CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Why does the rate of a reaction increase with rise in temperature? [NCERT Exemplar]
Answer:
At higher temperatures, larger fraction of colliding particles can cross the energy barrier (i.e., the activation energy), which leads to faster rate.

Teacher's Note:
1. The increase in the number of collisions alone is small; the big effect comes from the larger fraction of molecules with energy ≥ E_a.

Question 2. Define ‘order of a reaction’. [CBSE 2019 (56/2/1)]
Answer:
Order of a reaction may be defined as the sum of the powers of the concentration terms of the reactants in the rate law expression.

Teacher's Note:
1. Mention “rate law expression” in the definition – order comes from the experimental rate law, not from the balanced equation.

Question 3. Identify the order of reaction from the following unit for its rate constant:
L mol⁻¹ s⁻¹ [CBSE (F) 2010]
Answer:
Second order.

Teacher's Note:
1. Compare with (mol L⁻¹)¹⁻ⁿ s⁻¹: here 1 − n = −1, so n = 2.

Question 4. For the reaction A → B, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction? [HOTS]
Answer:
Let, r = k [A]ⁿ   …(i)
Given, 3r = k [9A]ⁿ   …(ii)
Dividing equation (ii) by (i), we get
\( \frac{3r}{r} \) = \( \frac{k[9A]^n}{k[A]^n} \) = \( \frac{9^n k[A]^n}{k[A]^n} \)
3 = 9ⁿ or 3¹ = 3²ⁿ
⇒ 2n = 1 or n = ½
∴ Rate, r = k[A]^1/2
Order of reaction = ½

Teacher's Note:
1. Writing both sides as powers of the same base (3 here) makes the exponent comparison immediate.

Question 5. For which type of reactions, order and molecularity have the same value? [NCERT Exemplar]
Answer:
If the reaction is an elementary reaction, order is same as molecularity.

Teacher's Note:
1. For complex reactions the two can differ, because order is decided by the slow step and found by experiment.

Question 6. Why is the probability of reaction with molecularity higher than three very rare? [NCERT Exemplar]
Answer:
The probability of more than three molecules colliding simultaneously is very small. Hence, possibility of molecularity being more than three is very low.

Teacher's Note:
1. Even termolecular reactions are uncommon; reactions that seem to involve many molecules usually go through several bimolecular steps.

Question 7. For a reaction, A + B → Products, the rate law is Rate = k[A][B]^3/2. Can the reaction be an elementary reaction? Explain. [HOTS]
Answer:
During an elementary reaction, the number of atoms or ions colliding to react is referred to as molecularity. Had this been an elementary reaction the order of reaction with respect to B would have been 1, but in the given rate law it is \( \frac{3}{2} \). This indicates that the reaction is not an elementary reaction.

Teacher's Note:
1. A fractional power in a rate law is a quick sign that the reaction is complex (multi-step).

Question 8. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is the order of the reaction? [HOTS]
Answer:
As t_75% = 2t_50%
Therefore, it is a first order reaction.

Teacher's Note:
1. Only for a first order reaction is the half-life constant, so the second 25% takes the same time as the first 50%.

Question 9. What is the effect of adding a catalyst on
(i) Activation energy (E_a), and
(ii) Gibbs energy (ΔG) of a reaction? [CBSE Delhi 2017; (AI) 2017]
Answer:
(i) Decreases
(ii) No effect

Teacher's Note:
1. A catalyst cannot make a non-spontaneous reaction spontaneous, because it does not change ΔG.

Question 10. Define threshold energy of a reaction.
Answer:
Threshold energy is the minimum energy which must be possessed by reacting molecules in order to undergo effective collision which leads to formation of product molecules.

Teacher's Note:
1. Threshold energy = average energy of reactants + activation energy.

Question 11. Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example. [NCERT Exemplar]
Answer:
Thermodynamically the conversion of diamond to graphite is highly feasible but this reaction is very slow because its activation energy is high.

Teacher's Note:
1. Thermodynamics (ΔG) tells whether a reaction can occur; kinetics (E_a) tells how fast it occurs.

Short Answer Questions (2 Marks)

Question 12. Define order of reaction. Predict the order of reaction in the given graphs:
where [R]₀ is the initial concentration of reactant and t_1/2 is half life. [CBSE Outside Delhi Set 2, 2019]

Answer:
Order of reaction is defined as the sum of powers to which the concentration terms are raised in the rate law equation.
(a) First order (b) Zero order

Teacher's Note:
1. Constant t_½ (independent of [R]₀) means first order; t_½ rising in proportion to [R]₀ means zero order, since t_½ = [R]₀/2k.

