CBSE Class 12 Chemistry Chemical Kinetics NEET JEE Questions Set 03

Single Correct Answer

Question 1. H₂O₂ is formed in the upper atmosphere through the following mechanism: H₂O + (O) → 2OH → H₂O₂. The overall enthalpy change and activation energy for the forward reaction are 72 kJ mol⁻¹ and 77 kJ mol⁻¹ respectively. The activation energy for the decomposition of H₂O₂ to give back H₂O and (O) will be
(a) 5 kJ mol⁻¹
(b) −5 kJ mol⁻¹
(c) 149 kJ mol⁻¹
(d) −149 kJ mol⁻¹
Answer:
Ans. (a)
Activation energy for backward reaction = E_a(forward) − ΔH = 77 − 72 = 5 kJ mol⁻¹.
(H₂O + O lies at the bottom; the peak is 77 kJ above it and H₂O₂ lies 72 kJ above it, so the peak is 5 kJ above H₂O₂.)

Teacher's Note:
1. Use E_a(f) − E_a(b) = ΔH; an activation energy can never be negative.

Question 2. When a catalyst increases the rate of a chemical reaction, the rate constant (AIPMT 2013)
(a) remains constant
(b) increases
(c) decreases
(d) may increase or decrease depending on the order of reaction
Answer:
Ans. (b)
Rate = k [Reactant]. In the presence of catalyst, at the same concentration of the reactant, as rate increases, k increases.
Alternatively, k = A e^−E_a/RT. In presence of catalyst at the same temperature, E_a decreases. Hence, k increases.

Teacher's Note:
1. A catalyst changes k (by lowering E_a) but not the equilibrium constant or ΔH.

Question 3. For a reaction taking place in three steps, the rate constants are k₁, k₂ and k₃. The overall rate constant is \( k = \frac{k_1k_2}{k_3} \). If the energy of activation values for the first, second and third stages are 40, 50 and 60 kJ mol⁻¹ respectively, then the overall energy of activation in kJ mol⁻¹ is (Kerala CEE 2008, AMU Engg. 2014)
(a) 30
(b) 150
(c) 50
(d) 60
Answer:
Ans. (a)
\( k = \frac{k_1k_2}{k_3} \) or \( Ae^{-E/RT} = \frac{Ae^{-E_1/RT}\,Ae^{-E_2/RT}}{Ae^{-E_3/RT}} \)
or e^−E/RT = e^{(−E₁ − E₂ + E₃)/RT} or E = E₁ + E₂ − E₃ = 40 + 50 − 60 = 30 kJ mol⁻¹

Teacher's Note:
1. Multiplying rate constants adds their activation energies and dividing subtracts them.

Question 4. The rate constants k₁ and k₂ for two different reactions are 10¹⁶ · e^−2000/T and 10¹⁵ · e^−1000/T respectively. The temperature at which k₁ = k₂ is (AIPMT 2008)
(a) 1000 K
(b) \( \frac{2000}{2.303} \) K
(c) 2000 K
(d) \( \frac{1000}{2.303} \) K
Answer:
Ans. (d)
When k₁ = k₂, 10¹⁶ e^−2000/T = 10¹⁵ e^−1000/T or 10 e^−2000/T = e^−1000/T
Taking natural logarithm of both sides, \( \ln10 - \frac{2000}{T} = \frac{-1000}{T} \) or \( 2.303 - \frac{2000}{T} = \frac{-1000}{T} \)
or \( \frac{1000}{T} = 2.303 \) or \( T = \frac{1000}{2.303} \) K

Teacher's Note:
1. Divide out the common powers of 10 first and then take natural logs; ln 10 = 2.303.

Question 5. For a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow the equation \( \log k = -(2000)\frac{1}{T} + 6.0 \). The pre-exponential factor A and the activation energy E_a, respectively, are (IIT 2009)
(a) 1.0 × 10⁶ s⁻¹ and 9.2 kJ mol⁻¹
(b) 6.0 s⁻¹ and 16.6 kJ mol⁻¹
(c) 1.0 × 10⁻¹ s⁻¹ and 16.6 kJ mol⁻¹
(d) 1.0 × 10⁶ s⁻¹ and 38.3 kJ mol⁻¹
Answer:
Ans. (d)
Given \( \log k = 6 - \frac{2000}{T} \). Comparing with \( \log k = \log A - \frac{E_a}{2.303RT} \)
log A = 6, i.e., A = 10⁶ s⁻¹ and \( \frac{E_a}{2.303R} = 2000 \)
or E_a = 2000 × 2.303 × 8.314 J mol⁻¹ = 38294 J mol⁻¹ = 38.3 kJ mol⁻¹

Teacher's Note:
1. With log (base 10), the slope of log k vs 1/T is −E_a/(2.303R); with ln it is −E_a/R.

