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A-Level · Chemistry

AC9 Chemistry: Rate Equations and Equilibrium

AQA 7405 · Calculator allowed · about 145 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A reaction between three species, P, Q and R, was investigated using the initial rates method. The table shows the results of four experiments carried out at the same constant temperature.

Experiment	[P] (mol dm-3)	[Q] (mol dm-3)	[R] (mol dm-3)	Initial rate (mol dm-3 s-1)
1	0.020	0.010	0.010	4.0 x 10^-4
2	0.040	0.010	0.010	8.0 x 10^-4
3	0.040	0.020	0.010	8.0 x 10^-4
4	0.040	0.020	0.020	1.6 x 10^-3

(a)Using experiments 1 and 2, determine the order of reaction with respect to P. Show your reasoning.(2)

(b)Using experiments 2 and 3, determine the order of reaction with respect to Q. Show your reasoning.(2)

(c)Using experiments 3 and 4, determine the order of reaction with respect to R. Show your reasoning.(2)

(d)Deduce the overall order of reaction and write the rate equation for this reaction.(2)

(e)Using the data from Experiment 1, calculate the rate constant, k, for this reaction, including its units.(3)

(f)Predict the initial rate of reaction when [P] = 0.060 mol dm-3, [Q] = 0.050 mol dm-3 and [R] = 0.030 mol dm-3, at the same temperature.(2)

(Total for Question 1 is 13 marks)

Dinitrogen pentoxide decomposes according to the equation:
2N2O5(g) -> 4NO2(g) + O2(g)
The concentration of N2O5 was monitored during this decomposition at a constant temperature. The results are shown in the table.

Time (min)	0	15	30	45
[N2O5] (mol dm-3)	0.640	0.320	0.160	0.0800

You may use: for a first order reaction, k = ln2 / t(1/2).

(a)Explain, using the data, why this reaction is first order with respect to N2O5.(2)

(b)State the half-life of this reaction.(1)

(c)Calculate the rate constant, k, for this reaction, in units of s-1. Give your answer to 3 significant figures.(3)

(d)Calculate the total time, from t = 0, taken for the concentration of N2O5 to fall to 0.0400 mol dm-3.(2)

(Total for Question 2 is 8 marks)

2-bromo-2-methylpropane, (CH3)3CBr, undergoes hydrolysis with aqueous sodium hydroxide:
(CH3)3CBr + OH- -> (CH3)3COH + Br-
The rate equation for this reaction, found by experiment, is:
rate = k[(CH3)3CBr]

(a)State the overall order of this reaction and the order with respect to hydroxide ions, OH-. Justify both answers using the rate equation given.(3)

(b)Deduce whether the mechanism for this reaction is SN1 or SN2. Explain your reasoning.(2)

(c)Write an equation for the rate-determining step of this mechanism.(2)

(Total for Question 3 is 7 marks)

The overall equation for a reaction is:
2E(g) + F(g) -> 2EF(g)
The experimentally determined rate equation is rate = k[E][F].
Two possible mechanisms have been suggested.
Mechanism 1:
Step 1 (slow): E + F -> EF + G
Step 2 (fast): E + G -> EF
Mechanism 2:
Step 1 (slow): 2E -> E2
Step 2 (fast): E2 + F -> 2EF

(a)Show that Mechanism 1 is consistent with both the overall equation and the experimental rate equation.(2)

(b)Deduce the rate equation that Mechanism 2 predicts, and explain why Mechanism 2 can be ruled out.(3)

(c)State the role of species G (in Mechanism 1) and of species E2 (in Mechanism 2), and explain how you can identify each as playing this role.(2)

(Total for Question 4 is 7 marks)

It is sometimes stated, as a rough rule of thumb, that the rate of a reaction roughly doubles for every 10 degC (10 K) rise in temperature. This question tests whether that rule is generally reliable, for a reaction with an activation energy, Ea, of 50.0 kJ/mol.
You may use: ln(k2/k1) = -(Ea/R)(1/T2 - 1/T1); R = 8.31 J K-1 mol-1.