Question 13. Explain the following terms:
(i) Rate constant (k)
(ii) Half-life period of reaction (t_1/2). [CBSE Comptt. OD 2015]
Answer:
(i) Rate constant (k): Rate constant is the rate of the reaction when the concentration of reactants is unity.
(ii) Half-life period of a reaction (t_1/2): Half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its original value.

Teacher's Note:
1. This is why k is also called the specific rate constant; it depends on temperature but not on concentration.

Question 14. Write units of rate constants for zero order and for the second order reactions if the concentration is expressed in mol L⁻¹ and time in second. [CBSE Comptt. OD 2015]
Answer:
Zero order: mol L⁻¹ s⁻¹
Second order: L mol⁻¹ s⁻¹

Teacher's Note:
1. Use (mol L⁻¹)¹⁻ⁿ s⁻¹: n = 0 gives mol L⁻¹ s⁻¹, n = 2 gives mol⁻¹ L s⁻¹.

Question 15. (i) What is the order of the reaction whose rate constant has same units as the rate of reaction?
(ii) For a reaction A + H₂O → B; Rate ∝ [A]. What is the order of this reaction? [CBSE Comptt. OD Set-3 2017]
Answer:
(i) Zero order
(ii) Pseudo-first order

Teacher's Note:
1. For (ii), the reaction looks second order (two reactants) but water is in large excess, so only [A] matters.

Question 16. For a chemical reaction R → P, variation in ln[R] vs time (t) plot is given below:
For this reaction:
(i) Predict the order of reaction.
(ii) What is the unit of rate constant (k)? [CBSE Comptt. Delhi Set-1, 2 2017]

Answer:
(i) First order.
(ii) s⁻¹/time⁻¹

Teacher's Note:
1. A straight line for ln[R] vs t (slope −k, intercept ln[R]₀) is the graphical test for a first order reaction.

Question 17. For a certain chemical reaction variation in concentration [A] vs. time (s) plot is given below:
(i) Predict the order of the given reaction.
(ii) What does the slope of the line and intercept indicate?
(iii) What is the unit of rate constant k?

Answer:
(i) Zero order reaction
(ii) Slope represents −k; Intercept represents [A]₀
(iii) mol L⁻¹ s⁻¹

Teacher's Note:
1. Concentration itself (not its log) falling in a straight line with time is the signature of zero order: [A] = [A]₀ − kt.

Question 18. (i) Explain why H₂ and O₂ do not react at room temperature.
(ii) Write the rate equation for the reaction A₂ + 3B₂ → 2C, if the overall order of the reaction is zero.
Answer:
(i) Due to high activation energy of the reaction between O₂ and H₂.
(ii) Rate = k[A₂]⁰[B₂]⁰

Teacher's Note:
1. A reaction can be thermodynamically favourable and still not happen at room temperature if its activation energy is too high – a spark supplies that energy.

Question 19. Derive integrated rate equation for rate constant of a first order reaction. [CBSE Comptt. OD Set-1, 2 2017]
Answer:
R → P
Rate = −\( \frac{d[R]}{dt} \) = k[R] or \( \frac{d[R]}{[R]} \) = −k dt
Integrating this equation, we get ln[R] = −kt + I …(i)
When t = 0, R = [R]₀, where [R]₀ is the initial concentration of the reactant. Therefore, equation (i) can be written as ln[R]₀ = −k × 0 + I ⇒ ln[R]₀ = I
Substituting the value of I in equation (i): ln[R] = −kt + ln[R]₀
Rearranging this equation: ln \( \frac{[R]}{[R]_0} \) = −kt
k = \( \frac{1}{t} \) ln \( \frac{[R]_0}{[R]} \)
k = \( \frac{2.303}{t} \) log \( \frac{[R]_0}{[R]} \)

Teacher's Note:
1. Do not skip the step that finds the integration constant (I = ln[R]₀) from t = 0 – it carries a mark in the board scheme.

Question 20. State a condition under which a bimolecular reaction is kinetically first order reaction.
Answer:
Let us take a bimolecular reaction: A + B → Product
Rate = k[A][B]
When concentration of [B] is taken in excess then rate law will become: Rate = k[A], where k = constant.
The rate depends only on one of the reactants as there is negligible change in its concentration, so it is bimolecular but is of first order.

