CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), if Δ[NH₃]/Δt = 4 × 10⁻⁸ mol L⁻¹ s⁻¹, what is the value of −Δ[H₂]/Δt ?
Answer:
Rate = \( -\frac{1}{3}\frac{\Delta[H_2]}{\Delta t} = +\frac{1}{2}\frac{\Delta[NH_3]}{\Delta t} \)
or \( -\frac{\Delta[H_2]}{\Delta t} = \frac{3}{2}\frac{\Delta[NH_3]}{\Delta t} \)
= \( \frac{3}{2} \) × 4 × 10⁻⁸ mol L⁻¹ s⁻¹
= 6 × 10⁻⁸ mol L⁻¹ s⁻¹

Teacher's Note:
1. Quick check: H₂ (coefficient 3) must disappear faster than NH₃ (coefficient 2) appears, so the answer must be larger than 4 × 10⁻⁸.

Question 2. Write the names of four factors affecting the rate of a reaction. (Raj. Board 2011)
Answer:
(i) Concentration of reactants (ii) Temperature (iii) Surface area of reactants (iv) Presence of catalyst.

Teacher's Note:
1. Nature of reactants and exposure to light (for photochemical reactions) are two more factors you can quote if asked for more.

Question 3. Define specific reaction rate or rate constant.
Answer:
Specific reaction rate is the rate of reaction when the molar concentration of each of the reactants is unity.

Teacher's Note:
1. Because it is the rate at unit concentrations, the rate constant changes with temperature but not with concentration.

Question 4. Define : Rate of reaction and rate constant. (CBSE 2010)
Answer:
Rate of reaction is defined as the change in the concentration of any one of the reactants or products per unit time.
Rate constant of a reaction is equal to the rate of reaction when concentration of each reactant is taken as 1 mol L⁻¹.

Teacher's Note:
1. Rate depends on concentration and falls as the reaction proceeds; the rate constant stays fixed at a given temperature.

Question 5. What is the difference between Rate law and Law of Mass Action?
Answer:
Rate law is an experimental law whereas law of mass action is a theoretical law based on the balanced chemical reaction.

Teacher's Note:
1. In the rate law the powers are found by experiment and may differ from the coefficients; in the law of mass action the powers are always the stoichiometric coefficients.

Question 6. The rate law for the decomposition of N₂O₅ is : rate = k [N₂O₅]. What is the significance of k in this equation ?
Answer:
k represents the rate constant or specific reaction rate for the reaction and is equal to the rate of reaction when concentration of N₂O₅ is 1 mol L⁻¹.

Teacher's Note:
1. Since this is a first order rate law, the unit of k here is s⁻¹ (time⁻¹).

Question 7. Define the term ‘order of reaction’ for chemical reactions. (CBSE 2008, 2010; Pb. Board 2011; AP Board 2012)
Answer:
Order of a reaction is the sum of the powers of the concentration terms of the reactants as they appear in the experimentally determined rate law expression.
For example, if Rate = k[A]^x[B]^y, then order of reaction = x + y.

Teacher's Note:
1. Order is found from experiment, not from the balanced equation, and it can be zero, a whole number or a fraction.

Question 8. Give an example of first order reaction. (HP Board 2013)
Answer:
2N₂O₅ → 4NO₂ + O₂, Rate = k [N₂O₅].

Teacher's Note:
1. Radioactive decay and the decomposition of H₂O₂ are other common examples of first order reactions.

Question 9. Rate of a reaction is given by the equation : Rate = k [A]² [B]
What are the units for the rate and the rate constant for this reaction ?
Answer:
Units of rate = mol L⁻¹ s⁻¹
Units of k = \( \frac{\text{Rate}}{[A]^2[B]} = \frac{\text{mol L}^{-1}\text{ s}^{-1}}{(\text{mol L}^{-1})^2(\text{mol L}^{-1})} \)
= L² mol⁻² s⁻¹.

Teacher's Note:
1. Shortcut: for an nth order reaction the unit of k is (mol L⁻¹)¹⁻ⁿ s⁻¹; here n = 3.

Question 10. The decomposition reaction of ammonia gas on platinum surface has a rate constant = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. What is the order of the reaction ?
Answer:
Zero order (on the basis of units of k)

Teacher's Note:
1. A rate constant with the same unit as rate (mol L⁻¹ s⁻¹) always means zero order.

