Chapter 3: Chemical Kinetics Long Questions | chemistry Class 12 CBSE

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Chapter 3: Chemical Kinetics Long Questions | chemistry Class 12 CBSE | ByteNotes.in

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Compare the integrated rate laws for zero order and first order reactions with respect to their mathematical form, graphical representation, half-life, and units of rate constant.

Zero order: [R] = [R]₀ – kt; plot of [R] vs t is a straight line with slope = –k; t₁/₂ = [R]₀/2k (depends on initial concentration); units of k: mol L⁻¹s⁻¹.
First order: ln[R] = –kt + ln[R]₀ or [R] = [R]₀e^(–kt); plot of ln[R] vs t is a straight line with slope = –k; t₁/₂ = 0.693/k (independent of initial concentration); units of k: s⁻¹.
Key distinction in half-life: for zero order reactions, each successive half-life is shorter (as [R]₀ decreases), while for first order reactions, every half-life is equal and constant.
Zero order reactions occur under special conditions (e.g., surface saturation), while first order reactions are common (e.g., radioactive decay, hydrogenation of ethene).
The independence of first order half-life from concentration is a defining diagnostic feature used to identify first order reactions experimentally.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 2

Long Form

Analyse the difference between molecularity and order of a reaction using at least two examples to support each point of distinction.

Order is experimentally determined and can be 0, 1, 2, 3, or a fractional value (e.g., 3/2 for CH₃CHO decomposition); molecularity is theoretical and can only be 1, 2, or 3 (positive integers).
Order applies to both elementary and complex reactions; molecularity is defined only for elementary reactions and has no meaning for complex multi-step reactions.
For complex reactions, the overall order is determined by the rate-determining (slowest) step; the molecularity of that slowest step equals the overall order of the reaction.
Example: KClO₃ + 6FeSO₄ + 3H₂SO₄ → products appears to be 10th order from stoichiometry but is actually second order, showing order ≠ sum of stoichiometric coefficients.
Example: for 2NO + O₂ → 2NO₂, experimentally Rate = k[NO]²[O₂], order = 3; this matches the molecularity only because this is an elementary reaction.

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Long Answer

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Medium

Question 3

Long Form

Explain the Arrhenius equation and describe how a graph of ln k vs 1/T can be used to calculate activation energy and the frequency factor.

The Arrhenius equation k = Ae^(–Ea/RT) shows that rate constant k depends exponentially on temperature T, where A is the frequency factor and Ea is the activation energy.
Taking natural log: ln k = –(Ea/R)(1/T) + ln A. This is of the form y = mx + c, where y = ln k, x = 1/T, slope m = –Ea/R, and intercept = ln A.
A plot of ln k vs 1/T gives a straight line with a negative slope; from slope = –Ea/R, activation energy Ea = –slope × R.
The intercept of the line on the y-axis (when 1/T = 0) equals ln A, so A = e^(intercept), giving the frequency factor.
This graphical method is used experimentally: rate constants are measured at several temperatures, ln k values are plotted against 1/T, and the best-fit line yields both Ea and A simultaneously.

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Question 4

Long Form

Describe collision theory of chemical reactions, including the concept of effective collision, steric factor, and the limitations of the theory.

Collision theory (Max Trautz and William Lewis, 1916–18) treats reactant molecules as hard spheres; reaction occurs when molecules collide. The rate = Z_AB × e^(–Ea/RT) for a bimolecular reaction, where Z_AB is the collision frequency and e^(–Ea/RT) is the fraction of molecules with energy ≥ Ea.
Not all collisions lead to products; only effective collisions do — those with both sufficient kinetic energy (≥ threshold energy) and proper molecular orientation.
To account for orientation, a steric factor P is introduced, modifying the equation to: Rate = PZ_AB e^(–Ea/RT). P takes into account that molecules must be properly aligned for bonds to break and reform.
The equation predicts rate constants accurately for reactions involving atomic species or simple molecules but shows significant deviations for complex molecules.
Limitation: collision theory treats molecules as featureless hard spheres, ignoring their actual structure and electronic properties, which is a major drawback for complex organic molecules.

