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Form 5 Chemistry Practice Paper: Reaction Rate, Chemical Equilibrium and Industrial Processes
Form 5 Chemistry – 12 questions – 100 marks – 75 minutes
Form 5 Chemistry paper: 12 questions, 100 marks, 75 minutes on rates, collision theory, Kc and Kp, Le Chatelier and industrial processes; all answers included.
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Paper details Form 5 Chemistry practice paper with 12 structured questions on reaction rate, chemical equilibrium and industrial processes; full answers and two printable PDFs.
Section A: Reaction Rate (Q1-Q4)
Answer all questions; show the expression, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1.00 g of zinc granules is added to excess 2.0 M hydrochloric acid and the hydrogen collected at room temperature (molar volume 24.0 cubic decimetres per mole). Data: 0 s, 0 cm3; 60 s, 120 cm3; 120 s, 210 cm3; 180 s, 280 cm3; 240 s, 330 cm3; 300 s, 360 cm3. (a) Find the average rate of hydrogen production between 0 s and 120 s in cm3 per second. (b) Find the average rate between 180 s and 300 s and explain the difference. (c) Calculate the maximum volume of hydrogen from 1.00 g of zinc and state whether the reaction is complete at 300 s. |
(a) Rate = (210 – 0) divided by 120 = 1.75 cm3 per second. (b) Rate = (360 – 280) divided by 120 = 0.67 cm3 per second. The later rate is lower because the acid concentration falls as it is used up and the surface area of zinc decreases. (c) n(Zn) = 1.00 divided by 65.4 = 0.0153 mol, so n(H2) = 0.0153 mol. V = 0.3670 dm3 = 367 cm3. Only 360 cm3 is collected, so the reaction is not complete. Marking notes: each average rate 2 marks; explanation 2 marks; moles and volume 2 marks; conclusion 1 mark. |
| 2 | Data for the reaction A + B that produces a single product, at constant temperature: Experiment 1: [A] = 0.10 M, [B] = 0.10 M, initial rate = 2.0e-3 mol dm-3 s-1; Experiment 2: [A] = 0.20 M, [B] = 0.10 M, initial rate = 4.0e-3 mol dm-3 s-1; Experiment 3: [A] = 0.10 M, [B] = 0.20 M, initial rate = 2.0e-3 mol dm-3 s-1. (a) Deduce the order with respect to A and to B, and explain your reasoning. (b) Write the rate equation and calculate the rate constant k with its unit. (c) Find the initial rate when [A] = 0.30 M and [B] = 0.50 M. |
(a) Comparing 1 and 2: doubling [A] doubles the rate, so the reaction is first order in A. Comparing 1 and 3: doubling [B] leaves the rate unchanged, so it is zero order in B. (b) Rate = k[A], so k = 2.0e-3 divided by 0.10 = 0.020 per second (first order, unit s-1). (c) Rate = 0.020 times 0.30 = 6.00e-03 mol dm-3 s-1, independent of [B]. Marking notes: each order 2 marks; reasoning 2 marks; rate equation and k 2 marks; unit 1 mark; calculation 2 marks. |
| 3 | A reaction has an activation energy of 50.0 kJ per mole and R = 8.314 J K-1 mol-1. (a) Using ln(k2/k1) = (Ea/R)(1/T1 – 1/T2), calculate the ratio of the rate constants when the temperature rises from 300 K to 310 K. (b) Explain why a rise of only 10 K increases the rate markedly, using the distribution of molecular energies. (c) If the activation energy were 60.0 kJ per mole instead, would the ratio be larger or smaller? Give a brief reason. |
(a) Ea/R = 50000 divided by 8.314 = 6013.95 K. (1/300 – 1/310) = 1.075269e-04 per K. ln(k2/k1) = 0.6467, so k2/k1 = 1.91, about 1.9 times. (b) At the higher temperature the mean kinetic energy rises, so a much larger fraction of molecules has energy at least equal to Ea; the number of effective collisions per second rises. (c) The ratio would be smaller, because a higher activation energy makes the rate less sensitive to a fixed temperature rise. Marking notes: substitution 3 marks; value 2 marks; energy distribution 2 marks; judgement 1 mark. |
| 4 | Without a catalyst a reaction has an activation energy of 75.0 kJ per mole; with a catalyst it is 55.0 kJ per mole. (a) At 298 K, calculate the factor by which the rate constant increases (R = 8.314 J K-1 mol-1). (b) State how the catalyst affects the enthalpy change and the position of equilibrium. (c) Explain why a catalyst cannot make a thermodynamically unfavourable reaction (positive delta G) feasible. |