Question 11. The rate constant of a reaction is 3 × 10² min⁻¹. What is the order of the reaction? (Assam Board 2012)
Answer:
On the basis of units of rate constant, order = 1.

Teacher's Note:
1. Only a first order rate constant has the unit time⁻¹ with no concentration term.

Short Answer Questions (2 Marks)

Question 12. Why boiling of an egg or cooking of rice in an open vessel takes more time at a hill station?
Answer:
At a hill station, due to higher altitude, atmospheric pressure is low. Hence, water boils at a lower temperature.
At a lower temperature, the rate of reaction decreases. Hence, it takes longer time.

Teacher's Note:
1. Link the chain: lower pressure → lower boiling point → lower temperature → slower reaction.

Question 13. When is the rate of reaction equal to specific reaction rate? (Manipur Board 2012)
Answer:
Rate = k [A]^a [B]^b etc.
Rate = k when either concentration of each reactant is 1 mol L⁻¹ or order with respect to each reactant is zero, i.e., the reaction is of zero order.

Teacher's Note:
1. “Specific reaction rate” is another name for the rate constant k – the rate at unit concentration of each reactant.

Question 14. A reaction is first order in X and second order in Y.
(i) How is the rate affected on increasing the concentration of Y three times?
(ii) How is the rate affected if the concentration of both X and Y are doubled? (CBSE 2020)
Answer:
Rate = k [X] [Y]²
(i) If [Y] is increased three times, new rate = k [X] [3Y]² = 9 k [X] [Y]², i.e., increases 9 times.
(ii) If both X and Y are doubled, Rate = k [2X] [2Y]² = 8 k [X] [Y]², i.e., increases 8 times.

Teacher's Note:
1. Raise the factor of change to the power of the order for each reactant, then multiply.

Question 15. The rate expression for the reaction, (CH₃)₃C–Cl + OH⁻ → (CH₃)₃C–OH + Cl⁻ is Rate = k [(CH₃)₃C–Cl]. Propose the mechanism for the reaction.
Answer:
The rate of reaction depends on the slowest step. Hence, we shall have
Step 1: (CH₃)₃C–Cl → (CH₃)₃C⁺ + Cl⁻ (slow)
Step 2: (CH₃)₃C⁺ + OH⁻ → (CH₃)₃C–OH (fast)

Teacher's Note:
1. OH⁻ is absent from the rate law, so it must enter only in a fast step after the slow one.

Question 16. For the reaction: 2NH₃ (g) → N₂ (g) + 3H₂ (g) (on Pt surface), Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the units of k. (CBSE 2016)
Answer:
(i) Rate = k shows that the rate of the reaction is independent of the concentrations of the reactants. Hence, it is a reaction of zero order. As two molecules of the reactant are present in the reaction, molecularity is two.
(ii) Units of k will be same as that of the rate of reaction viz. mol L⁻¹ s⁻¹.

Teacher's Note:
1. Decomposition of NH₃ on hot platinum is zero order because the metal surface is fully covered at high pressure.

Question 17. What are the units of the rate constant of a pseudo unimolecular reaction?
Answer:
For a pseudo unimolecular reaction, Rate = k′[A], where k′ = k [H₂O]. The observed rate constant k′ has units of a first order reaction, i.e., min⁻¹ (or s⁻¹).
The true rate constant: k′ = k [H₂O] ∴ k = \( \frac{k'}{[H_2O]} \) = \( \frac{min^{-1}}{mol\ L^{-1}} \) = L mol⁻¹ min⁻¹
Thus, the true rate constant has the units of a second order reaction.

Teacher's Note:
1. The pseudo first order constant k′ contains the constant concentration of the excess reactant, which is why k′ and k have different units.

Question 18. The rate constant for a second order reaction is k = \( \frac{2.303}{t(a-b)} \) log \( \frac{b(a-x)}{a(b-x)} \) where a and b are initial concentrations of the two reactants A and B involved. If one of the reactants is present in excess, it becomes pseudo unimolecular. Explain how?
Answer:
Suppose B is in excess so that b >> a or x. Neglecting a and x in comparison to b, the equation reduces to
kb = k′ = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \)
which is same as for reactions of 1st order.