Teacher's Note:
1. Such reactions are called pseudo first order; acid hydrolysis of ethyl acetate and inversion of cane sugar are the textbook examples.

Question 21. The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees. How old is that piece of wood? (log 3 = 0.4771, log 7 = 0.8540, Half-life of C-14 = 5730 years)
Answer:
k = 0.693/t_1/2 = 0.693/5730 years⁻¹
t = \( \frac{2.303}{k} \) log \( \frac{C_0}{C_t} \)
Let C₀ = 1 and C_t = 3/10, so C₀/C_t = 1/(3/10) = 10/3
t = \( \frac{2.303}{0.693} \) × 5730 × log \( \frac{10}{3} \)
t = 9957 years

Teacher's Note:
1. Radioactive decay is always first order, so carbon dating uses exactly the first order integrated equation.
2. log(10/3) = 1 − log 3 = 0.5229; the log 7 value given is not needed.

Question 22. 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?
Answer:
Initial amount = 5 g, Final amount = 3 g, Rate constant = 1.15 × 10⁻³ s⁻¹
We know that for a first order reaction, t = \( \frac{2.303}{k} \) log \( \frac{[R_{initial}]}{[R_{final}]} \)
= \( \frac{2.303}{1.15\times10^{-3}} \) log \( \frac{5}{3} \) = \( \frac{2.303}{1.15\times10^{-3}} \) × 0.2219
= 444.379 s

Teacher's Note:
1. log(5/3) = log 5 − log 3 = 0.6990 − 0.4771 = 0.2219; keep four decimal places to land on the answer.

Question 23. (i) Why the molecularity of a reaction can not be zero?
(ii) Write the formula for expressing rate of the reaction N₂ + 3H₂ → 2NH₃
Answer:
(i) The molecularity of a reaction is the number of total molecules taking part in elementary step of reaction. So, minimum one molecule is required for a reaction to occur. Hence, the value of molecularity is never zero.
(ii) N₂ + 3H₂ → 2NH₃
Rate of reaction = \( \frac{-d[N_2]}{dt} \) = \( -\frac{1}{3}\frac{d[H_2]}{dt} \) = \( \frac{1}{2}\frac{d[NH_3]}{dt} \)

Teacher's Note:
1. Reactant terms carry a minus sign and product terms a plus sign; each is divided by its own coefficient.

Question 24. (i) The rate of reaction decreases with the progress of reaction. Why?
(ii) Reactions having molecularity more than three occur rarely. Why?
Answer:
(i) The rate of reaction depends on the concentration of reactants. Since the concentration of reactants decreases with time, so rate of reaction also decreases.
(ii) It is because of the fact that collision of more than three molecules is not possible at a time.

Teacher's Note:
1. For (ii), a simultaneous collision of four or more molecules with the right energy and orientation is so unlikely that such steps are practically never observed.

Question 25. (i) If the rate equation is given below: Rate = k[A]²[B]
then what will be the unit of its rate and rate constant?
(ii) Write the slope value obtained in the plot of log [R₀]/[R] vs. time for a first order reaction. [CBSE Delhi Set 1, 2020]
Answer:
(i) Unit of rate = mol L⁻¹ s⁻¹
Unit of rate constant (k) = \( \frac{\text{Unit of rate}}{\text{Unit of [A]}^2\times\text{Unit of [B]}} \) = \( \frac{mol\ L^{-1}\,s^{-1}}{(mol\ L^{-1})^{2}(mol\ L^{-1})} \) = mol⁻² L² s⁻¹
(ii)
Slope value = \( \frac{k}{2.303} \)

Teacher's Note:
1. From log([R]₀/[R]) = kt/2.303 the line goes through the origin with a positive slope; it is the log[R] vs t plot that has slope −k/2.303.

Question 26. (i) For which reaction, the rate of reaction does not decrease with time?
(ii) What is the order of photochemical reaction?
Answer:
(i) For zero order reaction the rate of reaction does not decrease with time because it does not depend on concentration of reactants.
(ii) Zero order reaction.

Teacher's Note:
1. In a photochemical reaction such as H₂ + Cl₂ →(hν) 2HCl the rate depends on the light intensity, not on the gas concentrations.

