CBSE Class 12 Chemistry HOTs Chemical Kinetics Set 02

Higher Order Thinking Questions

Question 1. Assertion (A): The rate of a reaction increases with an increase in temperature.
Reason (R): The number of collisions increases with an increase in temperature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (b) Both statements are true. A rise in temperature does increase the collision frequency, but only slightly (a few per cent per 10 K).
The large increase in rate comes from the much larger increase in the fraction of molecules with energy ≥ E_a. So R is not the correct explanation of A.
In simple words: More collisions are a minor effect; more energetic collisions are the main reason.

Teacher's Note:
1. Separate 'more collisions' from 'more effective collisions'.
2. Quote the e^−E_a/RT term when you explain.

Question 2. Assertion (A): The order and the molecularity of a reaction are always equal.
Reason (R): Complex reactions take place in several steps and the slowest step is the rate determining step.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (d) A is false: order is found by experiment and can be zero or a fraction, while molecularity is a whole number found from an elementary step. For a complex reaction the order is decided by the slowest step.
R is a true statement about complex reactions.
In simple words: The two quantities agree only for simple one-step reactions.

Teacher's Note:
1. Give one example in which they differ, such as a reaction of fractional order.
2. R is true but does not rescue A.

Question 3. Assertion (A): The rate of a zero order reaction remains constant until the reaction is over.
Reason (R): The rate of a zero order reaction is independent of the concentration of the reactants.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) For a zero order reaction rate = k[R]⁰ = k. Because the rate does not depend on concentration, it stays constant as the reactant is used up, until the reactant is exhausted.
So R explains A.
In simple words: No concentration term in the rate law means the rate never changes.

Teacher's Note:
1. Write the rate law first.
2. The rate drops to zero only when the reactant has been completely consumed.

Question 4. Assertion (A): All collisions between reactant molecules lead to the formation of products.
Reason (R): Only those collisions in which the molecules have the correct orientation and sufficient kinetic energy lead to the formation of products.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (d) A is false: only a small fraction of the collisions are effective.
R is true: effective collisions need energy ≥ E_a and proper orientation.
In simple words: Most collisions only bounce off; a few have the energy and orientation needed.

Teacher's Note:
1. Effective collisions are a small fraction of the total.
2. Mention the two conditions in a long answer.

Question 5. Assertion (A): The half-life of a first order reaction is independent of the initial concentration of the reactant.
Reason (R): The half-life of a first order reaction is given by t½ = 0.693/k.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) t½ = 0.693/k contains only the rate constant, and k does not depend on concentration. So t½ is constant whatever the initial concentration, and R is the reason for A.
In simple words: The formula for t½ has no concentration in it, which is exactly why the half-life does not depend on concentration.

Teacher's Note:
1. Quote the formula as the reason.
2. Contrast with zero order, where t½ = [R]₀/2k depends on [R]₀.

Question 6. Assertion (A): A catalyst increases the rate of a reaction.
Reason (R): A catalyst changes the enthalpy of the reaction so that the products become more stable.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (c) A is true: a catalyst provides a new path with a lower activation energy, so more molecules cross the barrier.
R is false: a catalyst lowers the forward and the reverse barriers by the same amount and does not change ΔH or the energies of the reactants and products.
In simple words: A catalyst lowers the barrier; it does not change where the reaction starts and ends.

Teacher's Note:
1. Draw the catalysed path below the uncatalysed one, with the same start and end levels.
2. A catalyst does not alter ΔH, ΔG or the equilibrium constant.

Question 7. Assertion (A): The rate constant of a reaction increases with an increase in temperature.
Reason (R): The fraction of molecules having energy equal to or greater than the activation energy increases with an increase in temperature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) From the Arrhenius equation k = Ae^−E_a/RT, a larger T makes e^−E_a/RT (the fraction of molecules with E ≥ E_a) larger, and so k increases. R gives the reason for A.
In simple words: Heating puts more molecules above the energy barrier, so k becomes larger.

Teacher's Note:
1. Link the Arrhenius factor to the energy distribution curve.
2. E_a and A are almost unchanged over a small temperature range.

Teacher's Note:
1. Match the sure pairs first and eliminate options.
2. Keep the first-order half-life property in mind: every half-life is constant.

Question 9. Which of the following statements are correct for a zero order reaction? (more than one may be correct)
(a) The unit of the rate constant is the same as the unit of the rate
(b) The half-life is directly proportional to the initial concentration
(c) A plot of [R] against time is a straight line with slope −k
(d) A plot of rate against [R] is a straight line passing through the origin
Answer:
Ans. (a), (b) and (c)
(a) For rate = k, k has the unit mol L⁻¹ s⁻¹ like the rate. (b) t½ = [R]₀/2k. (c) [R] = [R]₀ − kt is a straight line of slope −k.
(d) is wrong: rate = k does not change with [R], so a plot of rate against [R] is a horizontal line.
In simple words: For zero order, concentration falls in a straight line and the rate is flat.

