CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. In a reaction 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval.
Answer:
Average rate = \( \frac{-\Delta[A]}{2\Delta t} \) = −\( \frac{1}{2} \) \( \left( \frac{0.4 - 0.5}{10} \right) \) = 5 × 10⁻³ M min⁻¹

Teacher's Note:
1. Do not forget to divide by the coefficient 2 of A; the rate of disappearance of A itself is 1 × 10⁻² M min⁻¹.

Question 2. In some cases large number of colliding reactant molecules have energy more than threshold energy even then the reaction is slow. Why?
Answer:
Because reactant molecules do not collide in proper orientation.

Teacher's Note:
1. Effective collisions need both sufficient energy and proper orientation.

Question 3. Give an example of a reaction having fractional order.
Answer:
Decomposition of acetaldehyde (order = 1.5).
CH₃CHO \( \xrightarrow{723\,K} \) CH₄ + CO

Teacher's Note:
1. Fractional order is possible only for complex reactions, never for elementary ones.

Question 4. Decomposition reaction of ammonia on Pt surface has rate constant = 2.5 × 10⁻¹ mol L⁻¹ s⁻¹. What is order of reaction?
Answer:
Unit of k explains that it is zero order reaction.

Teacher's Note:
1. The unit mol L⁻¹ s⁻¹ is the same as the unit of rate, which marks zero order.

Question 5. What is order of radioactive decay?
Answer:
First order

Teacher's Note:
1. That is why the half-life of a radioactive element does not depend on the amount taken.

Question 6. A reaction proceeds with a uniform rate throughout. What do you conclude? or Is there any reaction whose rate does not decrease with time?
Answer:
Reaction is of zero order.

Teacher's Note:
1. Only for zero order is the rate independent of concentration, so it does not fall as the reactant is used up.

Question 7. For a reaction, the graph of the rate of reaction against molar concentration of the reactant is as shown. What is the order of the reaction? (Karnataka Board 2012)

Answer:
Zero order.

Teacher's Note:
1. A horizontal rate vs [A] line means the rate does not change with concentration – the definition of zero order.

Question 8. What is the fraction of molecules having energy equal to or greater than activation energy, E_a? What is this quantity called?
Answer:
It is equal to e^−E_a/RT at temperature T. This quantity is called ‘Boltzmann factor’.

Teacher's Note:
1. The Boltzmann factor rises sharply with temperature, which is why a small rise in T can double the rate.

Question 9. Write expression for rate of reaction in terms of each reactant and product for the reaction
N₂ + 3H₂ → 2NH₃.
Answer:
Rate = \( -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt} \)

Teacher's Note:
1. Divide each rate of change by its stoichiometric coefficient so that the rate has one single value; reactant terms carry a minus sign.

Question 10. For the straight reaction, X₂(g) + 2Y₂(g) → 2XY₂(g), write the rate equation in terms of disappearance of Y₂.
Answer:
Rate = \( -\frac{1}{2}\frac{d[Y_2]}{dt} \) = k[X₂][Y₂]².
Rate of disappearance of Y₂ \( \left(-\frac{d[Y_2]}{dt}\right) \) = 2k[X₂][Y₂]²

Teacher's Note:
1. Rate of reaction and rate of disappearance of Y₂ are different quantities: the second is twice the first because 2 molecules of Y₂ are used up per reaction event.

Question 11. Express the rate of the following reaction in terms of disappearance of hydrogen in the reaction
3H₂(g) + N₂(g) → 2NH₃(g) (CBSE 2007)
Answer:
Rate = \( -\frac{1}{3}\frac{d[H_2]}{dt} \)

Teacher's Note:
1. The factor 1/3 comes from the coefficient 3 of H₂ in the balanced equation.

Short Answer Questions (2 Marks)

Question 12. For the reaction N₂(g) + 3H₂(g) → 2NH₃(g)
How is the rate of formation of ammonia related to the rate of disappearance of H₂?
Answer:
Rate of reaction = −\( \frac{1}{3} \) \( \frac{\Delta[H_2]}{\Delta t} \) = \( \frac{1}{2} \) \( \frac{\Delta[NH_3]}{\Delta t} \)
Or \( \frac{\Delta[NH_3]}{\Delta t} \) = −\( \frac{2}{3} \) \( \frac{\Delta[H_2]}{\Delta t} \)

Teacher's Note:
1. A quick check: for every 3 mol H₂ used, 2 mol NH₃ form, so NH₃ forms at 2/3 the rate at which H₂ disappears.

