CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. In some cases, it is found that a large number of colliding molecules have energy more than threshold value, yet the reaction is slow. Why? [CBSE Delhi 2013]
Answer:
This is due to improper orientation of the colliding molecules at the time of collision.

Teacher's Note:
1. Both conditions are needed for an effective collision: energy ≥ threshold energy and proper orientation.

Teacher's Note:
1. Diamond → graphite is thermodynamically feasible but so slow that it is imperceptible – a reminder that feasibility and rate are different things.

Question 3. Define rate constant (k). [CBSE Comptt. OD 2016]
Answer:
Rate constant is the rate of reaction when the concentration of reactant is unity.

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

Question 4. Write the unit of rate constant for a zero order reaction. [CBSE Comptt. OD 2013]
Answer:
Unit is mol L⁻¹ s⁻¹.

Teacher's Note:
1. For a zero order reaction, the unit of k is the same as the unit of rate.

Question 5. What is the effect of catalyst on
(i) Gibbs energy (ΔG) and
(ii) activation energy of a reaction? [CBSE Delhi 2017/OD 2013]
Answer:
(i) No effect.
(ii) Decreases.

Teacher's Note:
1. Remember the pair: a catalyst changes the path (E_a) but not the start and end points (ΔG, ΔH).

Question 6. In the Arrhenius equation, what does the factor e^−E_a/RT correspond to? [CBSE SQP 2017]
Answer:
e^−E_a/RT corresponds to the fraction of molecules that have kinetic energy equal to or greater than E_a.

Teacher's Note:
1. This fraction rises sharply with temperature, which is why a 10° rise roughly doubles the rate of many reactions.

Question 7. On increasing temperature, activation energy of a reaction decreases, why? [CBSE SQP 2016]
Answer:
Activation energy itself does not depend on temperature. On increasing temperature, the kinetic energy of molecules increases, so a larger fraction of molecules have energy equal to or greater than the threshold energy; the rate increases as if the energy barrier had become smaller.

Teacher's Note:
1. E_a is a property of the reaction path; only a catalyst lowers it. Temperature changes how many molecules can cross the barrier, not the height of the barrier.

Question 8. The reactions taking place in one step are called ............... reactions.
Answer:
elementary

Teacher's Note:
1. For elementary reactions alone, order and molecularity are the same.

Question 9. The order of reaction is ............... determined.
Answer:
experimentally

Teacher's Note:
1. Order cannot be predicted from the balanced equation; it must come from rate measurements.

Question 10. The rate law for a reaction is Rate = K[A][B]^3/2. Can the reaction be an elementary process? Explain.
Answer:
No, an elementary process would have a rate law with orders equal to its molecularities and therefore must be in integral form.

Teacher's Note:
1. A fractional order is a sure sign of a complex (multi-step) reaction.

Question 11. Identify the order of a reaction from the following rate constant:
k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹
Answer:
Second order

Teacher's Note:
1. Unit of k = (mol L⁻¹)¹⁻ⁿ s⁻¹; L mol⁻¹ s⁻¹ fits n = 2.

Short Answer Questions (2 Marks)

Question 12. The rate law for the reaction: Ester + H⁺ → Acid + Alcohol is:
\( \frac{dx}{dt} \) = k [Ester] [H⁺]⁰
What would be the effect on the rate if (i) concentration of the ester is doubled? (ii) concentration of H⁺ is doubled?
Answer:
(i) The rate of reaction will be doubled.
(ii) No effect on rate.

Teacher's Note:
1. Any concentration raised to the power zero equals 1, so changing [H⁺] cannot change the rate.

Teacher's Note:
1. The rate constant does depend on temperature (and catalyst), even though it does not depend on concentration.

Question 14. A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of rate constant for such a reaction? [CBSE (AI) 2011]
Answer:
Consider the reaction nR → Products
As the reaction is of second order
∴ Rate, r = k [R]²   …(i)
If the concentration of the reactant is reduced to half, then
Rate, r′ = k\( \left[\frac{R}{2}\right]^2 \)   …(ii)
Dividing equation (ii) by (i), we get
\( \frac{r'}{r} \) = \( \frac{k[R]^2}{4k[R]^2} \) = \( \frac{1}{4} \)
r′ = \( \frac{1}{4} \)r, i.e., rate of reaction becomes \( \frac{1}{4} \)th of the initial rate.
The unit of rate constant is mol⁻¹ L s⁻¹.