(a) ln(factor) = (75000 – 55000) divided by (8.314 times 298) = 8.0724, so the factor is 3205, about 3.2e3 times. (b) The enthalpy change is unchanged because the reactants and products are unchanged; the equilibrium constant and the position of equilibrium are also unchanged. Both forward and reverse reactions are speeded up, so equilibrium is reached sooner. (c) A catalyst changes only the path and its activation energy, not the free energy difference. With a positive delta G the products have higher free energy, so the equilibrium lies on the reactant side and a catalyst cannot alter that thermodynamic conclusion. Marking notes: factor 4 marks; enthalpy and equilibrium 2 marks; delta G explanation 2 marks. |
Section B: Chemical Equilibrium (Q5-Q8)
Data: molar volume 24.0 cubic decimetres per mole; R = 8.314 J K-1 mol-1; 1 atm = 101 kPa.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | In a 1.00 dm3 vessel, 1.00 mol of hydrogen is mixed with 1.00 mol of iodine vapour and heated. H2(g) + I2(g) is in equilibrium with 2HI(g). At equilibrium there is 1.60 mol of HI. (a) Calculate the moles of H2 and of I2 at equilibrium. (b) Calculate the equilibrium constant Kc at that temperature. (c) In another experiment at the same temperature, [H2] = 0.50 M, [I2] = 0.50 M and [HI] = 1.00 M are mixed. Use the reaction quotient Qc to decide the direction of the net change. |
(a) Forming 1.60 mol of HI uses 0.80 mol of each reactant, so [H2] = [I2] = 1.00 – 0.80 = 0.20 mol in 1.00 dm3. (b) Kc = [HI] squared divided by ([H2] times [I2]) = 2.56 divided by 0.16 = 64.0. (c) Qc = 1.00 squared divided by (0.50 times 0.50) = 4.0. Since Qc is less than Kc, the net change is to the right until Qc equals Kc. Marking notes: equilibrium moles 3 marks; Kc expression 2 marks; Kc value 2 marks; Qc and direction 2 marks. |
| 6 | Calcium carbonate decomposes at high temperature: CaCO3(s) is in equilibrium with CaO(s) + CO2(g). At 1100 K the concentration of carbon dioxide at equilibrium is 0.035 mol dm-3. (a) Write the Kc expression and explain why the solids do not appear in it. (b) Calculate Kc with its unit. (c) Use p = cRT to calculate the partial pressure of CO2 in atm, and state the effect of adding more solid CaO to the closed vessel. |
(a) Kc = [CO2]. The concentrations of the solids are constant and are incorporated into the equilibrium constant, so they are omitted. (b) Kc = 0.035 mol dm-3. (c) p = cRT = 0.035 times 0.0821 times 1100 = 3.16 atm. Adding more solid CaO has no effect on the CO2 concentration or on the position of equilibrium. Marking notes: expression and solids 3 marks; Kc and unit 2 marks; partial pressure 2 marks; CaO 1 mark. |
| 7 | The Haber process: N2(g) + 3H2(g) is in equilibrium with 2NH3(g), delta H = -92 kJ per mole. (a) Write the Kc expression and its unit in terms of mol dm-3. (b) At 700 K, Kc = 0.50 in those units. If [N2] = 0.10 M and [H2] = 0.30 M at equilibrium, find [NH3]. (c) Industry uses about 450 degrees Celsius and 200 atm. Explain the use of high pressure and why a lower temperature is not used. |
(a) Kc = [NH3] squared divided by ([N2] times [H2] cubed); the unit is (mol dm-3) to the power -2, that is dm6 mol-2. (b) [NH3] squared = 0.50 times 0.10 times 0.027 = 0.00135, so [NH3] = 0.0367 M. (c) High pressure favours the side with fewer gas moles (4 mol to 2 mol), raising the yield. A lower temperature would raise the yield further because the forward reaction is exothermic, but the rate would be too slow, so a compromise temperature with an iron catalyst is used. Marking notes: expression and unit 3 marks; calculation 3 marks; pressure 1 mark; temperature 1 mark. |
| 8 | First step of the Contact process: 2SO2(g) + O2(g) is in equilibrium with 2SO3(g). 1.00 tonne of sulfur dioxide is converted (M(SO2) = 64.1, M(SO3) = 80.1). (a) Calculate the theoretical mass of sulfur trioxide in tonnes. (b) If the conversion is only 96.0 per cent, find the actual yield in tonnes. (c) Explain why excess air rather than excess sulfur dioxide is used, and name the catalyst. |
(a) n(SO2) = 1.00e6 divided by 64.1 mol; n(SO3) is the same. Mass = 1.00 times 80.1 divided by 64.1 = 1.250 tonnes. (b) Actual yield = 1.250 times 0.960 = 1.200 tonnes. (c) Excess air shifts the equilibrium towards the product and raises the conversion of SO2; oxygen is cheap and abundant, and unreacted SO2 emissions are reduced. The catalyst is vanadium(V) oxide. Marking notes: mole ratio and theoretical mass 3 marks; actual yield 2 marks; excess air 2 marks; catalyst 1 mark. |
