A city bus depot is comparing two catalytic converter materials before fitting them to diesel service vehicles. Carbon monoxide in exhaust gas reacts with o...

Assessment: Chemistry 9202 | Paper 2 Mock 01 | Written Paper 2 Subject: Chemistry - 9202

Question 1 Report

A city bus depot is comparing two catalytic converter materials before fitting them to diesel service vehicles. Carbon monoxide in exhaust gas reacts with oxygen to make carbon dioxide. Fig. 1 compares an uncatalysed route with a route using a platinum catalyst. The products have the same energy in both routes.

Fig. 1
energyreaction progresswithout catalystwith catalystreactantsproducts© EAGLE BEACON GLOBAL

(a) Give the balanced equation for carbon monoxide reacting with oxygen. [2]
(b) Use Fig. 1 to compare the activation energies for the two routes. [2]
(c) Explain why a catalyst increases the rate of this reaction in the converter. [3]
(d) Describe what happens to the catalyst after a reaction between carbon monoxide and oxygen has occurred. [2]
(e) Suggest two reasons why a converter is most useful when the bus engine has become hot. [2]
(f) Calculate the mass of carbon dioxide formed when 5.60 g of carbon monoxide reacts completely. Relative formula masses: CO = 28.0, CO2 = 44.0. [4]

Answer Details

(a) 2CO + O2 → 2CO2. [2]

(b) The catalysed route has a lower activation energy than the uncatalysed route. This is shown by its lower peak on the energy profile. [2]

(c) A catalyst provides an alternative reaction route with lower activation energy. At the same temperature, more collisions have sufficient energy to react, so there are more successful collisions each second and the reaction is faster. [3]

(d) The catalyst is not used up and remains chemically unchanged after the reaction. It can therefore be used again. [2]

(e) At higher temperature, particles have more kinetic energy and collide more often. More particles also have energy greater than the activation energy, making the reaction faster. Any two. [2]

(f) moles of CO = mass ÷ Mr = 5.60 g ÷ 28.0 g mol-1 = 0.200 mol.
The equation gives a 1:1 mole ratio of CO to CO2, so moles of CO2 = 0.200 mol.
mass of CO2 = 0.200 mol × 44.0 g mol-1 = 8.80 g. [4]

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