Chemical Kinetics — Questions & Answers

Sample questions from Chemical Kinetics

1. Q1. A reaction is found to be first order. Starting concentration is 0.80 mol/L, and after 30 minutes it has fallen to 0.20 mol/L. How many half-lives have elapsed?

   • A.One half-life
   • B.Two half-lives✓
   • C.Three half-lives
   • D.Four half-lives
   Solution

   0.80 -> 0.40 (after one half-life) -> 0.20 (after a second half-life); the concentration has fallen to one-quarter (1/2^2) of its initial value, so exactly two half-lives have elapsed in the 30-minute interval, meaning t1/2 = 15 minutes for this reaction.

2. Q2. What is ACTIVATION ENERGY (Ea)?

   • A.The total energy released by a reaction as it proceeds from reactants to products
   • B.The minimum extra energy that reactant molecules must possess (above their average energy) in order to undergo effective collision and successfully form products✓
   • C.The energy required to break all the bonds in the product molecules only, unrelated to the reactants
   • D.The exact energy of the reactant molecules at the very start of the reaction, before any collisions occur
   Solution

   ACTIVATION ENERGY (Ea) is the MINIMUM EXTRA energy (above the average energy of the reactant molecules) that colliding reactant molecules must possess in order to successfully overcome the energy barrier and form the TRANSITION STATE (activated complex) en route to products -- it can be visualized as an 'energy hill' or barrier that must be climbed for reactants to be converted to products, REGARDLESS of whether the overall reaction is exothermic or endothermic -- reactions with a LOWER activation energy proceed FASTER (since a larger fraction of molecular collisions will possess sufficient energy to overcome this smaller barrier, at a given temperature), while reactions with a HIGHER activation energy proceed SLOWER -- activation energy is a key quantity in the ARRHENIUS EQUATION, which quantitatively relates the rate constant k to activation energy and temperature.

3. Q3. What does the ARRHENIUS EQUATION, k=Ae^(-Ea/RT), describe?

   • A.The dependence of reaction rate purely on reactant concentration, with no temperature dependence at all
   • B.The quantitative dependence of the rate constant (k) on TEMPERATURE (T) and ACTIVATION ENERGY (Ea), where A is the pre-exponential (frequency) factor and R is the gas constant✓
   • C.A relationship that applies only to zero-order reactions
   • D.An equation used exclusively to calculate the order of a reaction from experimental data
   Solution

   The ARRHENIUS EQUATION, k=Ae^(-Ea/RT), quantitatively describes how the RATE CONSTANT (k) of a reaction depends on TEMPERATURE (T) and ACTIVATION ENERGY (Ea) -- here, A is the PRE-EXPONENTIAL (or FREQUENCY) FACTOR, related to the frequency of molecular collisions and the fraction of those collisions with proper orientation, R is the universal gas constant, and e^(-Ea/RT) represents the FRACTION of molecules possessing sufficient energy (at least Ea) to react successfully at a given temperature -- this equation explains WHY reaction rates generally INCREASE with increasing TEMPERATURE (higher T increases the exponential term e^(-Ea/RT), since a larger fraction of molecules then possess sufficient energy to overcome the activation barrier) and WHY reactions with LOWER activation energy proceed faster at a given temperature (a smaller Ea makes the exponential term larger, closer to 1) -- the Arrhenius equation is fundamental to understanding both temperature effects on reaction rate and the practical effect of catalysts (which work by providing an alternative pathway with LOWER activation energy).

4. Q4. Two parallel first-order paths consume A: A -> B with k1 = 2.0 x 10^-3 s^-1 and A -> C with k2 = 3.0 x 10^-3 s^-1. What percentage of A ultimately ends up as B?

   • A.20 per cent
   • B.50 per cent
   • C.40 per cent✓
   • D.60 per cent
   Solution

   For competing first-order channels the product ratio equals the ratio of rate constants, so the fraction appearing as B is k1/(k1 + k2) = 2.0/(2.0 + 3.0) = 0.40, that is 40 per cent. This branching ratio is independent of time and of the starting concentration.

5. Q5. A zero-order reaction A -> P has k = 1.0 x 10^-3 mol L^-1 s^-1. If [A]0 = 0.10 mol/L, what are t1/2 and t3/4 (time for 75 per cent completion)?

   • A.t1/2 = 50 s, t3/4 = 75 s✓
   • B.t1/2 = 50 s, t3/4 = 100 s
   • C.t1/2 = 693 s, t3/4 = 1386 s
   • D.t1/2 = 100 s, t3/4 = 200 s
   Solution

   For zero order t = ([A]0 - [A])/k. Half completion: t = (0.10 - 0.05)/1.0 x 10^-3 = 50 s. Three-quarter completion: t = (0.10 - 0.025)/1.0 x 10^-3 = 75 s. The ratio t3/4 : t1/2 is 1.5 for zero order, whereas it is 2 for first order, providing a way to distinguish the two orders.

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