Teacher's Note:
1. With (a − b) ≈ −b and (b − x) ≈ b, the b terms cancel and only the first order expression in a survives.

Question 19. The reaction 2NO (g) + O₂ (g) → 2NO₂ (g) and 2CO (g) + O₂ (g) → 2CO₂ (g) look to be similar. Yet the former is faster than the latter at the same temperature. Explain why?
Answer:
The former has lower activation energy than the latter reaction. Lower the activation energy, faster is the reaction.

Teacher's Note:
1. Similar-looking equations tell you nothing about speed; rate is decided by the activation energy barrier.

Question 20. Can a reaction have zero activation energy?
Answer:
If E_a = 0, then according to Arrhenius equation, k = Ae^−E_a/RT = Ae⁰ = A, i.e., Rate constant = collision frequency.
This means every collision results into a chemical reaction which cannot be true. Hence, E_a ≠ 0.

Teacher's Note:
1. Use the Arrhenius equation as a check: E_a = 0 would make every collision effective, which contradicts collision theory.

Question 21. Can a reaction have negative activation energy?
Answer:
According to Arrhenius eqn., k = Ae^−E_a/RT. If E_a were negative, the exponential factor can be written as e^x where x = E_a/RT (taken positive). If x << 1, we have
e^x = 1 + x + \( \frac{x^2}{2!} \) + \( \frac{x^3}{3!} \) + ... = 1 + x (Neglecting higher powers of x) = 1 + E_a/RT
∴ k = A (1 + E_a/RT), i.e., k >> A which is impossible. Hence, E_a cannot be negative.

Teacher's Note:
1. The rate constant can never exceed the collision frequency A, because not more than every collision can react.

Question 22. What is the value of rate constant at extremely high temperature (= ∞)? Is this rate constant feasible? Why or why not?
Answer:
At T = ∞, k = Ae^−E_a/RT = Ae⁰ = A, i.e., Rate constant = collision frequency
which is not feasible as explained in Q. 17 (it would mean every collision results in reaction).

Teacher's Note:
1. As T → ∞, E_a/RT → 0, so the exponential factor tends to 1 and k approaches its upper limit A.

Question 23. Why equilibrium constant of a reaction does not change in the presence of a catalyst?
Answer:
The catalyst increases the speed of forward reaction as well as backward reaction to the same extent. Hence, equilibrium is not disturbed but is attained quickly.

Teacher's Note:
1. K = k_f/k_b; a catalyst raises both by the same factor, so their ratio is unchanged.

Question 24. The rate of a reaction depends upon the temperature and quantitatively expressed as k = Ae^−E_a/RT
(i) If a graph is plotted between log k and 1/T, write the expression for the slope of the graph.
(ii) At different conditions, E_a1 and E_a2 are the activation energies of the two reactions. If E_a1 = 40 J/mol and E_a2 = 80 J/mol, which of the two has larger value of the rate constant? (CBSE Sample Paper 2019)
Answer:
(i) ln k = ln A − \( \frac{E_a}{RT} \) or log k = log A − \( \frac{E_a}{2.303RT} \)
∴ Plot of log k vs \( \frac{1}{T} \) will be linear with slope = −\( \frac{E_a}{2.303R} \).
(ii) Lower the activation energy, faster is the reaction and hence greater is the rate constant. Hence, k₁ > k₂.

Teacher's Note:
1. With log₁₀ the slope is −E_a/2.303R; with ln it is −E_a/R – state which one your graph uses.

Question 25. What is meant by reaction rate ? Give its symbolic expression and units for the reaction
CO(g) + NO₂(g) → CO₂(g) + NO(g)
Answer:
The change in concentration of any one of the reactants or products per unit time is called the rate of reaction.
For the given reaction, Rate = \( -\frac{d[CO]}{dt} = -\frac{d[NO_2]}{dt} = +\frac{d[CO_2]}{dt} = +\frac{d[NO]}{dt} \)
Units of rate = mol L⁻¹ s⁻¹ (for gases, atm s⁻¹ may also be used).

Teacher's Note:
1. All coefficients here are 1, so every species changes at the same rate; only the signs differ.

