CBSE Class 12 Chemistry Chemical Kinetics Worksheet Set 05

Chemical Kinetics Worksheet

Question 1. What is meant by the rate of a chemical reaction? Distinguish between the average rate and the instantaneous rate of a reaction.
Answer:
Rate of reaction is the change in the concentration of a reactant or a product per unit time. It is always positive, so a minus sign is used when a reactant is considered and a plus sign when a product is considered. Its unit is mol L⁻¹ s⁻¹.
Average rate is the change in concentration over a measurable interval of time: \( r_{av} = -\frac{\Delta[R]}{\Delta t} = +\frac{\Delta[P]}{\Delta t} \). It is the slope of the chord joining two points on the concentration–time curve.
Instantaneous rate is the rate at one particular instant, obtained when the interval becomes very small: \( r_{inst} = -\frac{d[R]}{dt} = +\frac{d[P]}{dt} \). It is the slope of the tangent to the concentration–time curve at that instant.
For most reactions the rate falls with time, so the average rate depends on the interval chosen, while the instantaneous rate has one value at each moment.
In simple words: Average rate is for a stretch of time; instantaneous rate is for a single moment.

Teacher's Note:
1. Draw the curve and mark a chord and a tangent to show the difference.
2. Write the sign convention: negative for reactants, positive for products.

Question 2. For the reaction N₂ + 3H₂ → 2NH₃, the rate of formation of NH₃ is 3.0 × 10⁻⁴ mol L⁻¹ s⁻¹. Calculate the rate of disappearance of H₂ and N₂ and the rate of the reaction.
Answer:
Rate of reaction = \( -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt} \).
Rate of reaction = ½ × 3.0 × 10⁻⁴ = 1.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
−d[N₂]/dt = rate of reaction = 1.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
−d[H₂]/dt = 3 × 1.5 × 10⁻⁴ = 4.5 × 10⁻⁴ mol L⁻¹ s⁻¹.
In simple words: H₂ disappears three times as fast as N₂ does, and NH₃ forms twice as fast as N₂ disappears.

Teacher's Note:
1. Divide each rate by its coefficient to get the rate of reaction.
2. Then multiply by the coefficient of the species you need.

Teacher's Note:
1. A table is the best way to present this in the board exam.
2. Always add one example to show order and molecularity can differ.

Teacher's Note:
1. Find the pair of experiments in which one concentration changes at a time.
2. Verify the rate law with the third experiment before calculating k.

Question 5. Write the units of the rate constant for (a) a zero order, (b) a first order and (c) a second order reaction. Derive the general expression for an nth order reaction.
Answer:
For rate = k[A]ⁿ: \( k = \frac{\text{rate}}{[A]^n} = \frac{\text{mol L}^{-1}\text{ s}^{-1}}{(\text{mol L}^{-1})^{n}} \), so the unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹.
(a) n = 0: mol L⁻¹ s⁻¹.
(b) n = 1: s⁻¹.
(c) n = 2: mol⁻¹ L s⁻¹.
For gaseous reactions with pressure in atm, the unit becomes atm¹⁻ⁿ s⁻¹.
In simple words: The unit of k tells you the order: s⁻¹ means first order and mol L⁻¹ s⁻¹ means zero order.

Teacher's Note:
1. Start from the unit of rate and divide by the unit of concentration raised to n.
2. Memorise the three common cases.

Question 6. Derive the integrated rate equation for a zero order reaction. Obtain its half-life and describe the graph of [R] against time.
Answer:
For R → P with rate = k[R]⁰ = k: \( -\frac{d[R]}{dt} = k \), so d[R] = −k dt.
Integrating, [R] = −kt + I, where I is the constant of integration. At t = 0, [R] = [R]₀, so I = [R]₀.
[R] = [R]₀ − kt, that is, \( k = \frac{[R]_0 - [R]}{t} \).
Half-life: at t = t½, [R] = [R]₀/2, so \( t_{1/2} = \frac{[R]_0}{2k} \). It is directly proportional to the initial concentration.
Graph: [R] against t is a straight line with intercept [R]₀ and slope −k. The reaction is complete at t = [R]₀/k.
In simple words: In a zero order reaction the reactant is used up at a steady rate, so the plot of concentration against time is a straight line.

Teacher's Note:
1. Show the limits of integration or the constant I clearly.
2. Remember that t½ depends on [R]₀ only for zero order.

Question 7. Derive the integrated rate equation for a first order reaction and show that its half-life is independent of the initial concentration.
Answer:
For R → P with rate = k[R]: \( \frac{d[R]}{[R]} = -k\,dt \).
Integrating, ln [R] = −kt + I. At t = 0, [R] = [R]₀, so I = ln [R]₀.
Hence ln [R] = ln [R]₀ − kt, or \( k = \frac{1}{t}\ln\frac{[R]_0}{[R]} \), or \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \).
At t = t½, [R] = [R]₀/2, so \( t_{1/2} = \frac{1}{k}\ln 2 = \frac{0.693}{k} \).
[R]₀ has cancelled out, so t½ is the same whatever the starting concentration. A plot of ln [R] against t is a straight line of slope −k.
In simple words: Every halving of the concentration takes the same time, no matter how much reactant you start with.

Teacher's Note:
1. Show the step where [R]₀ cancels to earn the mark for independence.
2. Quote t½ = 0.693/k as the final result.

Question 8. The half-life of a first order reaction is 40 minutes. Calculate (a) the rate constant and (b) the time required for 90% of the reaction to be completed.
Answer:
(a) k = 0.693 ÷ t½ = 0.693 ÷ 40 = 0.01733 min⁻¹ (1.73 × 10⁻² min⁻¹).
(b) For 90% completion, [R]₀/[R] = 100/10 = 10.
t = (2.303/k) log 10 = 2.303 ÷ 0.01733 = 133 minutes (about 2.2 hours).
In simple words: Find k from the half-life, then use the logarithmic equation with the remaining fraction.

