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Reactivity 2: How much, how fast and how far? practice questions.

202 exam-style questions on Reactivity 2: How much, how fast and how far? with answers and explanations. Six real samples below — the full set is free with an account.

Syllabus

Topics covered

Key terms

Definitions you need

overall order of a reaction
The sum of the powers (orders) of the concentration terms in the experimentally determined rate equation.
rate-determining step
The slowest step in a reaction mechanism, which limits the overall rate; only species involved up to and including this step appear in the rate equation.
dynamic equilibrium
The state of a reversible reaction in a closed system where the rates of the forward and reverse reactions are equal, so the concentrations of reactants and products stay constant.
activation energy
Activation energy (Ea) is the minimum energy that colliding reactant particles must possess for a collision to lead to a reaction.
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Sample questions

A compound contains 40.0% C, 6.7% H and 53.3% O by mass. (Ar: C = 12.0, H = 1.0, O = 16.0.) What is its empirical formula?

  • CH2O
  • C2H4O2
  • CHO
  • C2H6O

Answer: CH2O

Divide each percentage by its Ar: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33. Divide by the smallest (3.33): C = 1, H = 2, O = 1. The empirical formula is CH2O.

For N2 + 3H2 -> 2NH3, if 2.0 mol N2 reacts with 3.0 mol H2, which is the limiting reactant and how much NH3 forms?

  • N2 limiting, 4.0 mol NH3
  • H2 limiting, 2.0 mol NH3
  • H2 limiting, 3.0 mol NH3
  • N2 limiting, 2.0 mol NH3

Answer: H2 limiting, 2.0 mol NH3

The ratio needed is 1 N2 : 3 H2. 2.0 mol N2 would need 6.0 mol H2, but only 3.0 mol H2 is available, so H2 is limiting. 3.0 mol H2 x (2 NH3 / 3 H2) = 2.0 mol NH3.

At STP (0 degrees C, 100 kPa), the molar volume of an ideal gas is 22.7 dm3/mol. What volume does 0.50 mol of oxygen gas occupy at STP?

  • 11.35 dm3
  • 22.7 dm3
  • 45.4 dm3
  • 5.68 dm3

Answer: 11.35 dm3

At STP the volume of any ideal gas = moles x molar volume = 0.50 x 22.7 = 11.35 dm3. The identity of the gas does not matter for an ideal gas.

A fixed amount of gas at constant temperature has volume 2.0 dm3 at 100 kPa. What is its volume when the pressure is increased to 250 kPa?

  • 0.80 dm3
  • 1.25 dm3
  • 5.0 dm3
  • 0.50 dm3

Answer: 0.80 dm3

By Boyle's law, P1V1 = P2V2 at constant T. V2 = (P1V1)/P2 = (100 x 2.0)/250 = 200/250 = 0.80 dm3.

On a Maxwell-Boltzmann energy distribution, what happens when temperature is increased?

  • The curve shifts towards lower energies and its peak becomes taller and much narrower
  • The peak shifts to higher energy and more molecules exceed the activation energy
  • The total area under the curve increases
  • The activation energy decreases

Answer: The peak shifts to higher energy and more molecules exceed the activation energy

At higher temperature the distribution broadens and its peak shifts to higher energy (and becomes lower). A larger fraction of molecules has energy greater than or equal to the activation energy, so the reaction rate increases. The total area (number of molecules) stays constant.

For a reaction A + B -> products, doubling [A] doubles the rate and doubling [B] quadruples the rate. What is the overall order of reaction?

  • 1
  • 2
  • 3
  • 0

Answer: 3

Doubling [A] doubles the rate: first order in A. Doubling [B] quadruples the rate (2^2): second order in B. Rate = k[A][B]^2. Overall order = 1 + 2 = 3.

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