Teacher's Note:
1. For second order, rate changes as the square of the concentration factor: halving gives (½)² = ¼.

Question 15. What do you understand by the rate law and rate constant of a reaction? Identify the order of a reaction if the units of its rate constant are:
(i) L⁻¹ mol s⁻¹     (ii) L mol⁻¹ s⁻¹. [CBSE (AI) 2011]
Answer:
An experimentally determined expression which relates the rate of reaction with the concentration of reactants is called rate law while the rate of reaction when concentration of each reactant is unity in a rate law expression is called rate constant.
(i) Comparing power of mole in L⁻¹ mol s⁻¹ and (mol L⁻¹)¹⁻ⁿ s⁻¹, we get
1 = 1 − n ⇒ n = 0 i.e., zero order reaction
(ii) Again comparing power of mole in L mol⁻¹ s⁻¹ and (mol L⁻¹)¹⁻ⁿ s⁻¹, we get
−1 = 1 − n ⇒ n = 2, i.e., second order reaction

Teacher's Note:
1. The power of mol in the unit of k is always (1 − n); compare only that power to find n quickly.

Question 16. Calculate the overall order of a reaction which has the rate expression,
(i) Rate = k[A]^1/2 [B]^3/2     (ii) Rate = k[A]^3/2 [B]⁻¹
Answer:
(i) Order = \( \frac{1}{2} \) + \( \frac{3}{2} \) = 2, i.e., second order.
(ii) Order = \( \frac{3}{2} \) + (−1) = \( \frac{1}{2} \), i.e., half order.

Teacher's Note:
1. A negative power means that species slows the reaction down (inhibitor); still add it with its sign when finding the overall order.

Teacher's Note:
1. In a two-mark answer, a table with two clear contrasting points scores better than a long paragraph.

Question 18. Consider the decomposition of hydrogen peroxide in alkaline medium which is catalysed by iodide ions.
2H₂O₂ \( \xrightarrow{OH^-} \) 2H₂O + O₂
This reaction takes place in two steps as given below:
Step-I   H₂O₂ + I⁻ → H₂O + IO⁻ (slow)
Step-II   H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast)
(i) Write the rate law expression and determine the order of reaction w.r.t. H₂O₂.
(ii) What is the molecularity of each individual step?
Answer:
(i) Rate = k [H₂O₂]¹ [I⁻]¹
Order of reaction w.r.t. H₂O₂ = 1
(ii) Molecularity of step I = 2 and step II = 2.

Teacher's Note:
1. Write the rate law from the slow step only; I⁻ appears in it even though it is a catalyst and is regenerated in step II.

Question 19. The rate constant for a reaction of zero order in A is 0.0030 mol L⁻¹ s⁻¹. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M?
Answer:
k = 0.0030 mol L⁻¹ s⁻¹, [R]₀ = 0.10 M, [R] = 0.075 M
We know that
[R] = −kt + [R]₀
0.075 = −0.0030t + 0.10
⇒ 3t = 100 − 75 ⇒ t = 8.33 seconds

Teacher's Note:
1. For zero order, the time is simply (change in concentration)/k = 0.025/0.0030 s; no logarithm is involved.

Question 20. Define the following terms:
(i) Pseudo first order reaction
(ii) Half life period of reaction (t_1/2) [CBSE Delhi 2014]
Answer:
(i) A reaction which is not truly of first order but under certain conditions becomes a reaction of first order is called pseudo first order reaction, e.g., acid hydrolysis of ethyl acetate.
CH₃COOC₂H₅ + H₂O \( \xrightarrow{H^+} \) CH₃COOH + C₂H₅OH
Rate ∝ [CH₃COOC₂H₅] as H₂O is in excess.
(ii) The half life (t_1/2) of a reaction is the time in which the concentration of reactant is reduced to one half of its initial concentration [R]₀.
For a first order reaction, t_1/2 = \( \frac{0.693}{k} \), i.e., independent of [R]₀.
For a zero order reaction, t_1/2 = \( \frac{[R]_0}{2k} \), i.e., t_1/2 ∝ [R]₀.

Teacher's Note:
1. Inversion of cane sugar is another standard example of a pseudo first order reaction.

