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In a closed system a reversible reaction reaches equilibrium when forward and reverse rates become equal. Concentrations remain constant, but they need not be equal. Adding a reactant usually shifts the position toward products; increasing pressure favours the side with fewer gaseous moles. If gaseous mole counts are equal there is no such shift. Raising temperature favours the endothermic direction. A catalyst speeds both directions and reaches the same equilibrium sooner.
Video by Eliot Rintoul. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.
Viewing focus: Explain equilibrium responses to changing conditions
After watching, attempt these questions before opening the answers. You can also use the written lesson below.
After watching 1: At equilibrium, which rates are equal?
Forward and reverse rates. The reversible processes continue at equal rates.
After watching 2: Which change can change Kc?
Temperature. Kc depends on temperature.
What you'll be able to do
In a closed system a reversible reaction reaches equilibrium when forward and reverse rates become equal. Concentrations remain constant, but they need not be equal. Adding a reactant usually shifts the position toward products; increasing pressure favours the side with fewer gaseous moles. If gaseous mole counts are equal there is no such shift. Raising temperature favours the endothermic direction. A catalyst speeds both directions and reaches the same equilibrium sooner.
For a homogeneous equilibrium, Kc is the product of equilibrium concentrations of products divided by those of reactants, each raised to its stoichiometric coefficient. Use equilibrium values rather than the initial mixture. If necessary use changes in amount and the equation ratio to obtain equilibrium amounts, then divide by vessel volume. In the concentration convention used in school calculations, units follow from cancelling the concentration units; do not assume every Kc is dimensionless.
At a given temperature Kc is constant, so changing concentration or pressure changes the composition needed to satisfy it. Changing temperature changes Kc itself. For an exothermic forward reaction, heating reduces Kc. Industrial conditions balance equilibrium yield, reaction rate, separation costs, pressure costs and safety. A moderate yield on one pass can be acceptable when unreacted feed is recycled. Neither a catalyst nor a faster production rate by itself proves that equilibrium favours the products.
Attempt these problems on paper before opening the worked reasoning. Explain your steps as well as giving a final answer.
Common mistakes to avoid
Key takeaways
Question 1: An exothermic gas reaction has fewer product gas molecules. Predict the effects of increasing pressure and increasing temperature separately.
Pressure increases product yield; temperature decreases it.
Check the assessed scope and practical or mathematical requirements.
An external explanation or interactive activity to use alongside the lesson. It may include extension material.
Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.
Test yourself
Answer a few questions, then review anything you missed.
Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.