Kinetics & Rate Equations

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

• AMost molecules are too large to collide
• BMost collisions involve molecules with combined energy below the activation energy and/or the wrong orientation✓
• CReactant molecules repel each other and rarely collide at all
• DThe activation energy is always greater than the bond 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 position of equilibrium towards the products
• CIt increases the average kinetic energy of the reacting molecules
• DIt makes the forward reaction more exothermic

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.