Question 6. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is (IIT 2010)
(a) Plot (a)
(b) Plot (b)
(c) Plot (c)
(d) Plot (d)

Answer:
Ans. (a)
\( k = Ae^{-E_a/RT} = \frac{A}{e^{E_a/RT}} \). Thus, as T increases, the exponential factor in the denominator decreases or Ae^−E_a/RT increases. Hence, k increases exponentially as T increases.

Teacher's Note:
1. k rises with temperature, and the rise becomes steeper at higher T, which gives the upward curving plot.

Question 7. The activation energy for a reaction at temperature T K was found to be 2.303 RT J mol⁻¹. The ratio of the rate constant to Arrhenius factor is (Karnataka CET 2010, 2011)
(a) 10⁻¹
(b) 10⁻²
(c) 2 × 10⁻³
(d) 2 × 10⁻²
Answer:
Ans. (a)
\( 2.303\log k = 2.303\log A - \frac{E_a}{RT} \). Putting E_a = 2.303 RT (Given), we get \( 2.303\log k = 2.303\log A - \frac{2.303RT}{RT} \)
or log k = log A − 1 or \( \log\frac{k}{A} = -1 \) or \( \frac{k}{A} = 10^{-1} \)

Teacher's Note:
1. k/A = e^−E_a/RT is the fraction of molecules with energy at least E_a.

Question 8. Two reactions R₁ and R₂ have identical pre-exponential factors. Activation energy of R₁ exceeds that of R₂ by 10 kJ mol⁻¹. If k₁ and k₂ are the rate constants for reactions R₁ and R₂ respectively at 300 K, then ln (k₂/k₁) is equal to (R = 8.314 J mol⁻¹ K⁻¹) (JEE Main 2017)
(a) 6
(b) 4
(c) 8
(d) 12
Answer:
Ans. (b)
k₁ = A e^−E_a1/RT, k₂ = A e^−E_a2/RT
\( \therefore \frac{k_2}{k_1} = e^{\frac{1}{RT}(E_{a1}-E_{a2})} \) or \( \ln\frac{k_2}{k_1} = \frac{E_{a1}-E_{a2}}{RT} = \frac{10\times10^3}{8.314\times300} \) = 4

Teacher's Note:
1. Convert 10 kJ to 10000 J before dividing by RT (about 2494 J mol⁻¹ at 300 K).

Question 9. For a reaction, consider the plot of ln k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10⁻⁵ s⁻¹, then the rate constant at 500 K is (JEE Main 2019)
(a) 2 × 10⁻⁴ s⁻¹
(b) 10⁻⁴ s⁻¹
(c) 10⁻⁶ s⁻¹
(d) 4 × 10⁻⁴ s⁻¹

Answer:
Ans. (b)
Arrhenius equation is \( \ln k = \frac{-E_a}{RT} + \ln A \). Thus, plot of ln k vs 1/T is linear with slope \( = \frac{-E_a}{R} \)
\( \ln\frac{k_2}{k_1} = \frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] \quad \therefore 2.303\log\frac{k_2}{10^{-5}} = 4606\left(\frac{1}{400}-\frac{1}{500}\right) \)
\( \log\frac{k_2}{10^{-5}} = \frac{4606}{2.303}\left(\frac{1}{2000}\right) = 1 \) or log k₂ − log 10⁻⁵ = 1 or log k₂ + 5 = 1
or log k₂ = −4 or k₂ = 10⁻⁴ s⁻¹

Teacher's Note:
1. The slope of ln k vs 1/T equals −E_a/R, so E_a/R = 4606 K can be used directly.

Question 10. For a reaction A → B, enthalpy of reaction is −4.2 kJ mol⁻¹ and enthalpy of activation is 9.6 kJ mol⁻¹. The correct potential energy profile for the reaction is shown in the option. (NEET 2021)
(a) Profile (a)
(b) Profile (b)
(c) Profile (c)
(d) Profile (d)

Answer:
Ans. (b)
As Δ_rH = H_B − H_A and enthalpy of reaction is negative, it can be so only when H_A > H_B, i.e., PE of A is greater than PE of B.
This is so only in diagram (b). In such cases, activation energy of backward reaction (E_b) > activation energy of forward reaction (E_f). This is again so in diagram (b) only.
In the present case, E_f = 9.6 kJ mol⁻¹, E_b = 9.6 + 4.2 = 13.8 kJ mol⁻¹

Teacher's Note:
1. For an exothermic reaction the products lie below the reactants, and the hump measured from the products is the larger one.

