Chemical Kinetics NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
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Every Chemical Kinetics question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Chemical Kinetics previous year questions
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Q1NEET 2021
The slope of Arrhenius Plot (ln k v/s 1/T) of first order reaction is −5 × 10³ K. The value of Ea of the reaction is. Choose the correct option for your answer. [Given R = 8.314 JK⁻¹ mol⁻¹]
Not quite — the answer is A.
ln k = ln A − Ea/(RT), so the slope of ln k vs 1/T is −Ea/R. Ea = 5 × 10³ × 8.314 = 41,570 J mol⁻¹ ≈ 41.5 kJ mol⁻¹. Ea is positive since the slope is negative, and 83 kJ is double the correct value.
Q2NEET 2020 · Q109
An increase in the concentration of the reactants of a reaction leads to change in :
Not quite — the answer is D.
More reactant molecules per unit volume means more collisions per second, so the collision frequency rises and the rate increases. Activation energy, threshold energy and heat of reaction are properties of the reaction itself and do not depend on concentration.
Q3NEET 2020 · Q132
The rate constant for a first order reaction is 4.606×10⁻³ s⁻¹. The time required to reduce 2.0 g of the reactant to 0.2 g is :
Not quite — the answer is C.
t = (2.303/k) log([R]₀/[R]) = (2.303/4.606×10⁻³) × log(2.0/0.2) = 500 × 1 = 500 s. Leaving out the 2.303 conversion factor would give about 217 s, near the 200 s trap.
Q4NEET 2019 · Q63
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:
Not quite — the answer is C.
t = (2.303/k) log([A]₀/[A]) = (2.303/k) log(100/1) = 4.606/k. 0.693/k is only the half-life, and 2.303/k corresponds to 90% completion.
Q5NEET 2019 · Q67
For the chemical reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) the correct option is:
Not quite — the answer is C.
Rate = −d[N₂]/dt = −(1/3)d[H₂]/dt = +(1/2)d[NH₃]/dt. The relation with a negative sign on the NH₃ term is wrong because NH₃ is a product whose concentration increases.
Q6NEET 2018 · Q79
The correct difference between first- and second-order reactions is that
Not quite — the answer is B.
For first order t½ = 0.693/k, independent of [A]₀, while for second order t½ = 1/(k[A]₀). Rates of both orders depend on concentration, and either type can be catalysed.
Q7NEET 2018 · Q84
When initial concentration of the reactant is doubled, the half-life period of a zero order reaction
Not quite — the answer is B.
For a zero-order reaction t½ = [A]₀/2k, directly proportional to initial concentration, so doubling [A]₀ doubles t½. A half-life independent of concentration is the first-order case.
Q8NEET 2017 · Q40
Mechanism of a hypothetical reaction X₂ + Y₂ → 2XY is given below: (i) X₂ → X + X (fast) (ii) X + Y₂ ⇌ XY + Y (slow) (iii) X + Y → XY (fast). The overall order of the reaction will be:
Not quite — the answer is D.
The slow step gives rate = k[X][Y₂]. X comes from the fast pre-equilibrium X₂ ⇌ 2X, so [X] = √(K[X₂]) and rate = k′[X₂]^½[Y₂]. The overall order is ½ + 1 = 1.5; taking the slow step's molecularity of 2 as the order ignores that X is an intermediate.
Q9NEET 2017 (Gujarati paper) · Q8
For the reaction, XA + YB → ZC, if −d[A]/dt = −d[B]/dt = 1.5 d[C]/dt, then the correct statement among the following is:
Not quite — the answer is B.
Rate = −(1/X)d[A]/dt = −(1/Y)d[B]/dt = (1/Z)d[C]/dt. Equal rates of loss of A and B give X = Y, and −d[A]/dt = 1.5 d[C]/dt gives X = 1.5Z, so the smallest whole numbers are X = Y = 3 and Z = 2. Z cannot also be 3, because then the rates would be equal, not in the ratio 1.5.
Q10NEET 2017 (Gujarati paper) · Q29
Consider the reaction between chlorine and nitric oxide Cl₂(g) + 2NO(g) → 2NOCl(g). On doubling the concentration of both reactants, the rate of the reaction increases by a factor of 8. However, if only the concentration of Cl₂ is doubled, the rate increases by a factor of 2. The order of this reaction with respect to NO is:
Not quite — the answer is C.
Let rate = k[Cl₂]ˣ[NO]ʸ. Doubling Cl₂ alone doubles the rate, so x = 1. Doubling both gives 2¹ × 2ʸ = 8, so 2ʸ = 4 and y = 2. The overall order is 3, which is the trap.
Q11NEET 2016 Phase 2 · Q48
The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. It is because the
Not quite — the answer is A.