Question 13. The rate of a gaseous reaction becomes half when volume of the vessel is doubled. What is the order of reaction?
Answer:
Suppose, order of reaction is n and the reaction is A(g) → Products
Rate = k[A]ⁿ   ...(i)
When volume is doubled, molar conc. becomes half and rate of reaction gets halved.
\( \frac{Rate}{2} \) = k \( \left( \frac{[A]}{2} \right)^n \)   ...(ii)
Dividing equation (i) by equation (ii),
(2)¹ = (2)ⁿ
Or n = 1

Teacher's Note:
1. Doubling the volume of a gas halves every concentration; the rate changes by (1/2)ⁿ.

Question 14. A reaction which is first order with respect to A has rate constant 6 min⁻¹. If we start with [A] = 0.5 mol L⁻¹, when would [A] reach the value of 0.05 mol L⁻¹?
Answer:
k = \( \frac{2.303}{t} \) log \( \frac{[A]_0}{[A]} \)
k = 6 min⁻¹, [A]₀ = 0.5, [A] = 0.05, t = ?
t = \( \frac{2.303}{6} \) log \( \frac{0.5}{0.05} \) = \( \frac{2.303}{6} \) log 10 = 0.3838 min

Teacher's Note:
1. A fall to 1/10th of the initial concentration always takes t = 2.303/k for a first order reaction.

Question 15. The conversion of the molecules X to Y follows second order kinetics. If the concentration of X is increased to three times, how will it affect the rate of formation of Y?
Answer:
Rate = k[X]²; Rate′ = k(3[X])² = 9 k[X]²
The rate of formation of Y becomes 9 times.

Teacher's Note:
1. For second order, rate scales with the square of the concentration factor: 3² = 9.

Question 16. 4NH₃ + 5O₂ → 4NO + 6H₂O. If rate of formation of NO is 6 × 10⁻⁴ atm min⁻¹, calculate the rate of formation of H₂O.
Answer:
\( \frac{1}{4} \) \( \frac{d[NO]}{dt} \) = \( \frac{1}{6} \) \( \frac{d[H_2O]}{dt} \) ⇒ \( \frac{d[H_2O]}{dt} \) = \( \frac{6}{4} \) × 6 × 10⁻⁴
= 9.0 × 10⁻⁴ atm min⁻¹

Teacher's Note:
1. Ratio of rates of formation = ratio of stoichiometric coefficients (6 : 4 here).

Question 17. Consider a certain reaction A → Product with K = 2.0 × 10⁻² s⁻¹. Calculate the concentration of A remaining after 100 s, if the initial concentration of A is 1.0 mol L⁻¹.
Answer:
The unit of K (s⁻¹) shows a first order reaction.
log \( \frac{[A]_0}{[A]} \) = \( \frac{Kt}{2.303} \) = \( \frac{2.0 \times 10^{-2} \times 100}{2.303} \) = 0.8684
\( \frac{[A]_0}{[A]} \) = 7.39 ⇒ [A] = \( \frac{1.0}{7.39} \)
[A] = 0.135 M

Teacher's Note:
1. Always read the order from the unit of the rate constant before choosing the formula.

Question 18. Explain with an example, what is a pseudo first order reaction? The graphs (A and B) given below are plots of rate of reaction vs concentration of the reactant. Predict the order from the graphs.

Answer:
A pseudo first order reaction is a reaction which looks like a higher order reaction but actually follows first order kinetics, because one reactant is present in large excess and its concentration remains almost constant.
Example: acid hydrolysis of ethyl acetate, CH₃COOC₂H₅ + H₂O \( \xrightarrow{H^+} \) CH₃COOH + C₂H₅OH; Rate = k[CH₃COOC₂H₅] (water in excess).
Graph A: rate increases linearly with [R], i.e. rate ∝ [R]; the reaction is first order.
Graph B: rate does not change with [R], i.e. rate = k[R]⁰; the reaction is zero order.

Teacher's Note:
1. Note the axes: these are rate vs concentration plots, not concentration vs time plots – a horizontal line here means zero order.

