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Form 5 Chemistry Practice Paper: Reaction Rate, Chemical Equilibrium and Industrial Chemistry (12 questions, 100 marks, 75 minutes)
Form 5 Chemistry – 12 questions – 100 marks – 75 minutes
Form 5 Chemistry paper: 12 questions, 100 marks, 75 min; measuring rates, collision theory, orders of reaction, Kc and Q, Le Chatelier, Haber and Contact processes, yield and atom economy.
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Paper details Form 5 Chemistry practice paper with 12 structured questions, full worked answers and a mark scheme (100 marks, 75 minutes).
Section A: Reaction Rate
Answer all parts; show all working and units in calculations.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Zinc granules react with excess dilute hydrochloric acid and the hydrogen collected in a gas syringe has volume: 0 s 0 cm³; 30 s 20 cm³; 60 s 34 cm³; 90 s 44 cm³; 120 s 50 cm³; 150 s 52 cm³. (a) Calculate the average rate from 0 to 60 s in cm³ s⁻¹. (b) Calculate the average rate from 60 to 120 s and compare it with (a). (c) Explain why the rate falls with time, and state how the first 30 s of the curve would change if zinc powder were used. |
(a) Rate = (V2 – V1) / (t2 – t1) = (34 – 0) / (60 – 0) = 0.57 cm³ s⁻¹. (b) Rate = (50 – 34) / (120 – 60) = 0.27 cm³ s⁻¹, about 47% of the first stage. (c) The acid is used up, so its concentration falls and collisions become less frequent. Zinc powder has a much larger surface area, so more gas is collected in the first 30 s (steeper curve), but the final volume is the same because the acid is still the limiting reagent. |
| 2 | 0.40 mol dm⁻³ sodium thiosulphate solution is mixed with dilute acid and the time for a sulphur precipitate to hide a cross under the flask is timed: 0.40 mol dm⁻³ takes 25 s, 0.20 mol dm⁻³ takes 50 s and 0.10 mol dm⁻³ takes 100 s. (a) Use 1 / time as a relative rate and tabulate the three values. (b) Deduce the order with respect to thiosulphate and justify it. (c) Why must the total volume, the temperature and the flask be kept the same in every run? |
(a) Relative rates are 0.04, 0.02 and 0.01 s⁻¹: halving the concentration doubles the time and halves the rate. (b) Rate is directly proportional to concentration (0.40 to 0.20 halves the rate from 0.04 to 0.02), so the reaction is first order in thiosulphate. (c) A fixed total volume keeps the dilution identical so concentration is the only variable; temperature sets the average kinetic energy of the particles; the same flask keeps the cross size and viewing distance constant. Judging when the cross disappears is subjective, so one observer should time every run. |
| 3 | Initial rate data for A + B gives products: Experiment 1: [A] = 0.10 mol dm⁻³, [B] = 0.10 mol dm⁻³, initial rate = 2.0e-03 mol dm⁻³ s⁻¹; Experiment 2: [A] = 0.20 mol dm⁻³, [B] = 0.10 mol dm⁻³, initial rate = 8.0e-03 mol dm⁻³ s⁻¹; Experiment 3: [A] = 0.10 mol dm⁻³, [B] = 0.20 mol dm⁻³, initial rate = 4.0e-03 mol dm⁻³ s⁻¹. (a) Find the order with respect to A and to B. (b) Write the rate equation and calculate the rate constant k with units. (c) If both [A] and [B] are doubled, by what factor does the initial rate change? |
(a) Experiments 1 and 2 differ only in [A]: doubling it raises the rate from 2.0e-03 to 8.0e-03, a factor of 4, so the order in A is 2. Experiments 1 and 3 differ only in [B]: doubling it gives a factor of 2, so the order in B is 1. (b) Rate = k[A]²[B]. k = rate / ([A]²[B]) = 2.0e-03 / (0.10² x 0.10) = 2.00 mol⁻² dm⁶ s⁻¹. (c) Factor = 2² x 2¹ = 8, so the rate rises from 2.0e-03 to 1.6e-02 mol dm⁻³ s⁻¹. |
| 4 | A reaction has rate v at 25 °C. As a rule of thumb the rate doubles for every 10 °C rise. (a) By what factor does the rate rise from 25 °C to 65 °C? (b) Use collision theory to explain the effect of temperature, and state the two main contributions. (c) A catalyst also speeds up the reaction. How does its effect on the energy profile differ from that of heating, and what does each do to the position of equilibrium? |