At low pressure only a small part of the metal surface is covered, and the coverage is proportional to the pressure of PH₃. The rate depends on the coverage, so it is proportional to pressure: first order. At high pressure the surface is saturated and the rate becomes independent of coverage (zero order).
Q12NEET 2016 Phase 1 · Q49
The rate of a first-order reaction is 0.04 mol l⁻¹ s⁻¹ at 10 seconds and 0.03 mol l⁻¹ s⁻¹ at 20 seconds after initiation of the reaction. The half-life period of the reaction is:
Not quite — the answer is A.
For first order, rate ∝ concentration, so 0.04/0.03 = e^(k×10) and k = ln(4/3)/10 = 0.2877/10 ≈ 0.0288 s⁻¹. t½ = 0.693/0.0288 ≈ 24.1 s. Using the 20 s mark as the time interval instead of the 10 s gap gives a much smaller k and longer half-lives.
Q13NEET 2016 Phase 1 · Q76
The addition of a catalyst during a chemical reaction alters which of the following quantities?
Not quite — the answer is D.
A catalyst provides an alternative pathway with a lower activation energy, which speeds up the reaction. Entropy, internal energy and enthalpy are state functions that depend only on the initial and final states, so a catalyst cannot change them.
Q14AIPMT 2015
The activation energy of a reaction can be determined from the slope of which of the following graphs?
Not quite — the answer is D.
From the Arrhenius equation ln k = ln A − Ea/RT, a plot of ln k against 1/T is a straight line of slope −Ea/R, from which Ea is found. ln k against T is curved.
Q15AIPMT 2015
When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is:
Not quite — the answer is C.
t½ ∝ [A]₀^(1 − n). It is independent of the initial concentration only when n = 1, as in t½ = 0.693/k. For zero order t½ doubles when [A]₀ doubles.
Q16NEET 2013
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R = 8.314 J mol⁻¹ K⁻¹)
Not quite — the answer is C.
log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂) with T₁ = 293 K, T₂ = 308 K. Ea = 2.303 × 8.314 × 0.3010 × (293 × 308)/15 = 5.763 × 6016 ≈ 3.47 × 10⁴ J mol⁻¹ = 34.7 kJ mol⁻¹.
Q17NEET 2013
A reaction having equal energies of activation for forward and reverse reactions has
Not quite — the answer is C.
ΔH = Ea(forward) − Ea(reverse). If the two activation energies are equal, ΔH = 0. ΔG and ΔS need not be zero, since they also depend on entropy changes.
Q18AIPMT 2012 Prelims
In a reaction, A + B → Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as
Not quite — the answer is A.
Let rate = k[A]^x[B]^y. Doubling [B] doubles the rate, so y = 1. Doubling both gives 2^(x+1) = 8, so x = 2. Rate = k[A]²[B]. k[A][B]² would also give 8 for doubling both, but it would quadruple, not double, the rate on doubling [B].
Q19AIPMT 2012 Prelims
In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become:
Not quite — the answer is C.
From 10 °C to 100 °C is a 90° rise, i.e. nine 10° steps. Rate becomes 2⁹ = 512 times. 256 (2⁸) comes from miscounting the steps. The zero order of the reaction does not affect the temperature factor.
Q20AIPMT 2011 Mains
The unit of rate constant for a zero order reaction is
Not quite — the answer is C.
For a zero-order reaction rate = k[A]⁰ = k, so k has the units of rate, mol L⁻¹ s⁻¹. s⁻¹ is the first-order unit and L mol⁻¹ s⁻¹ is the second-order unit.
Q21AIPMT 2011 Mains
The rate of the reaction 2N₂O₅ → 4NO₂ + O₂ can be written in three ways: −d[N₂O₅]/dt = k[N₂O₅] d[NO₂]/dt = k′[N₂O₅] d[O₂]/dt = k″[N₂O₅] The relationship between k and k′ and between k and k″ are
Not quite — the answer is D.
Rate = −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt. So d[NO₂]/dt = 2 × (−d[N₂O₅]/dt), giving k′ = 2k, and d[O₂]/dt = ½ × (−d[N₂O₅]/dt), giving k″ = k/2. Setting all three constants equal ignores the stoichiometric coefficients.
Q22AIPMT 2011 Mains
The half life of a substance in a certain enzyme-catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L⁻¹ to 0.04 mg L⁻¹ is
Not quite — the answer is A.
A constant half-life means first-order kinetics. 1.28/0.04 = 32 = 2⁵, so five half-lives are needed: 5 × 138 = 690 s. 552 s corresponds to only four half-lives.
Q23AIPMT 2011 Prelims
Which one of the following statements for the order of a reaction is incorrect?
Not quite — the answer is A.
Order is an experimental quantity and can be zero, fractional or even negative, so the claim that it is always a whole number is incorrect. It is determined experimentally, does not have to follow stoichiometric coefficients, and equals the sum of the powers of concentration terms in the rate law.