Question 21. For a reaction: A + H₂O → B, Rate ∝ [A].
What is its (i) molecularity (ii) order of reaction?
Answer:
(i) Pseudo unimolecular
(ii) Order = 1.

Teacher's Note:
1. Water takes part in the reaction but is in large excess, so it does not appear in the rate law.

Question 22. The rate for the reaction R → P is rate = k[R]. It has been shown graphically alongside. What is rate constant for the reaction?

Answer:
From the graph
Case I: Rate = k[A]
1 × 10⁻² mol L⁻¹ s⁻¹ = k (0.1 mol L⁻¹)
∴ k = \( \frac{1\times10^{-2}\ mol\ L^{-1}\ s^{-1}}{0.1\ mol\ L^{-1}} \) = 0.1 s⁻¹
Case II: 3 × 10⁻² mol L⁻¹ s⁻¹ = k (0.3 mol L⁻¹)
k = \( \frac{3\times10^{-2}\ mol\ L^{-1}\ s^{-1}}{0.3\ mol\ L^{-1}} \) = 0.1 s⁻¹
Hence, k = 0.1 s⁻¹

Teacher's Note:
1. The rate constant is simply the slope of the rate vs [R] straight line; reading two points and getting the same k confirms first order.

Question 23. The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce the initial concentration of the reactant to \( \frac{1}{10} \)th of its initial value? [CBSE (F) 2013]
Answer:
t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \) as [R] = \( \frac{[R]_0}{10} \)
∴ t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]_0/10} \) = \( \frac{2.303}{60} \) × log 10 = \( \frac{2.303}{60} \) × 1 = 3.838 × 10⁻² s

Teacher's Note:
1. Reducing to 1/10th always gives log 10 = 1, so t = 2.303/k directly for a first order reaction.

Question 24. (i) For a reaction A + B → P, the rate law is given by, r = k[A]^1/2[B]². What is the order of this reaction?
(ii) A first order reaction is found to have a rate constant k = 5.5 × 10⁻¹⁴ s⁻¹. Find the half life of the reaction. [CBSE (AI) 2013]
Answer:
(i) Order of reaction = \( \frac{1}{2} \) + 2 = \( \frac{5}{2} \)
(ii) Radioactive decay follows first order kinetics.
t_1/2 = \( \frac{0.693}{k} \) = \( \frac{0.693}{5.5\times10^{-14}} \) s = 1.26 × 10¹³ s

Teacher's Note:
1. A very small k means a very long half-life; check the sign of the exponent when dividing by a negative power of 10.

Question 25. 87.5% of the substance disintegrated in 45 minutes (first order reaction). What is its half-life?
Answer:
\( \left(\frac{1}{2}\right)^n = \frac{12.5}{100} \) because 87.5% has disintegrated, amount left is 12.5%.
\( \left(\frac{1}{2}\right)^n = \left(\frac{1}{8}\right) \) ⇒ \( \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 \)
Number of half lives = 3
t_1/2 = \( \frac{45}{3} \) = 15 minutes

Teacher's Note:
1. Count half-lives: 100 → 50 → 25 → 12.5%, i.e., three half-lives for 87.5% completion.

Short Answer Questions (3 Marks)

Question 26. The decomposition of NH₃ on platinum surface is a zero order reaction. What will be rate of production of N₂ and H₂ when the value of k is 2.5 × 10⁻⁹ mol L⁻¹ s⁻¹?
Answer:
2NH₃ → N₂ + 3H₂
Rate of reaction = −\( \frac{1}{2} \)\( \left(\frac{d[NH_3]}{dt}\right) \) = \( \frac{d[N_2]}{dt} \) = \( \frac{1}{3}\frac{d[H_2]}{dt} \)
∵ Reaction is zero order, so, the rate of reaction = k[NH₃]⁰ = k = 2.5 × 10⁻⁹ mol L⁻¹ s⁻¹
Rate of production of N₂: ∵ Rate of reaction = Rate of production of N₂
∴ Rate of production of N₂ = 2.5 × 10⁻⁹ mol L⁻¹ s⁻¹
Rate of production of H₂: ∵ Rate of reaction = \( \frac{1}{3} \) × Rate of production of H₂
∴ Rate of production of H₂ = 3 × Rate of reaction = 3 × 2.5 × 10⁻⁹ = 7.5 × 10⁻⁹ mol L⁻¹ s⁻¹

Teacher's Note:
1. For a zero order reaction the rate is simply equal to k; then use the coefficients (1 for N₂, 3 for H₂) to get each product’s rate.