(a) The rise is 65 – 25 = 40 °C, that is 4 intervals of 10 °C, so the factor is 2^4 = 16. (b) Heating raises the average kinetic energy, so particles move faster and collide more often. Far more important is that a much larger fraction of molecules now has energy at or above the activation energy (the Maxwell-Boltzmann distribution shifts right), so the number of effective collisions rises sharply. (c) Heating shifts the whole distribution to higher energy while the activation energy stays the same. A catalyst opens a new pathway with a lower activation energy, so the threshold drops but the shape of the distribution is unchanged. A catalyst lowers the activation energy of both forward and reverse reactions equally, so it does not change the equilibrium position or Kc, only the time taken to reach it; heating shifts the position of an exothermic reaction to the left and lowers Kc. |
Section B: Chemical Equilibrium
Answer all parts; give numerical answers to the stated precision.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | A reaction has an activation energy of 50 kJ mol⁻¹. (a) At 50 °C it takes 8 minutes. Using the rule that the rate doubles for each 10 °C, estimate the time needed at 100 °C. (b) Describe in words the energy distribution curves at 25 °C and 75 °C, mark the activation energy position, and explain what the area under each curve means. (c) Explain why reactions with a high activation energy are especially sensitive to temperature. |
(a) The rise is 50 °C, that is 5 intervals, so the rate becomes 2^5 = 32 times larger and the time falls to 8 / 32 = 0.25 minutes (about 15.0 seconds). (b) Both curves start at zero, peak at low energy and have a long tail to the right; the hotter curve has a lower peak and lies further right, with a fatter tail beyond the activation energy. The activation energy is a fixed threshold on the axis, and the area under the curve gives the fraction of molecules with that energy (total area 1). The area to the right of the threshold is the fraction able to collide effectively. (c) With a high activation energy only a tiny fraction of molecules (the far tail) can react, so a small temperature rise multiplies that tail area many times. For a low activation energy most molecules are already above the threshold, so the relative gain from heating is small. |
| 6 | At 430 °C, 0.500 mol dm⁻³ of hydrogen and an equal concentration of iodine vapour are sealed in a vessel: H2(g) + I2(g) ⇌ 2HI(g). At equilibrium the concentration of hydrogen iodide is 0.780 mol dm⁻³. (a) Calculate the equilibrium concentrations of H2 and I2 and the percentage conversion of H2. (b) Calculate Kc at this temperature, stating its units or why it is dimensionless. (c) If a little HI is added to the equilibrium mixture, state how the position shifts, whether Kc changes, and how the final HI concentration compares with the original. |
(a) Two moles of HI are formed per mole of H2 used, so H2 reacted = 0.780 / 2 = 0.390 mol dm⁻³. Hence [H2] = [I2] = 0.500 – 0.390 = 0.110 mol dm⁻³ and the conversion is 0.390 / 0.500 = 78%. (b) Kc = [HI]² / ([H2][I2]) = 0.780² / 0.110² = 50.3. There are two gas molecules on each side, so the concentration units cancel and Kc is dimensionless. (c) Adding HI raises the product concentration so the reaction quotient exceeds Kc and the net change runs backwards, decomposing some of the added HI. The final HI concentration is therefore higher than before but lower than the sum of the original and the amount added. Kc depends only on temperature and is unchanged; only a temperature change alters it. |
| 7 | The Haber process: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -92 kJ mol⁻¹. (a) 10 mol of N2 is mixed with 30 mol of H2. If the reaction went to completion, how many moles and how many grams of NH3 (M = 17) would form? At a 35% yield, what mass is obtained? (b) Explain from equilibrium why high pressure and low temperature favour ammonia, and why industry nevertheless uses about 450 °C. (c) Suggest two practical measures that raise the ammonia yield and explain each. |