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Question 5

Long Form

Explain the mechanism of decomposition of H₂O₂ catalysed by I⁻ ions in alkaline medium. Identify the intermediate, rate-determining step, and justify the experimental rate law.

The overall reaction is: 2H₂O₂ → 2H₂O + O₂ (catalysed by I⁻ in alkaline medium). The experimental rate law is: Rate = k[H₂O₂][I⁻], showing first order with respect to both H₂O₂ and I⁻.
The two-step mechanism proposed: Step 1 (slow, rate-determining): H₂O₂ + I⁻ → H₂O + IO⁻; Step 2 (fast): H₂O₂ + IO⁻ → H₂O + I⁻ + O₂.
The species IO⁻ is called an intermediate — it is formed during the course of the reaction but does not appear in the overall balanced equation.
Since step 1 is slow and rate-determining, the rate of the overall reaction equals the rate of step 1: Rate = k[H₂O₂][I⁻], which matches the experimental rate law.
Both steps are bimolecular elementary reactions; I⁻ is regenerated in step 2, confirming its role as a catalyst, not a reactant in the overall equation.

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Long Answer

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Hard

Question 6

Long Form

Analyse how a catalyst lowers activation energy and affects the rate of reaction without altering equilibrium. Use an energy profile diagram in your explanation.

A catalyst provides an alternate reaction pathway with a lower activation energy (Ea') compared to the uncatalysed pathway (Ea). According to the Arrhenius equation, lower Ea leads to a larger e^(–Ea/RT) factor, hence a higher rate constant k.
On the potential energy vs reaction coordinate diagram: the uncatalysed path shows a high energy barrier (activated complex C at high energy), while the catalysed path shows a lower peak C', representing the lower activation energy.
The catalyst does not alter the energies of reactants or products; hence the enthalpy change (ΔH) and Gibbs energy change (ΔG) of the reaction remain unchanged.
Since ΔG is unchanged, the equilibrium constant K is also unchanged (ΔG = –RT ln K). The catalyst speeds up both forward and backward reactions equally, so equilibrium is attained faster but the equilibrium position is the same.
A catalyst cannot make a thermodynamically non-spontaneous reaction (ΔG > 0) proceed; it only accelerates spontaneous reactions. A small quantity of catalyst is sufficient since it is regenerated after the reaction.

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Long Answer

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Hard

Question 7

Long Form

Derive the integrated rate equation for a first order reaction and show that time required for 99.9% completion is 10 times the half-life.

For R → P, rate = –d[R]/dt = k[R]. Rearranging: d[R]/[R] = –k dt. Integrating: ln[R] = –kt + ln[R]₀, or k = (2.303/t) log([R]₀/[R]).
For t₁/₂: [R] = [R]₀/2, so k = (2.303/t₁/₂) log 2 = (2.303 × 0.301)/t₁/₂, giving t₁/₂ = 0.693/k.
For 99.9% completion: [R] = [R]₀ – 0.999[R]₀ = 0.001[R]₀. So k = (2.303/t) log([R]₀/0.001[R]₀) = (2.303/t) log(10³) = (2.303 × 3)/t.
Therefore t = 6.909/k.
Ratio t/t₁/₂ = (6.909/k)/(0.693/k) = 6.909/0.693 = 10. Hence, time for 99.9% completion = 10 × t₁/₂.

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Long Answer

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Hard

Question 8

Long Form

Explain the Maxwell-Boltzmann energy distribution and analyse how temperature increase leads to a doubling of the rate constant for every 10°C rise.