Teacher's Note:
1. Choose pairs of experiments in which only one concentration changes: (ii) and (iv) for B, (i) and (iii) for A.
2. Remember the unit of k for a third order reaction: mol⁻² L² time⁻¹.

Question 28. A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
(Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [CBSE OD Set-1, 2, 3 2017]
Answer:
t = \( \frac{2.303}{k} \) log \( \frac{[A]_0}{[A]} \)
20 min = \( \frac{2.303}{k} \) log \( \frac{100}{75} \) …(i)
t = \( \frac{2.303}{k} \) log \( \frac{100}{25} \) …(ii)
Divide equation (i) by (ii): \( \frac{20}{t} \) = \( \frac{\log 4/3}{\log 4} \)
20/t = 0.1250/0.6021
t = 96.3 min

Teacher's Note:
1. Dividing the two equations cancels k, so you never need its value; just be careful that 25% decomposed leaves 75 and 75% decomposed leaves 25.

Question 29. 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 the reaction (ii) Rate of change in concentration of A and C?
Answer:
(i) As \( -\frac{d[B]}{dt} \) = 1 × 10⁻² mol L⁻¹ s⁻¹
∴ Rate = \( -\frac{1}{2}\frac{d[B]}{dt} \) = \( \frac{1}{2} \) × 1 × 10⁻² = 0.5 × 10⁻² mol L⁻¹ s⁻¹
(ii) Rate = \( -\frac{d[A]}{dt} = -\frac{1}{2}\frac{d[B]}{dt} = +\frac{1}{3}\frac{d[C]}{dt} \)
Rate of change in concentration of A
= \( -\frac{d[A]}{dt} = -\frac{1}{2}\frac{d[B]}{dt} \)
= 0.5 × 10⁻² mol L⁻¹ s⁻¹
Rate of change in concentration of C
= \( +\frac{d[C]}{dt} = -\frac{3}{2}\frac{d[B]}{dt} \) = \( \frac{3}{2} \) × 1 × 10⁻²
= 1.5 × 10⁻² mol L⁻¹ s⁻¹

Teacher's Note:
1. Rate of reaction is a single value; rates of change of individual species differ by their coefficients. A is consumed (decreases) while C is formed (increases).

Question 30. The rate of reaction, 2NO + Cl₂ → 2NOCl is doubled when concentration of Cl₂ is doubled and it becomes eight times when concentration of both NO and Cl₂ are doubled. Deduce the order of the reaction. [HOTS]
Answer:
Let r = k [NO]^x [Cl₂]^y   …(i)
2r = k [NO]^x [2Cl₂]^y   …(ii)
8r = k [2NO]^x [2Cl₂]^y   …(iii)
Dividing (iii) by (ii), we get
\( \frac{8r}{2r} \) = \( \frac{k[2NO]^x[2Cl_2]^y}{k[NO]^x[2Cl_2]^y} \)
2² = [2]^x, x = 2
Putting the value of x in (i) and (ii), we get
r = k [NO]² [Cl₂]^y; 2r = k [NO]² [2Cl₂]^y
\( \frac{2r}{r} \) = \( \frac{[2Cl_2]^y}{[Cl_2]^y} \), 2 = [2]^y, y = 1
Rate = k [NO]² [Cl₂]¹
Overall order of reaction = x + y = 2 + 1 = 3

Teacher's Note:
1. Divide the equations so that only one unknown power remains each time; pick the pair of experiments where only one concentration changes.