Short Answer Questions (3 Marks)

Question 27. A first order reaction takes 160 minutes time for 20% completion. Calculate time required for half completion of reaction.
Answer:
Given: [R]₀ = 1, [R] = 0.80, t = 160 min, t_1/2 = ?
For first order reaction, k = \( \frac{2.303}{t} \) log₁₀ \( \frac{[R]_0}{[R]} \)
or, k = \( \frac{2.303}{160} \) log₁₀ \( \frac{1}{0.8} \) or, k = \( \frac{2.303}{160\ \text{min}} \) × log 1.25
or, k = \( \frac{2.303\times0.0969}{160\ \text{min}} \) = 1.39 × 10⁻³ min⁻¹
Again t_1/2 = \( \frac{0.693}{k} \) = \( \frac{0.693}{1.39\times10^{-3}} \) = 498 min

Teacher's Note:
1. 20% completion leaves 80%, so use [R]₀/[R] = 1/0.8 = 1.25.

Question 28. Hydrogen peroxide, H₂O_2(aq) decomposes to H₂O_(l) and O_2(g) in a reaction that is first order in H₂O₂ and has a rate constant k = 1.06 × 10⁻³ min⁻¹.
(i) How long will it take for 15% of a sample of H₂O₂ to decompose?
(ii) How long will it take for 85% of the sample to decompose?
Answer:
(i) t = \( \frac{2.303}{k} \) log \( \frac{[A]_0}{[A]} \); given k = 1.06 × 10⁻³ min⁻¹, \( \frac{[A]_0}{[A]} \) = \( \frac{100}{85} \)
t = \( \frac{2.303}{1.06\times10^{-3}\ \text{min}^{-1}} \) log \( \frac{100}{85} \)
t = \( \frac{2303}{1.06} \) [2 log 10 − log 85] min
t = \( \frac{2303}{1.06} \) [2 × 1 − 1.9294] = \( \frac{2303\times0.0706}{1.06} \)
t = 153.39 min = 153.4 min
(ii) Given k = 1.06 × 10⁻³ min⁻¹, \( \frac{[A]_0}{[A]} \) = \( \frac{100}{15} \)
t = \( \frac{2.303}{1.06\times10^{-3}\ \text{min}^{-1}} \) log \( \frac{100}{15} \) = \( \frac{2303}{1.06} \) [2 log 10 − log 15] min
= \( \frac{2303}{1.06} \) [2 × 1 − 1.1761] = \( \frac{2303\times0.8239}{1.06} \) min = 1790 min

Teacher's Note:
1. In (ii), 85% decomposed means only 15% remains, so the ratio is 100/15, not 100/85.

Question 29. When inversion of sucrose is studied at pH = 5, the half-life period is always found to be 500 minutes irrespective of any initial concentration but when it is studied at pH = 6, the half-life period is found to be 50 minutes. Derive the rate law expression for the inversion of sucrose.
Answer:
At pH = 5, as half-life period is found to be independent of initial concentration of sucrose, this means with respect to sucrose, it is a reaction of first order, i.e., Rate = k[Sucrose].
If n is the order with respect to H⁺ ion, t_1/2 ∝ [H⁺]¹⁻ⁿ,
i.e., 500 ∝ (10⁻⁵)^{1 − n} [pH = 5 means [H⁺] = 10⁻⁵ M] ...(i)
and 50 ∝ (10⁻⁶)^{1 − n} [pH = 6 means [H⁺] = 10⁻⁶ M] ...(ii)
Dividing (i) by (ii), 10 = (10)^{1 − n} i.e. 1 − n = 1 or n = 0, i.e., order with respect to H⁺ ion = 0. Hence, overall rate law is Rate = k [Sucrose] [H⁺]⁰.

Teacher's Note:
1. Use t_1/2 ∝ [A]₀¹⁻ⁿ for the H⁺ dependence; a tenfold change in [H⁺] giving a tenfold change in t_1/2 fixes 1 − n = 1.

Question 30. What will be the rate of decomposition of N₂O₅ and rate of formation of NO₂ and O₂ when [N₂O₅] = 0.40 M for the reaction 2N₂O₅ → 4NO₂ + O₂. The rate constant for this reaction is 3.1 × 10⁻⁴ min⁻¹.
Answer:
The given reaction is 2N₂O₅ → 4NO₂ + O₂
Unit of rate constant suggests that rate of reaction is first order.
Hence, rate of reaction = k[N₂O₅] = 3.1 × 10⁻⁴ min⁻¹ × 0.40 M = 1.24 × 10⁻⁴ M min⁻¹
Rate of reaction = −\( \frac{1}{2} \) \( \frac{d[N_2O_5]}{dt} \) = +\( \frac{1}{4} \) \( \frac{d[NO_2]}{dt} \) = +\( \frac{d[O_2]}{dt} \)
∴ −\( \frac{d[N_2O_5]}{dt} \) = 1.24 × 10⁻⁴ M min⁻¹ × 2 = 2.48 × 10⁻⁴ M min⁻¹
Similarly \( \frac{d[NO_2]}{dt} \) = 4 × 1.24 × 10⁻⁴ M min⁻¹ = 4.96 × 10⁻⁴ M min⁻¹
+\( \frac{d[O_2]}{dt} \) = 1.24 × 10⁻⁴ M min⁻¹

