Reaction Rates & Kinetics — Practice · Grade 11 Science

Chapter: Rates & Equilibrium

What this practice covers

By the end of Reaction Rates & Kinetics you should be able to:

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

Vocabulary used: reaction rate, collision theory, effective collision, catalyst, surface area, inhibitor.

Problem 1

For a reaction whose rate increases with this reactant’s concentration, which change usually speeds it up while other conditions stay fixed?
  1. removing the catalyst
  2. cooling the reactants
  3. using large chunks instead of a fine powder
  4. increasing the concentration of a reactant
Answer and explanation

increasing the concentration of a reactant

Increasing that reactant’s concentration can increase rate under the stated conditions. Concentration effects depend on the actual reaction and rate law; this is not a universal rule for every reaction.

Problem 2

According to collision theory, an effective collision requires…
  1. enough energy and the correct orientation
  2. that a catalyst be present
  3. that the reaction be exothermic
  4. only that the particles touch
Answer and explanation

enough energy and the correct orientation

A collision only produces a reaction if the particles collide with at least the activation energy and are properly oriented.

Problem 3

A reaction releases 48 mL of gas in 12 s. What is the average rate of gas production in mL/s?
Answer and explanation

4 mL/s

Rate = 48 mL ÷ 12 s = 4.0 mL/s.

Problem 4

Explain, using collision theory, why powdered zinc reacts with acid much faster than a single lump of zinc of the same mass.
Answer and explanation

Powder has far more exposed surface area than a single lump, so many more zinc atoms are in contact with the acid at once. That allows far more effective collisions per second between zinc and acid particles, which speeds up the reaction.

Problem 5

In a reaction, 0.36 mol of a reactant is consumed in 45 s. What is the average rate of consumption in mol/s?
Answer and explanation

0.008 mol/s

Rate = 0.36 mol ÷ 45 s = 0.008 mol/s.

Problem 6

Doubling B at fixed A doubles rate. Find order in B. Show the reasoning, not just the conclusion.
Hints

Reaction coefficients do not generally supply rate-law exponents.

Answer and explanation

First order, since 2^n=2 gives n=1.

Problem 7

Why does a catalyst not increase equilibrium yield solely by lowering activation barriers?
Hints

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

Answer and explanation

It accelerates forward and reverse processes without changing the equilibrium constant at fixed temperature.

Problem 8

Temperature changes the population of particles above a barrierTemperature changes the population of particles above a barrierProbability density (inverse energy unit)00.10.20.30.40.5024681012Translational energy (selected units)Lower T modelHigher T model
Original normalized ideal translational-energy distributions with kBT values of 1 and 2 selected energy units. The dashed threshold is 4 units; the finite plot truncates the tails.
Interpret the supplied visual. Does the lower peak of the higher-temperature distribution mean fewer total particles?
Answer and explanation

No. These are normalized distributions. The population is spread over a wider energy range; compare areas and the threshold tail rather than peak height alone.

At thermal equilibrium a gas has a distribution of translational energies, not one common energy per molecule. Raising temperature shifts and broadens the ideal distribution so a greater fraction can exceed a selected activation threshold. This helps explain many temperature effects on rates, though orientation, mechanism and transport also matter. A catalyst changes the pathway/barrier rather than simply raising every molecule’s energy or changing equilibrium. The shaded-free curves here encode a probability density; an area, not peak height alone, represents a fraction.

No. These are normalized distributions. The population is spread over a wider energy range; compare areas and the threshold tail rather than peak height alone.

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 selected rate law is r=k[A]²[B]. Doubling A and halving B at the same temperature changes rate by what factor? If r uses M/s, give k units and one condition needed for the prediction.
Answer and explanation

The factor is 2²×0.5=2. For total order three, k units are M⁻²s⁻¹. Assume the same law, temperature, medium and relevant pathway remain valid; the change does not establish a new mechanism.

The factor is 2²×0.5=2. For total order three, k units are M⁻²s⁻¹. Assume the same law, temperature, medium and relevant pathway remain valid; the change does not establish a new mechanism.

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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