Teacher's Note:
1. For 90% completion the remaining fraction is 10%.
2. log 10 = 1 makes the working short.

Question 9. A first order reaction has a rate constant of 2.0 × 10⁻² s⁻¹. If the initial concentration is 0.50 mol L⁻¹, calculate (a) the concentration after 60 s and (b) the time taken for 75% of the reactant to react.
Answer:
(a) ln ([R]₀/[R]) = kt = 2.0 × 10⁻² × 60 = 1.2.
[R]₀/[R] = e^1.2 = 3.32, so [R] = 0.50 ÷ 3.32 = 0.151 mol L⁻¹ (about 0.15 mol L⁻¹).
(b) 75% completion means [R]₀/[R] = 4, so t = (2.303/k) log 4 = 2.303 × 0.6021 ÷ 2.0 × 10⁻² = 69.3 s.
This equals two half-lives (2 × 34.65 s), as expected.
In simple words: Use ln or log in the first order equation; 75% completion always takes two half-lives.

Teacher's Note:
1. Check (b) against two half-lives: t½ = 0.693/k = 34.65 s.
2. Do not forget to convert percentage completion into [R]₀/[R].

Question 10. What are pseudo first order reactions? Give two examples.
Answer:
A reaction that is truly of higher order but behaves as a first order reaction because one reactant is present in large excess, so that its concentration remains almost constant, is called a pseudo first order reaction.
Example 1: acid hydrolysis of an ester, CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH. Rate = k[CH₃COOC₂H₅][H₂O]. Water is in large excess, so rate = k′[CH₃COOC₂H₅], where k′ = k[H₂O].
Example 2: inversion of cane sugar, C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ (glucose and fructose). Rate = k′[C₁₂H₂₂O₁₁], with water in large excess.
In simple words: When one reactant is in huge excess its concentration hardly changes, so only the other reactant seems to control the rate.

Teacher's Note:
1. An excess reagent does not change the order found from the rate law but behaves as a constant.
2. Both examples are hydrolysis reactions in which the solvent is the excess reactant.

Question 11. State the collision theory of chemical reactions. What are the two conditions for a collision to be effective?
Answer:
According to collision theory, reactant molecules must collide for a reaction to take place, and the rate depends on the number of collisions per second (collision frequency, Z). Only a small fraction of the collisions lead to products; these are called effective collisions.
A collision is effective only when (i) the colliding molecules have a minimum energy, equal to or greater than the activation energy E_a, and (ii) they collide with the proper orientation, so that the bonds can break and form.
These ideas are expressed in k = PZe^−E_a/RT, where P is the steric (probability) factor for orientation and e^−E_a/RT is the fraction of molecules with energy ≥ E_a.
In simple words: Molecules must hit hard enough and in the right way for a reaction to happen.

Teacher's Note:
1. Name both the energy barrier and the orientation barrier in the answer.
2. Mention that only a small fraction of collisions are effective.

Question 12. (a) For an exothermic reaction A → B, the activation energy of the forward reaction is 17 kJ mol⁻¹ and the enthalpy change is −40 kJ mol⁻¹. Find the activation energy of the reverse reaction. (b) For an endothermic reaction with ΔH = +20 kJ mol⁻¹ and E_a(forward) = 85 kJ mol⁻¹, find E_a(reverse).
Answer:
The enthalpy change is the difference of the two activation energies: ΔH = E_a(f) − E_a(b).
(a) −40 = 17 − E_a(b), so E_a(b) = 17 + 40 = 57 kJ mol⁻¹.
(b) +20 = 85 − E_a(b), so E_a(b) = 85 − 20 = 65 kJ mol⁻¹.
In an exothermic reaction the reverse barrier is higher than the forward one; in an endothermic reaction it is lower.
In simple words: The reverse barrier is the forward barrier plus the energy released (exothermic) or minus the energy absorbed (endothermic).

Teacher's Note:
1. Sketch the potential energy diagram to avoid sign errors.
2. ΔH is the same for both parts of the question; only its sign changes.

Question 13. The rate of a reaction doubles when the temperature is raised from 25 °C to 35 °C. Calculate the activation energy of the reaction (R = 8.314 J K⁻¹ mol⁻¹).
Answer:
T₁ = 298 K, T₂ = 308 K and k₂/k₁ = 2.
Use \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) \).
log 2 = E_a ÷ (2.303 × 8.314) × (10 ÷ (298 × 308)).
E_a = 0.3010 × 2.303 × 8.314 × 298 × 308 ÷ 10 = 52898 J mol⁻¹ ≈ 52.9 kJ mol⁻¹.
In simple words: A rise of 10 °C that doubles the rate corresponds to an activation energy of about 53 kJ per mole.

Teacher's Note:
1. Convert °C to K before substituting.
2. Give the final answer in kJ mol⁻¹ and check the power of ten.

Question 14. Why does the rate of a reaction increase so much on raising the temperature by only 10 °C, although the collision frequency increases by only about 2%?
Answer:
At a higher temperature the kinetic energy of the molecules increases and the energy distribution curve flattens and shifts to the right.
The fraction of molecules having energy equal to or greater than the activation energy, e^−E_a/RT, increases very sharply; for a 10 °C rise it roughly doubles.
The total number of collisions rises only slightly, but many more of the collisions are effective. Hence the rate (and k) rises by a factor of about 2 to 3 for every 10 °C rise.
In simple words: Heating does not make many more collisions; it makes many more of the collisions energetic enough to react.

Teacher's Note:
1. Mention the Maxwell–Boltzmann energy distribution curve in the answer.
2. State that E_a itself does not change with temperature.