(a)Calculate the factor by which the rate constant increases (that is, the ratio k(310 K) / k(300 K)) for a temperature rise from 300 K to 310 K, given Ea = 50.0 kJ/mol.(4)

(b)Calculate the equivalent factor, k(510 K) / k(500 K), by which the rate constant increases for the same 10 K rise, but starting from 500 K instead, using the same value of Ea.(3)

(c)Using your answers to (a) and (b), evaluate the statement that the rate of a reaction roughly doubles for every 10 degC rise in temperature.(2)

(Total for Question 5 is 9 marks)

Sulfur dioxide is oxidised to sulfur trioxide in the Contact process:
2SO2(g) + O2(g) ≤> 2SO3(g) deltaH(forward) = -197 kJ/mol
1.00 mol of SO2 and 0.60 mol of O2 were mixed in a sealed 2.0 dm3 container and allowed to reach equilibrium at a constant temperature. At equilibrium, 0.20 mol of O2 remained.

(a)Construct an ICE (initial, change, equilibrium) summary for this reaction, and hence determine the amount, in mol, of SO2 and of SO3 at equilibrium.(3)

(b)Calculate the equilibrium concentration of SO2, O2 and SO3.(2)

(c)Write the expression for Kc for this equilibrium and use your answer to (b) to calculate its value, stating units.(3)

(d)State and explain the effect on the value of Kc of increasing the temperature of this system.(2)

(Total for Question 6 is 10 marks)

Nitrosyl chloride decomposes according to the equation:
2NOCl(g) ≤> 2NO(g) + Cl2(g)
2.00 mol of NOCl(g) was introduced into an evacuated container and allowed to reach equilibrium at a constant temperature; the total pressure at equilibrium was 100 kPa. At equilibrium, 1.00 mol of NOCl remained unreacted.

(a)Calculate the amount, in mol, of NO and of Cl2 present at equilibrium, and hence the total number of moles of gas present.(3)

(b)Calculate the mole fraction of each of the three gases at equilibrium.(2)

(c)Hence calculate the partial pressure of each gas at equilibrium.(2)

(d)Write the expression for Kp for this equilibrium and calculate its value, stating units.(3)

(e)Explain the effect, if any, of increasing the total pressure on this system, at constant temperature, on (i) the position of equilibrium and (ii) the value of Kp.(3)

(Total for Question 7 is 13 marks)

Increasing the temperature of a reversible, exothermic reaction increases the rate at which equilibrium is reached, but decreases the equilibrium yield of product.

(a)Explain, in terms of rate constants and activation energy, why both of these observations occur.(6)

(Total for Question 8 is 6 marks)

Propanone reacts with iodine in the presence of a hydrochloric acid catalyst:
CH3COCH3 + I2 -(H+ catalyst)-> CH3COCH2I + HI
Iodine solution is coloured, but propanone, hydrochloric acid and the organic products are colourless, so the reaction can be monitored using a colorimeter. A calibration curve for this colorimeter gives the relationship: Absorbance = 20.0 x [I2], where [I2] is in mol dm-3.
In one experiment, the absorbance of the reaction mixture was recorded every 60 seconds:

Time (s)	0	60	120	180	240
Absorbance	0.800	0.680	0.560	0.440	0.320

(a)Explain why a colorimeter can be used to monitor the concentration of iodine during this reaction, but cannot be used to monitor the concentration of propanone or hydrochloric acid.(2)

(b)State and explain what the shape of the absorbance-time graph shows about the order of reaction with respect to iodine.(3)

(c)Using the calibration relationship and the gradient of the absorbance-time graph, calculate the rate of this reaction, in mol dm-3 s-1.(3)

(d)Given that this reaction is also known to be first order with respect to propanone and first order with respect to H+ ions, write the overall rate equation for this reaction.(2)

(e)Suggest why it is convenient, when using this method to find the rate, that the reaction is zero order with respect to iodine.(2)

(Total for Question 9 is 12 marks)

In a follow-up investigation of the same acid-catalysed iodination of propanone (CH3COCH3 + I2 -(H+ catalyst)-> CH3COCH2I + HI), the initial rate was found (using the colorimetry method described previously) for four different starting concentrations of hydrochloric acid, with [propanone] kept constant at 0.500 mol dm-3 and [I2] also kept constant throughout.

