Chapter 3: Chemical Kinetics Case Study Questions | chemistry Class 12 CBSE

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

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A student studying reaction kinetics measured the concentration of butyl chloride (C4H9Cl) during its hydrolysis reaction with water. Starting with an initial concentration of 0.100 mol/L, she recorded concentrations at intervals: 0.0905 mol/L at 50 s, 0.0820 mol/L at 100 s, 0.0741 mol/L at 150 s, and 0.0671 mol/L at 200 s. She plotted concentration versus time and drew tangents at various points to find instantaneous rates. She noticed that as the reaction progressed, the rate consistently decreased even though external conditions remained unchanged. Her teacher explained that this was due to the decreasing concentration of the reactant over time, a classic feature of reactions that follow specific rate laws.

Question 1: Calculate the average rate of hydrolysis of C4H9Cl between t = 50 s and t = 100 s.

Average rate = -Δ[C4H9Cl]/Δt
= -(0.0820 - 0.0905)/(100 - 50)
= 0.0085/50 = 1.70 × 10⁻⁴ mol L⁻¹ s⁻¹

Question 2: The student observes that average rate decreases with time. Why can average rate not be used to predict the rate at a specific instant?

Average rate remains constant over the entire time interval for which it is calculated.
It cannot represent the continuously changing rate at a particular moment.
Instantaneous rate (slope of the tangent at a given point on the concentration-time graph) must be used instead.

Question 3: Describe how you would graphically determine the instantaneous rate of hydrolysis of C4H9Cl at t = 600 s. What mathematical expression represents the instantaneous rate?

Plot concentration of C4H9Cl on the y-axis versus time on the x-axis.
Draw a tangent line to the curve at the point corresponding to t = 600 s.
Calculate the slope of this tangent: r_inst = -d[C4H9Cl]/dt
The negative sign is used because concentration of reactant decreases with time; rate must be positive.

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A chemist studied the oxidation of nitrogen monoxide: 2NO(g) + O2(g) → 2NO2(g). She measured initial rates at different starting concentrations. When [NO] = 0.30 mol/L and [O2] = 0.30 mol/L, the rate was 0.096 mol L⁻¹ s⁻¹. Doubling [NO] to 0.60 mol/L while keeping [O2] constant gave a rate of 0.384 mol L⁻¹ s⁻¹. Doubling [O2] to 0.60 mol/L while keeping [NO] constant gave a rate of 0.192 mol L⁻¹ s⁻¹. From this data, she derived the rate law and rate constant. She also noted that the exponents in the rate law were identical to the stoichiometric coefficients in the balanced equation, though she cautioned her students that this is not always the case.

Question 1: What is the order of the reaction with respect to NO and O2? State the overall order.

When [NO] is doubled (O2 constant), rate increases 4 times → order with respect to NO = 2.
When [O2] is doubled (NO constant), rate doubles → order with respect to O2 = 1.
Overall order = 2 + 1 = 3 (third order).

Question 2: Write the rate law expression and calculate the rate constant k with proper units.

Rate = k[NO]²[O2]
k = Rate/([NO]²[O2]) = 0.096/((0.30)² × 0.30) = 0.096/0.027
k = 3.56 mol⁻² L² s⁻¹

Question 3: The chemist cautioned that rate law cannot always be predicted from the balanced equation. Explain this with a suitable example, and state what determines the rate law experimentally.

Rate law must be determined experimentally; it cannot be inferred from stoichiometric coefficients alone.
Example: CH3COOC2H5 + H2O → CH3COOH + C2H5OH has Rate = k[CH3COOC2H5]¹[H2O]⁰, where the exponent for water is 0, not 1.
The exponents (orders) reflect the mechanism of the reaction, specifically the slowest (rate-determining) step.
The rate constant k and the order are both determined from experimental data of initial rates at varying concentrations.