Question 11. The slope of the Arrhenius plot \( \left(\ln k\ \text{vs}\ \frac{1}{T}\right) \) of first order reaction is −5 × 10³ K. The value of E_a of the reaction is: Choose the correct option for your answer [Given R = 8.314 J K⁻¹ mol⁻¹] (NEET 2021)
(a) 41.5 kJ mol⁻¹
(b) 83.0 kJ mol⁻¹
(c) 166 kJ mol⁻¹
(d) −83 kJ mol⁻¹
Answer:
Ans. (a)
According to Arrhenius equation, k = Ae^−E_a/RT or \( \ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right) \). Hence, slope of the plot of ln k vs 1/T \( = -\frac{E_a}{R} \)
or \( -5\times10^3\ \text{K} = -\frac{E_a}{8.314\ \text{J K}^{-1}\text{mol}^{-1}} \) or E_a = (5 × 10³ K)(8.314 J K⁻¹ mol⁻¹) = 41.57 × 10³ J mol⁻¹ = 41.5 kJ mol⁻¹

Teacher's Note:
1. Do not multiply by 2.303 here: the plot uses ln k, not log k.

Question 12. The rate constant (k) of a reaction is measured at different temperatures (T) and the data are plotted in the given figure. The activation energy of the reaction in kJ mol⁻¹ is (R is gas constant) (JEE Main 2020)
(a) R
(b) 2R
(c) \( \frac{2}{R} \)
(d) \( \frac{1}{R} \)

Answer:
Ans. (b)
k = Ae^−E_a/RT, \( \ln k = \ln A - \frac{E_a}{RT} \), \( \ln k = \ln A - \frac{E_a}{R\times10^3}\times\frac{10^3}{T} \) (Multiplying numerator and denominator by 10³)
Slope of the graph \( = -\frac{E_a}{R\times10^3} = -\frac{10}{5} \)
∴ E_a = 2R × 10³ J = 2R kJ

Teacher's Note:
1. When the x-axis is 10³/T, the slope read from the graph must be divided by 10³ to get −E_a/R.

Question 13. In collision theory of chemical reaction, Z_AB represents (NEET 2020, Phase 2)
(a) the fraction of the molecules with energies greater than E_a
(b) the collision frequency of reactants A and B
(c) steric factor
(d) the fraction of molecules with energies equal to E_a
Answer:
Ans. (b)
In collision theory of chemical reaction, Z_AB represents the collision frequency of the reactants A and B.

Teacher's Note:
1. In rate = P Z_AB e^−E_a/RT: P is the steric factor, Z_AB the collision frequency and the exponential term the fraction of energetic collisions.

Question 14. One mole of N₂O₄ gas at 300 K is kept in a closed container at 1 atm. It is heated to 600 K when 20% by mass of N₂O₄ decomposes to NO₂(g). The resultant pressure in the container would be
(a) 1.2 atm
(b) 2.4 atm
(c) 2.0 atm
(d) 1.0 atm
Answer:
Ans. (b)
N₂O₄ → 2NO₂. Initial moles: 1, 0. Moles after dissociation: 1 − 0.2 = 0.8, 0.4
Total moles after dissociation = 0.8 + 0.4 = 1.2
Initial temp. = 300 K ∴ P₁V = n₁RT₁ or 1 × V = 1 × R × 300 ...(i)
Temp. after dissociation = 600 K ∴ P₂V₂ = n₂RT₂ or P₂ × V = 1.2 × R × 600 ...(ii)
Dividing eqn. (ii) by eqn. (i), \( P_2 = \frac{1.2\times600}{300} \) = 2.4 atm

Teacher's Note:
1. Both the increase in moles and the doubling of temperature raise the pressure at constant volume.

Question 15. Consider the following statements:
(i) increase in concentration of reactant increases the rate of zero order reaction.
(ii) rate constant k is equal to collision frequency A if E_a = 0.
(iii) rate constant k is equal to collision frequency A if E_a = ∞.
(iv) ln k vs T is a straight line.
(v) ln k vs 1/T is a straight line.
Correct statements are: (Kerala PET 2010)
(a) (i) and (iv)
(b) (ii) and (v)
(c) (iii) and (iv)
(d) (ii) and (iii)
(e) (i) and (v)
Answer:
Ans. (b)
Rate of a zero order reaction is independent of concentration. Hence, (i) is wrong.
k = A e^−E_a/RT. When E_a = 0, k = A. When E_a = ∞, k ≠ A.
Plot of ln k vs 1/T is a straight line.
Hence, only (ii) and (v) are correct.

Teacher's Note:
1. Put the limiting values into k = A e^−E_a/RT: E_a = 0 gives k = A, while E_a = ∞ gives k = 0.

Question 16. The oxidation of a certain metal is found to obey the equation A² = αt + β, where A is the thickness of the oxide film at time t, α and β are constants. The order of this reaction is
(a) 0
(b) 1
(c) −1
(d) 2
Answer:
Ans. (c)
A² = αt + β. Differentiating this equation, we get 2A dA = α dt or \( \frac{dA}{dt} = \frac{\alpha}{2A} = \frac{\alpha}{2}[\mathrm{A}]^{-1} \)
\( \frac{d[\mathrm{A}]}{dt} \) represents the rate of increase of thickness of oxide film with time. Hence, order of reaction = −1.