[HCl] (mol dm-3)	0.50	1.00	1.50	2.00
Initial rate (x10^-5 mol dm-3 s-1)	2.50	5.00	7.50	10.00

(a)Using the data, deduce the order of reaction with respect to H+ (the catalyst). Explain your reasoning.(2)

(b)Given that [propanone] was 0.500 mol dm-3 throughout, and that the reaction is zero order with respect to iodine, write the overall rate equation and use the data to calculate the rate constant, k, stating its units.(4)

(c)State two variables, other than the concentrations already discussed, that must be kept constant across this series of experiments for the comparison to be valid.(2)

(Total for Question 10 is 8 marks)

A reversible reaction can be catalysed to speed up the rate at which it reaches equilibrium.

(a)State how a catalyst affects the rate of a reaction, and explain why, in terms of activation energy.(2)

(b)Explain why the addition of a catalyst does not change the value of Kc for a reversible reaction.(2)

(Total for Question 11 is 4 marks)

X(g) + Y(g) ≤> 2W(g)
At a certain temperature, Kc for this equilibrium = 4.0 (no units). A mixture initially containing 1.00 mol dm-3 of X and 1.00 mol dm-3 of Y, with no W present, is allowed to reach equilibrium at constant volume and temperature.

(a)Write the expression for Kc for this equilibrium.(1)

(b)Using x mol dm-3 as the fall in concentration of X at equilibrium, complete an ICE table for this reaction in terms of x.(2)

(c)Use the value of Kc to calculate x, and hence the equilibrium concentration of X, Y and W.(4)

(d)Verify your answer to part (c) by substituting your equilibrium concentrations back into the Kc expression.(1)

(Total for Question 12 is 8 marks)

The concentration of a reactant, V, was monitored during a reaction at a constant temperature:

Time (min)	0	1.000	3.000	7.000
[V] (mol dm-3)	1.000	0.500	0.250	0.125

You may use, for a second order reaction: 1/[V] = 1/[V]0 + kt

(a)Using the data, calculate the time taken for each of the three successive halvings of [V] shown (that is, from 1.000 to 0.500, from 0.500 to 0.250, and from 0.250 to 0.125 mol dm-3).(2)

(b)Describe the pattern in these successive half-lives.(1)

(c)Deduce the order of reaction with respect to V. Justify your answer using the pattern identified in part (b).(2)

(d)Using two of the data points and the integrated rate equation for a second order reaction, calculate the rate constant, k, for this reaction, stating its units.(3)

(Total for Question 13 is 8 marks)

Mark scheme · AC9 Chemistry: Rate Equations and Equilibrium

Question 1

(a) M1 [P] doubles (0.020 to 0.040) with [Q] and [R] both constant, and the initial rate doubles (4.0x10^-4 to 8.0x10^-4) oe
(a) A1 order with respect to P = 1 cao
(a) Answer: Order with respect to P = 1
(b) M1 [Q] doubles (0.010 to 0.020) with [P] and [R] both constant, but the initial rate is unchanged (8.0x10^-4 in both experiments) oe
(b) A1 order with respect to Q = 0 (zero order) cao
(b) Answer: Order with respect to Q = 0
(c) M1 [R] doubles (0.010 to 0.020) with [P] and [Q] both constant, and the initial rate doubles (8.0x10^-4 to 1.6x10^-3) oe
(c) A1 order with respect to R = 1 cao
(c) Answer: Order with respect to R = 1
(d) M1 overall order = 1 + 0 + 1 = 2, ft from (a) to (c)
(d) A1 rate = k[P][R] (Q does not appear, since it is zero order)
(d) Answer: Overall order = 2; rate = k[P][R]
(e) M1 k = rate / ([P][R]) = 4.0x10^-4 / (0.020 x 0.010), ft from (d)
(e) A1 k = 2.0 cao
(e) A1 units: dm3 mol^-1 s^-1 (mol^-1 dm3 s-1)
(e) Answer: k = 2.0 dm3 mol^-1 s^-1
(f) M1 rate = k[P][R] = 2.0 x 0.060 x 0.030, ft from (e) ([Q] is not used, since the reaction is zero order in Q)
(f) A1 rate = 3.6 x 10^-3 mol dm-3 s-1 cao
(f) Answer: 3.6 x 10^-3 mol dm-3 s-1