(a) 10 mol of N2 needs 30 mol of H2, so the gases are in the exact 3:1 ratio. One mole of N2 gives two moles of NH3, so 20 mol forms, with mass 20 x 17 = 340 g. At a 35% yield the mass is 340 x 0.35 = 119.0 g. (b) Four moles of gas become two, so raising the pressure shifts the equilibrium right. The forward reaction is exothermic, so lowering the temperature also shifts it right. However, at low temperature the rate is far too slow and the iron catalyst is ineffective, so about 450 °C is a compromise that accepts a lower equilibrium yield in return for a workable rate, with high pressure and recycling of unreacted gas making up the loss. (c) Continuously liquefying and removing the ammonia lowers the product concentration, so Q falls below Kc and the equilibrium keeps shifting right; recycling the unreacted nitrogen and hydrogen at high pressure raises the reactant partial pressures, shifting the equilibrium right and saving raw material. Using a catalyst also shortens the time needed to reach equilibrium. |
| 8 | The key step of the Contact process: 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = -197 kJ mol⁻¹, with a V2O5 catalyst at about 450 °C and slightly above 1 atm. (a) How many moles of oxygen are needed to oxidise 100 mol of SO2 completely, and how many moles and grams of SO3 (M = 80) are formed? (b) Explain why industry does not use the theoretically best combination of very low temperature and very high pressure. (c) Compare the principles behind the choice of conditions in the Haber and Contact processes. |
(a) The equation gives a 2:1:2 ratio, so 100 mol of SO2 needs 100 / 2 = 50 mol of O2 and produces 100 mol of SO3, that is 100 x 80 = 8000 g. (b) A very low temperature gives an almost complete shift to the right, but the rate becomes far too slow and the catalyst loses activity, so equilibrium cannot be reached in a sensible time. Very high pressure needs thick-walled vessels, raising capital cost, and hot acidic gases corrode the equipment. A compromise of moderate temperature, modest pressure and a catalyst is used, with inter-stage cooling and an absorption tower that removes SO3 so the equilibrium keeps shifting right. (c) Both are exothermic reversible reactions that reduce the number of gas moles, so high pressure and lower temperature favour the product. In practice both are limited by kinetics and economics (equipment strength, energy cost, catalyst activity), so a moderate temperature is chosen and the yield is recovered by catalysis and product removal. Thermodynamics sets the ceiling; kinetics and cost set the operating point. |
Section C: Industrial Chemistry and Applications
Answer all parts; support explanations with principles or data.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | In the laboratory 25.0 g of calcium carbonate (M = 100) is heated: CaCO3(s) gives CaO(s) + CO2(g). The mass of calcium oxide (M = 56) actually obtained is 12.6 g. (a) Calculate the theoretical yield, the percentage yield and the mass of carbon dioxide released. (b) Calculate the atom economy for CaO and explain how it differs from percentage yield. (c) Give two reasons why the actual yield is below 100% and suggest improvements. |
(a) Moles of CaCO3 = 25.0 / 100 = 0.25 mol. The equation is 1:1, so the theoretical mass of CaO is 0.25 x 56 = 14.0 g and the yield is 12.6 / 14.0 = 90%. The CO2 released is 0.25 x 44 = 11.0 g, consistent with the mass loss 25.0 – 14.0 = 11.0 g. (b) Atom economy = molar mass of the desired product / total molar mass of reactants = 56 / 100 = 56%. It is fixed by the equation and is the inherent ceiling of the reaction. Percentage yield compares what was actually obtained with the theoretical amount, so it depends on incomplete reaction, side reactions and handling losses and can be improved by better technique. (c) The temperature or heating time may be too low, leaving some carbonate undecomposed, and powder is lost on transfer or sticks to the crucible. Improvements: heat to constant mass at a higher temperature, cool in a covered crucible before weighing, and minimise transfers. Impure samples or absorbed water also lower the result. |
| 10 | For the reversible reaction A(g) ⇌ B(g), Kc = 4.0 at 298 K. At one instant a vessel contains [A] = 0.250 mol dm⁻³ and [B] = 0.250 mol dm⁻³. (a) Calculate the reaction quotient Q and state the direction of the net change. (b) Explain how the size of Kc relates to the position of equilibrium, and whether a very large Kc means a fast reaction. (c) On raising the temperature, Kc falls to 1.0. Deduce whether the forward reaction is exothermic or endothermic and state how [B] changes. |
