CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. Define ‘molecularity of a reaction’. (Pb. Board 2011)
Answer:
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction, is called the molecularity of the reaction.

Teacher's Note:
1. Molecularity is a theoretical idea defined only for an elementary step, and it is always a whole number (1, 2 or rarely 3), never zero or a fraction.

Question 2. What is the rate determining step of a reaction ? (Chhattisgarh Board 2011)
Answer:
The slowest step is the rate determining step.

Teacher's Note:
1. The overall rate law is written from the slow step, which is why the rate law of a complex reaction can differ from its balanced equation.

Question 3. What is the molecularity of the reaction, Cl → \( \frac{1}{2} \) Cl₂(g) ?
Answer:
Two. Molecularity cannot be fractional; the elementary step actually involves the collision of two chlorine atoms, 2Cl → Cl₂(g).

Teacher's Note:
1. Never read molecularity off a fractional coefficient; write the step with whole numbers of colliding particles first.

Question 4. For the reaction NO₂ + CO → CO₂ + NO, the rate law is : Rate = k [NO₂]². Propose the probable mechanism of this reaction.
Answer:
NO₂ + NO₂ \( \xrightarrow{\text{slow}} \) NO + NO₃ ;
NO₃ + CO \( \xrightarrow{\text{fast}} \) CO₂ + NO₂

Teacher's Note:
1. Check a mechanism two ways: the steps must add up to the overall equation, and the slow step must give the observed rate law (here two NO₂ and no CO).

Question 5. State any one condition under which a bimolecular reaction may be kinetically of first order.
Answer:
A bimolecular reaction may become kinetically of first order if one of the reactants is present in excess.

Teacher's Note:
1. Acid hydrolysis of an ester in a large excess of water is the standard example of this situation.

Question 6. The reaction A + B → C has zero order. What is the rate equation ?
Answer:
Rate = \( \left(\frac{dx}{dt}\right) \) = k[A]⁰[B]⁰ = k (rate const.)

Teacher's Note:
1. For a zero order reaction the rate stays constant throughout, so the concentration falls linearly with time.

Question 7. A substance with initial concentration ‘a’ follows zero order kinetics. In how much time will the reaction go to completion ?
Answer:
\( \frac{dx}{dt} \) = k or dx = k dt ∴ x = kt + I.
When t = 0, x = 0. ∴ I = 0.
Hence, x = kt or t = x/k. For completion,
x = a ∴ t = a/k

Teacher's Note:
1. Unlike a first order reaction, a zero order reaction does reach completion in a finite time, which is twice its half-life (a/2k).

Question 8. How is half-life period (t_1/2) of a first order reaction related to its rate constant (k)? (MP Board 2011, J & K Board 2012)
Answer:
\( t_{1/2} = \frac{\ln 2}{k} = \frac{2.303}{k}\log 2 = \frac{0.693}{k} \).

Teacher's Note:
1. The relation contains no concentration term, which is why the half-life of a first order reaction is independent of initial concentration.

Question 9. How is half-life period related to initial concentration for a second order reaction ?
Answer:
\( t_{1/2} \propto a^{1-n} \). For n = 2, \( t_{1/2} \propto a^{-1} \), i.e., \( t_{1/2} \propto \frac{1}{a} \).
Thus, half-life period of a second order reaction is inversely proportional to the initial concentration of the reactant.

Teacher's Note:
1. Learn the pattern: zero order t_1/2 ∝ a, first order independent of a, second order ∝ 1/a.

Question 10. Write expression for half-life in case of a reaction between hydrogen and chlorine to form hydrochloric acid gas.
Answer:
Given reaction is a reaction of zero order. Hence, t_1/2 = [A]₀/2k.

Teacher's Note:
1. The photochemical reaction of H₂ and Cl₂ over water is a standard example of a zero order reaction.

Question 11. Draw a schematic graph showing how the rate of a first order reaction changes with change in concentration of the reactant.

Answer:
As for a first order reaction, rate ∝ molar concentration, graph between Rate vs Molar conc. will be a straight line passing through the origin.

Teacher's Note:
1. The slope of this straight line is equal to the rate constant k; for a zero order reaction the same plot would be a horizontal line.