Teacher's Note:
1. Differentiate to get the rate, then read the order as the power of A in the rate expression; orders can be negative.

Question 17. The initial rate of hydrolysis of methyl acetate (1 M) by a weak acid (HA, 1 M) is 1/100th of that of a strong acid (HX, 1 M) at 25°C. The K_a of HA is (JEE Advanced 2013)
(a) 1 × 10⁻⁴
(b) 1 × 10⁻⁵
(c) 1 × 10⁻⁶
(d) 1 × 10⁻³
Answer:
Ans. (a)
Rate w.r.t. weak acid, R₁ = k [H⁺]_WA [ester]; Rate w.r.t. strong acid, R₂ = k [H⁺]_SA [ester]
\( \therefore \frac{R_1}{R_2} = \frac{[\mathrm{H^+}]_{WA}}{[\mathrm{H^+}]_{SA}} = \frac{1}{100} \) (Given) \( \therefore [\mathrm{H^+}]_{WA} = \frac{[\mathrm{H^+}]_{SA}}{100} = \frac{1}{100}\times1\ \text{M} \) = 0.01 M = 10⁻² M
HA ⇄ H⁺ + A⁻. Initial: C mol L⁻¹. At eqm.: C − Cα = C(1 − α), Cα, Cα
[H⁺] = Cα or 10⁻² = 1 × α or α = 10⁻²
\( K_a = \frac{C\alpha\cdot C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \simeq C\alpha^2 \) = 1 × (10⁻²)² = 10⁻⁴

Teacher's Note:
1. Acid-catalysed ester hydrolysis has rate proportional to [H⁺], so the rate ratio gives the [H⁺] ratio directly.

Question 18. The decomposition of PH₃ on tungsten at low pressure is a first order reaction. It is because the (NEET Phase II 2016)
(a) rate is independent of surface coverage
(b) rate of decomposition is very low
(c) rate is proportional to the surface coverage
(d) rate is inversely proportional to the surface coverage
Answer:
Ans. (c)
The decomposition of PH₃ on tungsten at low pressure is a first order reaction because rate is proportional to the surface coverage.

Teacher's Note:
1. At low pressure the surface coverage is proportional to the gas pressure, so rate ∝ p (first order); at high pressure the surface is full and the reaction becomes zero order.

Question 19. The correct difference between first and second order reaction is that (NEET 2018)
(a) the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
(b) the half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀
(c) a first-order reaction can be catalysed; a second-order reaction cannot be catalysed
(d) the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
Answer:
Ans. (b)
The half-life of a first-order reaction is independent of [A]₀ whereas half-life of a second-order reaction is inversely proportional to [A]₀.

Teacher's Note:
1. Only a zero order reaction has a rate independent of concentration; both first and second order rates depend on it.

Question 20. For the reaction scheme, A → B → C (rate constants k₁ and k₂), if all the reactions are of 1st order and if the rate of formation of B is set to be zero, then the concentration of B is given by (JEE Main 2019)
(a) (k₁ + k₂) [A]
(b) \( \left(\frac{k_1}{k_2}\right)[\mathrm{A}] \)
(c) k₁k₂ [A]
(d) (k₁ − k₂) [A]
Answer:
Ans. (b)
Rate of formation of B = Rate of formation of B from A − Rate of decomposition of B to C, i.e., \( \frac{d[\mathrm{B}]}{dt} = k_1[\mathrm{A}] - k_2[\mathrm{B}] \)
Setting \( \frac{d[\mathrm{B}]}{dt} = 0 \) means 0 = k₁[A] − k₂[B] or k₂[B] = k₁[A] or \( [\mathrm{B}] = \frac{k_1}{k_2}[\mathrm{A}] \)

Teacher's Note:
1. This is the steady-state approximation: an intermediate forms and is used up at the same rate.

Question 21. In the reaction xA → yB, \( \log\left[-\frac{d[\mathrm{A}]}{dt}\right] = \log\left[\frac{d[\mathrm{B}]}{dt}\right] + 0.3010 \). 'A' and 'B' respectively can be (JEE Main 2019)
(a) C₂H₄ and C₄H₈
(b) n-Butane and iso-Butane
(c) N₂O₄ and NO₂
(d) C₂H₂ and C₆H₆
Answer:
Ans. (a)
The given expression can be written as \( \log\left[-\frac{d[\mathrm{A}]}{dt}\right] = \log\left[\frac{d[\mathrm{B}]}{dt}\right] + \log 2 \) ∴ \( \log\left[\frac{-d[\mathrm{A}]}{dt}\right] = \log\left[2\times\frac{d[\mathrm{B}]}{dt}\right] \)
or \( \frac{-d[\mathrm{A}]}{dt} = 2\times\frac{d[\mathrm{B}]}{dt} \) ...(i)
For xA → yB, Rate \( = -\frac{1}{x}\frac{d[\mathrm{A}]}{dt} = \frac{1}{y}\frac{d[\mathrm{B}]}{dt} \) or \( -\frac{d[\mathrm{A}]}{dt} = \frac{x}{y}\frac{d[\mathrm{B}]}{dt} \) ...(ii)
Comparing eqns. (i) and (ii), \( \therefore \frac{x}{y} = 2 \). This is so for the reaction 2C₂H₄ → C₄H₈