Question 26. Show graphically the average and instantaneous rate of a reaction. (Hr. Board 2011)

Answer:
When the concentration of a reactant is plotted against time, a curve is obtained (see figure).
Average rate between times t₁ and t₂ is the slope of the chord joining the two points on the curve:
r_av = \( -\frac{[R]_2 - [R]_1}{t_2 - t_1} = -\frac{\Delta [R]}{\Delta t} \)
Instantaneous rate at any time t is the slope of the tangent drawn to the curve at that point: r_inst = \( -\frac{d[R]}{dt} \).

Teacher's Note:
1. Average rate uses a chord between two points; instantaneous rate uses a tangent at one point – do not mix up the two constructions.

Short Answer Questions (3 Marks)

Question 27. The rate of most of reactions double when their temperature is raised from 298 K to 308 K. Calculate the activation energy of such a reaction.
Answer:
log \( \frac{K_2}{K_1} \) = \( \frac{E_a}{2.303R} \) \( \left[ \frac{1}{T_1} - \frac{1}{T_2} \right] \)
log 2 = \( \frac{E_a}{2.303 \times 8.314} \) \( \left[ \frac{308 - 298}{298 \times 308} \right] \)
E_a = \( \frac{2.303 \times 8.314 \times 298 \times 308 \times 0.3010}{10} \) J mol⁻¹
E_a = 52.89 kJ/mol

Teacher's Note:
1. This ~53 kJ mol⁻¹ value is the activation energy behind the rule of thumb that rates double for every 10 K rise near room temperature.

Question 28. A first order reaction takes 69.3 min for 50% completion. Set up an equation for determining the time needed for 80% completion.
Answer:
K = \( \frac{0.693}{t_{1/2}} \) = \( \frac{0.693}{69.3} \) min⁻¹ = 10⁻² min⁻¹
t = \( \frac{2.303}{K} \) log \( \frac{[R_0]}{[R]} \)
For 80% completion, [R] = 0.2[R]₀, so \( \frac{[R_0]}{[R]} \) = 5
= \( \frac{2.303}{10^{-2}} \) log 5
= 160.9 min

Teacher's Note:
1. 80% completion means 20% is left, so the ratio is 100/20 = 5 (not 100/80).

Question 29. The activation energy of a reaction is 94.14 kJ/mol and the value of rate constant at 40°C is 1.8 × 10⁻¹ s⁻¹. Calculate the frequency factor A.
Answer:
Given, E_a = 94.14 × 10³ J mol⁻¹, T = 40 + 273 = 313 K, K = 1.8 × 10⁻¹ s⁻¹
By using, K = Ae^−E_a/RT ⇒ ln K = ln A − \( \frac{E_a}{RT} \)
Or log K = log A − \( \frac{E_a}{2.303RT} \)
Or log (1.8 × 10⁻¹) + \( \frac{94.14 \times 10^3}{2.303 \times 8.314 \times 313} \) = log A
Or −0.7447 + 15.7082 = log A = 14.9635
Or A = antilog (14.9635) = 9.194 × 10¹⁴ s⁻¹

Teacher's Note:
1. A has the same unit as k (s⁻¹ here); a large A like 10¹⁴ s⁻¹ is typical for first order reactions.

Question 30. The rate constants of a reaction at 500 K and 700 K are 0.02 s⁻¹ and 0.07 s⁻¹ respectively. Calculate the values of E_a and A.
Answer:
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \) \( \left[ \frac{T_2 - T_1}{T_1 T_2} \right] \)
log \( \frac{0.07}{0.02} \) = \( \frac{E_a}{2.303 \times 8.314} \) \( \left[ \frac{700 - 500}{700 \times 500} \right] \)
0.5441 = \( \frac{E_a}{19.147} \) × 5.714 × 10⁻⁴
E_a = 18.23 kJ mol⁻¹
log A = log k + \( \frac{E_a}{2.303RT} \) = log 0.02 + \( \frac{18230}{2.303 \times 8.314 \times 500} \) = −1.699 + 1.904 = 0.205
A = 1.603 s⁻¹

Teacher's Note:
1. Find E_a first from the two rate constants, then put either (k, T) pair back into the Arrhenius equation to get A.