Question 2

(a) B1 the time taken for [N2O5] to halve is the same (15 minutes) for each interval shown in the table oe
(a) B1 a constant half-life, independent of concentration, is characteristic of a first order reaction oe
(a) Answer: The half-life is constant at 15 minutes throughout, which is characteristic of first order kinetics.
(b) B1 15 minutes cao
(b) Answer: 15 minutes
(c) M1 t(1/2) = 15 min = 900 s
(c) M1 k = ln2 / 900
(c) A1 k = 7.70 x 10^-4 s-1 awrt
(c) Answer: k = 7.70 x 10^-4 s-1
(d) M1 recognises that 0.0400 mol dm-3 is reached after one further half-life beyond t = 45 minutes (0.0800 halves to 0.0400)
(d) A1 60 minutes cao
(d) Answer: 60 minutes

Question 3

(a) B1 overall order = 1 (first order overall)
(a) B1 order with respect to OH- = 0 (zero order)
(a) B1 because [OH-] does not appear in the rate equation, so changing [OH-] has no effect on the rate oe
(a) Answer: Overall order = 1; order with respect to OH- = 0, because [OH-] is absent from the rate equation.
(b) B1 SN1 (unimolecular nucleophilic substitution)
(b) B1 because the rate depends only on the concentration of the halogenoalkane, independent of [OH-], showing the rate-determining step involves only one species (the halogenoalkane) oe
(b) Answer: SN1, because the rate is independent of [OH-], showing the rate-determining step involves only the halogenoalkane.
(c) M1 (CH3)3CBr -> (CH3)3C+ + Br-
(c) A1 correct species and charges shown, with this identified as the (slow) rate-determining step, involving only the halogenoalkane (OH- is not involved until the second, fast step)
(c) Answer: (CH3)3CBr -> (CH3)3C+ + Br- (rate-determining step; OH- reacts with the carbocation in a second, fast step)

Question 4

(a) M1 adding the two steps of Mechanism 1 together, species G cancels (formed in step 1, used up in step 2), giving the overall equation 2E + F -> 2EF oe
(a) A1 the rate-determining step (step 1) involves one molecule of E and one molecule of F reacting, matching the experimental rate equation, rate = k[E][F] cso
(a) Answer: Adding the steps of Mechanism 1 gives 2E + F -> 2EF (G cancels); step 1 (the RDS) involves one E and one F, matching rate = k[E][F].
(b) M1 the rate-determining step in Mechanism 2 (2E -> E2) involves two molecules of E only, with no F
(b) A1 Mechanism 2 predicts rate = k[E]² (second order in E, zero order in F)
(b) A1 this does not match the experimental rate equation, rate = k[E][F] (which is first order in F), so Mechanism 2 can be ruled out cso
(b) Answer: Mechanism 2 predicts rate = k[E]², which does not match the experimental rate = k[E][F], so it is ruled out.
(c) B1 G is a reaction intermediate in Mechanism 1, since it is formed in step 1 and used up in step 2, so it does not appear in the overall equation
(c) B1 E2 is (similarly) a reaction intermediate in Mechanism 2, formed in step 1 and used up in step 2, so it also does not appear in the overall equation oe
(c) Answer: Both G and E2 are reaction intermediates: each is formed in the first step of its mechanism and consumed in the second, so neither appears in the overall equation.