Teacher's Note:
1. Recognise 0.3010 as log 2; A must be consumed twice as fast as B is formed.

Question 22. In the study of oxidation of iodide ions by hydrogen peroxide in presence of an acid and starch as a clock reaction,
(a) thiosulphate added should be in large amount
(b) thiosulphate added should be in small amount
(c) amount of thiosulphate added has no importance
(d) thiosulphate is added to speed up the reaction
Answer:
Ans. (b)
Thiosulphate added should be in small amount so that it reacts away with iodine and the iodine further liberated can give blue colour with starch.

Teacher's Note:
1. Thiosulphate acts as a timer: the blue colour appears only after all of it has been used up.

Question 23. In the study of oxidation of iodide ions by hydrogen peroxide in presence of dilute H₂SO₄ and starch as a clock reaction, if excess of thiosulphate is added
(a) blue colour will appear immediately
(b) blue colour will not appear at all
(c) blue colour will appear only on heating the reaction mixture
(d) blue colour will appear after a very long time
Answer:
Ans. (b)
If thiosulphate is present in excess, all the iodine liberated will react with the thiosulphate and no blue colour will appear at all.

Teacher's Note:
1. 2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻ removes iodine as fast as it forms.

Question 24. In the study of oxidation of iodide ions by hydrogen peroxide in presence of dilute H₂SO₄, thiosulphate and starch as a clock reaction, if concentration of iodide ions is increased
(a) less time will be taken for blue colour to appear
(b) more time will be taken for blue colour to appear
(c) same time will be taken for blue colour to appear
(d) any one of the above can happen depending upon the amount of H₂SO₄ present
Answer:
Ans. (a)
On increasing the concentration of I⁻ ions, the rate of reaction will increase. After reaction with thiosulphate, less time will be taken for blue colour to appear.

Teacher's Note:
1. In a clock reaction, a shorter time for the colour means a faster reaction (rate ∝ 1/time).

One or More Correct Answers

Question 25. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is (are) (JEE Advanced 2017)
(a) The activation energy of the reaction is unaffected by the value of the steric factor
(b) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation
(c) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally
(d) Since P = 4.5, the reaction will not proceed unless an effective catalyst is added
Answer:
Ans. (a, b)
Actual k = PAe^−E_a/RT. Theoretical k = Ae^−E_a/RT
E_a is independent of steric factor. Hence, option (a) is correct.
Further, \( P = \frac{k_{\text{actual}}}{k_{\text{theoretical}}} \) (actual means observed or experimental value, theoretical means predicted by Arrhenius equation)
As P = 4.5, k_actual > k_theoretical. This can be so when A_actual > A_theoretical. Hence, option (b) is correct.

Teacher's Note:
1. P = (observed A)/(calculated A); P > 1 simply means the observed frequency factor is larger than predicted.

Question 26. For a first order reaction, A(g) → 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t = 0) and at time t are P₀ and P_t, respectively. Initially only A is present with concentration [A]₀ and t_1/3 is the time required for the partial pressure of A to reach 1/3rd of its initial value. The correct option(s) is (are) (Assume that all gases behave as ideal gases) (JEE Advanced 2018)
(a) Graph (a)
(b) Graph (b)
(c) Graph (c)
(d) Graph (d)

Answer:
Ans. (a, d)
A(g) → 2B(g) + C(g), V constant, T = 300 K
t = 0: P₀; t = t: P₀ − x, 2x, x ∴ P_t = P₀ − x + 2x + x = P₀ + 2x or 2x = P_t − P₀
\( t = \frac{1}{k}\ln\frac{P_0}{P_0-x} = \frac{1}{k}\ln\frac{P_0}{P_0-\frac{(P_t-P_0)}{2}} = \frac{1}{k}\ln\frac{2P_0}{3P_0-P_t} \)
or kt = ln 2P₀ − ln (3P₀ − P_t) or ln (3P₀ − P_t) = −kt + ln 2P₀
Thus, graph between ln (3P₀ − P_t) vs 't' is a straight line with −ve slope. So, (a) is correct.
\( t_{1/3} = \frac{1}{k}\ln\frac{P_0}{P_0/3} = \frac{1}{k}\ln 3 \). Thus, t_1/3 is independent of initial concentration. So (b) is wrong. As rate constant is a constant quantity and independent of initial concentration, so graph (d) is correct.