The Maxwell-Boltzmann distribution plots the fraction of molecules (N_E/N_T) against kinetic energy. The peak represents the most probable kinetic energy; fewer molecules exist at very high or low energies.
At temperature T, only molecules with energy ≥ Ea can react. The area under the curve to the right of Ea represents this fraction, corresponding to e^(–Ea/RT) in the Arrhenius equation.
When temperature is raised by 10°C (to T+10), the curve shifts right and broadens; the area beyond Ea effectively doubles. This is because the exponent –Ea/R(T+10) is less negative than –Ea/RT, so e^(–Ea/R(T+10)) ≈ 2 × e^(–Ea/RT).
Since k = Ae^(–Ea/RT) and the exponential factor doubles, the rate constant k also approximately doubles for every 10°C rise in temperature.
This doubling rule is empirical and holds approximately for reactions with activation energies around 50–60 kJ/mol near room temperature; deviations are observed at very high or low temperatures.

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Long Answer

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Question 9

Long Form

For a first order gas phase reaction A(g) → B(g) + C(g), derive the integrated rate equation in terms of initial pressure and total pressure at time t.

Let pᵢ be the initial pressure of A and p_t be the total pressure at time t. If x atm is the decrease in pressure of A at time t, then at time t: p_A = pᵢ – x; p_B = x; p_C = x.
Total pressure: p_t = (pᵢ – x) + x + x = pᵢ + x, so x = p_t – pᵢ.
Therefore p_A = pᵢ – x = pᵢ – (p_t – pᵢ) = 2pᵢ – p_t.
Substituting into the first order rate equation k = (2.303/t) log([R]₀/[R]) in terms of pressure: k = (2.303/t) log(pᵢ/p_A) = (2.303/t) log[pᵢ/(2pᵢ – p_t)].
This equation allows calculation of the rate constant k from experimentally measured values of total pressure p_t at various times t, without needing individual partial pressures directly.

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Long Answer

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Question 10

Long Form

Explain the concept of rate constant k and analyse how its value and units change with the order of the reaction.

The rate constant k is the proportionality constant in the rate law: Rate = k[A]ˣ[B]ʸ. It is specific to a reaction at a given temperature and does not depend on the concentration of reactants.
Physically, k equals the rate of reaction when all reactant concentrations are unity (1 mol L⁻¹). A larger k indicates a faster reaction at any given concentration.
Units of k depend on reaction order: k = Rate/([A]ˣ[B]ʸ) = (mol L⁻¹s⁻¹)/(mol L⁻¹)ⁿ = (mol L⁻¹)^(1–n) s⁻¹. For zero order: mol L⁻¹s⁻¹; first order: s⁻¹; second order: mol⁻¹ L s⁻¹.
The unit of k can be used as a diagnostic tool to identify reaction order: if k has units of L mol⁻¹s⁻¹, the reaction is second order (as in Example 3.4 of the chapter: k = 2.3 × 10⁻⁵ L mol⁻¹s⁻¹ is second order).
k increases exponentially with temperature (Arrhenius equation), while changing concentration has no effect on k — distinguishing k from rate itself.

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Chapter 3: Chemical Kinetics | Long Questions

Compare the integrated rate laws for zero order and first order reactions with respect to their mathematical form, graphical representation, half-life, and units of rate constant.

Analyse the difference between molecularity and order of a reaction using at least two examples to support each point of distinction.

Explain the Arrhenius equation and describe how a graph of ln k vs 1/T can be used to calculate activation energy and the frequency factor.

Describe collision theory of chemical reactions, including the concept of effective collision, steric factor, and the limitations of the theory.

Explain the mechanism of decomposition of H₂O₂ catalysed by I⁻ ions in alkaline medium. Identify the intermediate, rate-determining step, and justify the experimental rate law.

Analyse how a catalyst lowers activation energy and affects the rate of reaction without altering equilibrium. Use an energy profile diagram in your explanation.

Derive the integrated rate equation for a first order reaction and show that time required for 99.9% completion is 10 times the half-life.

Explain the Maxwell-Boltzmann energy distribution and analyse how temperature increase leads to a doubling of the rate constant for every 10°C rise.

For a first order gas phase reaction A(g) → B(g) + C(g), derive the integrated rate equation in terms of initial pressure and total pressure at time t.

Explain the concept of rate constant k and analyse how its value and units change with the order of the reaction.

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