Question 19. Differentiate between:
(a) Average rate and instantaneous rate of a chemical reaction.
(b) Molecularity and order of reaction.
Answer:
(a) Average rate is the change in concentration of a reactant or product divided by the time interval over which the change occurs (−Δ[R]/Δt); it is measured over a long time interval. Instantaneous rate is the rate at a particular moment of time (−d[R]/dt); it is obtained from the slope of the tangent to the concentration–time curve at that instant.
(b) Molecularity is the number of reacting species taking part in an elementary reaction which must collide simultaneously; it is a theoretical concept and is always a whole number (never zero). Order is the sum of the powers of the concentration terms in the experimental rate law; it is determined experimentally and can be zero or fractional.

Teacher's Note:
1. Instantaneous rate becomes equal to average rate only when the time interval Δt approaches zero.

Question 20. Show that in case of first order reaction, the time required for 99.9% of the reaction to take place is about ten times than that required for half the reaction.
Answer:
For a first order reaction, t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \)
For 99.9% completion, [R] = [R]₀ − 0.999[R]₀ = 0.001[R]₀
t_99.9% = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{0.001[R]_0} \) = \( \frac{2.303}{k} \) log 10³ = \( \frac{6.909}{k} \)
For half the reaction, t_1/2 = \( \frac{0.693}{k} \)
\( \frac{t_{99.9\%}}{t_{1/2}} \) = \( \frac{6.909}{0.693} \) ≈ 10
Hence, t_99.9% ≈ 10 × t_1/2.

Teacher's Note:
1. The rate constant cancels in the ratio, so this result holds for every first order reaction.

Question 21. For the reaction NO₂ + CO → CO₂ + NO, the experimentally determined rate expression below 400 K is rate = k[NO₂]². What mechanism can be proposed for this reaction?
Answer:
Since CO does not appear in the rate law, it must take part only after the rate determining step, and the slow step must involve two molecules of NO₂. A possible mechanism is:
Step 1. NO₂ + NO₂ → NO₃ + NO (slow)
Step 2. NO₃ + CO → NO₂ + CO₂ (fast)
Rate = k[NO₂]² (from the slow step); adding the two steps gives the overall reaction NO₂ + CO → CO₂ + NO.

Teacher's Note:
1. Any proposed mechanism must satisfy two checks: the steps must add up to the overall equation, and the slow step must give the observed rate law.

Question 22. The half life period of a first order reaction is 60 min. What % will be left after 240 min?
Answer:
Number of half-lives, n = \( \frac{240}{60} \) = 4
Fraction left = \( \left( \frac{1}{2} \right)^4 \) = \( \frac{1}{16} \)
% left = 6.25%

Teacher's Note:
1. When the time is an exact multiple of t_1/2, counting half-lives is faster than using the log formula.

Question 23. Time for half change for a first order reaction is 25 min. What time will be required for 99% reaction?
Answer:
k = \( \frac{0.693}{25} \) = 0.02772 min⁻¹
t = \( \frac{2.303}{k} \) log \( \frac{100}{1} \) = \( \frac{2.303 \times 2}{0.02772} \)
t = 166.16 min

Teacher's Note:
1. For 99% reaction, [R]₀/[R] = 100/1, not 100/99 – a very common slip.

Question 24. For a reaction A → B, the rate of reaction can be denoted by −\( \frac{dA}{dt} \) or +\( \frac{dB}{dt} \). State the significance of plus and minus signs in this case.
Answer:
Rate of reaction is always positive. Minus sign indicates decrease in the concentration of the reactant, i.e., d[A] is −ve so that (−) × (−) = +ve.
Plus sign indicates increase in the concentration of the product with time during the reaction, i.e., d[B] is +ve.

Teacher's Note:
1. The signs are only a convention so that the rate comes out positive whether it is written for a reactant or a product.

Question 25. Why in general a reaction does not proceed with a uniform rate throughout or why instantaneous rate is preferred over average rate?
Answer:
This is because the rate of reaction at any time depends upon the concentrations of the reactants at that time which keep on decreasing with time.

Teacher's Note:
1. Average rate hides this continuous change; the instantaneous rate (slope of the tangent to the concentration-time curve) gives the rate at one moment.

