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Chemical Equilibrium

Pressure, catalysts and investigating a shift in equilibrium

  • 2 things to know
  • Components 01 and 03

Overview

Recall questions on the effect of pressure in a gaseous system, why a catalyst does not move the position, and how a shift can be investigated experimentally.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Numbers

What does lowering the pressure do to a gaseous equilibrium, what does changing the pressure do to H₂(g) + I₂(g) ⇌ 2HI(g), and how does a catalyst affect the forward and reverse rates?

Numbers

What does a catalyst do to the time taken to reach equilibrium, and how can the effects of concentration and of temperature on equilibrium position be investigated?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    How does increasing pressure affect the position of equilibrium in a gaseous system?

    Model answer

    The equilibrium position shifts towards the side with fewer moles of gas.

  2. 2

    What happens to the equilibrium position when pressure is decreased in a gaseous equilibrium?

    Model answer

    The equilibrium position shifts towards the side with more moles of gas.

  3. 3

    For the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g), what effect does changing pressure have on the position of equilibrium?

    Model answer

    No effect, because both sides have equal moles of gas (2 moles on each side).

  4. 4

    How does a catalyst affect the rates of the forward and reverse reactions at equilibrium?

    Model answer

    A catalyst increases the rate of both the forward and reverse reactions by the same amount.

  5. 5

    Does adding a catalyst change the position of equilibrium?

    Model answer

    No, a catalyst does not change the position of equilibrium.

  6. 6

    What is the effect of a catalyst on the time taken to reach equilibrium and on the equilibrium concentrations?

    Model answer

    A catalyst allows equilibrium to be reached more quickly but does not alter the equilibrium concentrations.

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AThe first step is Cl• + O₃ → ClO• + O₂.
BThe second step is ClO• + O → Cl• + O₂.
CThe overall equation is O₃ + O → 2O₂.

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In ClO⁻, chlorine has oxidation number +1.
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