Topic 9

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Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, go back to the full page.

Definitions to learn word for word

Measuring a rate

Method Rate
Disappearing cross (a fixed amount of product forms) rate ∝ 1 ÷ time, in s⁻¹
Gas collected volume change ÷ time, in cm³ s⁻¹
Concentration against time gradient of a tangent, in mol dm⁻³ s⁻¹

Collision theory and the factors

Reaction needs a collision with energy of at least EaE_\text{a} and the right orientation.

Change What changes Effect on rate
Higher concentration or pressure more particles per unit volume, so more frequent collisions up
Larger surface area (solid) more exposed particles up
Higher temperature more frequent collisions and a much greater proportion with energy of at least EaE_\text{a} up (the second is the main effect)
Catalyst lower EaE_\text{a}, so a greater proportion of collisions are successful up

Write "frequency of successful collisions", not "more collisions" alone. A gas at pressure pp has concentration c=p÷RTc = p \div RT (SI units).

The Maxwell–Boltzmann distribution

Reaction profiles

Catalysts in industry

Less energy: lower EaE_\text{a}, so a lower temperature, so less fuel and less CO₂. Higher atom economy: a better route with fewer by-products. Atom economy = MrM_\text{r} of the desired product ÷ sum of MrM_\text{r} of all products × 100%.

Le Chatelier at a glance

Change Position shifts
Add a reactant or remove a product to the right
Add a product or remove a reactant to the left
Raise the pressure towards the side with fewer moles of gas
Lower the pressure towards the side with more moles of gas
Raise the temperature in the endothermic direction
Lower the temperature in the exothermic direction
Add a catalyst no shift (equilibrium is reached faster)

Justify in four parts: the change, "shifts to oppose it", the direction, the result. Count only gas moles. Equal moles of gas on both sides means pressure has no effect on the position.

The compromise

For an exothermic reversible reaction, a low temperature gives a high yield but a slow rate, and a high temperature gives a fast rate but a low yield.

Process Conditions
Haber (NH₃) about 450 °C, 200 atm, iron catalyst, NH₃ liquefied and removed, unreacted gases recycled
Contact (SO₃) about 450 °C, 1–2 atm, V₂O₅ catalyst
Hydration of ethene about 300 °C, 60–70 atm, H₃PO₄ catalyst, unreacted gases recycled

Evaluate data in four steps: describe with numbers, explain with chemistry, spot the conflict, recommend a compromise (catalyst, recycling, cost and safety).

The ten traps

  1. Writing "more collisions" with no word for successful or frequent.
  2. Saying that temperature works only by more frequent collisions. The main effect is the proportion with enough energy.
  3. Drawing a Maxwell–Boltzmann curve that touches the x-axis, starts away from the origin, or has a different area at a different temperature.
  4. Moving the curve for a catalyst, instead of the EaE_\text{a} line.
  5. Putting the intermediate at the top of a hump, or changing the product level for a catalysed route.
  6. Saying that equilibrium means the reaction has stopped or that the concentrations are equal.
  7. Applying a pressure argument to a reaction with equal gas moles, or counting solids and liquids.
  8. Shifting to the exothermic side when the temperature is raised.
  9. Saying a catalyst increases the yield or shifts the position.
  10. Saying a high temperature is used in the Haber process "to raise the yield". It raises the rate at the cost of the yield.

