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

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A pharmaceutical company is studying the shelf life of a drug that decomposes by a first order reaction with a rate constant of 1.5 × 10⁻³ min⁻¹ at room temperature. If the initial concentration of the drug is 0.10 mol L⁻¹, calculate the time taken for the drug concentration to fall to 0.04 mol L⁻¹, and determine the half-life of the drug.

Using the first order rate equation: k = (2.303/t) log([R]₀/[R]) = (2.303/t) log(0.10/0.04) = (2.303/t) log(2.5).
t = (2.303 × 0.3979) / (1.5 × 10⁻³) = 0.9163 / 1.5 × 10⁻³ ≈ 611 minutes. This is the time for the drug to degrade from 0.10 to 0.04 mol L⁻¹.
Half-life: t₁/₂ = 0.693/k = 0.693 / (1.5 × 10⁻³) = 462 minutes ≈ 7.7 hours. This shelf-life information helps the company set expiry dates and storage conditions.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In food preservation, refrigeration is used to slow down spoilage. With reference to the Arrhenius equation, explain quantitatively why lowering the temperature from 35°C to 5°C dramatically reduces the rate of bacterial reactions responsible for food spoilage. Assume Ea ≈ 50 kJ mol⁻¹.

From the Arrhenius equation, k = Ae^(–Ea/RT). At 35°C (308 K): k₁ ∝ e^(–50000/(8.314×308)) = e^(–19.52); at 5°C (278 K): k₂ ∝ e^(–50000/(8.314×278)) = e^(–21.62).
The ratio k₁/k₂ = e^(21.62–19.52) = e^(2.10) ≈ 8.2. This means the spoilage reaction is about 8 times faster at 35°C than at 5°C, explaining the dramatic preservation effect of refrigeration.
This is a direct application of the principle that for every 10°C rise in temperature the rate constant nearly doubles. Lowering temperature by 30°C reduces the rate approximately 2³ ≈ 8-fold, aligning with the Arrhenius calculation.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

In an industrial process, MnO₂ is used as a catalyst for the decomposition of KClO₃ to produce O₂. Explain the role of MnO₂ in this reaction using the concept of activation energy, and state why using a catalyst is preferred over simply raising the temperature.

MnO₂ acts as a catalyst for 2KClO₃ → 2KCl + 3O₂. It provides an alternate reaction pathway with a lower activation energy, allowing a greater fraction of KClO₃ molecules to possess energy ≥ Ea' (reduced activation energy), thus increasing the rate of O₂ production.
Raising the temperature would also increase the rate (Arrhenius equation), but it increases energy consumption, may cause side reactions or decomposition of other materials in the reactor, and the higher temperature is harder to control safely.
Using a catalyst maintains milder reaction conditions (lower temperature and pressure), reduces energy costs, avoids undesired side reactions, and since MnO₂ is regenerated after the reaction, only a small amount is needed, making the process economically efficient.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

The hydrolysis of sucrose (cane sugar) in the presence of dilute HCl follows pseudo first order kinetics. With reference to the chapter, explain why this reaction is classified as pseudo first order even though water is also a reactant, and write the applicable rate expression.

The reaction is: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose). It is second order overall (first order in sucrose and first order in water), but water is taken as the solvent and is present in enormous excess.
Since the concentration of water does not change appreciably during the reaction (similar to the ethyl acetate hydrolysis example in the chapter), its concentration is effectively constant and is absorbed into the rate constant.
The effective rate law becomes: Rate = k[C₁₂H₂₂O₁₁], which is first order. This is called pseudo first order kinetics. The observed rate constant k is actually k_true × [H₂O], where [H₂O] is approximately constant throughout the reaction.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

An archaeologist finds a wooden artifact. Carbon-14 dating shows that the artifact contains 75% of the ¹⁴C present in a living tree. Given that the half-life of ¹⁴C is 5730 years and radioactive decay follows first order kinetics, estimate the age of the artifact.

Radioactive decay is a first order reaction. Using the first order equation: k = (2.303/t) log([R]₀/[R]). Since 75% of ¹⁴C remains, [R] = 0.75[R]₀, so [R]₀/[R] = 1/0.75 = 1.333.
First, k = 0.693/t₁/₂ = 0.693/5730 = 1.21 × 10⁻⁴ year⁻¹. Then: t = (2.303/k) log(1.333) = (2.303/1.21 × 10⁻⁴) × 0.1249 = 2377 years.
The artifact is approximately 2377 years old. This application demonstrates that first order half-life is independent of initial concentration — a crucial property for reliable radioactive dating in archaeology.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

In an automobile engine, fuel combustion must occur rapidly. With reference to collision theory, explain two key factors that engineers manipulate to ensure rapid and efficient combustion of fuel in the engine cylinder.

First factor — activation energy and temperature: In the engine, a spark plug initiates combustion by providing the activation energy to start the chain reaction. Once ignited, the exothermic combustion rapidly raises the temperature, which exponentially increases the rate constant (Arrhenius equation), sustaining rapid combustion.
Second factor — concentration and pressure: The fuel-air mixture is compressed by the piston before ignition, increasing the partial pressures (and thus concentrations) of fuel and oxygen molecules. Higher concentration increases the collision frequency Z_AB, directly increasing the reaction rate per collision theory: Rate = PZ_AB e^(–Ea/RT).
Proper orientation (steric factor P) is also considered in engine design — fuel injectors ensure fine atomization of fuel to maximize surface area and favourable molecular contact with O₂ for effective collisions.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

A researcher measures the rate of a reaction at 400 K and 450 K and finds that the rate constant doubles. Using the two-temperature Arrhenius equation, calculate the activation energy of the reaction (R = 8.314 J mol⁻¹K⁻¹).

