Chemical Equilibrium — Practice · Grade 11 Science

Chapter: Rates & Equilibrium

What this practice covers

By the end of Chemical Equilibrium you should be able to:

9 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: reversible reaction, dynamic equilibrium, Le Châtelier's principle, equilibrium shift, equilibrium constant (Keq), Haber process.

Problem 1

At dynamic equilibrium…
  1. the reaction has completely stopped
  2. the forward and reverse reactions occur at equal rates
  3. reactants and products are always present in equal amounts
  4. no more collisions occur
Answer and explanation

the forward and reverse reactions occur at equal rates

Equilibrium is dynamic: both reactions still run, but at equal rates, so concentrations stay constant.

Problem 2

Adding more reactant to a system at equilibrium causes the equilibrium to shift…
  1. toward the reactants
  2. toward the products
  3. so the reaction stops
  4. in no direction at all
Answer and explanation

toward the products

By Le Châtelier's principle, the system shifts toward the products to consume the added reactant.

Problem 3

At equilibrium, the forward reaction has a rate of 0.020 mol/(L·s). What is the rate of the reverse reaction at that same moment?
Answer and explanation

0.02 mol/(L·s)

At equilibrium the forward and reverse rates are equal by definition, so the reverse rate is also 0.020 mol/(L·s).

Problem 4

A student says, 'At equilibrium, the reaction has stopped.' Explain why this statement is incorrect.
Answer and explanation

The forward and reverse reactions are both still happening — they have not stopped. They simply run at equal rates, so the concentrations stay constant. Equilibrium is dynamic, not static.

Problem 5

For the exothermic reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + heat, increasing the temperature will…
  1. cause the reaction to stop entirely
  2. shift the equilibrium right, making more SO₃
  3. shift the equilibrium left, making less SO₃
  4. have no effect on the equilibrium
Answer and explanation

shift the equilibrium left, making less SO₃

Heat behaves like a product in an exothermic reaction, so adding heat shifts the equilibrium in the reverse (endothermic) direction — left, reducing the amount of SO₃.

Problem 6

What does adding a catalyst do to an established equilibrium composition in the ideal model? Show the reasoning, not just the conclusion.
Hints

Equilibrium concentrations change, while K remains fixed at that temperature.

Answer and explanation

It does not shift equilibrium; it changes rates and how quickly equilibrium is reached.

Problem 7

Why can changing temperature change K while adding reactant at fixed temperature does not?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

K depends on thermodynamics and temperature; concentration changes alter Q and the response, not fixed-temperature K.

Problem 8

Not every intervention changes the equilibrium constantNot every intervention changes the equilibrium constantChange at same reaction conditionsK effect / responseConcentration disturbance at fixed TQ changes; K does notCatalyst at fixed TRates change; K does notTemperature changeK can changeInert gas at fixed V, ideal mixtureReactant partial pressuresunchangedPressure effects require the actual volume/amount constraint.
Original condition-comparison table; ideal mixture assumptions and unchanged reaction identity are explicit.
Interpret the supplied visual. Why is “raising total pressure always shifts equilibrium toward fewer gas molecules” incomplete?
Answer and explanation

The method matters. Adding inert gas at fixed volume does not change ideal reacting-species partial pressures; compression is a different disturbance.

For a specified reaction and standard-state convention, K depends on temperature rather than how much material was initially added. A catalyst changes approach rates but not the equilibrium constant for the same conditions. Temperature changes can alter K; their direction requires the appropriate thermodynamic account. Adding an inert gas to an ideal mixture at fixed volume can raise total pressure without changing reacting-species partial pressures. Distinguish total pressure from the quantities entering Q before predicting a composition shift.

The method matters. Adding inert gas at fixed volume does not change ideal reacting-species partial pressures; compression is a different disturbance.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 9

A⇌B has K=[B]/[A]=4 under a selected ideal concentration convention. Initially at equilibrium [A]=0.20 M and [B]=0.80 M. Add A to 0.70 M at unchanged volume/T. Find immediate Q and predict net direction without claiming original concentrations return.
Answer and explanation

Q=0.80/0.70≈1.14<4, so net forward conversion is favored. Under the closed 1:1 model total concentration becomes 1.50 M and final values are A=0.30 M, B=1.20 M; the ratio returns, not the original amounts.

Q=0.80/0.70≈1.14<4, so net forward conversion is favored. Under the closed 1:1 model total concentration becomes 1.50 M and final values are A=0.30 M, B=1.20 M; the ratio returns, not the original amounts.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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