Question 31. The rate constants of a reaction at 700 K and 760 K are 0.011 M⁻¹ s⁻¹ and 0.105 M⁻¹ s⁻¹ respectively. Calculate the values of Arrhenius parameters.
Answer:
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \) \( \left[ \frac{T_2 - T_1}{T_1 T_2} \right] \)
log \( \frac{0.105}{0.011} \) = \( \frac{E_a}{2.303 \times 8.314} \) \( \left[ \frac{760 - 700}{700 \times 760} \right] \)
0.9798 = \( \frac{E_a}{19.147} \) × 1.128 × 10⁻⁴
E_a = 166.3 kJ mol⁻¹
log A = log k + \( \frac{E_a}{2.303RT} \) = log 0.011 + \( \frac{166340}{2.303 \times 8.314 \times 700} \) = −1.9586 + 12.4108 = 10.4522
A = 2.83 × 10¹⁰ M⁻¹ s⁻¹

Teacher's Note:
1. "Arrhenius parameters" means both E_a and A – give both, with units.

Question 32. The initial concentration of N₂O₅ in the first order reaction N₂O₅ → 2NO₂ + ½O₂ was 1.24 × 10⁻² mol L⁻¹ at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 × 10⁻² mol L⁻¹. Calculate the rate constant of the reaction at 318 K.
Answer:
K = \( \frac{2.303}{t} \) log \( \frac{[A]_0}{[A]} \) = \( \frac{2.303}{t} \) log \( \frac{[N_2O_5]_0}{[N_2O_5]_t} \) = \( \frac{2.303}{60} \) log \( \frac{1.24 \times 10^{-2}}{0.2 \times 10^{-2}} \)
= \( \frac{2.303}{60} \) log 6.2 = \( \frac{2.303}{60} \) × 0.7924 min⁻¹
= 0.0304 min⁻¹

Teacher's Note:
1. The time is in minutes, so k comes out in min⁻¹; convert only if the question asks for s⁻¹.

Question 33. A first order reaction is 20% complete in 20 minutes. Calculate the time taken for the reaction to go to 80% completion.
Answer:
K = \( \frac{2.303}{20} \) log \( \frac{100}{80} \) = \( \frac{2.303}{20} \) × 0.0969 = 1.116 × 10⁻² min⁻¹
t = \( \frac{2.303}{K} \) log \( \frac{100}{20} \) = \( \frac{2.303 \times 0.6990}{1.116 \times 10^{-2}} \)
t = 144.3 minutes

Teacher's Note:
1. At 20% completion 80% is left, and at 80% completion 20% is left – keep the two ratios (100/80 and 100/20) straight.

Question 34. For a first order reaction, calculate the ratio between the time taken to complete ¾ of the reaction and the time taken to complete half of the reaction.
Answer:
t_3/4 = \( \frac{2.303}{k} \) log 4 = \( \frac{2.303 \times 2 \times 0.3010}{k} \);   t_1/2 = \( \frac{2.303}{k} \) log 2 = \( \frac{2.303 \times 0.3010}{k} \)
\( \frac{t_{3/4}}{t_{1/2}} \) = Two

Teacher's Note:
1. log 4 = 2 log 2, which is why 75% completion takes exactly two half-lives.

Teacher's Note:
1. Always compare two experiments in which only one concentration changes; then check the rate law against the remaining experiment.

Long Answer Questions (5 Marks)

Question 36. (a) Define order of reaction.
(b) Rates of reaction double with every 10° rise in temperature. If this generalization holds for a reaction in the temperature range 298 K to 308 K, what would be the value of activation energy for this reaction? R = 8.314 J K⁻¹ mol⁻¹.
Answer:
(a) Order of Reaction: It is the sum of powers to which the conc. terms are raised in rate law expression.
(b) log \( \frac{K_2}{K_1} \) = \( \frac{E_a}{2.303R} \) \( \left[ \frac{1}{T_1} - \frac{1}{T_2} \right] \)
Here, T₁ = 298 K, T₂ = 308 K, R = 8.314 J K⁻¹ mol⁻¹, \( \frac{K_2}{K_1} \) = 2
log 2 = \( \frac{E_a}{2.303 \times 8.314} \) \( \left[ \frac{1}{298} - \frac{1}{308} \right] \)
0.3010 = \( \frac{E_a}{2.303 \times 8.314} \) \( \left[ \frac{10}{298 \times 308} \right] \)
E_a = \( \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10} \)
= 52898 J mol⁻¹
= 52.898 kJ mol⁻¹

Teacher's Note:
1. Write (1/T₁ − 1/T₂) as (T₂ − T₁)/(T₁T₂) to avoid rounding errors from tiny reciprocals.

