For a first order reaction A → B, the half life is 30 min. The time taken for 75% completion of the reaction is: (log 2 = 0.3010)
A30 min
B45 min
C60 min
D90 min
✅ Correct: C
75% completion → 25% remains = (1/2)². Requires 2 half-lives. t = 2×30 = \(\mathbf{60\text{ min}}\). For 87.5% completion (1/2)³: 3 half-lives. For 99.9%: about 10 half-lives.
Q2
A molecule undergoes two independent first order reactions with half lives 12 min and 3 min. Both reactions occur simultaneously. Time taken for 50% consumption of reactant:
In a first order decomposition reaction, the time for decomposition to 1/4 and 1/8 of initial concentration are t₁ and t₂ respectively. The ratio t₁/t₂ is:
A4/3
B3/2
C3/4
D2/3
✅ Correct: D
For first order: \(t_{1/4} = 2t_{1/2}\) (two half-lives) and \(t_{1/8} = 3t_{1/2}\) (three half-lives). \(t_1/t_2 = 2t_{1/2}/3t_{1/2} = \mathbf{2/3}\).
Q4
For a first order reaction A → B with half life 0.3010 min. The ratio of initial concentration to concentration at time 2.0 min is: (log 2 = 0.3010)
A and B radioactive substances have half lives 15 min and 5 min. Initial concentration of B is 4 times that of A. How long until concentrations are equal?
A5 min
B10 min
C15 min
D20 min
✅ Correct: C
Let initial: A₀ and 4A₀. At time t: A₀(1/2)^(t/15) = 4A₀(1/2)^(t/5). (1/2)^(t/15-t/5) = 4 = (1/2)^(−2). So t/5 − t/15 = 2 → 2t/15 = 2 → t = \(\mathbf{15\text{ min}}\).
Q6
For elementary reaction A(g) + B(g) → C(g) + D(g), if volume is suddenly reduced to 1/3, the reaction rate becomes x times original. Value of x:
A3
B9
C1/3
D1/9
✅ Correct: B
Rate = k[A][B]. Reducing volume to 1/3 → concentrations become 3× each. New rate = k(3[A])(3[B]) = 9k[A][B] = 9× original. x = \(\mathbf{9}\).
Q7
Time required for 99.9% completion of a first order reaction is ______ times the time required for 90% completion:
A2
B3
C4
D5
✅ Correct: B
For first order: t = (2.303/k)log(C₀/C). t₁ for 90%: log(100/10) = 1. t₂ for 99.9%: log(100/0.1) = log(1000) = 3. Ratio \(= 3/1 = \mathbf{3}\).
Q8
The activation energy of a reaction is 40 kJ/mol. The rate constant at 300 K is 1.5 × 10⁻³ s⁻¹. The rate constant at 400 K is approximately: (R = 8.314 J/mol·K)
The rate of a reaction doubles for every 10°C rise in temperature. If rate at 20°C is r, rate at 60°C is:
A4r
B8r
C16r
D32r
✅ Correct: C
Rate doubles for every 10°C. From 20°C to 60°C is 4 intervals of 10°C. Rate \(= r\times2^4 = \mathbf{16r}\). Temperature coefficient Q₁₀ = 2 for this reaction.
Q10
In a zero order reaction A → products, the rate is:
AProportional to [A]
BProportional to [A]²
CIndependent of [A]
DInversely proportional to [A]
✅ Correct: C
Zero order reaction: rate = k[A]⁰ = k = constant, independent of concentration. Rate does not change with changing concentration. Example: decomposition of NH₃ on Pt surface (heterogeneous catalysis).
Q11
Half life of a first order reaction is independent of:
ARate constant
BInitial concentration
CTemperature
DActivation energy
✅ Correct: B
For first order: \(t_{1/2} = 0.693/k\). Half-life depends on k (and hence T and Ea) but is independent of initial concentration. This is a defining characteristic of first order reactions.
Q12
Which of the following is the molecularity of the reaction: O₃ + hν → O₂ + O?
AUnimolecular
BBimolecular
CTermolecular
DZero
✅ Correct: A
Molecularity = number of molecules colliding in the elementary step. Here only O₃ (1 molecule) decomposes → molecularity = 1 = unimolecular. Note: hν (photon) is not counted as a reactant molecule for molecularity.