Teacher's Note:
1. First find the rate of reaction from k[N₂O₅], then multiply by each coefficient to get the individual species rates.

Teacher's Note:
1. Convert total pressure to reactant pressure first: here p_SO2Cl2 = 2P₀ − P_t = 0.8 − P_t.

Teacher's Note:
1. Always pair experiments where only one concentration changes; here 1 and 2 isolate A, and 2 and 3 isolate B.

Teacher's Note:
1. Here P_t in the formula means the pressure of C₂H₅Cl left (0.10 atm), not the total pressure 0.50 atm.

Question 34. For the reaction R → P, the concentration of a reactant changes from 0.03 to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:
R → P; [R₁] = 0.03 M, [R₂] = 0.02 M, Δt = 25 min
Average rate = −\( \frac{\Delta[R]}{\Delta t} \) = −\( \left[\frac{[R_2]-[R_1]}{\Delta t}\right] \) = −\( \left[\frac{0.02-0.03}{25}\right] \) = −\( \left[\frac{-0.01}{25}\right] \)
= 4 × 10⁻⁴ mol L⁻¹ min⁻¹
= \( \frac{4\times10^{-4}}{60} \) mol L⁻¹ s⁻¹ = 6.66 × 10⁻⁶ mol L⁻¹ s⁻¹

Teacher's Note:
1. To change per-minute into per-second, divide by 60 (a rate per second must be a smaller number).

Question 35. For the reaction 2A + B → A₂B, Rate = k[A][B]², k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹. Calculate the initial rate of the reaction when [A] = 0.1 mol L⁻¹ and [B] = 0.2 mol L⁻¹. Calculate the rate of reaction after [A] is reduced to 0.06 mol L⁻¹.
Answer:
2A + B → A₂B
Initial rate = k[A][B]² = 2.0 × 10⁻⁶ × 0.1 × (0.2)² = 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹
When concentration of [A] is reduced from 0.10 mol L⁻¹ to 0.06 mol L⁻¹, i.e., 0.04 mol L⁻¹ of A has been used in the reaction. Therefore, the concentration of B used = \( \frac{1}{2} \) × 0.04 = 0.02 mol L⁻¹
Hence, [B] = 0.2 − 0.02 = 0.18 mol L⁻¹
Rate = k[A][B]² = 2.0 × 10⁻⁶ × 0.06 × (0.18)² = 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹

Teacher's Note:
1. The key step is the stoichiometry: 2 mol of A react with 1 mol of B, so B falls by only half as much as A.

Long Answer Questions (5 Marks)

Teacher's Note:
1. Find the orders first, then k from any one experiment, and only then substitute the new concentrations.

Question 37. A reaction is of first order with respect to A and second order with respect to B.
(i) Write differential rate equation.
(ii) How is the rate affected when the concentration of B is tripled?
(iii) How is the rate affected when the concentration of both A and B is doubled?
Answer:
(i) Differential rate equation: \( \frac{dx}{dt} \) = k[A][B]²
(ii) Let [A] = a, [B] = b. If [B] increases three times, [B] = 3b
∴ Rate = k[A][B]²; Rate₁ = k × a × b² …(i); Rate₂ = k × a × (3b)² …(ii)
From eq. (i) and (ii): \( \frac{Rate_2}{Rate_1} \) = \( \frac{k\times a\times(3b)^2}{k\times a\times b^2} \) ⇒ Rate₂ = 9 × Rate₁
∴ The rate becomes nine times when the concentration of B is tripled.
(iii) If [A] and [B] are doubled then [A] = 2a, [B] = 2b
Rate₁ = k × a × b² …(i); Rate₂ = k × (2a) × (2b)² …(ii)
From eq. (i) and (ii): \( \frac{Rate_2}{Rate_1} \) = \( \frac{k\times(2a)\times(2b)^2}{k\times a\times b^2} \) = 8 ⇒ Rate₂ = 8 Rate₁
∴ The rate becomes eight times when the concentration of both A and B is doubled.