Short Answer Questions (2 Marks)

Question 12. Express the relationship between the rate of production of HCl and the rate of disappearance of H₂ in the reaction : H₂ + Cl₂ → 2HCl
Answer:
Rate = \( -\frac{d[H_2]}{dt} = -\frac{d[Cl_2]}{dt} = +\frac{1}{2}\frac{d[HCl]}{dt} \)
Hence, \( \frac{d[HCl]}{dt} = 2\left(-\frac{d[H_2]}{dt}\right) \), i.e., HCl is produced at twice the rate at which H₂ disappears.

Teacher's Note:
1. The factor 2 comes straight from the coefficient of HCl in the balanced equation.

Question 13. What is meant by ‘rate constant, k’ of a reaction ? If the concentration be expressed in mol L⁻¹ units and time in seconds, what would be the units for k (i) for a zero order reaction (ii) for a first order reaction. (CBSE 2008, Manipur Board 2012)
Answer:
Rate constant is the rate of reaction when the concentration of each reactant is unity (1 mol L⁻¹). It is the proportionality constant in the rate law.
(i) Zero order: k = Rate = mol L⁻¹ s⁻¹
(ii) First order: k = \( \frac{\text{mol L}^{-1}\text{ s}^{-1}}{\text{mol L}^{-1}} \) = s⁻¹

Teacher's Note:
1. Use the general unit (mol L⁻¹)¹⁻ⁿ s⁻¹ to check any order quickly.

Question 14. What is order of reaction? Write units of the rate constant k for zero order, first order and second order reaction. (MP Board 2012)
Answer:
Order of a reaction is the sum of the powers of the concentration terms of the reactants in the rate law expression.
Units of k: zero order = mol L⁻¹ s⁻¹; first order = s⁻¹; second order = L mol⁻¹ s⁻¹ (mol⁻¹ L s⁻¹).

Teacher's Note:
1. The power of mol L⁻¹ in the unit of k goes 1, 0, −1 as the order goes 0, 1, 2.

Question 15. The gas phase decomposition of acetaldehyde, CH₃CHO(g) → CH₄(g) + CO(g), at 680 K is observed to follow the rate expression :
Rate = −d[CH₃CHO]/dt = k[CH₃CHO]^3/2.
If the rate of decomposition is followed by monitoring the partial pressure of acetaldehyde, we can express the rate as :
−dP_CH3CHO/dt = k[P_CH3CHO]^3/2
If the pressure is measured in atmospheres and the time in minutes, then
(a) What are the units of the rate of reaction ?
(b) What are the units of the rate constant k ?
Answer:
(a) Units of rate = atm min⁻¹
(b) k = \( \frac{\text{Rate}}{(P_{CH_3CHO})^{3/2}} = \frac{\text{atm min}^{-1}}{\text{atm}^{3/2}} \) = atm^−1/2 min⁻¹

Teacher's Note:
1. With pressure units the same rule works: replace mol L⁻¹ by atm, so k has unit atm¹⁻ⁿ time⁻¹.

Question 16. Answer the following questions (Do any two) :
(a) Identify the order of reaction from the following units of its rate constant : mol⁻¹ L s⁻¹.
(b) The conversion of molecule A to B follows second order kinetics. If concentration of A is increased to three times, how will it affect the rate of formation of B ?
(c) Write the expression of integrated rate equation for zero order reaction. (CBSE 2022)
Answer:
(a) 2nd order.
(b) Rate = k [A]². If [A] is increased 3 times, rate will become 9 times and so rate of formation of B will become 9 times.
(c) For a zero order reaction R → P, [R] = [R]₀ − kt or \( k = \frac{[R]_0 - [R]}{t} \).

Teacher's Note:
1. In (b), square the factor (3² = 9) because the order is 2; a common slip is to answer 6 times.

Question 17. Observe the graph shown in the figure and answer the following questions :
(a) What is the order of reaction ?
(b) What is the slope of the curve ?
(c) Write the relationship between k and t_1/2 (half-life period). (CBSE 2022)

Answer:
(a) First order reaction (a plot of log [R]₀/[R] vs time is a straight line passing through the origin).
(b) From \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \), \( \log\frac{[R]_0}{[R]} = \frac{k}{2.303}\,t \). Hence, slope = \( \frac{k}{2.303} \).
(c) \( t_{1/2} = \frac{0.693}{k} \)

Teacher's Note:
1. If the graph were ln [R]₀/[R] vs time, the slope would be k itself; the 2.303 appears only with log to base 10.