Short Answer Questions (3 Marks)

Question 26. A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.
(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 J K⁻¹ mol⁻¹) [CBSE 2018]
Answer:
For a first order reaction, k = \( \frac{0.693}{t_{1/2}} \)
So, k₁ = \( \frac{0.693}{40} \) min⁻¹ and k₂ = \( \frac{0.693}{20} \) min⁻¹
\( \frac{k_2}{k_1} \) = \( \frac{0.693\ min^{-1}/20}{0.693\ min^{-1}/40} \) = 2
Here, T₁ = 300 K, T₂ = 320 K and R = 8.314 J K⁻¹ mol⁻¹
Substituting these values in the expression
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303\times R} \)\( \left(\frac{T_2 - T_1}{T_1 T_2}\right) \), we get
log 2 = \( \frac{E_a}{2.303\times8.314\ J K^{-1} mol^{-1}} \)\( \left(\frac{320\ K - 300\ K}{300\ K\times320\ K}\right) \)
0.3010 = \( \frac{E_a\times20}{19.147\ J mol^{-1}\times300\times320} \)
or E_a = \( \frac{0.3010\times19.147\times300\times320\ J mol^{-1}}{20} \) = 27663.58 J mol⁻¹
E_a = 27.66 kJ mol⁻¹

Teacher's Note:
1. The ratio of half-lives gives the ratio of rate constants in reverse: t₁/t₂ = k₂/k₁.

Question 27. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentration of both A and B are doubled? [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) Rate = \( -\frac{d[R]}{dt} \) = k[A][B]²
(ii) Rate becomes 9 times.
(iii) Rate becomes 8 times.

Teacher's Note:
1. Raise each factor to its own order: (3)² = 9 for B alone, and 2 × 2² = 8 when both are doubled.

Question 28. Half-life for a first order reaction is 693 s. Calculate the time required for 90% completion of this reaction. [CBSE Comptt. OD 2017]
Answer:
t_1/2 = \( \frac{0.693}{k} \)
k = \( \frac{0.693}{693\text{ s}} \) = 0.001 s⁻¹
k = \( \frac{2.303}{t} \) log [R]₀/[R]
t = \( \frac{2.303}{k} \) log [R]₀/[R] = \( \frac{2.303}{0.001} \) log 100/10
t = 2303 s

Teacher's Note:
1. For 90% completion, 10% is left, so [R]₀/[R] = 100/10 = 10 and log 10 = 1.
2. Write the unit of the final answer: it is a time (s), not s⁻¹.

Question 29. The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce 1 g of the reactant to 0.0625 g? [CBSE SQP 2017]
Answer:
t = \( \frac{2.303}{k} \) log \( \frac{[R]_0}{[R]} \)
t = \( \frac{2.303}{60} \) log \( \frac{1}{0.0625} \)
t = 0.0462 s

Teacher's Note:
1. 1/0.0625 = 16 = 2⁴, so the answer is exactly four half-lives (4 × 0.693/60 s) – a handy cross-check.

Teacher's Note:
1. SO₂Cl₂ → SO₂ + Cl₂ turns one mole of gas into two, which is why p_A = 2p₀ − p_t applies.

Question 31. Rate constant k for a first order reaction has been found to be 2.54 × 10⁻³ s⁻¹. Calculate its 3/4^th life. (log 4 = 0.6020) [CBSE Comptt. OD 2013]
Answer:
k = 2.54 × 10⁻³ s⁻¹
For a first order reaction, t = \( \frac{2.303}{k} \) log \( \frac{a}{a-x} \)
For 3/4^th life, a − x = a − \( \frac{3}{4} \)a
On substitution of these values,
t_3/4 = \( \frac{2.303}{2.54\times10^{-3}} \) log \( \frac{a}{a-\frac{3}{4}a} \) = \( \frac{2.303}{2.54\times10^{-3}} \) log \( \frac{a}{\frac{1}{4}a} \)
= \( \frac{2.303}{2.54\times10^{-3}} \) log 4 = \( \frac{2.303}{2.54\times10^{-3}} \) × 0.6020
= 545.82 s

Teacher's Note:
1. 3/4^th life is simply two half-lives: 2 × 0.693/(2.54 × 10⁻³) ≈ 546 s.

Question 32. A first order gas phase reaction: A₂B₂(g) → 2A(g) + 2B(g) at the temperature 400°C has the rate constant k = 2.0 × 10⁻⁴ s⁻¹. What percentage of A₂B₂ is decomposed on heating for 900 seconds? [Antilog 0.0781 = 1.197] [CBSE Comptt. OD 2013]
Answer:
k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \)
⇒ 2 × 10⁻⁴ = \( \frac{2.303}{900} \) log \( \frac{100}{100-x} \)
log \( \frac{100}{100-x} \) = \( \frac{9}{115.15} \) = 0.0781
⇒ \( \frac{100}{100-x} \) = Antilog (0.0781)
⇒ 100 = 1.197 (100 − x)
⇒ 100 = 119.7 − 1.197x
⇒ 1.197x = 19.7
∴ x = \( \frac{19.7}{1.197} \) = 16.45%

Teacher's Note:
1. Take a = 100 so that x comes out directly as the percentage decomposed; [A] in the formula is what is left, 100 − x.