Teacher's Note:
1. Quick check: the rate factor is (factor for A)¹ × (factor for B)² – here 3² = 9 and 2 × 2² = 8.

Question 38. (a) Reaction 2A + B + C → D + 2E shows first order with respect to A, second order with respect to B and zero order with respect to C. Determine:
(i) Rate law of reaction
(ii) What will be the rate of reaction on doubling of concentration of A, B and C?
(b) For the reaction, N₂O₅ → 2NO₂ + ½O₂
The rate of dissociation of N₂O₅ is 5.65 × 10⁻⁵ mol L⁻¹ s⁻¹. Determine:
(i) Rate of reaction
(ii) Rate of formation of NO₂
(iii) Rate of formation of O₂
Answer:
(a) (i) 2A + B + C → D + 2E. Rate law: Rate (r) = k[A]¹[B]²[C]⁰
(ii) If concentration of A, B and C is doubled: r₁ = k[A]¹[B]²[C]⁰ …(i); r₂ = k[2A]¹[2B]²[2C]⁰ …(ii)
From equation (i) and (ii): \( \frac{r_1}{r_2} \) = \( \frac{k[A]^1[B]^2[C]^0}{k[2A]^1[2B]^2[2C]^0} \) = \( \frac{1}{8} \) ∴ r₂ = 8 × r₁
Hence rate of reaction becomes eight times.
(b) (i) N₂O₅ → 2NO₂ + ½O₂
Rate of reaction = \( \frac{-d[N_2O_5]}{dt} \) = \( \frac{1}{2}\left(\frac{d[NO_2]}{dt}\right) \) = \( 2\left(\frac{d[O_2]}{dt}\right) \)
Rate of reaction = Rate of dissociation of N₂O₅ = 5.65 × 10⁻⁵ mol L⁻¹ s⁻¹
(ii) Rate of dissociation of N₂O₅ = \( \frac{1}{2} \) × Rate of formation of NO₂
Rate of formation of NO₂ = 2 × 5.65 × 10⁻⁵ = 11.3 × 10⁻⁵ mol L⁻¹ s⁻¹
(iii) Rate of dissociation of N₂O₅ = 2 × Rate of formation of O₂
Rate of formation of O₂ = \( \frac{1}{2} \) × Rate of dissociation of N₂O₅ = \( \frac{1}{2} \) × 5.65 × 10⁻⁵ = 2.825 × 10⁻⁵ mol L⁻¹ s⁻¹

Teacher's Note:
1. A zero order reactant has no effect: doubling [C] multiplies the rate by 2⁰ = 1.
2. With a coefficient of ½ for O₂, its rate of formation is half the rate of loss of N₂O₅.

Question 39. (a) A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?
(b) Define order of reaction. Write the condition under which a bimolecular reaction follows first order kinetics. [CBSE Delhi Set 1, 2020]
Answer:
(a) t = \( \frac{2.303}{k} \) log \( \frac{[A]_0}{[A]} \)
(i) 40 min = \( \frac{2.303}{k} \) log \( \frac{100}{75} \) …(i)
(ii) t = \( \frac{2.303}{k} \) log \( \frac{100}{20} \) …(ii)
Dividing (i) by (ii): \( \frac{40}{t} \) = \( \frac{\log 4/3}{\log 5} \) = \( \frac{0.1250}{0.6990} \)
t = \( \frac{0.6990\times40}{0.1250} \) = 223.7 min
k = \( \frac{2.303}{t} \) log \( \frac{100}{100-25} \) = \( \frac{2.303}{40} \) log \( \frac{100}{75} \) = \( \frac{2.303}{40} \) (log 4 − log 3)
= \( \frac{2.303}{40} \) (0.6021 − 0.4771) = \( \frac{2.303}{40} \) × 0.125 = 0.007196
= 7.196 × 10⁻³ min⁻¹
(b) Order of reaction: The sum of the powers of the concentration of the reacting species that are involved in the rate equation for the reaction is called order of reaction.
The condition under which a bimolecular reaction follows first order kinetics is when one of the reactants is taken in large excess so that its concentration hardly changes.

Teacher's Note:
1. 80% complete means 20% of the reactant is left, so the ratio [A]₀/[A] = 100/20 = 5, not 100/80.