Question 18. Comment on the statement that molecularity of the slowest step in a reaction gives the overall molecularity of the reaction.
Answer:
A complex reaction takes place in a number of elementary steps and the overall rate of the reaction is controlled by the slowest step (the rate determining step).
Hence, the molecularity of the slowest step is taken as the molecularity of the overall reaction, and the rate law (order) of the overall reaction is also written from this step.
Strictly speaking, molecularity has a meaning only for an elementary step; for a complex reaction as a whole it has no direct meaning.

Teacher's Note:
1. Remember the companion idea: since the slow step decides the rate, the rate law contains only the species taking part in (or before) that step.

Question 19. What is meant by the rate controlling step in a reaction ?
Answer:
In a complex reaction which takes place in a series of steps, the slowest step is called the rate controlling (rate determining) step, because the overall rate of the reaction cannot be faster than the rate of this slowest step.

Teacher's Note:
1. Think of it as the narrowest point of a funnel: however fast the other steps are, the slowest one sets the overall speed.

Question 20. Explain with suitable example, how the molecularity of a reaction is different from the order of a reaction ? (UP Board 2012)
Answer:
Consider the hydrolysis of ethyl acetate in presence of acid:
CH₃COOC₂H₅ + H₂O \( \xrightarrow{H^+} \) CH₃COOH + C₂H₅OH
Two molecules (ester and water) take part in the reaction, so its molecularity is 2.
But water is present in large excess, so its concentration remains practically constant. Hence, Rate = k[CH₃COOC₂H₅], i.e., order = 1.
Thus molecularity (theoretical, from the number of colliding species) and order (experimental, from the rate law) can be different.

Teacher's Note:
1. Pseudo first order reactions are the easiest examples to quote whenever order and molecularity differ.

Question 21. Define and explain the term ‘molecularity of a reaction’ with suitable example. (Maharashtra Board 2012)
Answer:
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction, is called the molecularity of the reaction.
Examples: NH₄NO₂ → N₂ + 2H₂O (unimolecular, molecularity = 1); 2HI → H₂ + I₂ (bimolecular, molecularity = 2); 2NO + O₂ → 2NO₂ (termolecular, molecularity = 3).
Reactions with molecularity more than three are rare because the chance of more than three molecules colliding at the same time is very small.

Teacher's Note:
1. Molecularity can only be 1, 2 or 3 and is never zero or fractional.

Question 22. Show that for the reactions of first order
(i) Half-life period is independent of initial concentrations.
(ii) Units of rate constant do not depend upon the units of concentration.
Answer:
(i) For a first order reaction, \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \). At t_1/2, [R] = [R]₀/2, so \( t_{1/2} = \frac{2.303}{k}\log 2 = \frac{0.693}{k} \). This expression does not contain [R]₀; hence half-life is independent of initial concentration.
(ii) In \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \), the concentration term appears as a ratio, so its units cancel. Hence the unit of k is time⁻¹ (e.g., s⁻¹), whatever units are used for concentration.

Teacher's Note:
1. Because only a ratio of concentrations is needed, first order calculations can use moles, grams, pressure or even percentage directly.

Question 23. Rate of reaction is the change in concentration of any one of the reactants or any one of the products in unit time
(i) Express the rate of the following reaction in terms of reactants and products : 2HI → H₂ + I₂
(ii) If the expression for the above reaction is, rate = k [HI]², what is the order of the reaction.
(iii) Are the order and molecularity of the above reaction same? Give reason. (Kerala Board 2012)
Answer:
(i) Rate = \( -\frac{1}{2}\frac{d[HI]}{dt} = \frac{d[H_2]}{dt} = \frac{d[I_2]}{dt} \)
(ii) Order = 2
(iii) Yes. The reaction takes place in a single step (elementary reaction) in which two HI molecules collide, so molecularity = 2, which is the same as the order. (However, order is an experimental quantity while molecularity is a theoretical concept.)

Teacher's Note:
1. Order and molecularity are equal only for elementary reactions, and the HI decomposition is the standard example of this.

Question 24. (a) How does the rate of a reaction depend upon the concentration of reactants? (J & K Board 2012)
(b) What is rate constant? Write ‘two applications’ of rate law. (Maharashtra Board 2012)
Answer:
(a) The rate of a reaction is proportional to the concentrations of the reactants, each raised to some power (found by experiment): Rate = k[A]^x[B]^y. In general, rate increases with increase in concentration of the reactants (except for a zero order reaction).
(b) Rate constant is the rate of reaction when the concentration of each reactant is unity.
Applications of rate law: (i) it helps to find the order of reaction; (ii) it helps to calculate the rate of reaction at any concentration of reactants (and to propose the mechanism of the reaction).

