1.7.6Physical Chemistry
Catalysts — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 4 parts Lesson 45 of 102
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 4How Catalysts Work
Alternative route
- Alternative route A catalyst provides an alternative reaction pathway that has a lower activation energy.
More successful collisions
- More successful collisions At the same temperature a greater proportion of collisions has enough energy to react, so the rate increases.
Not used up
- Not used up The catalyst is regenerated at the end of the reaction, so it is chemically unchanged and a small amount works repeatedly.
No change in ΔH
- No change in ΔH The energy of reactants and products are unchanged, so the enthalpy change of reaction is the same.
Equilibrium
- Equilibrium A catalyst speeds up the forward and reverse reactions equally, so it does not alter the position of equilibrium.
Part 2 of 4Types
Types
Homogeneous
- Homogeneous A homogeneous catalyst is in the same phase as the reactants, such as aqueous ions catalysing a reaction in solution.
Heterogeneous
- Heterogeneous A heterogeneous catalyst is in a different phase, usually a solid with gaseous or liquid reactants adsorbed on its surface.
Examples
- Examples Iron is used in the Haber process, vanadium(V) oxide in the Contact process and platinum or rhodium in catalytic converters.
Benefits
- Benefits Catalysts allow lower temperatures and pressures, so they reduce energy use and costs, and often reduce carbon dioxide emissions.
| Catalyst | Process | Phase relation |
|---|---|---|
| Iron | Haber process | Heterogeneous |
| Vanadium(V) oxide | Contact process | Heterogeneous |
| Platinum/rhodium | Catalytic converter | Heterogeneous |
Part 3 of 4Maxwell–Boltzmann Explanation
1
Draw the distribution curve for the reaction at a fixed temperature.
2
Mark the original activation energy on the energy axis.
3
Mark a second, lower activation energy for the catalysed pathway.
4
Show that the area to the right of the lower Ea is much larger than the area to the right of the original Ea.
5
Conclude that a greater fraction of collisions are successful, so the rate rises without raising the temperature.
Part 4 of 4Exam Checklist & Tips
1.7.6 – Key Points for the Exam
- Catalyst A substance that increases the rate of reaction without being used up, by lowering the activation energy.
- Homogeneous catalyst A catalyst in the same physical state as the reactants.
- Heterogeneous catalyst A catalyst in a different physical state from the reactants.
- Adsorption The attachment of reactant molecules to the surface of a solid catalyst.
Common Mistakes
- Saying that a catalyst changes the position of equilibrium or the yield at equilibrium.
- Stating that a catalyst increases the energy of the particles.
- Describing the catalyst as being used up in the reaction.
- Showing a change in ΔH when drawing the catalysed pathway.
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