Question 37. (a) What are pseudo order reactions? Give example.
(b) Rate constant K of a reaction varies with temperature ‘T’ according to the equation:
log K = log A − \( \frac{E_a}{2.303R} \) \( \left( \frac{1}{T} \right) \)
where E_a is the activation energy. When a graph is plotted for log K vs 1/T, a straight line with a slope of −4250 K is obtained. Calculate E_a for the reaction.
Answer:
(a) The chemical reactions which look like higher order reactions but in real they follow lower order kinetics.
For example, CH₃COOC₂H₅ + H₂O (excess) \( \xrightarrow{H^+} \) CH₃COOH + C₂H₅OH
Rate = K[CH₃COOC₂H₅], Order = 1
(b) Slope = −\( \frac{E_a}{2.303R} \) = −4250 K
So, E_a = −2.303 × R × Slope
= −2.303 × 8.314 J K⁻¹ mol⁻¹ × (−4250 K)
= 81375.3 J mol⁻¹
= 81.375 kJ mol⁻¹

Teacher's Note:
1. The slope of log K vs 1/T is negative; E_a must come out positive, which is a quick check on the sign.

Question 38. (a) Determine the units of rate constant for first and zero order reaction.
(b) Show that time required for the completion of 99% of the first order reaction is twice the time required for 90% completion of the reaction.
Answer:
(a) K = (mol)¹⁻ⁿ Lⁿ⁻¹ s⁻¹
For zero order, n = 0
So, K = (mol)¹⁻⁰ L⁰⁻¹ s⁻¹ = mol L⁻¹ s⁻¹
For first order, n = 1
So, K = (mol)¹⁻¹ L¹⁻¹ s⁻¹ = s⁻¹
(b) For a first order reaction, t = \( \frac{2.303}{K} \) log \( \frac{[A]_0}{[A]} \)
[A]₀ = a, [A] = a − \( \frac{a \times 99}{100} \) = 0.01a
t(99%) = \( \frac{2.303}{K} \) log \( \frac{a}{0.01a} \) = \( \frac{2.303}{K} \) log 100 = \( \frac{2.303}{K} \) × 2   ...(i)
For 90% completion of reaction, [A] = a − \( \frac{a \times 90}{100} \) = 0.1a
t(90%) = \( \frac{2.303}{K} \) log \( \frac{a}{0.1a} \) = \( \frac{2.303}{K} \) × 1   ...(ii)
Dividing equation (i) by equation (ii), we get
t(99%) = 2 × t(90%)

Teacher's Note:
1. Each tenfold fall in concentration takes the same time (2.303/K) in a first order reaction: 100% → 10% and 10% → 1% take equal times.

Question 39. (a) Define rate constant of reaction.
(b) A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Answer:
(a) Rate constant: It is the rate of chemical reaction when the concentration of reactant is taken as unity at a given temperature.
(b) Let initial conc. = a
Conc. after 40 min = a − \( \frac{a \times 30}{100} \) = 0.70a
K = \( \frac{2.303}{t} \) log \( \frac{[A]_0}{[A]} \) = \( \frac{2.303}{40} \) log \( \frac{a}{0.70a} \) = \( \frac{2.303}{40} \) log \( \frac{1}{0.70} \)
= \( \frac{2.303}{40} \) × 0.1549 = 8.92 × 10⁻³ min⁻¹
t_1/2 = \( \frac{0.693}{K} \) = \( \frac{0.693}{8.92 \times 10^{-3}} \) = 77.7 min

Teacher's Note:
1. 30% decomposition leaves 70%, so use 100/70 in the log, not 100/30.

Teacher's Note:
1. For A(g) → B(g) + C(g), the reactant pressure is 2P_i − P_t; derive it once (P_t = P_i + x) rather than memorising it.