8.1. Rate of reaction
A subsection of Chemistry, 9701, through 8. Reaction kinetics
Listing 10 of 86 questions
Propanone, CH3COCH3 , reacts with iodine, I2 , in the presence of an acid catalyst. CH3COCH3 + I2 CH3COCH2I + H+ + I– The rate equation for this reaction is shown. rate = k[H+] Complete Table 1.1 to describe the order of the reaction. Table 1.1 order of the reaction with respect to order of the reaction with respect to order of the reaction with respect to [H+] overall order of the reaction An experiment is performed using a large excess of CH3COCH3 and a large excess of H+. The initial concentration of I2 is 1.00 × 10–5 mol dm–3. The initial rate of decrease in the I2 concentration is 2.27 × 10–7 mol dm–3 s–1. Use the axes to draw a graph of against time for the first 10 seconds of the reaction. 1.00 × 10–5 0.90 × 10–5 0.80 × 10–5 0.70 × 10–5 0.60 × 10–5 0.50 × 10–5 0.40 × 10–5 0.30 × 10–5 0.20 × 10–5 0.10 × 10–5 / mol dm–3 time / s State whether it is possible to calculate the numerical value of the rate constant, k, for this reaction from your graph. Explain your answer. The experiment is repeated at a different temperature. The initial concentrations of H+ ions, I2 and CH3COCH3 are all 0.200 mol dm–3. The value of k at this temperature is 2.31 × 10–5 mol–1 dm3 s–1. Calculate the initial rate of this reaction. rate = mol dm–3 s–1 The experiment is repeated using an excess of H+. The new rate equation is shown. rate = k1 The value of k1 is 1.1 × 10–3 s–1. Calculate the value of the half-life, t1 2 . t1 2 = s Use your answer to to draw a graph of against time for this reaction. The initial value of on your graph should be 0.200 mol dm–3. The final value of on your graph should be 0.0250 mol dm–3. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200X / mol dm–3 time / s A four-step mechanism is suggested for the overall reaction. CH3COCH3 + I2 CH3COCH2I + H+ + I– rate = k [H+ ] Part of this mechanism is shown. step 1: CH3COCH3 + H+ CH3C+(OH)CH3 step 2: CH3C+(OH)CH3 CH3C(OH)=CH2 + H+ step 3: step 4: CH3C+(OH)CH2I CH3COCH2I + H+ Write an equation for step 3. Suggest the slowest step of the mechanism. Explain your answer. Identify one conjugate acid-conjugate base pair in the mechanism. conjugate acid conjugate base
9701_w23_qp_42
THEORY
2023
Paper 4, Variant 2
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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_43
THEORY
2023
Paper 4, Variant 3
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A and B react together to give product AB. A + B AB When the concentrations of A and B are both 0.0100 mol dm–3, the rate of formation of AB is 7.62 × 10–4 mol dm–3 s–1. When the concentrations of A and B are both 0.0200 mol dm–3, the rate of formation of AB is 3.05 × 10–3 mol dm–3 s–1. Complete the three possible rate equations that are consistent with these data. rate = rate = rate = Choose one of the rate equations you have written in , and calculate the value of the rate constant, k. Include the units of k. k = units Explain why it is not possible to calculate a value for the half-life, t1 , of this reaction using the value of the rate constant k calculated in and the equation k = 0.693 / t1 . Catalysts may be homogeneous or heterogeneous. Identify two metals that act as heterogeneous catalysts in the removal of NO2 from the exhaust gases of car engines. and Iron acts as a heterogeneous catalyst in the Haber process. Describe the mode of action of this iron catalyst. Fe2+ ions act as a homogeneous catalyst in the reaction between I–and S2O8 2–. Write equations for the two reactions that occur when Fe2+is added to a mixture of I–and S2O8 2–. equation 1 S2O8 2– + equation 2 Explain the difference between a homogeneous catalyst and a heterogeneous catalyst. Fe2+ ions can be oxidised to Fe3+ ions under alkaline conditions by suitable oxidising agents. Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states. The half-equation for the reduction of Fe3+ under alkaline conditions, and its E o value, are shown. Fe(OH)3 + e– Fe(OH)2 + OH– E o = – 0.56 V Four more half-equations for reactions under alkaline conditions, and their E o values, are shown. Al (OH)4 – + 3e– Al + 4OH– E o = –2.35 V Cl O– + H2O + 2e– Cl – + 2OH– E o = +0.89 V O2 + 2H2O + 4e– 4OH– E o = +0.40 V Zn(OH)4 2– + 2e– Zn + 4OH– E o = –1.22 V Select two oxidising agents that can oxidise Fe2+ ions to Fe3+ ions under alkaline conditions. Write an equation, and give the E cell value, for each of the two reactions that occur. oxidising agent 1: equation: E cell = V oxidising agent 2: equation: E cell = V o o o
9701_w24_qp_41
THEORY
2024
Paper 4, Variant 1
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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
