Kinetics & Rate Equations

According to collision theory, why does only a small fraction of collisions between reactant molecules lead to a reaction?

• AMost reactant molecules are simply too large to collide at ordinary concentrations
• BMost collisions involve molecules with combined energy below the activation energy and/or the wrong orientation✓
• CReactant molecules repel one another so strongly that they hardly ever collide at all, even in a concentrated solution
• DThe activation energy is always greater than the bond or lattice enthalpies involved

Answer: For a collision to be successful the colliding molecules must have a combined energy equal to or greater than the activation energy AND collide with the correct orientation. Most collisions fail one or both criteria, so only a small fraction are effective.

A small increase in temperature can cause a large increase in the rate of a reaction. The primary reason for this is that increasing temperature:

• Alowers the activation energy of the reaction
• Bincreases the enthalpy change of the reaction
• Cgreatly increases the proportion of molecules with energy greater than or equal to the activation energy✓
• Donly increases the frequency of collisions, which alone explains the large effect

Answer: Raising the temperature shifts the Maxwell–Boltzmann distribution so that a much larger proportion of molecules have energy ≥ Ea. This rise in the fraction of successful collisions dominates; the increase in collision frequency is comparatively small. Temperature does not change Ea or ΔH.

How does a catalyst increase the rate of a chemical reaction?

• AIt provides an alternative reaction pathway with a lower activation energy✓
• BIt shifts the equilibrium towards the products, raising the overall yield
• CIt raises the average kinetic energy of the reacting molecules present
• DIt makes the forward reaction considerably more exothermic overall

Answer: A catalyst provides an alternative route with a lower activation energy, so a greater proportion of collisions are successful at a given temperature. It does not change ΔH or the position of equilibrium, and it speeds up forward and reverse reactions equally.

The following initial-rate data were obtained for the reaction A + B → products. Experiment 1: [A] = 0.10, [B] = 0.10, rate = 0.20 mol dm⁻³ s⁻¹ Experiment 2: [A] = 0.20, [B] = 0.10, rate = 0.80 mol dm⁻³ s⁻¹ Experiment 3: [A] = 0.10, [B] = 0.20, rate = 0.40 mol dm⁻³ s⁻¹ What is the overall order of the reaction?

• AFirst order
• BZero order
• CSecond order
• DThird order✓

Answer: Comparing experiments 1 and 2: [A] doubles, rate ×4, so order in A = 2. Comparing 1 and 3: [B] doubles, rate ×2, so order in B = 1. Overall order = 2 + 1 = 3 (third order).

A reaction has the rate equation rate = k[X][Y]². If the concentration of Y is tripled while the concentration of X is kept constant, by what factor does the rate change?

• A×3
• B×6
• C×27
• D×9✓

Answer: The order with respect to Y is 2, so the rate depends on [Y]². Tripling [Y] multiplies the rate by 3² = 9. X is unchanged, so it has no effect here.

For a multi-step reaction, the species that appear in the experimentally determined rate equation are those involved in or before the:

• Arate-determining (slowest) step✓
• Bfastest step in the mechanism
• Cfinal product-forming step
• Dstep with the largest enthalpy change

Answer: The overall rate is governed by the slowest step, the rate-determining step. Only species taking part up to and including this step (i.e. its reactants, including those from prior fast equilibria) appear in the rate equation; substances added after it do not.

The Arrhenius equation is k = A e^(−Ea/RT). According to this relationship, which change would cause the largest increase in the rate constant k for a given reaction?

• AIncreasing the activation energy, Ea
• BDecreasing the temperature, T
• CDecreasing the activation energy, Ea (e.g. by adding a catalyst)✓
• DDecreasing the pre-exponential factor, A

Answer: In k = A e^(−Ea/RT), lowering Ea makes the exponent less negative, so e^(−Ea/RT) increases and k rises sharply — this is how a catalyst works. Raising Ea or lowering T or A all decrease k.

What is the order of reaction with respect to a reactant?

• AThe coefficient of that reactant in the balanced equation
• BThe number of its molecules in the rate-determining step product
• CThe ratio of its concentration to that of the other reactants
• DThe power of its concentration in the rate equation✓

Answer: Order comes only from experiment and cannot be read off the balanced equation. It reflects how many particles of that species appear in the rate-determining step.

What does a zero-order reactant tell you?

• AIts concentration affects the rate in direct proportion
• BIts concentration affects the rate as a squared term
• CIt is not present in the reaction mixture at all
• DIts concentration does not affect the rate at all✓

Answer: A zero-order species is absent from the rate equation, so changing its concentration changes nothing. That means it takes no part in the rate-determining step, although it is certainly involved in the overall reaction.

Why does a small rise in temperature produce a large increase in reaction rate?

• AIt sharply increases the proportion of molecules with E ≥ Ea✓
• BIt sharply increases the total number of molecules colliding
• CIt lowers the activation energy needed for a successful hit
• DIt changes the orientation at which the molecules collide

Answer: Raising the temperature shifts the Maxwell-Boltzmann distribution so a disproportionately large fraction of molecules exceed the activation energy. Collision frequency rises too, but that effect on its own is far too small to account for the size of the rate change.

What does the rate-determining step govern?

• AThe overall rate, being the fastest step in the mechanism
• BThe overall rate, being the slowest step in the mechanism✓
• CThe overall yield obtained from the reaction mixture
• DThe position of equilibrium reached by the reaction

Answer: A multi-step mechanism can proceed no faster than its slowest step, so only species involved up to and including that step appear in the rate equation. Comparing an experimental rate equation with a proposed mechanism is how mechanisms are tested.

Why does raising temperature increase rate substantially?

• AFewer particles exceed the activation energy on collision
• BThe activation energy of the reaction is lowered
• CMore particles exceed the activation energy on collision✓
• DThe particles become physically larger and collide more

Answer: Heating shifts the Maxwell-Boltzmann distribution so a much greater proportion of particles has energy above Ea. Collisions also become more frequent, but the energy effect is by far the larger of the two.

What are the units of the rate constant for a second-order reaction?

• Amol dm⁻³ s⁻¹
• Bs⁻¹
• Cmol⁻¹ dm³ s⁻¹✓
• Dmol⁻² dm⁶ s⁻¹

Answer: Rate has units mol dm⁻³ s⁻¹, so for rate = k[A]² the units of k must cancel one extra power of concentration. First order gives s⁻¹ and third order gives mol⁻² dm⁶ s⁻¹.

What does the Arrhenius equation relate?

• AThe rate constant to concentration and reaction order
• BThe rate constant to temperature and activation energy✓
• CThe equilibrium constant to temperature and pressure
• DThe enthalpy change to entropy and free energy

Answer: k = Ae^(−Ea/RT), so plotting ln k against 1/T gives a straight line whose gradient is −Ea/R. It is the standard experimental route to an activation energy.

Why does increasing the surface area of a solid increase rate?

• AFewer particles are exposed but each one collides much harder
• BMore particles are exposed and available for collision✓
• CThe activation energy of the reaction is substantially reduced
• DThe concentration of the solid reactant increases

Answer: Reaction happens at the surface, so grinding a solid exposes far more particles to the other reactant and collision frequency rises. The concentration of a pure solid does not change and is not part of the rate equation.

What is measured in a clock reaction?

• ATotal volume of gas produced by the reaction
• BMass lost by the reaction vessel over time
• CTime taken to reach a fixed visible change✓
• DFinal temperature reached by the mixture

Answer: The time to a sudden colour change is measured, and 1/t is taken as proportional to the initial rate. It is a neat method because a single easily-spotted endpoint replaces continuous monitoring.