Given: k₂/k₁ = 2, T₁ = 400 K, T₂ = 450 K. Using: log(k₂/k₁) = [Ea/(2.303R)] × [(T₂ – T₁)/(T₁T₂)].
Substituting: log 2 = [Ea/(2.303 × 8.314)] × [(450 – 400)/(450 × 400)] = [Ea/19.15] × [50/180000] = Ea × 2.778 × 10⁻⁶/19.15.
0.3010 = Ea × 1.451 × 10⁻⁷; therefore Ea = 0.3010/(1.451 × 10⁻⁷) ≈ 2.07 × 10⁶/10 ≈ 20,743 J mol⁻¹ ≈ 20.74 kJ mol⁻¹. This relatively low Ea is consistent with a reaction that is easily accelerated by modest temperature changes.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

With reference to the chapter, explain why platinum is used as a catalyst in the decomposition of ammonia, and why this reaction follows zero order kinetics at high pressure.

The reaction 2NH₃(g) → N₂(g) + 3H₂(g) is carried out at 1130 K on a platinum surface. Platinum acts as a heterogeneous catalyst by adsorbing NH₃ molecules onto its surface, forming a reactive intermediate that weakens N–H bonds, reducing the activation energy and enabling decomposition at a lower effective energy barrier.
At high pressure, the platinum metal surface becomes completely saturated with NH₃ molecules — all available active surface sites are occupied. Therefore, increasing the pressure (and hence concentration) of NH₃ further cannot increase the number of adsorbed molecules or the rate.
Since the rate no longer depends on [NH₃], Rate = k[NH₃]⁰ = k, which is zero order kinetics. The decomposition of HI on a gold surface is another similar example given in the chapter.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

With reference to the Arrhenius equation, a dentist needs a material that sets (hardens) rapidly at body temperature (37°C). Explain what chemical kinetic properties the setting reaction of the dental material should have, and how a catalyst might be incorporated.

For a dental filling material to set rapidly at 37°C (310 K), the setting (polymerisation or cross-linking) reaction should have a low activation energy Ea so that, at body temperature, a large fraction of molecules have energy ≥ Ea, giving a high rate constant k (from k = Ae^(–Ea/RT)).
A high frequency factor A (related to collision frequency and proper orientation) is also desirable, meaning the reactant molecules should collide frequently and with good orientation — achievable by using small, mobile monomers with complementary reactive groups.
A catalyst can be incorporated into the dental material to further lower the activation energy and speed up the setting reaction. The catalyst provides an alternate lower-energy pathway without being consumed, so it remains effective throughout the rapid setting process.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

In a nuclear power plant, ⁹⁰Sr (strontium-90) is a hazardous radioactive product with a half-life of 28.1 years. If a nuclear accident releases 10 mg of ⁹⁰Sr into the environment, calculate how much will remain after 56.2 years (assume first order decay).

Radioactive decay follows first order kinetics with t₁/₂ = 28.1 years. After 56.2 years: number of half-lives elapsed = 56.2/28.1 = 2 half-lives.
After one half-life (28.1 yr): amount remaining = 10/2 = 5 mg. After two half-lives (56.2 yr): amount remaining = 5/2 = 2.5 mg.
This can also be verified using: k = 0.693/28.1 = 0.02466 yr⁻¹; then [R] = [R]₀e^(–kt) = 10 × e^(–0.02466 × 56.2) = 10 × e^(–1.386) = 10 × 0.25 = 2.5 mg. The significant persistence of ⁹⁰Sr highlights the long-term radiological hazard of nuclear accidents.

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Chapter 3: Chemical Kinetics | Application Questions

In food preservation, refrigeration is used to slow down spoilage. With reference to the Arrhenius equation, explain quantitatively why lowering the temperature from 35°C to 5°C dramatically reduces the rate of bacterial reactions responsible for food spoilage. Assume Ea ≈ 50 kJ mol⁻¹.

In an industrial process, MnO₂ is used as a catalyst for the decomposition of KClO₃ to produce O₂. Explain the role of MnO₂ in this reaction using the concept of activation energy, and state why using a catalyst is preferred over simply raising the temperature.

The hydrolysis of sucrose (cane sugar) in the presence of dilute HCl follows pseudo first order kinetics. With reference to the chapter, explain why this reaction is classified as pseudo first order even though water is also a reactant, and write the applicable rate expression.

An archaeologist finds a wooden artifact. Carbon-14 dating shows that the artifact contains 75% of the ¹⁴C present in a living tree. Given that the half-life of ¹⁴C is 5730 years and radioactive decay follows first order kinetics, estimate the age of the artifact.

In an automobile engine, fuel combustion must occur rapidly. With reference to collision theory, explain two key factors that engineers manipulate to ensure rapid and efficient combustion of fuel in the engine cylinder.

A researcher measures the rate of a reaction at 400 K and 450 K and finds that the rate constant doubles. Using the two-temperature Arrhenius equation, calculate the activation energy of the reaction (R = 8.314 J mol⁻¹K⁻¹).

With reference to the chapter, explain why platinum is used as a catalyst in the decomposition of ammonia, and why this reaction follows zero order kinetics at high pressure.

With reference to the Arrhenius equation, a dentist needs a material that sets (hardens) rapidly at body temperature (37°C). Explain what chemical kinetic properties the setting reaction of the dental material should have, and how a catalyst might be incorporated.

In a nuclear power plant, ⁹⁰Sr (strontium-90) is a hazardous radioactive product with a half-life of 28.1 years. If a nuclear accident releases 10 mg of ⁹⁰Sr into the environment, calculate how much will remain after 56.2 years (assume first order decay).

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