CBSE Important Questions Class 12 Chemistry Chapter 3 – Chemical Kinetics

Class 12 Chemistry Chapter 3 Chemical Kinetics is part of the CBSE Class 12 Chemistry syllabus. It studies how fast a chemical reaction goes and what affects its speed. Students learn the rate of a reaction, rate law and order of a reaction, the integrated rate equations for zero and first order reactions, half-life, the effect of temperature and catalyst, the Arrhenius equation, activation energy and collision theory.

These CBSE important questions for Class 12 Chemistry Chapter 3 cover the rate of reaction from changing concentrations, the effect of a catalyst on activation energy, how to decide whether a reaction is exothermic or endothermic, filling in a rate table for a first order reaction, the rate constant, average rate, order with respect to each reactant, and threshold energy. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. For a reaction A + B → C + D, the concentration of C changes from 5 mmol to 10 mmol in 5 seconds. The rate of appearance of D is 1.5 times the rate of disappearance of B, which is twice the rate of disappearance of A. The rate of appearance of D is 9 mmol. Find the rate of reaction.

Answer: Rate of appearance of C = (10 − 5)/5 = 1 mmol/s.
Working through the given relations:
Rate of appearance of D = 9 mmol
Rate of disappearance of B = 9/1.5 = 6 mmol
Rate of disappearance of A = 6/2 = 3 mmol
For a reaction, the rate of reaction is defined with respect to the stoichiometric coefficients. Taking the relation through, the rate of reaction works out to
1 mmol per second

Q2. For a reaction at 300 K, the activation energy of the catalysed reaction is lower by 0.314 kJ than that of the uncatalysed reaction. By what factor does the rate increase?
(a) 2 times
(b) 1.5 times
(c) 1.13 times
(d) 3.14 times

Answer: (c) 1.13 times
Using the Arrhenius equation, the ratio of the two rate constants is
kcat/kuncat = eΔEa/RT
ΔEa = 0.314 kJ = 314 J, R = 8.314 J/K/mol, T = 300 K
ΔEa/RT = 314 / (8.314 × 300) = 314/2494 = 0.1259
kcat/kuncat = e0.1259 = 1.13
So the catalyst makes the reaction about 1.13 times faster.

Q3. Define the rate of a reaction. How do average rate and instantaneous rate differ?

Answer: The rate of a reaction is the change in concentration of a reactant or product per unit time. Its usual unit is mol L⁻¹ s⁻¹.
• Average rate is measured over a time interval: −Δ[R]/Δt or +Δ[P]/Δt.
• Instantaneous rate is the rate at a particular instant, found by making the time interval very small: −d[R]/dt or +d[P]/dt. On a concentration-time graph it is the slope of the tangent at that point.
Note that the rate of a reaction is always taken as a positive quantity, which is why the minus sign is used for reactants.

Q4. How can we decide whether a chemical reaction is exothermic or endothermic?

Answer: A chemical reaction happens in two stages:
1. The bonds in the reactant molecules are broken to give free atoms. Breaking bonds needs energy, so this step is endothermic.
2. The free atoms then join to form the product molecules. Forming bonds releases energy, so this step is exothermic.
The decision: Compare the two energies.
• If the energy released in forming bonds is more than the energy absorbed in breaking them, the overall reaction is exothermic (ΔH is negative).
• If the energy absorbed is more, the reaction is endothermic (ΔH is positive).

Q5. A biochemical reaction is carried out in a laboratory outside the human body in the absence of an enzyme. If the rate obtained is 10⁻⁶ times, then the activation energy of the reaction in the presence of an enzyme is
(a) 6/RT
(b) P is required
(c) different from the Ea obtained in the laboratory
(d) same as obtained in the laboratory

Answer: (c) Different from the Ea obtained in the laboratory
An enzyme is a biological catalyst. A catalyst works by providing an alternative path with a lower activation energy.
Since the reaction is a million times slower without the enzyme, the activation energy without it must be much higher. So the Ea in the presence of the enzyme is different from (and lower than) the laboratory value.

Q6. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the table.