Q13
The integrated rate law for a second order reaction 2A → products (if 2A → products, rate = k[A]²) is:
Aln[A] = ln[A]₀ − kt
B[A] = [A]₀ − kt
C1/[A] = 1/[A]₀ + kt
D[A]² = [A]₀² − kt
✅ Correct: C
For second order (rate = k[A]²): \(\dfrac{1}{[A]} = \dfrac{1}{[A]_0} + kt\). Graph of 1/[A] vs t gives a straight line with slope k. Half-life: \(t_{1/2} = 1/(k[A]_0)\) — depends on initial concentration.
Q14
Catalyst increases the rate of reaction by:
AIncreasing activation energy
BDecreasing activation energy
CIncreasing temperature
DChanging equilibrium constant
✅ Correct: B
Catalyst provides an alternative reaction pathway with lower activation energy. It does not change ΔH, ΔG, or the equilibrium constant — only the rate of achieving equilibrium. Both forward and reverse reactions are accelerated equally.
Q15
The rate constant of a reaction has units of mol⁻¹ L s⁻¹. The order of this reaction is:
AZero order
BFirst order
CSecond order
DThird order
✅ Correct: C
Rate constant units: zero order = mol L⁻¹ s⁻¹; first order = s⁻¹; second order = mol⁻¹ L s⁻¹; third order = mol⁻² L² s⁻¹. Units mol⁻¹ L s⁻¹ correspond to second order.
Q16
For a reversible reaction at equilibrium, the rate of forward reaction is 10⁻³ mol/L·s and backward reaction is also 10⁻³ mol/L·s. The equilibrium constant K is:
A10⁻³
B1
C10³
D0
✅ Correct: B
At equilibrium, rate of forward = rate of backward. \(K = k_f/k_b = 10^{-3}/10^{-3} = \mathbf{1}\). When K = 1, concentrations of reactants and products are equal (approximately, depending on stoichiometry).
Q17
The slope of the line in the plot of log k vs 1/T (Arrhenius plot) is:
AEa/R
B−Ea/R
CEa/2.303R
D−Ea/2.303R
✅ Correct: D
Arrhenius equation: \(\log k = \log A - \dfrac{E_a}{2.303R}\cdot\dfrac{1}{T}\). Plot of log k vs 1/T: slope = \(\mathbf{-E_a/2.303R}\). From this slope, activation energy can be calculated.
Q18
The rate of reaction: 2NO + Cl₂ → 2NOCl was studied and the following data was obtained. If [NO] doubles (keeping [Cl₂] constant) rate becomes 4×. If [Cl₂] doubles (keeping [NO] constant) rate becomes 2×. Rate law is:
Ar = k[NO][Cl₂]
Br = k[NO]²[Cl₂]
Cr = k[NO]²[Cl₂]²
Dr = k[NO][Cl₂]²
✅ Correct: B
Order with respect to NO: rate ∝ [NO]² (doubling [NO] → 4× rate → order 2). Order with respect to Cl₂: rate ∝ [Cl₂]¹ (doubling → 2× rate → order 1). Rate law: \(r = \mathbf{k[NO]^2[Cl_2]}\). Overall order = 3.
Q19
Which of the following statements about enzyme catalysis is incorrect?
AEnzymes are highly specific
BEnzymes lower the activation energy
CEnzyme-substrate complex is formed
DEnzymes shift the equilibrium constant
✅ Correct: D
Enzymes are biological catalysts. They are highly specific, form enzyme-substrate complex, and lower activation energy. However, like all catalysts, enzymes do not change the equilibrium constant — they only speed up the attainment of equilibrium.
Q20
The time required for completion of 90% of a first order reaction is 2.303 s. The rate constant k is: (log 2 = 0.301)
A1 s⁻¹
B2.303 s⁻¹
C0.1 s⁻¹
D0.01 s⁻¹
✅ Correct: A
\(t = \dfrac{2.303}{k}\log\dfrac{100}{10} = \dfrac{2.303}{k}\times1\). Given t = 2.303 s: \(k = 2.303/2.303 = \mathbf{1\text{ s}^{-1}}\).
R
Roshan
Expert · 5 Years Experience
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