Teacher's Note:
1. Express the pressure of A in terms of measurable P_t first; any fractional life of a first order reaction is independent of [A]₀.

Question 27. For the reaction 2X + Y → P (rate constant k), the rate of reaction is \( \frac{dP}{dt} = k[\mathrm{X}] \). Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is (are) (Use ln 2 = 0.693) (JEE Advanced 2021)
(a) The rate constant, k, of the reaction is 13.86 × 10⁻⁴ s⁻¹
(b) Half-life of X is 50 s
(c) At 50 s, \( \frac{-d[\mathrm{X}]}{dt} \) = 13.86 × 10⁻³ mol L⁻¹ s⁻¹
(d) At 100 s, \( \frac{-d[\mathrm{Y}]}{dt} \) = 3.46 × 10⁻³ mol L⁻¹ s⁻¹
Answer:
Ans. (b, c, d)
2X + Y → P. t = 0: 2, 1, 0. t = 50 s: 2 − 2 × 0.5 = 1, 0.5
Rate \( = -\frac{1}{2}\frac{d[\mathrm{X}]}{dt} = -\frac{d[\mathrm{Y}]}{dt} = \frac{dP}{dt} = k[\mathrm{X}] \) (Given), so \( -\frac{d[\mathrm{X}]}{dt} = 2k[\mathrm{X}] \), i.e., X decays by first order kinetics with rate constant 2k.
In 50 s, [X] falls from 2 to 1 mol L⁻¹, therefore t_1/2 of X = 50 s. So (b) is correct.
\( 2k = \frac{0.693}{50}\ \text{s}^{-1} = 13.86\times10^{-3}\ \text{s}^{-1} \) or k = 6.93 × 10⁻³ s⁻¹. So (a) is wrong.
At 50 s, \( -\frac{d[\mathrm{X}]}{dt} = 2k[\mathrm{X}] \) = (13.86 × 10⁻³ s⁻¹)(1 mol L⁻¹) = 13.86 × 10⁻³ mol L⁻¹ s⁻¹. So (c) is correct.
At 100 s (two half-lives), [X] = 0.5 mol L⁻¹. \( -\frac{d[\mathrm{Y}]}{dt} = k[\mathrm{X}] \) = (6.93 × 10⁻³)(0.5) = 3.46 × 10⁻³ mol L⁻¹ s⁻¹. So (d) is correct.

Teacher's Note:
1. When the rate is defined with a stoichiometric coefficient, X disappears at 2k[X]; the half-life of X is 0.693/2k, not 0.693/k.

Passage-Based Questions

Question 28. Comprehension 2: The progress of the reaction A ⇄ nB with time is represented in the Fig. below:
The value of n is
(a) 1
(b) 2
(c) 3
(d) 1.5

Answer:
Ans. (b)
From time t = 1 hour to t = 5 hour, Δt = 4 hr
Decrease in conc. of A = 0.5 − 0.3 = 0.2 mol L⁻¹; Increase in conc. of B = 0.6 − 0.2 = 0.4 mol L⁻¹.
Thus, in the same time interval, increase in conc. of B is double than decrease in the conc. of A. Hence, n = 2.

Teacher's Note:
1. Compare changes over the same time interval; the ratio of changes gives the stoichiometric ratio.

Question 29. Comprehension 2: The progress of the reaction A ⇄ nB with time is represented in the Fig. below:
The equilibrium constant K will be
(a) 2
(b) 1.2
(c) 0.5
(d) 6.67

Answer:
Ans. (b)
At equilibrium, there is no further change in conc. Hence, concentrations at equilibrium are [A] = 0.3 mol L⁻¹, [B] = 0.6 mol L⁻¹
\( \therefore \) Eqm. const. \( K = \frac{[\mathrm{B}]^2}{[\mathrm{A}]} = \frac{(0.6)^2}{0.3} \) = 1.2 mol L⁻¹

Teacher's Note:
1. Use the flat parts of the curves (5 h onwards) for equilibrium concentrations, and raise [B] to the power n = 2.

Question 30. Comprehension 2: The progress of the reaction A ⇄ nB with time is represented in the Fig. below:
The initial rate of conversion of A will be
(a) 0.1 mol L⁻¹ hr⁻¹
(b) 0.2 mol L⁻¹ hr⁻¹
(c) 0.4 mol L⁻¹ hr⁻¹
(d) 0.8 mol L⁻¹ hr⁻¹

Answer:
Ans. (a)
From t = 0 to t = 1 hour, for A, Δx = 0.6 − 0.5 = 0.1 mol L⁻¹
\( \therefore \) Initial rate of conversion of A \( = \frac{\Delta x}{\Delta t} = \frac{0.1\ \text{mol L}^{-1}}{1\ \text{hr}} \) = 0.1 mol L⁻¹ hr⁻¹

Teacher's Note:
1. Read the initial concentration of A (0.6 mol L⁻¹) where its curve meets the y-axis.