Question 31. Following reaction takes place in one step:
2NO(g) + O₂(g) → 2NO₂(g)
How will the rate of the above reaction change if the volume of the reaction vessel is reduced to one-third of its original volume? Will there be any change in the order of the reaction with reduced volume? [HOTS]
Answer:
Rate = k [NO]² [O₂]
Let initially, moles of NO = a, moles of O₂ = b, volume of the vessel = V. Then
[NO] = \( \frac{a}{V} \) M, [O₂] = \( \frac{b}{V} \) M
∴ Rate (r₁) = \( k\left(\frac{a}{V}\right)^2\left(\frac{b}{V}\right) = k\frac{a^2 b}{V^3} \)   …(i)
Now, new volume = \( \frac{V}{3} \)
∴ New concentrations: [NO] = \( \frac{a}{V/3} \) = \( \frac{3a}{V} \), [O₂] = \( \frac{b}{V/3} \) = \( \frac{3b}{V} \)
∴ New rate (r₂) = \( k\left(\frac{3a}{V}\right)^2\left(\frac{3b}{V}\right) = \frac{27ka^2 b}{V^3} \)   …(ii)
∴ \( \frac{r_2}{r_1} \) = 27 or r₂ = 27r₁, i.e., rate becomes 27 times.
Thus, there is no effect on the order of reaction.

Teacher's Note:
1. Since it is a one-step (elementary) reaction, the rate law can be written from the equation; order is a property of the rate law, so changing volume cannot change it.

Question 32. Nitric oxide, NO, reacts with oxygen to produce nitrogen dioxide.
2NO(g) + O₂(g) → 2NO₂(g)
The rate law for this reaction is:
Rate = k [NO]² [O₂]
Propose a mechanism for the reaction. [HOTS]
Answer:
The probable proposed mechanism may be,
Step I: NO + O₂ ⇄ NO₃ (fast)
Step II: NO₃ + NO \( \xrightarrow{k_1} \) NO₂ + NO₂ (slow)
Since slowest reaction is the rate determining step, therefore
Rate = k₁ [NO₃] [NO]
K = \( \frac{[NO_3]}{[NO][O_2]} \)
[NO₃] = K [NO] [O₂]
Rate = k₁K [NO] [O₂] [NO] = K′[NO]² [O₂], where K′ = k₁.K

Teacher's Note:
1. An intermediate (NO₃) cannot appear in the final rate law; replace it using the equilibrium constant of the fast step.

Question 33. Define order of reaction. How does order of a reaction differ from molecularity for a complex reaction? [CBSE 2019 (56/3/2), 2020 (56/5/1)]
Answer:
(i) The sum of powers of the concentration of the reactants in the rate law expression is known as order of that reaction.
Consider the general reaction, aA + bB → cC + dD
Let Rate = k[A]^x[B]^y
where x and y represent the order w.r.t. the reactants A and B respectively. Overall order of reaction = x + y.
(ii) Order is applicable to elementary as well as complex reactions whereas molecularity is applicable only for elementary reactions.
For a complex reaction molecularity has no meaning and order is given by the slowest step. Molecularity of the slowest step is same as the order of the overall reaction.

Teacher's Note:
1. Always add the illustration (aA + bB → products, Rate = k[A]^x[B]^y) to a definition answer – it shows the examiner you know where x and y come from.

Question 34. Observe the graph in diagram and answer the following questions.
(i) If slope is equal to −2.0 × 10⁻⁶ sec⁻¹, what will be the value of rate constant?
(ii) How does the half-life of zero order reaction relate to its rate constant? [CBSE Sample Paper 2017]

Answer:
(i) Slope = −\( \frac{k}{2.303} \) or k = −2.303 × Slope
∴ k = −2.303 × (−2.0 × 10⁻⁶ s⁻¹)
k = 4.606 × 10⁻⁶ s⁻¹
(ii) For a zero order reaction
t = \( \frac{[R]_0 - [R]}{k} \)
At t = t_1/2, [R] = \( \frac{[R]_0}{2} \)
∴ t_1/2 = \( \frac{[R]_0 - \frac{[R]_0}{2}}{k} \) or t_1/2 = \( \frac{[R]_0}{2k} \)

Teacher's Note:
1. A straight line on a log[R] vs t plot means first order; its slope is −k/2.303, so do not forget the 2.303 factor.