Teacher's Note:
1. For a zero order reaction the rate does not change with concentration at all – mention this exception in (a).

Question 25. Define half-life period. How is it related to the order of a reaction? (Karnataka Board 2012)
Answer:
The time in which half of the reactant is converted into products (i.e., concentration falls to half of its initial value) is called the half-life period, t_1/2.
For a reaction of nth order, \( t_{1/2} \propto \frac{1}{[A]_0^{\,n-1}} \)
Zero order: \( t_{1/2} = \frac{[A]_0}{2k} \) (∝ [A]₀); first order: \( t_{1/2} = \frac{0.693}{k} \) (independent of [A]₀); second order: t_1/2 ∝ 1/[A]₀.

Teacher's Note:
1. By measuring t_1/2 at two different initial concentrations you can find the order of a reaction.

Short Answer Questions (3 Marks)

Question 26. The decomposition of A into product has value of K as 4.5 × 10³ s⁻¹ at 10°C and energy of activation 60 kJ/mol. At what temperature would K be 1.5 × 10⁴ s⁻¹?
Answer:
log \( \frac{K_2}{K_1} \) = \( \frac{E_a}{2.303R} \) \( \left[ \frac{1}{T_1} - \frac{1}{T_2} \right] \), T₁ = 283 K
log \( \frac{1.5 \times 10^4}{4.5 \times 10^3} \) = \( \frac{60000}{2.303 \times 8.314} \) \( \left[ \frac{1}{283} - \frac{1}{T_2} \right] \)
0.5229 = 3133.6 \( \left[ \frac{1}{283} - \frac{1}{T_2} \right] \) ⇒ \( \frac{1}{T_2} \) = 3.5336 × 10⁻³ − 1.669 × 10⁻⁴ = 3.3667 × 10⁻³
T₂ = 297 K
= 24°C

Teacher's Note:
1. Convert 10°C to 283 K before using the equation, and convert the answer back to °C if asked.

Question 27. (a) Write rate law and order of the following reaction:
AB₂ + C₂ → AB₂C + C (slow)
AB₂ + C → AB₂C (fast)
(b) Define energy of activation of a reaction.
(c) What is the relationship between rate constant and activation energy of a reaction?
Answer:
(a) Rate = K[AB₂][C₂], Order = 1 + 1 = 2
(b) Energy of activation is the minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to the threshold energy.
(c) K = Ae^−E_a/RT

Teacher's Note:
1. The rate law comes from the slow step; the higher the activation energy, the smaller the rate constant at a given temperature.

Question 28. For a chemical reaction R → P, the variation in the concentration [R] vs time (t) plot is given below.
(a) Predict the order of reaction.
(b) Write down its rate law.
(c) What is the slope of the curve?

Answer:
(a) [R] decreases linearly with time, so the reaction is of zero order.
(b) Rate = k[R]⁰ = k
(c) [R] = [R]₀ − kt; comparing with y = c + mx, slope = −k (intercept = [R]₀).

Teacher's Note:
1. A straight line on a [R] vs t plot means zero order; for first order you need ln[R] vs t to get a straight line.

Question 29. What do you understand by a first order reaction? Show that for a first order reaction time required to complete a definite fraction of the reaction is independent of initial concentration.
Answer:
A reaction in which the rate of reaction is directly proportional to the first power of the concentration of the reactant is called a first order reaction. Rate = k[R].
For a first order reaction, t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \)
Let a definite fraction x of the reaction be completed. Then [R] = [R]₀ − x[R]₀ = (1 − x)[R]₀
t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{(1-x)[R]_0} \) = \( \frac{2.303}{k} \) log \( \frac{1}{1-x} \)
Since [R]₀ cancels out, the time required to complete a definite fraction of the reaction is independent of the initial concentration. For example, for half the reaction (x = ½), t_1/2 = \( \frac{0.693}{k} \).

Teacher's Note:
1. This is the general result behind the constant half-life of a first order reaction.

