10–4 mol dm–3 and the initial concentration of oxygen is 1.90×10–3 mol dm–3.  rate of reaction = mol dm–3 s–1 The drug cisplatin, Pt(NH3)2Cl 2, hydrolyses in water. Pt(NH3)2Cl 2 + H2O → [Pt(NH3)2Cl (H2O)]+ + Cl – The rate equation is shown. rate = k [Pt(NH3)2Cl 2] The value of k is 2.50×10–5 s–1 under certain conditions. This reaction has a constant half-life. Explain why this is the case. Use the information in this question to show that the half-life of this reaction is 2.77×104 s.  8.00×10–6 moles of Pt(NH3)2Cl 2 are added to 100 cm3 of water. Calculate the time taken for the concentration of Pt(NH3)2Cl 2 to fall to 2.50×10–6 mol dm–3.  time taken = s 
9701_w22_qp_43
THEORY
2022
Paper 4, Variant 3
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Fluorine reacts with chlorine dioxide, Cl O2, as shown. F2+ 2Cl O22FCl O2The rate of the reaction is first order with respect to the concentration of F2 and first order with respect to the concentration of Cl O2. No catalyst is involved. Suggest a two-step mechanism for this reaction. step 1 step 2 Identify the rate-determining step in this mechanism. Explain your answer. When the rate of the reaction is measured in mol dm−3 s−1 the numerical value of the rate constant, k, is 1.22 under certain conditions. Complete the rate equation for this reaction, stating the overall order of the reaction. rate = overall order of reaction = Use your rate equation in to calculate the rate of the reaction when the concentrations of F2 and Cl O2 are both 2.00 × 10−3 mol dm−3. rate = mol dm−3 s−1 Under different conditions, and in the presence of a large excess of ClO2, the rate equation is as shown. rate = k1 The half-life, t1 2, of the concentration of F2 is 4.00 s under these conditions. Calculate the numerical value of k1, giving its units. Give your answer to three significant figures. k1 = units An experiment is performed under these conditions in which the starting concentration of F2 is 0.00200 mol dm–3. Draw a graph on the grid in to show how the concentration of F2 changes over the first 12 s of the reaction. 0.0002 0.0004 0.0006 0.0008 0.0010 0.0012 0.0014 0.0016 0.0018 0.0020 / mol dm–3 time / s Use your graph in to find the rate of the reaction when the concentration of F2 is 0.00100 mol dm–3. Show your working on the graph. rate = mol dm–3 s−1
9701_w23_qp_41
THEORY
2023
Paper 4, Variant 1
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