Section C: Industrial Processes and Calculations (Q9-Q12)
Give numerical answers to three significant figures and state the unit of every equilibrium constant.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | In a 2.00 dm3 vessel, 2SO2(g) + O2(g) is in equilibrium with 2SO3(g); at equilibrium the vessel contains 0.40 mol SO2, 0.20 mol O2 and 1.60 mol SO3. (a) Calculate the equilibrium concentration of each gas. (b) Calculate Kc at that temperature. (c) If the volume is halved to 1.00 dm3 at constant temperature, state the direction of change in the concentrations and in the moles of SO3. |
(a) [SO2] = 0.20 M, [O2] = 0.10 M, [SO3] = 0.80 M. (b) Kc = 0.80 squared divided by (0.20 squared times 0.10) = 0.64 divided by 0.0040 = 160.0. (c) Halving the volume doubles every concentration at once, so Qc falls to one half of its former value. Qc is then less than Kc and the equilibrium shifts to the right, so the moles of SO3 increase. Marking notes: concentrations 3 marks; Kc expression 2 marks; Kc value 2 marks; compression 2 marks. |
| 10 | Hydrogen peroxide decomposes as follows: 2H2O2(aq) gives 2H2O(l) + O2(g). 50.0 cm3 of hydrogen peroxide solution is mixed with a catalyst and 12.0 cm3 of oxygen is collected in 20 s (molar volume 24.0 cubic decimetres per mole). (a) Calculate the moles of oxygen formed in 20 s. (b) Calculate the decrease in concentration of H2O2 in mol dm-3. (c) Find the average rate over the first 20 s in mol dm-3 s-1 and explain why it is not the initial rate. |
(a) n(O2) = 0.0120 divided by 24.0 = 5.0e-04 mol. (b) From the equation, H2O2 consumed = 2 times 5.0e-04 = 1.0e-03 mol, so the concentration falls by 1.0e-03 divided by 0.0500 = 0.020 mol dm-3. (c) Average rate = 0.020 divided by 20 = 1.0e-03 mol dm-3 s-1. The initial rate is the instantaneous rate at t = 0, obtained from the tangent to the concentration-time curve at t = 0; the average rate describes the whole interval instead. Marking notes: moles of oxygen 2 marks; concentration change 3 marks; average rate 2 marks; distinction 1 mark. |
| 11 | Without a catalyst a reaction reaches equilibrium in 60.0 minutes. With a catalyst the rate constant is 100 times larger, while concentrations and temperature are unchanged. (a) Calculate the time needed to reach equilibrium with the catalyst. (b) State whether the catalyst changes the equilibrium concentrations and explain why. (c) Give two practical benefits of catalysts in industrial processes. |
(a) Time is inversely proportional to rate: t = 60.0 times 60 divided by 100 = 36 s, about 0.60 minutes. (b) It does not change them. The catalyst lowers the activation energy of the forward and reverse reactions by the same amount, so both rates rise equally, Kc is unchanged and the equilibrium concentrations are the same as without a catalyst. (c) Shorter production cycles and lower energy demand (operation at a lower temperature), hence a higher output per unit time. Marking notes: time 3 marks; judgement 2 marks; explanation 2 marks; two benefits 1 mark. |
| 12 | For the same zinc and acid experiment the volume-time curve gives 90 cm3 at 40 s and 210 cm3 at 80 s, while the average rate from 0 to 120 s is 1.75 cm3 per second. (a) Use a chord to find the average rate between 40 s and 80 s in cm3 per second. (b) Explain why this rate is greater than the average over 0 to 120 s. (c) Describe how to obtain a more accurate instantaneous rate at 60 s, and state between which two values that instantaneous rate must lie. |
(a) Rate = (210 – 90) divided by (80 – 40) = 3.00 cm3 per second. (b) The reactant concentration is highest early in the reaction and the rate falls with time, so an average over the early interval must exceed the average over the whole 120 s, which includes the slower later stage. (c) Draw the tangent to the curve at t = 60 s and measure its gradient. The instantaneous rate must lie between 1.75 and 3.00 cm3 per second. Marking notes: chord calculation 3 marks; explanation 3 marks; tangent method 1 mark; range 1 mark. |
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