6.2. Rate of reaction
A subsection of Chemistry, 0620, through 6. Chemical reactions
Listing 10 of 319 questions
Hydrogenperoxide, H2O2, breaks down slowly at 40 °C to produce oxygen gas and water. 2H2O2 → 2H2O + O2 A student investigates the breakdown of hydrogenperoxide at 40 °C in the presence of a catalyst. Fig.6.1 shows the volume of oxygen gas released as the reaction proceeds. time / s volume of oxygen gas / cm3 Deduce the volume of oxygen gas released after 15seconds.  volume of oxygen = cm3 The student repeats the experiment at 20 °C. All other conditions stay the same. Draw a line on the grid in Fig.6.1 to show how the volume of oxygen changes when a temperature of 20 °C is used. The student repeats the experiment without a catalyst. All other conditions stay the same. Describe how the rate of reaction differs when no catalyst is used. The student repeats the experiment using a lower concentration of hydrogenperoxide. All other conditions stay the same. Describe how the rate of reaction differs when a lower concentration of hydrogenperoxide is used. Hydrogenperoxide can act as a reducing agent in the presence of an alkali. State the meaning of the term alkali. Give the formula of the ion that is present in all alkaline solutions. State the colour of methyl orange in an alkaline solution. Aqueous ammonia is an alkali. Complete the word equation for the reaction of aqueous ammonia with hydrochloricacid. ammonia + hydrochloric acid →  
0620_w23_qp_33
THEORY
2023
Paper 3, Variant 3
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Aqueous hydrogen peroxide, H2O2, slowly forms water and oxygen at room temperature and pressure, r.t.p. This reaction is catalysed by manganese(oxide. The equation is shown. 2H2O2→ 2H2O+ O2State the test for oxygen gas. test observations  A student investigates the rate of formation of oxygen gas when manganese(oxide is added to aqueous hydrogenperoxide. The volume of oxygen gas formed is measured at regular time intervals at r.t.p. The results are plotted onto the graph in Fig.4.1. volume of oxygen gas time t2 t1 State how the graph in Fig.4.1 shows the rate of reaction at time t2, is lower than at time t1. Explain, using collision theory, why the rate of reaction at time t2 is lower than at time t1. On Fig.4.1, sketch the graph obtained when the experiment is repeated using aqueous hydrogenperoxide at a higher temperature. All other conditions remain the same. Manganese(oxide is added to 20 cm3 of aqueous hydrogenperoxide. The total volume of oxygen gas produced is 72 cm3 at r.t.p. 2H2O2→ 2H2O+ O2Calculate the concentration of the aqueous hydrogen peroxide in g / dm3 using the following steps. ● Calculate the number of moles of oxygen gas produced.  mol ● Determine the number of moles of hydrogen peroxide which reacts.  mol ● Calculate the concentration of aqueous hydrogenperoxide in mol / dm3.  mol / dm3 ● Calculate the concentration of aqueous hydrogen peroxide in g / dm3.  g / dm3  Suggest the identity of one other metal oxide which also catalyses this reaction. 
0620_w23_qp_41
THEORY
2023
Paper 4, Variant 1
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This question is about rate of reaction and equilibrium. A student investigates the rate of decomposition of aqueous hydrogen peroxide, H2O2, using manganese(oxide as a catalyst. The equation for the reaction is shown. 2H2O2→ 2H2O+ O2The student uses the apparatus shown in . loosely fitting cotton wool aqueous hydrogen peroxide balance manganese(oxide catalyst container The student: ● adds the catalyst to the aqueous hydrogen peroxide ● replaces the container on the balance ● starts a stop-watch ● records the mass at regular time intervals. Table 5.1 shows the mass recorded at regular time intervals. Table 5.1 time / s mass / g 50.64 49.80 49.38 49.17 49.07 49.02 48.99 48.97 48.97 48.97 Suggest why the mass decreases as time increases. After a certain time the reaction stops. Explain why the reaction stops. Suggest why it is not possible to use the results in Table 5.1 to determine the exact time when the reaction stops. shows a graph of the mass against time. mass / g time / s The experiment is repeated at a higher temperature. All other conditions remain the same. Explain, in terms of collision theory, why the rate of reaction is higher at a higher temperature. On , sketch the line expected when the experiment is repeated at a higher temperature. Manganese(oxide is the catalyst in this reaction. Explain the meaning of (in manganese(oxide. State how the mass of the catalyst has changed, if at all, at the end of the experiment. Nitrogen monoxide gas, NO, and oxygen gas, O2, react to produce nitrogen dioxide gas, NO2, at room temperature. The reaction can reach equilibrium. The equation is shown. 2NO+ O22NO2∆H = –113 kJ / mol NOand O2are passed into a beaker as shown in . beaker O2NOExplain why the method shown in will not allow the reaction to reach equilibrium. The apparatus is changed and equilibrium is reached. The temperature of the equilibrium system is then increased and the position of equilibrium shifts to the left. Explain why the position of equilibrium shifts to the left. The pressure of the equilibrium system is then increased. State the direction, if any, in which the position of equilibrium shifts. Explain your answer. direction explanation
0620_w24_qp_43
THEORY
2024
Paper 4, Variant 3
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