Q24AIPMT 2010 Mains · Q110
The rate of the reaction 2NO + Cl₂ → 2NOCl is given by the rate equation rate = k[NO]²[Cl₂]. The value of the rate constant can be increased by
Not quite — the answer is A.
The rate constant depends on temperature (Arrhenius, k = A e^(−Ea/RT)), not on concentrations. Raising concentrations of NO or Cl₂ raises the rate but leaves k unchanged.
Q25AIPMT 2010 Prelims · Q58
For the reaction N₂O₅(g) → 2NO₂(g) + (1/2)O₂(g) the value of rate of disappearance of N₂O₅ is given as 6.25 × 10⁻³ mol L⁻¹ s⁻¹. The rate of formation of NO₂ and O₂ is given respectively as:
Not quite — the answer is C.
Rate = −d[N₂O₅]/dt = ½ d[NO₂]/dt = 2 d[O₂]/dt. So d[NO₂]/dt = 2 × 6.25 × 10⁻³ = 1.25 × 10⁻² and d[O₂]/dt = 6.25 × 10⁻³/2 = 3.125 × 10⁻³ mol L⁻¹ s⁻¹.
Q26AIPMT 2010 Prelims · Q60
During the kinetic study of the reaction 2A + B → C + D, following results were obtained: Run I: [A] = 0.1, [B] = 0.1, rate = 6.0 × 10⁻³ Run II: [A] = 0.3, [B] = 0.2, rate = 7.2 × 10⁻² Run III: [A] = 0.3, [B] = 0.4, rate = 2.88 × 10⁻¹ Run IV: [A] = 0.4, [B] = 0.1, rate = 2.40 × 10⁻² ([A], [B] in mol L⁻¹; initial rate of formation of D in mol L⁻¹ min⁻¹) Based on the above data which one of the following is correct?
Not quite — the answer is A.
Runs II and III: [A] fixed, [B] doubled, rate × 4, so order in B = 2. Runs I and IV: [B] fixed, [A] × 4, rate × 4, so order in A = 1. Rate = k[A][B]²; check: Run II/Run I = (3 × 4) = 12 = 7.2 × 10⁻²/6.0 × 10⁻³.
Q27AIPMT 2009 Prelims · Q159
For the reaction N₂ + 3H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴ mol L⁻¹ s⁻¹, the value of −d[H₂]/dt would be :
Not quite — the answer is D.
Rate = −(1/3)d[H₂]/dt = (1/2)d[NH₃]/dt. So −d[H₂]/dt = (3/2) × 2 × 10⁻⁴ = 3 × 10⁻⁴ mol L⁻¹ s⁻¹. Multiplying by 3 alone gives 6 × 10⁻⁴.
Q28AIPMT 2009 Prelims · Q163
In the reaction BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(l) + 3H₂O(l), the rate of appearance of bromine (Br₂) is related to rate of disappearance of bromide ions as following:
Not quite — the answer is D.
Rate = (1/3)d[Br₂]/dt = −(1/5)d[Br⁻]/dt, so d[Br₂]/dt = −(3/5)d[Br⁻]/dt. The minus sign is needed because [Br⁻] decreases while [Br₂] increases, and 5/3 inverts the stoichiometric ratio.
Q29AIPMT 2009 Prelims · Q169
Half life period of a first-order reaction is 1386 seconds. The specific rate constant of the reaction is:
Not quite — the answer is B.
For a first-order reaction k = 0.693/t½ = 0.693/1386 = 5.0 × 10⁻⁴ s⁻¹ = 0.5 × 10⁻³ s⁻¹. 5.0 × 10⁻³ s⁻¹ is ten times too large.
Q30AIPMT 2009 Prelims · Q199
For the reaction A + B → products, it is observed that: (a) on doubling the initial concentration of A only, the rate of reaction is also doubled and (b) on doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction. The rate of this reaction is given by:
Not quite — the answer is A.
Doubling [A] alone doubles the rate, so the order in A is 1. Doubling both raises the rate 8 times: 2¹ × 2ⁿ = 8, so n = 2 for B. Rate = k[A][B]².
Q31AIPMT 2008 Prelims · Q58
The bromination of acetone that occurs in acid solution is represented by this equation CH₃COCH₃(aq) + Br₂(aq) → CH₃COCH₂Br(aq) + H⁺(aq) + Br⁻(aq) These kinetic data were obtained for given reaction concentrations. Initial concentrations (M) [CH₃COCH₃], [Br₂], [H⁺] → initial rate of disappearance of Br₂ (M s⁻¹): 0.30, 0.05, 0.05 → 5.7 × 10⁻⁵ 0.30, 0.10, 0.05 → 5.7 × 10⁻⁵ 0.30, 0.10, 0.10 → 1.2 × 10⁻⁴ 0.40, 0.05, 0.20 → 3.1 × 10⁻⁴ Based on these data, the rate equation is
Not quite — the answer is B.