(a) Q = [B] / [A] = 0.250 / 0.250 = 1.00. Since Q is below Kc (1.00 is less than 4.0), the net change runs forward: [B] rises and [A] falls until Q equals Kc. (b) A large Kc means products predominate at equilibrium, so the position lies to the right; a small Kc means it lies to the left. Kc is a thermodynamic quantity and says nothing about rate: a very large Kc may still be reached extremely slowly if the activation energy is high, in which case a catalyst is needed. (c) Raising the temperature lowers Kc from 4.0 to 1.0, so the equilibrium shifts left. By Le Chatelier’s principle heating favours the endothermic direction, so the reverse reaction is endothermic and the forward reaction is exothermic with a negative ΔH. Some B reverts to A, so at the new equilibrium [B] is lower and [A] higher than before. |
| 11 | The equilibrium yield of an exothermic reversible reaction is 78% at 400 °C and 63% at 500 °C. (a) Calculate the fall in percentage points and the relative fall when the temperature rises from 400 °C to 500 °C. (b) Describe how the rate-time graph and the yield-time graph change when a catalyst is added at the same temperature. (c) A student claims that a catalyst raises the yield. Identify the error and correct it. |
(a) The yield falls by 78 – 63 = 15 percentage points, a relative fall of 15 / 78 = 19.2%. (b) Rate-time graph: with a catalyst the curve jumps to a much higher peak at once and then falls as reactants are used up, approaching zero; without it the curve is flatter with a lower peak. Yield-time graph: both reach exactly the same plateau, but the catalysed curve reaches it sooner, that is equilibrium is attained faster. (c) The error is to confuse rate with position of equilibrium. A catalyst speeds up the forward and reverse reactions equally, so Kc and the equilibrium yield are unchanged; it only raises the amount produced per unit time and lets the mixture reach equilibrium within a practical time. To raise the equilibrium yield, use a lower temperature or a higher pressure (for a reaction that reduces the number of gas moles) or remove the product. |
| 12 | A chemical plant must allocate limited capital between the Haber and Contact processes, and both involve a trade-off between the theoretically best conditions and those that are practical. (a) Using the Haber process, explain how thermodynamics, kinetics and economics together fix the operating conditions. (b) State two concrete measures for saving energy and cutting emissions in either process and explain why each works. (c) An engineer proposes lowering the reactor to 300 °C to raise the yield. Evaluate the proposal using principles or data. |
(a) Thermodynamics: the forward reaction is exothermic and reduces the number of gas moles, so low temperature and high pressure raise the equilibrium yield; this sets the ceiling. Kinetics: the lower the temperature the slower the reaction, and the catalyst loses activity, so about 450 °C is needed for a practical rate. Economics: high pressure demands thick-walled vessels and expensive compression, and recycling and cooling also cost energy. The chosen compromise is a moderate temperature, a high enough pressure, an iron catalyst and continuous removal of ammonia, giving an acceptable yield at the lowest overall cost. (b) Inter-stage cooling with heat exchangers recovers the heat of reaction to preheat the feed, cutting fuel use. Continuously removing the product (liquefying ammonia, absorbing SO3 in concentrated sulphuric acid) keeps Q below Kc so the equilibrium keeps shifting right and less gas has to be recycled. A more active and longer-lived catalyst allows a practical rate at a lower temperature, saving both energy and emissions. (c) Lowering to 300 °C does raise the equilibrium yield, but the rate collapses: with a doubling every 10 °C, 150 °C is 15 intervals, so the rate falls to roughly 1 / 2^15 of its value at 450 °C, about one thirty-thousandth, and the catalyst may be inactive. Contact time would have to be enormous and output per hour tiny unless the reactor and catalyst bed were greatly enlarged. The proposal therefore looks good only if yield alone is judged; decisions must be based on output per unit time and per unit cost. |
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