Long Answer Questions (5 Marks)

Question 35. (i) Write the rate law for a first order reaction. Justify the statement that half life for a first order reaction is independent of the initial concentration of the reactant.
(ii) For a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%. [CBSE Comptt. OD Set-1, 2 2017]
Answer:
(i) For a first order reaction, k = \( \frac{2.303}{t} \) log \( \frac{[R]_0}{[R]} \), where [R]₀ = initial concentration, [R] = conc. after time t
When half of the reaction is completed, [R] = [R]₀/2. Representing the time taken for half of the reaction to be completed by t_1/2, the equation becomes:
k = \( \frac{2.303}{t_{1/2}} \) log \( \frac{[R]_0}{[R]_0/2} \) ⇒ t_1/2 = \( \frac{2.303}{k} \) log 2 = \( \frac{2.303}{k} \) × 0.3010 ⇒ t_1/2 = \( \frac{0.693}{k} \)
The above equation shows that half-life of a first order reaction is independent of the initial concentration of the reactant.
(ii) For a first order reaction, t = \( \frac{2.303}{k} \) log \( \frac{a}{a-x} \)
t_99% = \( \frac{2.303}{k} \) log \( \frac{100}{1} \) = \( \frac{2.303}{k} \) log 100 = \( \frac{2.303\times2}{k} \) = \( \frac{4.606}{k} \)
and t_90% = \( \frac{2.303}{k} \) log \( \frac{100}{10} \) = \( \frac{2.303}{k} \) log 10 = \( \frac{2.303}{k} \)
\( \frac{t_{99\%}}{t_{90\%}} \) = 2 ⇒ t_99% = 2 × t_90%

Teacher's Note:
1. The rate law asked in (i) is Rate = k[R]; the [R]₀ cancels inside the log when [R] = [R]₀/2, which is the whole justification.

Question 36. (i) A first order reaction is 75% completed in 40 minutes. Calculate its t_1/2.
(ii) Predict the order of the reaction in the given plots:
where [R]₀ is the initial concentration of reactant. (Given: log 2 = 0.3010, log 4 = 0.6021) [CBSE Foreign Set-1, 2, 3 2017]

Answer:
(i) k = \( \frac{2.303}{t} \) log \( \frac{[A]_0}{[A]} \) = \( \frac{2.303}{40} \) log \( \frac{100}{25} \) = \( \frac{2.303}{40} \) log 4 = \( \frac{2.303}{40} \) × 0.6021
k = 0.0347 min⁻¹
t_1/2 = \( \frac{0.693}{k} \) = \( \frac{0.693}{0.0347\ min^{-1}} \) = 19.98 min = 20 min
(ii) (a) First order reaction (b) Zero order reaction

Teacher's Note:
1. Shortcut for (i): 75% completion is exactly two half-lives (100 → 50 → 25), so t_½ = 40/2 = 20 min.

Question 37. (i) For a reaction A + B → P, the rate is given by Rate = k[A][B]²
(a) How is the rate of reaction affected if the concentration of B is doubled?
(b) What is the overall order of reaction if A is present in large excess?
(ii) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 = 0.3010)
Answer:
(i) A + B → P; Rate = k[A][B]²
(a) When concentration of B is doubled it means concentration of B becomes 2 times. Thus, Rate = k[A]¹[2B]² = k[A][4B²]. So, the rate becomes 4 times.
(b) Order of reaction is the number of molecules whose concentration alters after the reaction. If A is present in excess i.e., its concentration is unaffected. So, rate depends only on the concentration of B, as Rate = k′[B]². Thus, the reaction is of second order.
(ii) Step 1: For the 1st order reaction: k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \), t = 30 × 60 = 1800 s
k = \( \frac{2.303}{1800} \) log \( \frac{100}{100-50} \) = \( \frac{2.303}{1800} \) log 2 = \( \frac{2.303}{1800} \) × 0.3010
Step 2: and t = \( \frac{2.303}{k} \) log \( \frac{a}{a-x} \) = \( \frac{2.303}{k} \) log \( \frac{100}{100-90} \) = \( \frac{2.303}{k} \) log 10 = \( \frac{2.303}{k} \)
Step 3: By putting the value of k, we get t = \( \frac{2.303\times1800}{2.303\times0.3010} \) = 5.98 × 10³ s

Teacher's Note:
1. Useful check: t_90% = t_½/log 2 ≈ 3.32 × t_½, i.e. about 99.7 min here.

Teacher's Note:
1. Real data rarely give exact ratios: 3.94 ≈ 4 = 2² and 0.493 ≈ 0.495, so round to the nearest simple order.

Teacher's Note:
1. Pick pairs of experiments in which only one concentration changes; the order for that reactant follows from the ratio of rates.