Chemical Kinetics

1 exercises12 questions solved

Exercise 3.1Rate of Reaction, Rate Law and Integrated Rate Equations

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Solution

Δ[R] = 0.02 – 0.03 = –0.01 M, Δt = 25 min Average rate = –Δ[R]/Δt In minutes: Rate = 0.01 M / 25 min = 4 × 10⁻⁴ M min⁻¹ In seconds (25 min = 1500 s): Rate = 0.01 M / 1500 s = 6.67 × 10⁻⁶ M s⁻¹

In a reaction 2A → Products, the concentration of A decreases from 0.5 mol/L to 0.4 mol/L in 10 minutes. Calculate the rate during this interval.

Solution

For 2A → Products: Rate = –(1/2) × Δ[A]/Δt Δ[A] = 0.4 – 0.5 = –0.1 mol/L, Δt = 10 min Rate = –(1/2) × (–0.1)/10 = 0.1/20 = 0.005 mol L⁻¹ min⁻¹ = 5 × 10⁻³ mol L⁻¹ min⁻¹

For a reaction, A + B → Product; the rate law is given by r = k[A]^(1/2)[B]². What is the order of the reaction?

Solution

Rate law: r = k[A]^(1/2)[B]² Order with respect to A = 1/2 Order with respect to B = 2 Overall order = 1/2 + 2 = 5/2 = 2.5 The reaction is 2.5th order (or 5/2 order) overall.

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Solution

For second order kinetics: r = k[X]² Initial rate: r₁ = k[X]² When [X] is tripled: r₂ = k[3X]² = k × 9[X]² = 9r₁ The rate of formation of Y will increase by 9 times (becomes 9-fold).

A first order reaction has a rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?

Solution

k = 1.15 × 10⁻³ s⁻¹, [A]₀ = 5 g, [A] = 3 g For first order reaction: t = (2.303/k) × log([A]₀/[A]) = (2.303/1.15 × 10⁻³) × log(5/3) = 2002.6 × log(1.667) = 2002.6 × 0.2219 = 444.4 s ≈ 444 s

Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Solution

For first order: t₁/₂ = 0.693/k t₁/₂ = 60 min k = 0.693/t₁/₂ = 0.693/60 min = 0.01155 min⁻¹ In seconds: k = 0.01155/60 = 1.925 × 10⁻⁴ s⁻¹ Rate constant = 1.925 × 10⁻⁴ s⁻¹

What will be the effect of temperature on rate constant?

Solution

The rate constant increases with temperature. This is described by the Arrhenius equation: k = A × e^(–Ea/RT) where: • k = rate constant • A = pre-exponential (frequency) factor • Ea = activation energy • R = gas constant (8.314 J mol⁻¹ K⁻¹) • T = temperature in Kelvin Key observations: 1. As T increases, e^(–Ea/RT) increases → k increases 2. For a 10°C rise in temperature, the rate roughly doubles (rule of thumb) 3. Higher activation energy → more sensitive to temperature change 4. The logarithmic form: ln k = ln A – Ea/RT A plot of ln k vs 1/T gives a straight line with slope = –Ea/R

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate the energy of activation.

Solution

T₁ = 298 K, T₂ = 308 K, k₂/k₁ = 2 Using the Arrhenius equation in two-temperature form: log(k₂/k₁) = [Ea/(2.303R)] × [(T₂ – T₁)/(T₁T₂)] log 2 = [Ea/(2.303 × 8.314)] × [(308 – 298)/(298 × 308)] 0.3010 = [Ea/19.147] × [10/91784] 0.3010 = Ea × (5.353 × 10⁻⁶) / 19.147 0.3010 = Ea × 2.796 × 10⁻⁷ Ea = 0.3010 / (2.796 × 10⁻⁷) = 1.076 × 10⁶ J/mol ≈ 52.9 kJ/mol

The activation energy of a reaction is 75 kJ mol⁻¹ in the absence of a catalyst and 50 kJ mol⁻¹ with a catalyst. How many times will the rate of reaction grow in the presence of the catalyst if the reaction proceeds at 25°C?

Solution

Ea₁ = 75 kJ/mol = 75000 J/mol (without catalyst) Ea₂ = 50 kJ/mol = 50000 J/mol (with catalyst) T = 25°C = 298 K, R = 8.314 J mol⁻¹ K⁻¹ Using Arrhenius equation (same A): k_cat/k_uncat = e^(–Ea₂/RT) / e^(–Ea₁/RT) = e^((Ea₁ – Ea₂)/RT) = e^((75000 – 50000)/(8.314 × 298)) = e^(25000/2477.6) = e^10.09 = 2.41 × 10⁴ The rate will be approximately 2.4 × 10⁴ times faster with the catalyst.

Q10

A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, how will the rate of reaction change?

Solution

For second order with respect to CO: Rate = k[CO]² Initial rate: r₁ = k[CO]² When [CO] is doubled: r₂ = k[2CO]² = 4k[CO]² = 4r₁ The rate of reaction will increase 4 times (quadruple).

Q11

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law with t₁/₂ = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Solution

t₁/₂ = 3.00 h, t = 8 h For first order: k = 0.693/3.00 = 0.231 h⁻¹ Using: [A]/[A]₀ = e^(–kt) = e^(–0.231 × 8) = e^(–1.848) = 0.158 Alternatively using logarithmic form: log([A]/[A]₀) = –kt/2.303 = –0.231 × 8/2.303 = –0.8024 [A]/[A]₀ = 10^(–0.8024) = 0.1576 Fraction remaining ≈ 0.158 (about 15.8%)

Q12

The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.

Solution

t₁/₂ = 5730 years k = 0.693/5730 = 1.209 × 10⁻⁴ year⁻¹ [A]/[A]₀ = 80% = 0.80 Using first order: t = (2.303/k) × log([A]₀/[A]) = (2.303/1.209 × 10⁻⁴) × log(1/0.80) = 19049.6 × log(1.25) = 19049.6 × 0.0969 = 1845.9 years ≈ 1846 years The sample is approximately 1846 years old.

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Chemical Kinetics

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