Question 30. The decomposition of NH₃ on platinum surface is a zero order reaction. What are the rates of production of N₂ and H₂? [K = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹]
Answer:
2NH₃ → N₂ + 3H₂
−\( \frac{1}{2} \) \( \frac{d[NH_3]}{dt} \) = \( \frac{d[N_2]}{dt} \) = \( \frac{1}{3} \) \( \frac{d[H_2]}{dt} \)
Rate = −\( \frac{d[NH_3]}{dt} \) = k × [NH₃]⁰ = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹
\( \frac{d[N_2]}{dt} \) = −\( \frac{1}{2} \) \( \frac{d[NH_3]}{dt} \) = \( \frac{1}{2} \) × 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹
\( \frac{d[H_2]}{dt} \) = −\( \frac{3}{2} \) \( \frac{d[NH_3]}{dt} \) = \( \frac{3}{2} \) × 2.5 × 10⁻⁴
Rate of production of N₂ = 1.25 × 10⁻⁴ mol L⁻¹ s⁻¹
Rate of production of H₂ = 3.75 × 10⁻⁴ mol L⁻¹ s⁻¹

Teacher's Note:
1. For zero order the rate of disappearance of NH₃ equals k; then scale by the coefficients (1/2 for N₂, 3/2 for H₂).

Question 31. Define the following :
(i) Order of a reaction (ii) Molecularity of a reaction (iii) Rate law (iv) Law of mass action
Answer:
(i) Order of a reaction is the sum of the powers of the concentration terms of the reactants in the experimentally determined rate law.
(ii) Molecularity is the number of reacting species (atoms, ions or molecules) taking part in an elementary reaction which must collide simultaneously to bring about the reaction.
(iii) Rate law is the expression, found by experiment, which relates the rate of reaction to the concentrations of the reactants, each raised to some power, e.g., Rate = k[A]^x[B]^y.
(iv) Law of mass action states that the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants, each raised to the power equal to its stoichiometric coefficient in the balanced equation.

Teacher's Note:
1. The difference between (iii) and (iv) is the powers: in the rate law they come from experiment, in the law of mass action from the balanced equation.

Teacher's Note:
1. Both rate and rate constant increase with temperature – that is a similarity, so do not list it as a difference.

Question 33. The rate expressions of some reactions are given below :
(a) 2N₂O₅ → 4NO₂ + O₂, Rate = k [N₂O₅]
(b) 2NO₂ + F₂ → 2NO₂F, Rate = k [NO₂] [F₂]
(c) NO₂ + CO → CO₂ + NO, Rate = k [NO₂]²
Propose the probable mechanism of each of the above reactions.
Answer:
(a) N₂O₅ \( \xrightarrow{\text{slow}} \) N₂O₃ + O₂ ; N₂O₃ + N₂O₅ \( \xrightarrow{\text{fast}} \) 4NO₂
(b) NO₂ + F₂ \( \xrightarrow{\text{slow}} \) NO₂F + F ; NO₂ + F \( \xrightarrow{\text{fast}} \) NO₂F
(c) NO₂ + NO₂ \( \xrightarrow{\text{slow}} \) NO + NO₃ ; NO₃ + CO \( \xrightarrow{\text{fast}} \) CO₂ + NO₂
In each case the slow step gives the observed rate law and the steps add up to the overall reaction.

Teacher's Note:
1. Always add the steps to check that intermediates (N₂O₃, F, NO₃) cancel and the overall equation is recovered.

Teacher's Note:
1. The zero/fraction point is the one examiners look for most often in this comparison.

Long Answer Questions (5 Marks)

Question 35. (a) The activation energy of a reaction is 100 kJ/mol. In the presence of catalyst the activation energy is decreased by 75%. What is the effect on rate constant of the reaction at 20°C?
(b) A + 2B → 3C + 2D
The rate of disappearance of B is 1 × 10⁻² mol L⁻¹ s⁻¹. What will be (i) rate of reaction (ii) rate of change in the concentration of A and C?
Answer:
(a) log \( \frac{K_2}{K_1} \) = \( \frac{E_a - E_p}{2.303RT} \) ⇒ log \( \frac{K_2}{K_1} \) = \( \frac{75 \times 10^3}{2.303 \times 8.314 \times 293} \) = 13.37
The rate constant increases 2.35 × 10¹³ times.
(b) Rate = −\( \frac{d[A]}{dt} \) = −\( \frac{1}{2} \) \( \frac{d[B]}{dt} \) = \( \frac{1}{3} \) \( \frac{d[C]}{dt} \)
(i) Rate of reaction = \( \frac{1}{2} \) × 1 × 10⁻² = 5 × 10⁻³ mol L⁻¹ s⁻¹
(ii) Rate of disappearance of A = 5 × 10⁻³ mol L⁻¹ s⁻¹; rate of formation of C = 3 × 5 × 10⁻³ = 15 × 10⁻³ mol L⁻¹ s⁻¹

Teacher's Note:
1. A catalyst lowers E_a by the same amount for the forward and backward reactions; here the 75 kJ drop (not 25 kJ) goes into the formula, since 75% of 100 kJ is removed.