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Case Set 3

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During a laboratory activity, students investigated the decomposition of N2O5 in CCl4 at 318 K: 2N2O5(g) → 4NO2(g) + O2(g). The initial concentration was 2.33 mol/L; after 184 minutes it dropped to 2.08 mol/L. A second group studied the thermal decomposition of H2O2 in alkaline medium using iodide ions as catalyst. They found the rate equation to be: Rate = k[H2O2][I⁻], implying first order in both reactants. They identified an intermediate IO⁻ formed in the first slow step, confirming it as the rate-determining step. Both groups compared their results and discussed how the mechanism of a reaction determines its experimentally observed rate law.

Question 1: Calculate the average rate of decomposition of N2O5 in mol L⁻¹ min⁻¹.

Average Rate = -(1/2) × Δ[N2O5]/Δt
= -(1/2) × (2.08 - 2.33)/184
= -(1/2) × (-0.25/184) = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹

Question 2: In the H2O2 decomposition, IO⁻ is described as an intermediate. What is a reaction intermediate, and how does it differ from a catalyst?

A reaction intermediate is a species formed during the course of a reaction but not present in the overall balanced equation.
IO⁻ is generated in step 1 and consumed in step 2, so it does not appear in the net equation.
A catalyst is a substance that increases the rate without being permanently consumed; it appears in the mechanism but is regenerated by the end.

Question 3: Explain the concept of rate-determining step using the H2O2 decomposition mechanism. How does the slowest step control the overall rate?

The H2O2 decomposition proceeds in two steps: (1) H2O2 + I⁻ → H2O + IO⁻ (slow), and (2) H2O2 + IO⁻ → H2O + I⁻ + O2 (fast).
The first step is slow and controls how quickly products form — analogous to the slowest runner in a relay team determining the team's overall time.
The rate law derived from the slow step: Rate = k[H2O2][I⁻] matches the experimentally observed rate equation.
For complex reactions, the order equals the molecularity of the slowest step.

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A pharmaceutical researcher studied the degradation of a drug molecule A in solution. She found it followed zero order kinetics under saturated enzyme conditions. Starting with 0.80 mol/L, the concentration fell to 0.60 mol/L in 200 seconds. She also studied a related drug B that underwent first order degradation with rate constant 0.0231 min⁻¹. For regulatory approval, she needed to determine shelf life — the time for the drug concentration to fall to 50% (half-life). She plotted concentration vs time for A and ln(concentration) vs time for B, obtaining straight lines in each case, confirming the respective reaction orders.

Question 1: For drug A (zero order), calculate the rate constant and the half-life.

k = ([R]0 - [R])/t = (0.80 - 0.60)/200 = 0.001 mol L⁻¹ s⁻¹
t½ = [R]0/2k = 0.80/(2 × 0.001) = 400 s

Question 2: Calculate the half-life of drug B and explain why the half-life of a first order reaction is independent of initial concentration.

t½ = 0.693/k = 0.693/0.0231 = 30 min
For first order: t½ = 0.693/k — there is no [R]0 term, so it is concentration-independent.
This is a key difference from zero order kinetics where t½ ∝ [R]0.

Question 3: Compare the integrated rate laws, graphical plots, and units of rate constant for zero order and first order reactions in a tabular manner.

Zero order: Integrated law: kt = [R]0 - [R]; Graph: [R] vs t (straight line, slope = -k); Units of k: mol L⁻¹ s⁻¹.
First order: Integrated law: kt = ln([R]0/[R]); Graph: ln[R] vs t (straight line, slope = -k); Units of k: s⁻¹.
Zero order half-life depends on initial concentration; first order half-life is constant and independent of [R]0.

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Case Set 5

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In an industrial setup, engineers studied the effect of temperature on the rate of a chemical reaction used in polymer synthesis. They found that at 300 K, the rate constant was 2.0 × 10⁻³ s⁻¹ and at 320 K it was 8.0 × 10⁻³ s⁻¹. They used this data to compute activation energy using the Arrhenius equation. They also noted that even at high temperatures, not all molecular collisions led to product formation. A separate factor P (steric factor) needed to be incorporated, giving the modified rate equation: Rate = PZ_AB × e^(-Ea/RT), where Z_AB is the collision frequency. The engineers emphasized that both adequate energy AND correct molecular orientation were essential for effective collisions.