Question 5

(a) M1 correctly rearranged expression, e.g. ln(k2/k1) = (Ea/R)(1/T1 - 1/T2), with T1 = 300 K and T2 = 310 K
(a) M1 1/300 - 1/310 = 1.075 x 10^-4 K-1 awrt
(a) M1 ln(k2/k1) = (50000/8.31) x 1.075x10^-4 = 0.647 awrt
(a) A1 k2/k1 = e^0.647 = 1.91 awrt cao (the rate constant increases by a factor of about 1.9)
(a) Answer: k(310)/k(300) = 1.91 (awrt)
(b) M1 1/500 - 1/510 = 3.92 x 10^-5 K-1 awrt
(b) M1 ln(k2/k1) = (50000/8.31) x 3.92x10^-5 = 0.236 awrt
(b) A1 k2/k1 = e^0.236 = 1.27 awrt cao (ecf from method)
(b) Answer: k(510)/k(500) = 1.27 (awrt)
(c) B1 the statement is only a rough approximation and is not generally reliable oe
(c) B1 the actual factor depends on the absolute temperature (as well as on Ea); here it is about 1.9x (closer to doubling) for the rise from 300 K to 310 K, but only about 1.3x (well below doubling) for the same size rise from 500 K to 510 K, so the 'doubling' rule does not hold consistently oe
(c) Answer: The rule is only a rough approximation; the true factor (1.9x at 300-310 K but only 1.3x at 500-510 K here) depends on the absolute temperature, so it does not hold consistently.

Question 6

(a) M1 change in moles of O2 = -0.40 mol; change in moles of SO2 = -0.80 mol (using 2:1 stoichiometry with O2)
(a) M1 change in moles of SO3 = +0.80 mol (using 2:1 stoichiometry with O2)
(a) A1 n(SO2) at equilibrium = 0.20 mol and n(SO3) at equilibrium = 0.80 mol cao
(a) Answer: n(SO2) = 0.20 mol; n(SO3) = 0.80 mol at equilibrium
(b) M1 concentration = moles / volume (dividing each amount by 2.0 dm3), ft from (a)
(b) A1 [SO2] = [O2] = 0.10 mol dm-3 and [SO3] = 0.40 mol dm-3 cao
(b) Answer: [SO2] = 0.10 mol dm-3; [O2] = 0.10 mol dm-3; [SO3] = 0.40 mol dm-3
(c) B1 Kc = [SO3]² / ([SO2]² [O2])
(c) M1 Kc = (0.40)² / ((0.10)² x 0.10), ft from (b)
(c) A1 Kc = 160 dm3 mol^-1 awrt cao
(c) Answer: Kc = 160 dm3 mol^-1
(d) B1 Kc decreases
(d) B1 because the forward reaction is exothermic, increasing temperature favours the endothermic (reverse) reaction, shifting the position of equilibrium to the left (reducing [SO3] and increasing [SO2] and [O2]), so the value of Kc decreases oe
(d) Answer: Kc decreases, because increasing temperature favours the endothermic reverse reaction, shifting equilibrium away from SO3.

Question 7

(a) M1 moles of NOCl reacted = 2.00 - 1.00 = 1.00 mol
(a) M1 moles NO formed = 1.00 mol (1:1 with NOCl reacted); moles Cl2 formed = 0.50 mol (1:2 with NOCl reacted)
(a) A1 total moles at equilibrium = 1.00 + 1.00 + 0.50 = 2.50 mol cao
(a) Answer: n(NO) = 1.00 mol; n(Cl2) = 0.50 mol; total = 2.50 mol
(b) M1 mole fraction = moles of component / total moles, ft from (a)
(b) A1 x(NOCl) = 0.400; x(NO) = 0.400; x(Cl2) = 0.200 cao
(b) Answer: x(NOCl) = 0.400; x(NO) = 0.400; x(Cl2) = 0.200
(c) M1 partial pressure = mole fraction x total pressure, ft from (b)
(c) A1 p(NOCl) = 40.0 kPa; p(NO) = 40.0 kPa; p(Cl2) = 20.0 kPa cao
(c) Answer: p(NOCl) = 40.0 kPa; p(NO) = 40.0 kPa; p(Cl2) = 20.0 kPa
(d) B1 Kp = (p(NO))² x p(Cl2) / (p(NOCl))²
(d) M1 Kp = (40.0)² x 20.0 / (40.0)², ft from (c)
(d) A1 Kp = 20.0 kPa cao
(d) Answer: Kp = 20.0 kPa
(e) B1 the position of equilibrium shifts towards NOCl (to the left), the side with fewer moles of gas (2 mol compared with 3 mol) oe
(e) B1 because this opposes the increase in pressure (Le Chatelier's principle) by reducing the total number of gas particles oe
(e) B1 the value of Kp is unaffected (unchanged), because Kp depends only on temperature oe
(e) Answer: Equilibrium shifts left, towards NOCl (fewer gas moles), opposing the pressure increase; Kp itself is unchanged, since Kp depends only on temperature.