Mixed test

  1. A tangent to a graph of [A] against time at 20 s passes through (0 s, 0.80 mol dm⁻³) and (40 s, 0.20 mol dm⁻³). Find the rate of reaction at 20 s.
  2. In a disappearing cross experiment the cross vanishes after 40 s at 20 °C and after 20 s at 30 °C. Calculate the rate at each temperature, and the ratio.
  3. 0.50 g of calcium carbonate reacts with an excess of hydrochloric acid. Calculate the total volume of carbon dioxide at room temperature and pressure. If 60 cm³ is collected in the first 30 s, find the mean rate over that time. (MrM_\text{r} of CaCO₃ = 100.1)
  4. Explain, using collision theory, why powdered marble reacts faster than marble chips.
  5. Draw and describe the Maxwell–Boltzmann distribution at two temperatures, and explain why a small rise in temperature can double the rate.
  6. A catalyst is added to a reaction at constant temperature. State what happens to the Maxwell–Boltzmann curve and to EaE_\text{a}, and explain the effect on the rate.
  7. State two features of a system at dynamic equilibrium.
  8. For 2 NO(g)+OX2(g)⇌2 NOX2(g)\ce{2NO(g) + O2(g) <=> 2NO2(g)}, ΔH = −114 kJ mol⁻¹, predict and justify the effect on the position of (a) raising the temperature, (b) raising the pressure, (c) adding a catalyst, (d) adding more O₂.
  9. For the Haber process, equilibrium yield falls from 89% at 200 °C to 8% at 600 °C, while the relative rate rises from 1 to 3500. Calculate the factors and explain why about 450 °C is used.
  10. Explain two ways in which a catalyst makes an industrial process more sustainable.

Answers cover these until you have tried the questions

  1. Gradient = (0.20 − 0.80) ÷ (40 − 0) = −0.015 mol dm⁻³ s⁻¹. The rate is 0.015 mol dm⁻³ s⁻¹ (1.5 × 10⁻²), quoted as a positive number.
  2. 20 °C: 1 ÷ 40 = 0.025 s⁻¹. 30 °C: 1 ÷ 20 = 0.050 s⁻¹. The ratio is 0.050 ÷ 0.025 = 2.0: the rate doubles for the 10 °C rise.
  3. n(CaCO3)n(\text{CaCO}_3) = 0.50 ÷ 100.1 = 0.004995 mol. The equation CaCOX3+2 HCl→CaClX2+HX2O+COX2\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2} gives 1 : 1, so n(CO2)n(\text{CO}_2) = 0.004995 mol. VV = 0.004995 × 24.0 = 0.1199 dm³ = 120 cm³. Mean rate = 60 ÷ 30 = 2.0 cm³ s⁻¹.
  4. Powder has a much larger surface area for the same mass, so more acid particles can collide with exposed marble particles at once. The frequency of successful collisions rises, so the rate rises.
  5. Sketch: two curves from the origin, the higher-temperature curve with a lower peak to the right, a higher tail and the same area, and a vertical EaE_\text{a} line to the right of both peaks. A small rise in temperature increases the proportion of molecules with energy of at least EaE_\text{a} greatly, because the tail is steep there. It also makes collisions slightly more frequent. Together they roughly double the rate.
  6. The curve does not change (same temperature). EaE_\text{a} moves to a lower value. More molecules have energy of at least the new EaE_\text{a}, so a greater proportion of collisions are successful, and the rate increases.
  7. The rates of the forward and backward reactions are equal, and the concentrations of reactants and products are constant, in a closed system.
  8. (a) Raising the temperature favours the endothermic direction, the backward reaction, so the position shifts left and less NO₂ forms. (b) 3 mol of gas on the left and 2 on the right, so higher pressure shifts it right. (c) A catalyst has no effect on the position; equilibrium is reached faster. (d) More O₂ is used up, so the position shifts right.
  9. Yield falls by 89 ÷ 8 = 11 times; rate rises by 3500 ÷ 1 = 3500 times. At 200 °C the yield is high but the reaction is far too slow; at 600 °C the rate is fast but the yield is very low (exothermic reaction). About 450 °C, with an iron catalyst to give a fast rate and recycling of unreacted gas, gives an acceptable yield at a workable rate.
  10. Less energy: the catalyst lowers the activation energy, so a lower temperature is enough, so less fuel is burned and less CO₂ is released. Higher atom economy: the catalyst allows a route with fewer steps and fewer by-products, so less waste.