A2-edexcel-answers-to-test-urself-1.pdf

1 H o w fa st? – rates A n sw ers

Answers to Topic 1 Test yourself questions 1 a) Concentrated hydrochloric acid reacts much faster with marble chips than dilute hydrochloric acid. b) Any reaction involving gases such as the manufacture of ammonia from nitrogen and hydrogen. c) Magnesium powder reacts much faster with dilute hydrochloric acid than magnesium ribbon. d) Catalytic converters are only effective in speeding up the reactions which remove pollutants from car exhausts once they are hot. e) A platinum–rhodium alloy catalyses the oxidation of ammonia to nitrogen oxide. 2 a) In a more concentrated solution there are more collisions per second between reactants and so an increased rate of reaction. b) Increasing the pressure forces the molecules closer together and increases the rate of collisions which lead to reaction. c) When a solid reacts with a liquid or gas the reaction takes place on the interface where the reactants meet. The larger the surface, the greater the area open to reaction. d) At a higher temperature the Maxwell– Boltzmann distribution shifts to the right so the proportion of molecules with energy greater than the activation energy increases, and so more collisions which are more energetic can lead to reaction. e) A catalyst provides an alternative reaction pathway with a lower activation energy. With a lower activation energy there are more collisions with enough energy to lead to reaction. 3

(0.55 − 0.42) moldm3 = 0.0087 mol dm−3  15s s−1

4 a) 2N2O5(g) → 4NO2(g) + O2(g) b) 2 mol of NO2 is formed from 1 mol of N2O5; 7.0 × 10−4 mol dm−3 s−1 5 a) Collect the gas in a graduated syringe. b) Remove samples at intervals, stop the reaction by cooling and then titrate against alkali the acid produced by the reaction. c) Measure the conductivity of the solution to follow the increase in the concentration of ions. d) Carry out the reaction in a flask, with a loose plug of cotton wool in its neck, on a balance and record the loss in mass at regular intervals. 6 Rate = k[peroxide] rate k= [peroxide] =

7.4× 10−6 moldm−3 s−1 0.02moldm− 3

= 3.7 × 10−4 s−1 7 Rate = k[ester][OH−] rate k= − [ester][OH ] 0.00069 moldm−3 s−1 0.05moldm− 3 × 0.10moldm− 3 = 0.138 dm3 mol−1 s−1 8 a) The graph is a straight line so the reaction is first order with respect to bromine. b) i) Half-lives are all close to 200 s wherever they are read from the graph. ii) This is consistent with the answer to a). The half-life for a first-order process is independent of the starting concentration. 9 For a first-order reaction the gradient of the rate–concentration graph gives the rate constant. 10 a) Rate = k[RBr][OH−] b) Second order =

1 of 4 © G. Hill and A. Hunt 2009 Edexcel Chemistry for A2

1 H o w fa st? – rates A n sw ers c) Units: dm3 mol−1 s−1 11 a) Graph similar to Figure 1.13. b) 2NH3(g) → N2(g) + 3H2(g) Rate = k 12 a) Rate = k[RBr][OH−] rate b) k = − [RBr][OH ] =

1.36moldm−3 s−1 0.02moldm− 3 × 0.02moldm− 3

= 3400 dm3 mol−1 s−1 13 a) Rate = k[R´Br] rate 40.40 moldm−3 s−1 b) k = = [R´Br] 0.02moldm−3 = 2020 s−1 14 a) 2H2(g) + 2NO(g) → N2(g) + 2H2O(g) b) Rate = k[H2(g)][NO(g)]2

2 of 4 © G. Hill and A. Hunt 2009 Edexcel Chemistry for A2

1 H o w fa st? – rates A n sw ers 15 Reaction between molecules involve the breaking of covalent bonds. The range of activation energies roughly corresponds to the range of values for covalent bond energies. 16 a) These are the units for a first-order reaction. b) As the value of k gets larger, the rate gets faster for a given concentration of the reactant. So the reaction speeds up as the temperature rises. c) The 10 degree rise from 298 K to 308 K brings about a 5-fold increase in rate. The 20 degree rise from 278  K to 298 K brings about just over a 25-fold increase in rate. 17 a) Ea/RT becomes smaller in magnitude as T rises. Because of the negative sign in the equation, this means that ln k becomes more positive. So k gets larger as T rises and the rate is faster. b) The larger the activation energy the larger the value of Ea/RT and so the smaller the magnitude of ln k. Hence k gets smaller and the rate less. Ea 18 ln (4.93 × 10−4) = constant − 8.314× 295 Ea ln (1.40 × 10−3) = constant − 8.314× 305 Subtract and solve for Ea Ea = 78 kJ mol−1 19 a) Heterogeneous b) homogeneous c) heterogeneous d) heterogeneous 20 In the presence of a catalyst, the reaction pathway has an activation energy which is much lower than when there is no catalyst. Tungsten metal adsorbs hydrogen into the upper layers of the crystal structure as single atoms. So the catalyst breaks the bonds between the atoms in one of the reactants. The pathway with a lower activation energy allows the reaction to proceed much faster.

21 There are several steps to picking up a meal from a canteen counter. The rate at which the queue moves can be slowed down if there is a ratedetermining step such as waiting for a toaster to make pieces of toast or for a coffee machine to deliver cups of coffee. Similarly the flow of traffic along a motorway slows down overall if there are lane closures and the traffic has to travel at 50 mph along a coned-off section of the road. 22 In the first step a strong covalent bond has to break. In the second step two oppositely charged ions, which attract each other, come together to form a bond. 23 a) NO2(g) + CO(g) → NO(g) + CO2(g) b) Rate = k[NO2(g)]2 c) Zero order 24 The intermediate has a double bond as in an alkene, and an –OH group as in an alcohol. Hence it is called an enol. 25 a) Rate = k[CH3COCH3][H+] b) Hydrogen ions act as the catalyst for the reaction. They are not used up in the reaction and so do not appear in the balanced equation. However, hydrogen ions are involved in the rate-determining step so that the concentration of hydrogen ions affects the rate of reaction. As a result, the concentration of hydrogen ions appears in the rate equation. 26 Hydrogen ions are used to form the enol intermediate in steps 1 and 2 but then as many hydrogen ions are released at the end of step 2. 27 In both reactions the rate is determined by the rate of formation of the enol, which depends on the concentrations of propanone and hydrogen ions. Bromine or iodine then react very quickly with the enol intermediate as soon as it is formed.

3 of 4 © G. Hill and A. Hunt 2009 Edexcel Chemistry for A2

1 H o w fa st? – rates A n sw ers

4 of 4 © G. Hill and A. Hunt 2009 Edexcel Chemistry for A2