Matrix Match Questions

Teacher's Note:
1. Units of k = (mol L⁻¹)¹⁻ⁿ s⁻¹; put n = 0, 1, 2 to check each unit.

Teacher's Note:
1. Temperature always speeds up a reaction, but it shifts K according to the sign of ΔH (Le Chatelier).

Integer and Numerical Answer Questions

Question 33. For a given chemical reaction, y₁A + y₂B → y₃C + y₄D, concentration of C changes from 10 mmol dm⁻³ to 20 mmol dm⁻³ in 10 seconds. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance of A. The rate of appearance of D has been experimentally determined to be 9 mmol dm⁻³ s⁻¹. Therefore, the rate of reaction is ...... mmol dm⁻³ s⁻¹. (Nearest integer) (JEE Main 2022)
Answer:
Ans. 1
\( -\frac{1}{y_1}\left(\frac{dA}{dt}\right) = \frac{-1}{y_2}\left(\frac{dB}{dt}\right) = +\frac{1}{y_3}\left(\frac{dC}{dt}\right) = +\frac{1}{y_4}\left(\frac{dD}{dt}\right) \)
\( \frac{dC}{dt} = \frac{20-10}{10} = \frac{10}{10} \) = 1 mmol dm⁻³ s⁻¹
\( \frac{dD}{dt} = 1.5\times\left(\frac{-dB}{dt}\right) = 2\times\left(\frac{-dA}{dt}\right) \). But \( \frac{dD}{dt} = \frac{y_4}{y_2}\left(\frac{-dB}{dt}\right) \therefore \frac{y_4}{y_2} = 1.5 \); \( \frac{dD}{dt} = \frac{y_4}{y_1}\left(\frac{-dA}{dt}\right) \therefore \frac{y_4}{y_1} = 2 \)
\( \frac{dD}{dt} = 9 = \frac{y_4}{y_3}\frac{dC}{dt} = \frac{y_4}{y_3}\times1 \) or \( \frac{y_4}{y_3} = 9 \)
Thus, y₄ = 1.5 y₂ = 2 y₁ = 9 y₃. Taking y₃ = 1, y₄ = 9.
Rate of reaction \( = \frac{1}{y_3}\frac{dC}{dt} = \frac{1}{y_4}\frac{dD}{dt} = \frac{1}{9}\times9 \) mmol dm⁻³ s⁻¹ = 1 mmol dm⁻³ s⁻¹

Teacher's Note:
1. Rate of reaction = (rate of change of any species)/(its coefficient); the ratios of rates give the ratios of the coefficients.

Question 34. At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.6 kPa, the half-life was found to be 170 s. The order of the reaction is ...... (integer answer) (JEE Main 2022)
Answer:
Ans. 0
For a gaseous reaction, \( t_{1/2} \propto \frac{1}{P_0^{\,n-1}} \)
\( \frac{(t_{1/2})_1}{(t_{1/2})_2} = \frac{(P_0^{\,n-1})_2}{(P_0^{\,n-1})_1} = \left[\frac{(P_0)_2}{(P_0)_1}\right]^{n-1} \)
\( \frac{340}{170} = \left(\frac{27.6}{55.5}\right)^{n-1} \) or \( 2^1 = \left(\frac{1}{2}\right)^{n-1} = 2^{1-n} \) or 1 − n = 1 or n = 0

Teacher's Note:
1. Halving the initial pressure halved the half-life: t_1/2 ∝ P₀, which is zero order.

Question 35. Consider the following reversible reaction A(g) + B(g) ⇄ AB(g). The activation energy of the backward reaction exceeds that of the forward reaction by 2RT (in J mol⁻¹). If the pre-exponential factor of the forward reaction is 4 times that of the reverse reaction, the absolute value of ΔG° (in J mol⁻¹) for the reaction at 300 K is ............ (Given: ln (2) = 0.7, RT = 2500 J mol⁻¹ at 300 K and G is Gibbs energy) (JEE Advanced 2018)
Answer:
Ans. 8500
Given: (E_a)_b − (E_a)_f = 2RT and A_f = 4A_b
Now, rate constant for forward reaction, k_f = A_f e^{−(E_a)_f/RT} and rate constant for backward reaction, k_b = A_b e^{−(E_a)_b/RT}
Equilibrium constant, \( K_{eq} = \frac{k_f}{k_b} = \frac{A_f}{A_b}e^{-(E_{a(f)}-E_{a(b)})/RT} \) = 4e^2RT/RT = 4e²
Now, ΔG° = −RT ln K_eq = −2500 ln (4e²) = −2500 (ln 4 + ln e²) = −2500 (1.4 + 2) = −2500 × 3.4 = −8500 J mol⁻¹
Absolute value of ΔG° = 8500 J mol⁻¹

Teacher's Note:
1. K = k_f/k_b links kinetics to thermodynamics; a K greater than 1 gives a negative ΔG°.