Teacher's Note:
1. Write [R] = [R]₀ − kt and test each statement.
2. A line through the origin is the rate plot of a first order reaction.

Question 10. Explain why the molecularity of a reaction cannot be zero or a fraction, although the order can be both. Why can the rate law of a complex reaction not be written from its balanced equation?
Answer:
Molecularity is the number of molecules, atoms or ions that actually collide in an elementary step. You cannot have zero or half a molecule taking part, so it is always a whole number (1, 2 or 3).
Order is the sum of the powers of concentration terms in an experimentally found rate law. The powers can be zero (the reactant does not affect the rate) or fractions.
A complex reaction proceeds in several steps, and the overall rate is controlled by the slowest step. The stoichiometric coefficients of the overall equation do not show the slow step, so the rate law can be found only by experiment.
In simple words: Molecularity counts real particles in one step; order is just a number from the experiment.

Teacher's Note:
1. Use the words 'elementary step' for molecularity and 'experiment' for order.
2. Give an example such as 2N₂O₅ → 4NO₂ + O₂, which is first order although two molecules appear.

Question 11. The decomposition of a compound A is of first order. A plot of ln [A] against time (in minutes) is shown. The line passes through the points (0, ln 0.80), (20, −1.61) and (40, −3.00).
(a) What is the slope of the line and the rate constant? (b) Find the half-life. (c) What is the concentration of A after 30 minutes?
Answer:
(a) Slope = (−3.00 − (−0.223)) ÷ 40 = −0.0693 min⁻¹. For a first order reaction the slope of ln [A] against t is −k, so k = 0.0693 min⁻¹.
(b) t½ = 0.693 ÷ 0.0693 = 10 minutes.
(c) 30 minutes = 3 half-lives, so [A] = 0.80 ÷ 2³ = 0.10 mol L⁻¹.
Check: ln [A] = −0.223 − 0.0693 × 30 = −2.30, and e^−2.30 = 0.10.
In simple words: A straight line on a ln [A] against t graph shows first order, and its slope is −k.

Teacher's Note:
1. The intercept of the line on the ln [A] axis is ln [A]₀.
2. Use the number of half-lives for a quick check.

Question 12. The Arrhenius plot of a first order reaction is a straight line of ln k against 1/T, as shown. It passes through ln k = 8 at 1/T = 0.0030 K⁻¹ and has a slope of −5000 K.
(a) Calculate the activation energy. (b) Find ln A and the pre-exponential factor A. (c) Calculate k at 400 K. (R = 8.314 J K⁻¹ mol⁻¹)
Answer:
(a) From ln k = ln A − E_a/RT, slope = −E_a/R. So E_a = 5000 × 8.314 = 41 570 J mol⁻¹ ≈ 41.6 kJ mol⁻¹.
(b) ln A = ln k + (E_a/R)(1/T) = 8 + 5000 × 0.0030 = 23, so A = e²³ ≈ 9.7 × 10⁹ s⁻¹ (the intercept at 1/T = 0).
(c) At 400 K, 1/T = 0.0025, so ln k = 23 − 5000 × 0.0025 = 10.5. Hence k = e^10.5 ≈ 3.6 × 10⁴ s⁻¹.
In simple words: The slope of the Arrhenius plot gives the activation energy, and the intercept gives ln A.

Teacher's Note:
1. Slope = −E_a/R (not −E_a/2.303R) when ln k is plotted.
2. If log₁₀ k is plotted, the slope is −E_a/2.303R.

Question 13. The rate of a certain reaction doubles for every 10 K rise in temperature. By what factor does the rate increase when the temperature is raised from 300 K to 330 K? Why is this rule only approximate?
Answer:
From 300 K to 330 K the rise is 30 K, which is three steps of 10 K.
Rate factor = 2³ = 8; the rate becomes 8 times.
The rule 'doubling for every 10 K' is only approximate. From the Arrhenius equation, log (k₂/k₁) = (E_a/2.303R)(T₂ − T₁)/(T₁T₂), so the ratio for a 10 K rise depends on both E_a and the temperature itself. It is about 2 only for E_a near 50 kJ mol⁻¹ at room temperature, and it becomes smaller at higher temperatures.
In simple words: Eight times faster after a 30 K rise if the rate doubles at each 10 K step.

Teacher's Note:
1. Count how many 10 K steps there are before using the power of 2.
2. Remember that a 10 K rule is an approximation, not a law.