Question 36. (a) Derive the general form of the expression for the half-life of first order reaction.
(b) The decomposition of NH₃ on platinum surface is a zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹ ? (CBSE 2008)
Answer:
(a) For a first order reaction R → P, \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \)
At t = t_1/2, [R] = \( \frac{[R]_0}{2} \)
\( \therefore\; t_{1/2} = \frac{2.303}{k}\log\frac{[R]_0}{[R]_0/2} = \frac{2.303}{k}\log 2 = \frac{2.303 \times 0.3010}{k} \)
or \( t_{1/2} = \frac{0.693}{k} \), which is independent of the initial concentration.
(b) 2NH₃(g) \( \xrightarrow{Pt} \) N₂(g) + 3H₂(g)
Rate = \( -\frac{1}{2}\frac{\Delta[NH_3]}{\Delta t} = \frac{\Delta[N_2]}{\Delta t} = \frac{1}{3}\frac{\Delta[H_2]}{\Delta t} \) = k (zero order)
\( \frac{\Delta[N_2]}{\Delta t} \) = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹
\( \frac{\Delta[H_2]}{\Delta t} \) = 3 × 2.5 × 10⁻⁴ = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹

Teacher's Note:
1. For a zero order reaction the rate equals k; then use the coefficients (1 for N₂, 3 for H₂) to get each rate of production.

Question 37. (a) List the factors on which the rate of a chemical reaction depends.
(b) The half-life for the decay of radioactive ¹⁴C is 5730 years. An archaeological artifact containing wood has only 80% of the ¹⁴C activity as found in living trees. Calculate the age of the artifact. (CBSE 2008)
Answer:
(a) The rate of a chemical reaction depends upon (i) nature of reactants (ii) concentration of reactants (iii) temperature (iv) presence of catalyst (v) surface area of reactants (vi) exposure to radiation (for photochemical reactions).
(b) Radioactive decay is a first order process. k = \( \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \) year⁻¹
\( t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{(0.693/5730)}\log\frac{100}{80} \) = 1845 years

Teacher's Note:
1. Activity is proportional to the amount of ¹⁴C, so 100 and 80 can be used directly for [A]₀ and [A].

Question 38. (a) Explain the following terms : (i) Rate of a reaction. (ii) Activation energy of a reaction.
(b) The decomposition of phosphine, PH₃, proceeds according to the following equation : 4PH₃(g) → P₄(g) + 6H₂(g)
It is found that the reaction follows the following rate equation : Rate = k [PH₃]
The half-life of PH₃ is 37.9 s at 120°C.
(i) How much time is required for 3/4th of PH₃ to decompose. (ii) What fraction of the original sample of PH₃ remains behind after 1 minute ? (CBSE 2010)
Answer:
(a) (i) Rate of a reaction is the change in concentration of any one of the reactants or products per unit time.
(ii) Activation energy is the minimum extra energy which must be absorbed by the reactant molecules so that their energy becomes equal to the threshold energy.
(b) The reaction is of first order, k = 0.693/t_1/2.
(i) \( t = \frac{2.303}{0.693/t_{1/2}}\log\frac{a}{a - \frac{3}{4}a} = \frac{2.303 \times 37.9}{0.693}\log 4 \) = 75.82 s
(ii) \( \log\frac{[A]_0}{[A]} = \frac{kt}{2.303} = \frac{(0.693/37.9)(60)}{2.303} \) = 0.4763
[A]₀/[A] = Antilog 0.4763 = 2.994
or [A]/[A]₀ = 1/2.994 = 0.334 (i.e., about 33.4% remains)

Teacher's Note:
1. Check: 1 minute is about 1.6 half-lives, so the fraction left must lie between 1/2 and 1/4; 0.334 fits.