Question 8

(a) L3 (5-6): A logical, coherent explanation that links the effect of temperature on the rate constants of both the forward and reverse reactions to the observed change in both rate and equilibrium yield, using the Arrhenius equation and/or the Maxwell-Boltzmann distribution. The answer shows clear understanding of why the reverse (endothermic) rate constant increases proportionally more than the forward (exothermic) rate constant, and links this correctly to the shift in equilibrium position.
(a) L2 (3-4): Some relevant, mostly correct explanation of why rate increases with temperature (e.g. more particles with energy greater than or equal to Ea) is given, with limited or no valid link to why the equilibrium yield decreases.
(a) L1 (1-2): Basic, isolated relevant statement(s), e.g. that rate increases with temperature or that yield decreases, without valid supporting explanation.
(a) L0 (0): No relevant content.
(a) Answer: Levels-of-response essay; see indicative content.

Question 9

(a) B1 iodine solution is coloured and absorbs visible light, so its concentration is related (proportional) to the absorbance measured oe
(a) B1 propanone, hydrochloric acid and the organic products are colourless, so they do not absorb visible light and their concentration cannot be found this way oe
(a) Answer: Iodine is coloured and absorbs light, so its concentration can be linked to absorbance; the other species are colourless and cannot be monitored this way.
(b) M1 the absorbance (and therefore [I2]) decreases by an equal amount, 0.120, in each successive 60 s interval, i.e. the graph is a straight line of constant negative gradient
(b) A1 this shows that the rate of reaction is constant and does not depend on (does not slow down as) [I2] falls
(b) A1 so the reaction is zero order with respect to iodine cao
(b) Answer: The absorbance falls linearly (constant gradient), showing the rate is independent of [I2], i.e. zero order with respect to iodine.
(c) M1 gradient of absorbance-time graph = (0.320 - 0.800) / (240 - 0) = -2.00 x 10^-3 s-1
(c) M1 converts to a concentration gradient by dividing by the calibration constant, 20.0: -2.00x10^-3 / 20.0
(c) A1 rate = 1.00 x 10^-4 mol dm-3 s-1 awrt cao
(c) Answer: Rate = 1.00 x 10^-4 mol dm-3 s-1
(d) B1 rate = k[CH3COCH3][H+]
(d) B1 no [I2] term is included, since the reaction is zero order with respect to iodine oe
(d) Answer: rate = k[CH3COCH3][H+]
(e) B1 because the rate (and therefore the gradient of the absorbance-time graph) is constant throughout the reaction, not just at t = 0 oe
(e) B1 so the rate can be found accurately using any two points on the (straight-line) graph, rather than needing to draw a tangent at t = 0 oe
(e) Answer: Because the rate is constant throughout (the graph is a straight line), the rate can be found from any two points, without needing to draw a tangent at t = 0.