Experiment [A] / mol L⁻¹ [B] / mol L⁻¹ Initial rate / mol L⁻¹ min⁻¹
I 0.1 0.1 2.0 × 10⁻²
II — 0.2 4.0 × 10⁻²
III 0.4 0.4 —
IV — 0.2 2.0 × 10⁻²

Answer: Since the reaction is first order in A and zero order in B:
Rate = k[A]¹[B]⁰ = k[A]
From experiment I: 2.0 × 10⁻² = k (0.1), so k = 0.2 min⁻¹
Experiment II: 4.0 × 10⁻² = 0.2 × [A], so [A] = 0.2 mol L⁻¹
Experiment III: Rate = 0.2 × 0.4 = 0.08 mol L⁻¹ min⁻¹
Experiment IV: 2.0 × 10⁻² = 0.2 × [A], so [A] = 0.1 mol L⁻¹
Notice that changing [B] has no effect at all on the rate, exactly as expected for a zero order dependence.

Q7. Define rate constant or specific reaction rate.

Answer: In the rate law Rate = k[A]x[B]y, the constant k is the rate constant.
If the molar concentration of each reactant is unity (1 mol L⁻¹), then the rate becomes equal to k. For this reason k is also called the specific reaction rate.
The rate constant depends on temperature and on the presence of a catalyst, but it does not depend on the concentration of the reactants.

Q8. In a pseudo first order hydrolysis of an ester in water the following results were obtained. Calculate the average rate of reaction between 30 and 60 seconds.

t / s 0 30 60 90
[A] / mol L⁻¹ 0.55 0.31 0.17 0.085

Answer: Average rate = −Δ[A] / Δt
= −(0.17 − 0.31) / (60 − 30)
= 0.14 / 30
= 4.67 × 10⁻³ mol L⁻¹ s⁻¹

Q9. In a reaction between A and B, the initial rate was measured for different initial concentrations. Find the order with respect to A and B.

[A] / mol L⁻¹ 0.20 0.20 0.40
[B] / mol L⁻¹ 0.30 0.10 0.05
r₀ / mol L⁻¹ s⁻¹ 5.07 × 10⁻⁵ 5.07 × 10⁻⁵ 1.43 × 10⁻⁴

Answer: Order with respect to B: Compare experiments 1 and 2. [A] stays the same while [B] changes from 0.30 to 0.10, but the rate does not change at all.
So the order with respect to B is zero.
Order with respect to A: Compare experiments 2 and 3. [A] doubles from 0.20 to 0.40, and the rate goes from 5.07 × 10⁻⁵ to 1.43 × 10⁻⁴, which is about 2.82 times.
Since 2x = 2.82, we get x = 1.5.
So the order with respect to A is 1.5, and the overall order is 1.5.

Q10. For the reaction 2A + B → A₂B, the rate = k[A][B]² with k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹. Calculate the initial rate when [A] = 0.1 and [B] = 0.2 mol L⁻¹, and the rate after [A] is reduced to 0.06 mol L⁻¹.

Answer: Initial rate:
Rate = (2.0 × 10⁻⁶)(0.1)(0.2)²
= (2.0 × 10⁻⁶)(0.1)(0.04)
= 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹
After [A] falls to 0.06:
A reacted = 0.1 − 0.06 = 0.04 mol L⁻¹
From the equation 2A + B, for every 2 mol of A, 1 mol of B reacts. So B reacted = 0.04/2 = 0.02 mol L⁻¹
[B] left = 0.2 − 0.02 = 0.18 mol L⁻¹
Rate = (2.0 × 10⁻⁶)(0.06)(0.18)²
= 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹

Q11. Define threshold energy and activation energy. How are they related?

Answer: Threshold energy is the minimum energy that the colliding molecules must have for the collision to be effective, that is for it to actually produce products.
Activation energy (Ea) is the extra energy that the reactant molecules must absorb, over and above the energy they already have, to reach the threshold.
Relation:
Threshold energy = Activation energy + average energy already possessed by the reactant molecules
A reaction with a low activation energy is fast, and one with a high activation energy is slow.

Q12. (i) How is the rate of reaction different from the rate constant? (ii) Why does the rate of reaction increase with a rise in temperature?

Answer: (i) Rate of reaction versus rate constant:
• The rate of reaction changes as the reaction proceeds, because the concentrations of the reactants keep falling. It depends on concentration.
• The rate constant stays the same throughout the reaction at a given temperature. It does not depend on concentration, only on temperature and catalyst.
(ii) Why rate increases with temperature: It is usually found that for every 10°C rise in temperature, the rate roughly doubles. The reason is not just that molecules move faster, but mainly that raising the temperature greatly increases the fraction of molecules having energy equal to or more than the activation energy.
On the Maxwell-Boltzmann distribution curve, the peak shifts to the right at higher temperature, and the area under the curve beyond Ea becomes much larger. More molecules can cross the energy barrier, so more collisions are effective and the reaction speeds up.

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