Question 33. The rate of most reactions becomes double when their temperature is raised from 298 K to 308 K. Calculate their activation energy.
[Given R = 8.314 J mol⁻¹ K⁻¹] [CBSE Comptt. OD 2016; DDE]
Answer:
T₁ = 298 K, T₂ = 308 K, R = 8.314 J mol⁻¹ K⁻¹
k₂ = 2k₁
log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \) \( \left[\frac{1}{T_1} - \frac{1}{T_2}\right] \)
⇒ log 2 = \( \frac{E_a}{19.15\text{ J K}^{-1}\text{ mol}^{-1}} \) \( \left[\frac{1}{298} - \frac{1}{308}\right] \)
E_a = \( \frac{0.3010\times19.15\times298\times308}{10} \) J mol⁻¹
E_a = 52905 J mol⁻¹ or 52.905 kJ mol⁻¹

Teacher's Note:
1. 1/298 − 1/308 = 10/(298 × 308); writing it this way avoids rounding errors with tiny decimals.
2. This is the “rate doubles for a 10° rise” rule of thumb – it corresponds to E_a of about 53 kJ mol⁻¹ near room temperature.

Question 34. Two reactions of the same order have equal pre exponential factors but their activation energies differ by 24.9 kJ mol⁻¹. Calculate the ratio between the rate constants of these reactions at 27°C. (Gas constant R = 8.314 J K⁻¹ mol⁻¹) [CBSE SQP 2018-2019]
Answer:
The Arrhenius equation: k = Ae^−E_a/RT
Taking log on both sides: log k = log A − \( \frac{E_a}{2.303RT} \)
For reaction (i): log k₁ = log A − \( \frac{E_a(1)}{2.303RT} \)
For reaction (ii): log k₂ = log A − \( \frac{E_a(2)}{2.303RT} \)
Subtracting (i) from (ii):
log \( \frac{k_2}{k_1} \) = \( \frac{E_a(1) - E_a(2)}{2.303RT} \) = \( \frac{24.9\times1000}{2.303\times8.314\times300} \) = 4.335
\( \frac{k_2}{k_1} \) = antilog (4.335) = 2.16 × 10⁴

Teacher's Note:
1. Convert kJ to J (24.9 kJ = 24900 J) and °C to K (27°C = 300 K) before substituting.
2. The reaction with the lower activation energy is the faster one, by a factor of about 2 × 10⁴ here.

Long Answer Questions (5 Marks)

Question 35. (i) For the reaction A → B, the rate of reaction becomes twenty seven times when the concentration of A is increased three times. What is the order of the reaction?
(ii) The activation energy of a reaction is 75.2 kJ mol⁻¹ in the absence of a catalyst and it lowers to 50.14 kJ mol⁻¹ with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25°C? [CBSE SQP 2016]
Answer:
(i) r = k[R]ⁿ
When concentration is increased three times, [R] = 3a
27r = k(3a)ⁿ
\( \frac{27r}{r} \) = \( \frac{k(3a)^n}{ka^n} \) or 27 = 3ⁿ or 3³ = 3ⁿ
⇒ n = 3
(ii) According to Arrhenius equation, log k = log A − \( \frac{E_a}{2.303RT} \)
For uncatalysed reaction: log k₁ = log A − \( \frac{E_a(1)}{2.303RT} \)   ...(i)
For catalysed reaction: log k₂ = log A − \( \frac{E_a(2)}{2.303RT} \)   ...(ii)
A is equal for both the reactions. Subtracting equation (i) from equation (ii):
log \( \frac{k_2}{k_1} \) = \( \frac{E_a(1) - E_a(2)}{2.303RT} \)
⇒ log \( \frac{k_2}{k_1} \) = \( \frac{(75.2 - 50.14)\text{ kJ mol}^{-1}}{2.303\times8.314\text{ J K}^{-1}\text{mol}^{-1}\times298\text{ K}} \)
⇒ log \( \frac{k_2}{k_1} \) = 4.39
⇒ \( \frac{k_2}{k_1} \) = antilog (4.39) = 2.45 × 10⁴
Rate of reaction increases by 2.45 × 10⁴ times.

