MCQz

Reaction Kinetics

Chemistry · 11th Class (Intermediate Part 1) · BISE Lahore

70 questions

Tests for this chapter

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Full Book Test

20 questions · Periodic Table and Periodic Properties × 1, Atomic Structure × 1, Chemical Bonding × 2, Stoichiometry × 1, States and Phases of Matter × 1, Chemical Energetics × 2, Reaction Kinetics × 1, Chemical Equilibrium × 1, Acid-Base Chemistry × 1, Electrochemistry × 2, Hydrocarbons × 2, Nitrogen and Sulfur × 1, Halogens × 1, Atmosphere × 1, Basic Separation Techniques × 1, Lab Safety and Practical Skills × 1

Reaction Kinetics Test

10 questions · Reaction Kinetics × 10

Sample questions

Showing 10 of 70 questions from this chapter. The full bank is in the chapter test below.

  1. 1

    The minimum amount of energy that colliding particles must possess in order to react is called the:

    • A.bond dissociation energy
    • B.lattice energy
    • C.activation energy (correct answer)
    • D.ionization energy
    Why: Activation energy is the minimum energy needed for a collision to be effective, that is, for the colliding species to be converted into products. Bond dissociation energy and ionization energy describe other changes and are not the collision-energy threshold.
  2. 2

    Molecules of two gases collide with energy greater than the activation energy, yet no reaction takes place. The most likely reason is that:

    • A.the colliding molecules are not properly oriented (correct answer)
    • B.the collisions are far too infrequent
    • C.the temperature of the mixture is too low
    • D.no catalyst is present in the mixture
    Why: An effective collision requires both sufficient energy and the proper orientation of the colliding species. Since the energy condition is already satisfied, the missing requirement is suitable orientation.
  3. 3

    The branch of chemistry that deals with the rates of chemical reactions, their orders and their mechanisms is called:

    • A.thermochemistry
    • B.electrochemistry
    • C.stoichiometry
    • D.chemical kinetics (correct answer)
    Why: Chemical kinetics is the study of reaction rates together with the experimental methods used to measure rates, orders and mechanisms. Thermochemistry deals with heat changes and electrochemistry with electrical effects in reactions.
  4. 4

    Which of the following is the slowest process under ordinary conditions?

    • A.precipitation of AgCl when AgNOX3\ce{AgNO3} solution is added to NaCl solution
    • B.rusting of iron (correct answer)
    • C.acid hydrolysis of an ester
    • D.explosion of a hydrogen and oxygen mixture
    Why: Rusting of iron takes days or months, while precipitation of AgCl is almost instantaneous, ester hydrolysis proceeds at a moderate rate and an explosion is over in a fraction of a second.
  5. 5

    The rate of a chemical reaction is defined as the:

    • A.change in the mass of a reactant divided by the volume of the vessel
    • B.change in the concentration of a reactant or product divided by the time taken for the change (correct answer)
    • C.total amount of product formed when the reaction is complete
    • D.time taken for the reaction to reach completion
    Why: Rate of reaction = change in concentration of the substance / time taken for the change. Mass, total yield and total time do not give a rate, because a rate always compares a change with the time over which it occurs.
  6. 6

    For a reaction in solution, concentration is measured in mol dm−3\pu{mol dm-3} and time in seconds. The rate of the reaction is therefore expressed in:

    • A.mol dm−3\pu{mol dm-3} s−1\pu{s-1} (correct answer)
    • B.mol dm−3\pu{mol dm-3}
    • C.s−1\pu{s-1}
    • D.dm3\pu{dm3} mol−1\pu{mol-1} s−1\pu{s-1}
    Why: Rate is a concentration change divided by a time, so its units are mol dm−3\pu{mol dm-3} divided by s, that is mol dm−3\pu{mol dm-3} s−1\pu{s-1}. For a slow reaction the same unit may be written with minutes or hours in place of seconds.
  7. 7

    For the reaction A→B\ce{A -> B}, the rate is written as -Δ[A]/Δt. The negative sign is used because:

    • A.the reaction rate is always numerically negative
    • B.A is used up faster than B is formed
    • C.concentrations are measured in mol dm−3\pu{mol dm-3}
    • D.the concentration of A decreases as the reaction proceeds (correct answer)
    Why: The concentration of the reactant A falls with time, so Δ[A] is negative; the minus sign is inserted to make the rate positive. The rate of a reaction is a positive quantity.
  8. 8

    The rate of a reaction at any one instant during a time interval is called the:

    • A.average rate
    • B.initial rate
    • C.instantaneous rate (correct answer)
    • D.specific rate
    Why: The rate at a single instant is the instantaneous rate and is obtained from the slope of the tangent to the concentration-time curve. The average rate is measured over two specified times and equals the instantaneous rate at only one instant of the interval.
  9. 9

    In the formation of ammonia, NX2(g)+3 HX2(g)→2 NHX3(g)\ce{N2(g) + 3H2(g) -> 2NH3(g)}, the concentration of ammonia is 3.5 mol dm−3\pu{mol dm-3} after 1.0 min and 6.2 mol dm−3\pu{mol dm-3} after 4.0 min. The average rate of formation of ammonia between 1.0 min and 4.0 min is:

    • A.3.5 mol dm−3\pu{mol dm-3} min-1
    • B.0.90 mol dm−3\pu{mol dm-3} min-1 (correct answer)
    • C.2.7 mol dm−3\pu{mol dm-3} min-1
    • D.0.68 mol dm−3\pu{mol dm-3} min-1
    Why: Δ[NHX3\ce{NH3}] = 6.2 - 3.5 = 2.7 mol dm−3\pu{mol dm-3} and Δt = 4.0 - 1.0 = 3.0 min, so the average rate is 2.7/3.0 = 0.90 mol dm−3\pu{mol dm-3} min-1. The value 2.7 is only the change in concentration, 3.5 is the instantaneous rate after 1.0 min, and 0.68 comes from dividing the change by 4.0 min instead of by the length of the interval.
  10. 10

    For the reaction HX2(g)+IX2(g)→2 HI(g)\ce{H2(g) + I2(g) -> 2HI(g)}, the concentration of iodine falls from 0.010 mol dm−3\pu{mol dm-3} to 0.0080 mol dm−3\pu{mol dm-3} in the first 100.0 s. The average rate of the reaction during this period is:

    • A.1.0×10−41.0 \times 10^{-4} mol dm−3\pu{mol dm-3} s−1\pu{s-1}
    • B.8.0×10−58.0 \times 10^{-5} mol dm−3\pu{mol dm-3} s−1\pu{s-1}
    • C.2.0×10−42.0 \times 10^{-4} mol dm−3\pu{mol dm-3} s−1\pu{s-1}
    • D.2.0×10−52.0 \times 10^{-5} mol dm−3\pu{mol dm-3} s−1\pu{s-1} (correct answer)
    Why: The iodine concentration falls by 0.010 - 0.0080 = 0.0020 mol dm−3\pu{mol dm-3} in 100.0 s, so the average rate is 0.0020/100.0 = 2.0×10−52.0 \times 10^{-5} mol dm−3\pu{mol dm-3} s−1\pu{s-1}. Dividing the whole initial or final concentration by 100 s gives 1.0×10−41.0 \times 10^{-4} and 8.0×10−58.0 \times 10^{-5}, while 2.0×10−42.0 \times 10^{-4} uses the correct change in concentration but a time of only 10 s.
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