Chemical Kinetics – JEE Advanced Previous Year Questions with Solutions

Chemical Kinetics JEE Advanced Previous Year Questions with Solutions are provided here to help JEE aspirants study productively for the exams. Students will find a set of IIT JEE Advanced Chemical Kinetics important questions that further comes with a detailed step by step solution along with a diagrammatic explanation. This will enable students to revise and understand the concepts more clearly. As students go through these questions they will further gain insight into the types of questions, difficulty level and pattern of the question paper. More significantly these will also help students in the last minute preparations as well.

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JEE Advanced Previous Year Questions on Chemical Kinetics

Question 1. 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 ΔGθ (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 the Gibbs energy)

Solution: (8500)

A(g) + B(g) ⇌ AB(g)

(E_a)_b − (E_a)_f = 2RT

A_f / A_b = 4

ΔG^o = −RTlnK_eq

Kf = A_fe^–(E_a)_f / RT

Kb = A_be^–(E_a)_b/ RT

K_eq = K_f/ K_b = A_f / A_b × ee^–(E_a)_f / RT × e⁺(E_a)_b/ RT = 4 × e(E_a)_b − (E_a)_f / RT

K_eq = 4 × e²

ΔG^o = −RT × ln(4 × e²)

ΔG^o = −RT(ln4 + 2lne)

ΔG^o = −RT(2 × 0.7 + 2)

ΔG^o = −RT(1.40 + 2)

ΔG^o = −RT(3.40)

ΔG^o = −2500 × 3.40

ΔG^o = −8500J.

Question 2. For a reaction, A ⇋ P, the plots of [A] and [P] with time at temperatures T1 and T2 are given below.

If T₂ > T₁, the correct statement(s) is (are) (Assume ΔH^θ and ΔS^θ are independent of temperature and ratio of lnK at T₁ to lnK at T₂ are greater than T₂/T1₁ Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

A. △H^θ < 0, △S < 0

B. △G^θ < 0, △H^θ > 0

C. △G^θ < 0, △S^θ < 0

D. △G^θ < 0, △S^θ > 0

Solution: (A and C)

Reaction: A ⇌ P

Given T₂ >T₁

lnK₁ / lnK₂ > T₁ / T₂

⇒ T₁ lnK₁ > T₂ lnK₂

⇒ −△G₁^o > −△G₂^o

⇒ (−△H^o + T₁△S^o) > (−△H^o + T₂△S^o)

⇒ T₁△S^o > T₂△S^o

⇒ △S^o < 0

Question 3. In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are)

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. Since P = 4.5, the reaction will not proceed unless an effective catalyst is used

D. The value of the frequency factor predicted by the Arrhenius equation is higher than that determined experimentally

Solution: (A and B)

ρ = 4.5

∴ ρ > 1

We know, K_exp / K_collison = ρ

K exp = ρAe^{-Ea / RT}

A_exp = ρA

∴ A_exp > A

∴ the experimentally determined value of the frequency factor is higher than that predicted by the Arrhenius equation.

∴ option A is correct and C is incorrect.

K_exp = ρAe^{-Ea / RT}

Since in this equation E_a is not changed, option B is correct.

Since K_exp > K, there is no need for an effective catalyst to reduce activation energy so that K_exp increases.

∴ option D is also not correct.

Question 4. According to the Arrhenius equation,

A. A high activation energy usually implies a fast reaction

B. Rate constant increases with an increase in temperature. This is due to a greater number of collisions whose energy exceeds the activation energy.

C. Higher the magnitude of activation energy, the stronger is the temperature dependence of the rate constant.

D. The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy.

Solution: (B, C and D)

According to Arrhenius Equation

K = Ae^{-Ea / RT}

Therefore at T → 0

e^{-Ea / RT} → 0

Therefore rate constant decreases with temperature.

And as T → ∞ K → A.

Thus, as temperature increases, the rate constant increases and approaches a value close to the Arrhenius constant.

Now

dK / dT = −EaA / RT e^{-Ea / RT}

Hence,

Higher the activation energy higher is the required temperature for reaction to occur, and hence greater is the temperature dependency of the rate constant.

Question 5. A closed vessel with rigid walls contains 1 mol of .

Solution: (9)

I nitial moles of glasses = 1

Total gaseous moles after decomposition = 8 + 1 = 9 moles

Ratio of pressures = P_f / P_i = n_f / n_i = 9

Question 6. In the decay sequence,

x₁, x₂, x₃ and x₄ are particles/radiation emitted by the respective isotopes. The correct option(s) is(are)

A. x₃ is γ-ray

B. Z is an isotope of uranium

C. x₁ will deflect towards the negatively charged plate

D. x₂ is β–

Solution: (B, C and D)

Z has the atomic number 92.

x₁ = α since positively charged so will be attracted by the negative plate.

x₂ = β negatively charged and attracted by positive plate

x₃ = β negatively charged and attracted by positive plate

x₄ = α since positively charged so will be attracted by the negative plate.

U and Z have the same atomic number so they are isotope

Question 7. A plot of the number of neutrons (n) against the number of protons (p) of stable nuclei exhibits upward deviation from linearity for atomic number, Z>20. For an unstable nucleus having an n / p n/p ratio less than 1, the possible mode (s) of decay is (are)

A. β^– − decay ( β – emission)

B. Orbital or K-electron capture

C. Neutron emission

D. β⁺ – decay (positron emission)

Solution: (B and D)

B. In this process, there is a decrement in p which increases the ratio and stability.

A + ^o_-1e → X

D. In this process, there is a decrement in p which increases the ratio and stability.

A → X + ^o_-1e

Question 8. In nuclear transmutation.

(X, Y) is (are)

A. (γ, n)

B. (p, D)

C. (n, D)

D. (γ, p)

Solution: (A, B)

As we can see from the above reaction, the atomic number of the element remains the same, but the mass number differs.

We start by applying the law conservation of mass and conservation of energy (separately).

⇒ 4 + a = 4 + c

⇒ 9 + b = 8 + d

Question 9. Which of the following plots is(are) correct for the given reaction? ([P]_o is the initial concentration of P)

Solution: (A)

As there is no inversion. Hence should be

SN1 → 1st order

c) x = a{1 – e–kt}

x/a = 1 – e–kt

x/a = Q/P0

Question 10. ln a first-order reaction the concentration of reactant decreases from 800 mol/dm³ to 50 mol/dm³ is 2 × 10⁴ sec. The rate constant of reaction in sec^-1 is:

A. 2 × 10⁴

B. 3.45 × 10^-5

C. 1.386 × 10^-4

D. 2 × 10^-4

Solution: (C)

1.386 × 10^-4

Assuming decay is happening in the above example, we can say it will take 4 half-lives in 2 × 10⁴ sec.

So, each half-life will take 5 × 10³ sec.

Now using, t_1/2 = 0.693 / k and putting values we get

k = 0.693 / 5 × 10³

k = 1.386 × 10^-4

Option C is correct.

Chemical Kinetics and Surface Chemistry