Teacher's Note:
1. In (ii) the difference 25.06 kJ must be converted to 25060 J before dividing by 2.303RT.
2. A catalyst does not change A, which is why the log A terms cancel on subtraction.

Question 36. (i) The decomposition of A into products has a value of k as 4.5 × 10³ s⁻¹ at 10°C and energy of activation 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?
(ii) (a) If half life period of a first order reaction is x and 3/4^th life period of the same reaction is y, how are x and y related to each other?
(b) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why? [CBSE Comptt. Delhi 2013]
Answer:
(i) We know that log \( \frac{k_2}{k_1} \) = \( \frac{E_a}{2.303R} \) \( \left(\frac{T_2 - T_1}{T_1 T_2}\right) \)
⇒ log \( \frac{1.5\times10^{4}}{4.5\times10^{3}} \) = \( \frac{60000}{2.303\times8.314} \) \( \left(\frac{T_2 - 283}{283\,T_2}\right) \)
⇒ 0.5228 = 3133.62 \( \left(\frac{T_2 - 283}{283\,T_2}\right) \)
⇒ \( \frac{0.5228}{3133.62} \) = \( \frac{T_2 - 283}{283\,T_2} \)
1.668 × 10⁻⁴ = \( \frac{T_2 - 283}{283\,T_2} \)
0.04721 T₂ = T₂ − 283
T₂ − 0.04721 T₂ = 283
0.95279 T₂ = 283
Hence, T₂ = \( \frac{283}{0.95279} \) = 297.0 K
(ii) (a) y = 2x. This is because for 3/4^th of the reaction to complete, time required = two half lives.
(b) This is because of improper orientation of the colliding molecules at the time of collision.

Teacher's Note:
1. Convert 10°C to 283 K first; a common error is to substitute the Celsius value.
2. Check the direction: k increases, so T₂ must come out higher than 283 K.

Question 37. (i) A first order reaction takes 100 minutes for completion of 60% of the reaction. Find the time when 90% of the reaction will be completed.
(ii) With the help of diagram explain the role of activated complex in a reaction. [CBSE Comptt. Delhi 2013]
Answer:
(i) (a) For the reaction of first order, k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \)
⇒ k = \( \frac{2.303}{100} \) log \( \frac{100}{40} \) = 0.00916 min⁻¹
(b) k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \)
⇒ 0.00916 = \( \frac{2.303}{t} \) log \( \frac{100}{10} \)
⇒ t = \( \frac{2.303}{0.00916} \) log 10
= 251.4 minutes
(ii) It is believed that when the reactant molecules absorb energy, their bonds are loosened and new loose bonds are formed between them. The intermediate thus formed is called an activated complex. It is unstable and immediately dissociates to form the stable products.
A–B + A–B (reactants) → A···B / A···B (activated complex) → A–A + B–B (products)

Teacher's Note:
1. For 60% completion, 40% is left, so the ratio is 100/40 – using 100/60 is the usual slip.
2. The activated complex sits at the top of the energy curve; its energy minus the reactant energy is the activation energy of the forward reaction.

Question 38. All energetically effective collisions do not result in a chemical change. Explain with the help of an example. [NCERT Exemplar]
Answer:
Effective collision leads to the formation of products, implying that during collisions molecules collide with sufficient kinetic energy called threshold energy (Threshold energy = Activation energy + Energy possessed by reacting species) and proper orientation, which leads to a chemical change because it enhances the old bond breaking between reactant molecules and formation of the new ones in products.
For example: formation of methanol from bromomethane depends upon the orientation of the reactant molecules. Proper orientation of reactant molecules leads to bond formation while improper orientation does not allow bond formation and hence no products are formed.
CH₃Br + OH⁻ → CH₃OH + Br⁻
Improper orientation: OH⁻ collides with the Br end of CH₃Br → no products.
Proper orientation: OH⁻ collides with the carbon atom from the side opposite to Br → [HO···CH₃···Br]⁻ → CH₃OH + Br⁻
To estimate effective collisions, another factor P which is called as probability or steric factor is introduced:
k = PZ_ABe^−E_a/RT

Teacher's Note:
1. Two conditions must both be met for a collision to be effective: energy at least equal to the threshold energy and proper orientation.