Teacher's Note:
1. Since the reaction is zero order in B, the value [B] = 0.25 is a distractor.

Question 37. The decomposition reaction 2N₂O₅(g) → 2N₂O₄(g) + O₂(g) (Δ) is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y × 10³ s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 × 10⁻⁴ s⁻¹, assuming ideal gas behaviour the value of Y is ............. (JEE Advanced 2019)
Answer:
Ans. 2.30
The units of the rate constant show that it is a reaction of 1st order. Suppose after time t, decrease in pressure of N₂O₅ is 2p atm.
2N₂O₅ → 2N₂O₄ + O₂. Initial pressure: 1 atm, 0, 0. After time t: 1 − 2p, 2p, p
Total pressure after time t (P_t): 1 − 2p + 2p + p = (1 + p) atm ∴ 1 + p = 1.45 (Given) or p = 0.45 atm, so p(N₂O₅) = 1 − 2 × 0.45 = 0.1 atm
For the reaction as written, rate \( = -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = k[\mathrm{N_2O_5}] \), so \( -\frac{d[\mathrm{N_2O_5}]}{dt} = 2k[\mathrm{N_2O_5}] \) and \( t = \frac{2.303}{2k}\log\frac{1}{1-2p} \)
\( \therefore Y\times10^3 = \frac{2.303}{2\times5\times10^{-4}}\log\frac{1}{0.1} \) = 2.30 × 10³ ∴ Y = 2.30

Teacher's Note:
1. When the rate constant refers to the reaction as written (2N₂O₅), N₂O₅ itself disappears with rate constant 2k.

Assertion–Reason Questions

Question 38. Assertion: The hydrolysis of methyl acetate by dilute HCl is a pseudo first order reaction.
Reason: HCl acts as a catalyst for the hydrolysis. (AIIMS 2007)
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. During hydrolysis of methyl acetate by dil HCl, water is taken in large excess and hence, its concentration does not change during hydrolysis.
HCl does act as a catalyst, but this is not why the reaction is pseudo first order.

Teacher's Note:
1. Pseudo first order behaviour is due to the large excess of water, not the catalyst.

Question 39. Assertion: Average life of a radioactive element is that period in which 63% of it is decayed.
Reason: Average life τ = 1.44 t_1/2. (AIIMS 2007)
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Average life τ = 1/λ. In time τ, N = N₀ e⁻¹ = 0.37 N₀, i.e., 63% has decayed. Hence, A is true.
τ = 1/λ = t_1/2/0.693 = 1.44 t_1/2. R is also true but it does not explain why 63% decays in time τ.

Teacher's Note:
1. The 63% figure comes from e⁻¹ = 0.37; the relation with t_1/2 is a separate fact.

Question 40. Assertion: By 10° rise of temperature, the rate of reaction increases by 100% to 200%.
Reason: By 10° rise of temperature, the collision frequency increases by 100 to 200%.
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct R. By 10° rise of temperature, the collision frequency increases by 2 to 3% only. The large rise in rate is due to the increase in the fraction of effective (energetic) collisions.

Teacher's Note:
1. Collision frequency varies only as √T; it is the fraction of molecules with energy ≥ E_a that roughly doubles.

Question 41. Assertion: A catalyst increases the rate of a reaction.
Reason: In presence of a catalyst, the activation energy of the reaction decreases. (AIIMS 2011)
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
A catalyst provides an alternative pathway with lower activation energy; hence more molecules can cross the energy barrier and the rate increases. R explains A.

Teacher's Note:
1. A catalyst lowers E_a for both forward and backward reactions equally, so it does not change ΔH or the equilibrium constant.

Question 42. Assertion: According to transition state theory, for the formation of an activated complex, one of the vibrational degree of freedom is converted into a translational degree of freedom.
Reason: Energy of the activated complex is higher than the energy of the reactant molecules. (AIIMS 2006)
(a) If both assertion and reason are true and reason is the true explanation of the assertion.
(b) If both assertion and reason are true but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
A + B → X* → Products
The activated complex X* has higher energy than reactants. It is a special molecule which decomposes such that one vibrational degree of freedom is converted into a translational degree of freedom along the reaction coordinate.

Teacher's Note:
1. The activated complex sits at the top of the energy barrier; its breaking along the reaction coordinate is what leads to products.