9701_w24_qp_42
THEORY
2024
Paper 4, Variant 2
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A and B react together to give product AB. A + B AB When the concentrations of A and B are both 0.0100 mol dm–3, the rate of formation of AB is 7.62 × 10–4 mol dm–3 s–1. When the concentrations of A and B are both 0.0200 mol dm–3, the rate of formation of AB is 3.05 × 10–3 mol dm–3 s–1. Complete the three possible rate equations that are consistent with these data. rate = rate = rate = Choose one of the rate equations you have written in , and calculate the value of the rate constant, k. Include the units of k. k = units Explain why it is not possible to calculate a value for the half-life, t1 , of this reaction using the value of the rate constant k calculated in and the equation k = 0.693 / t1 . Catalysts may be homogeneous or heterogeneous. Identify two metals that act as heterogeneous catalysts in the removal of NO2 from the exhaust gases of car engines. and Iron acts as a heterogeneous catalyst in the Haber process. Describe the mode of action of this iron catalyst. Fe2+ ions act as a homogeneous catalyst in the reaction between I–and S2O8 2–. Write equations for the two reactions that occur when Fe2+is added to a mixture of I–and S2O8 2–. equation 1 S2O8 2– + equation 2 Explain the difference between a homogeneous catalyst and a heterogeneous catalyst. Fe2+ ions can be oxidised to Fe3+ ions under alkaline conditions by suitable oxidising agents. Iron is a transition element. Explain why iron forms stable compounds in both the +2 and the +3 oxidation states. The half-equation for the reduction of Fe3+ under alkaline conditions, and its E o value, are shown. Fe(OH)3 + e– Fe(OH)2 + OH– E o = – 0.56 V Four more half-equations for reactions under alkaline conditions, and their E o values, are shown. Al (OH)4 – + 3e– Al + 4OH– E o = –2.35 V Cl O– + H2O + 2e– Cl – + 2OH– E o = +0.89 V O2 + 2H2O + 4e– 4OH– E o = +0.40 V Zn(OH)4 2– + 2e– Zn + 4OH– E o = –1.22 V Select two oxidising agents that can oxidise Fe2+ ions to Fe3+ ions under alkaline conditions. Write an equation, and give the E cell value, for each of the two reactions that occur. oxidising agent 1: equation: E cell = V oxidising agent 2: equation: E cell = V o o o
9701_w24_qp_43
THEORY
2024
Paper 4, Variant 3
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Chlorine dioxide undergoes the following reaction in aqueous solution. 2Cl O2 + 2OH– Cl O2 – + Cl O3 – + H2O The initial rate of the reaction was measured at different initial concentrations of Cl O2 and OH–. The table shows the results obtained. experiment [Cl O2] / mol dm–3 [OH–] / mol dm–3 initial rate / mol dm–3 s–1 1.25 × 10–2 1.30 × 10–3 2.33 × 10–4 2.50 × 10–2 1.30 × 10–3 9.34 × 10–4 2.50 × 10–2 2.60 × 10–3 1.87 × 10–3 Use the data in the table to determine the rate equation, showing the order with respect to each reactant. Show your reasoning. rate equation = Calculate the value of the rate constant, k, using the data from experiment 2. State its units. k = units Explain the difference between heterogeneous and homogeneous catalysts. Complete the table using ticks () to indicate whether the catalyst used in the reaction is heterogeneous or homogeneous. catalysed reaction heterogeneous homogeneous manufacture of ammonia in the Haber process removal of nitrogen oxides from car exhausts oxidation of sulfur dioxide in the atmosphere Some reactions are catalysed by one of the products of the reaction. This is called autocatalysis. An example of autocatalysis is the reaction between acidifiedmanganate(ions, MnO4 –, and ethanedioicacid, (CO2H)2. Mn2+ ions catalyse this reaction. The reaction is slow in the absence of a catalyst. Balance the equation for this reaction. MnO4 – + H+ + (CO2H)2 Mn2+ + CO2 + H2O The graph shown is a concentration-time graph for a typical reaction. concentration of reactant time On the axes below, sketch the curve you would expect for the autocatalysed reaction in . concentration of MnO4 – time Describe, with the aid of a reaction pathway diagram, the effect of a catalyst on a reversible reaction. Suggest why catalysts are used in industrial processes. The reaction for the Haber process to produce ammonia is shown. N2+ 3H22NH3ΔH o = –92 kJ mol–1 At 500 °C, when pressure is measured in atmospheres, the numerical value of Kp for this equilibrium is 1.45 × 10–5. ● Write the expression for Kp for this equilibrium. Kp = ● Calculate the partial pressure of NH3 at equilibrium at 500 °C, when the partial pressure of N2 is 20atm and that of H2 is 60atm. pNH3 = atm
9701_m17_qp_42
THEORY
2017
Paper 4, Variant 2
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