Question 10

(a) M1 the initial rate is directly proportional to [HCl] (e.g. doubling [HCl] from 0.50 to 1.00 mol dm-3 doubles the initial rate from 2.50x10^-5 to 5.00x10^-5 mol dm-3 s-1) oe
(a) A1 order with respect to H+ = 1 (first order) cao
(a) Answer: Order with respect to H+ = 1
(b) B1 rate = k[CH3COCH3][H+] (no [I2] term, since zero order in I2)
(b) M1 k = rate / ([CH3COCH3][H+]) = 2.50x10^-5 / (0.500 x 0.50), using any experiment
(b) A1 k = 1.00 x 10^-4 cao
(b) B1 units: dm3 mol^-1 s^-1 (mol^-1 dm3 s-1), consistent with an overall second order reaction
(b) Answer: rate = k[CH3COCH3][H+]; k = 1.00 x 10^-4 dm3 mol^-1 s^-1
(c) B1 temperature
(c) B1 the total volume of the reaction mixture (so the concentrations calculated remain accurate), or the colorimeter/calibration settings used to convert absorbance to concentration
(c) Answer: Temperature, and the total volume of the reaction mixture (or the colorimeter calibration used).

Question 11

(a) B1 a catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy oe
(a) B1 a greater proportion of particles now have energy greater than or equal to this lower activation energy, so more collisions are successful, increasing the rate constant, k, for both the forward and reverse reactions oe
(a) Answer: A catalyst provides an alternative pathway of lower activation energy, so more particles have sufficient energy to react, increasing k for both the forward and reverse reactions.
(b) B1 a catalyst increases the rate constants of the forward and reverse reactions, kf and kr, by the same factor oe
(b) B1 since Kc = kf/kr, this ratio is unchanged, so the value of Kc is unaffected; equilibrium is reached faster, but its position is unchanged oe
(b) Answer: Because the catalyst increases kf and kr by the same factor, the ratio Kc = kf/kr is unchanged; only the time taken to reach equilibrium is reduced.

Question 12

(a) B1 Kc = [W]² / ([X][Y])
(a) Answer: Kc = [W]² / ([X][Y])
(b) M1 equilibrium concentrations of X and Y in terms of x: [X] = [Y] = (1.00 - x) mol dm-3
(b) A1 equilibrium concentration of W in terms of x: [W] = 2x mol dm-3 (using 1:1:2 stoichiometry)
(b) Answer: [X] = [Y] = (1.00 - x); [W] = 2x, all in mol dm-3
(c) M1 substitutes into the Kc expression, ft from (a) and (b): 4.0 = (2x)² / (1.00 - x)²
(c) M1 takes the square root of both sides (valid since [X] = [Y]): 2.0 = 2x / (1.00 - x), or equivalent full expansion/solution
(c) M1 correct rearrangement and solution: x = 0.50
(c) A1 [X] = [Y] = 0.50 mol dm-3 and [W] = 1.00 mol dm-3 cao
(c) Answer: x = 0.50; [X] = [Y] = 0.50 mol dm-3; [W] = 1.00 mol dm-3
(d) B1 Kc = (1.00)² / (0.50 x 0.50) = 1.00/0.25 = 4.0, which matches the given value, confirming the answer cso
(d) Answer: Kc = (1.00)² / (0.50 x 0.50) = 4.0, matching the given value.

Question 13

(a) M1 correctly identifies each interval: 1.000 min, (3.000 - 1.000) = 2.000 min, (7.000 - 3.000) = 4.000 min
(a) A1 all three values correct: 1.000 min, 2.000 min, 4.000 min cao
(a) Answer: 1.000 min, 2.000 min, 4.000 min
(b) B1 each successive half-life is double the one before it; the half-life is not constant, it increases oe
(b) Answer: Each half-life is double the previous one (the half-life increases, it is not constant).
(c) B1 second order (with respect to V)
(c) B1 because the half-life doubling each time [V] halves (half-life inversely proportional to concentration) is characteristic of second order kinetics oe (a constant half-life would indicate first order; a halving half-life would indicate zero order)
(c) Answer: Second order, because the half-life doubles each time the concentration halves, which is characteristic of second order kinetics.
(d) M1 correct substitution using two data points, e.g. 1/0.500 = 1/1.000 + k(1.000)
(d) M1 correct rearrangement: k = (2.000 - 1.000)/1.000
(d) A1 k = 1.00 dm3 mol^-1 min^-1 cao
(d) Answer: k = 1.00 dm3 mol^-1 min^-1

Chemistry: Rate Equations and Equilibrium - Worksheets, Questions and Revision