Runs 1 and 2: doubling [Br₂] leaves the rate unchanged, so the reaction is zero order in Br₂. Runs 2 and 3: doubling [H⁺] doubles the rate, so it is first order in H⁺. Run 4 checks acetone: 5.7 × 10⁻⁵ × (0.40/0.30) × (0.20/0.05) ≈ 3.0 × 10⁻⁴, matching 3.1 × 10⁻⁴, so it is first order in acetone. Rate = k[CH₃COCH₃][H⁺]; any form containing [Br₂] contradicts runs 1 and 2.
Q32AIPMT 2008 Prelims · Q66
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
Not quite — the answer is A.
Setting 10¹⁶e^(−2000/T) = 10¹⁵e^(−1000/T) gives 10 = e^(1000/T), so 1000/T = ln 10 = 2.303 and T = 1000/2.303 K ≈ 434 K. 1000 K would come from wrongly taking ln 10 as 1.
Q33AIPMT 2007 Prelims · Q55
The reaction of hydrogen and iodine monochloride is given as: H₂(g) + 2ICl(g) → 2HCl(g) + I₂(g) The reaction is of first order with respect to H₂(g) and ICl(g), following mechanisms were proposed. Mechanism A: H₂(g) + 2ICl(g) → 2HCl(g) + I₂(g) Mechanism B: H₂(g) + ICl(g) → HI(g) ; slow HI(g) + ICl(g) → HCl(g) + I₂(g) ; fast Which of the above mechanism(s) can be consistent with the given information about the reaction?
Not quite — the answer is D.
The observed rate law is rate = k[H₂][ICl]. Mechanism A, as a single step, would be second order in ICl. In mechanism B the slow step H₂ + ICl gives exactly rate = k[H₂][ICl], so only B is consistent.
Q34AIPMT 2007 Prelims · Q62
In a first-order reaction A → B, if k is rate constant and initial concentration of the reactant A is 0.5 M, then the half-life is
Not quite — the answer is C.
For a first-order reaction t½ = ln 2/k = 0.693/k, independent of the initial concentration. log 2/k uses the base-10 log, and 0.693/(0.5k) wrongly brings in the 0.5 M concentration.
Q35AIPMT 2007 Prelims · Q75
If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately (log 4 = 0.60, log 5 = 0.69)
Not quite — the answer is A.
k = (2.303/60) log(100/40) = (2.303/60)(0.69 − 0.30) ≈ 0.0150 min⁻¹. t½ = 0.693/0.0150 ≈ 46 min, closest to 45 minutes. 50 minutes assumes a linear (zero-order) fall.
Q36AIPMT 2006 Prelims · Q68
For the reaction 2A + B → 3C + D which of the following does not express the reaction rate ?
Not quite — the answer is A.
Rate = −½ d[A]/dt = −d[B]/dt = +⅓ d[C]/dt = +d[D]/dt. Since C is a product its concentration rises, so −d[C]/3dt is negative and cannot represent the rate. The other three expressions are correct.
Q37AIPMT 2006 Prelims · Q77
Consider the reaction N₂(g) + 3H₂(g) → 2NH₃(g). The equality relationship between d[NH₃]/dt and −d[H₂]/dt is :
Not quite — the answer is B.
Rate = −⅓ d[H₂]/dt = ½ d[NH₃]/dt. Therefore d[NH₃]/dt = −(2/3) d[H₂]/dt: two moles of NH₃ form for every three moles of H₂ used. The ratio 3/2 inverts the stoichiometry.
Q38AIPMT 2005 Prelims · Q73
For a first order reaction A → B, the reaction rate at reactant concentration of 0.01 M is found to be 2.0 × 10⁻⁵ mol L⁻¹ s⁻¹. The half life period of the reaction is :
Not quite — the answer is D.
For first order, rate = k[A], so k = 2 × 10⁻⁵/0.01 = 2 × 10⁻³ s⁻¹. t½ = 0.693/k = 346.5 ≈ 347 s.
Q39AIPMT 2005 Prelims · Q83
The rate of reaction between two reactants A and B decreases by a factor of 4, if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is :
Not quite — the answer is B.
rate ∝ [B]ⁿ. Doubling [B] makes the rate one quarter: 2ⁿ = 1/4, so n = −2. Order +2 would make the rate four times larger.
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Questions from this chapter in each paper on this site.
- NEET 2021: 1
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- NEET 2018: 2
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 2
- NEET 2016 Phase 2: 1
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- NEET 2013: 2
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- AIPMT 2008 Prelims: 2
- AIPMT 2007 Prelims: 3
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- AIPMT 2005 Prelims: 2
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