Question 1: State the Arrhenius equation and identify each term. What does the factor e^(-Ea/RT) physically represent?

Arrhenius equation: k = A × e^(-Ea/RT)
A = frequency/pre-exponential factor (collision frequency); Ea = activation energy (J mol⁻¹); R = gas constant; T = absolute temperature.
e^(-Ea/RT) represents the fraction of molecules possessing kinetic energy equal to or greater than the activation energy Ea.

Question 2: Calculate the activation energy Ea using the given data (k1 = 2.0 × 10⁻³ s⁻¹ at 300 K, k2 = 8.0 × 10⁻³ s⁻¹ at 320 K). Use R = 8.314 J mol⁻¹ K⁻¹.

log(k2/k1) = Ea(T2 - T1)/(2.303R × T1 × T2)
log(8.0×10⁻³/2.0×10⁻³) = Ea × (320-300)/(2.303 × 8.314 × 300 × 320)
log 4 = Ea × 20/(2.303 × 8.314 × 96000)
0.602 = Ea × 20/1.793×10⁵; Ea = 0.602 × 1.793×10⁵/20 ≈ 5.40 × 10³ × 10 ≈ 53,900 J mol⁻¹ ≈ 53.9 kJ mol⁻¹

Question 3: Explain the Maxwell-Boltzmann distribution of energies and how increasing temperature affects the fraction of effective collisions. Why does the rate approximately double for every 10°C rise?

The Maxwell-Boltzmann distribution shows that molecules have a spread of kinetic energies with a peak at the most probable energy.
When temperature increases by 10°C, the curve shifts right and broadens; the fraction of molecules with energy ≥ Ea approximately doubles.
This causes the area under the curve beyond Ea to roughly double, hence the rate constant nearly doubles.
Because rate = k × f([A], [B]) and k doubles with a 10° rise, the overall rate also approximately doubles.

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Chapter 3: Chemical Kinetics | Case Study Questions

Passage

Question 1: Calculate the average rate of hydrolysis of C4H9Cl between t = 50 s and t = 100 s.

Question 2: The student observes that average rate decreases with time. Why can average rate not be used to predict the rate at a specific instant?

Question 3: Describe how you would graphically determine the instantaneous rate of hydrolysis of C4H9Cl at t = 600 s. What mathematical expression represents the instantaneous rate?

Passage

Question 1: What is the order of the reaction with respect to NO and O2? State the overall order.

Question 2: Write the rate law expression and calculate the rate constant k with proper units.

Question 3: The chemist cautioned that rate law cannot always be predicted from the balanced equation. Explain this with a suitable example, and state what determines the rate law experimentally.

Passage

Question 1: Calculate the average rate of decomposition of N2O5 in mol L⁻¹ min⁻¹.

Question 2: In the H2O2 decomposition, IO⁻ is described as an intermediate. What is a reaction intermediate, and how does it differ from a catalyst?

Question 3: Explain the concept of rate-determining step using the H2O2 decomposition mechanism. How does the slowest step control the overall rate?

Passage

Question 1: For drug A (zero order), calculate the rate constant and the half-life.

Question 2: Calculate the half-life of drug B and explain why the half-life of a first order reaction is independent of initial concentration.

Question 3: Compare the integrated rate laws, graphical plots, and units of rate constant for zero order and first order reactions in a tabular manner.

Passage

Question 1: State the Arrhenius equation and identify each term. What does the factor e^(-Ea/RT) physically represent?

Question 2: Calculate the activation energy Ea using the given data (k1 = 2.0 × 10⁻³ s⁻¹ at 300 K, k2 = 8.0 × 10⁻³ s⁻¹ at 320 K). Use R = 8.314 J mol⁻¹ K⁻¹.

Question 3: Explain the Maxwell-Boltzmann distribution of energies and how increasing temperature affects the fraction of effective collisions. Why does the rate approximately double for every 10°C rise?

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