25 questions in 25 minutes on the NTA computer-based test interface, all from Chemical Kinetics. +4 / -1 marking, instant score, full solutions. Free, no login.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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How many questions are in this Chemical Kinetics mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 18 of them are NEET previous-year questions.
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Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
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Chemistry
- Q1In a reaction A + B → Product, the rate is doubled when the concentration of B alone is doubled, and the rate increases by a factor of 8 when the concentrations of both A and B are doubled. The rate law for the reaction is:
- rate = k[A][B]
- rate = k[A]²[B]²
- rate = k[A]²[B]
- rate = k[A][B]²
Answer: (C) rate = k[A]²[B]
Doubling B alone doubles the rate ⇒ order in B = 1 (2¹ = 2). Doubling both gives 8× ⇒ 2ˣ·2¹ = 8 ⇒ 2ˣ = 4 ⇒ x = 2 (order in A). So rate = k[A]²[B].
- Q2A plot of log k versus 1/T for a reaction is a straight line. Its slope equals:
- −Eₐ/R
- −Eₐ/2.303R
- +Eₐ/2.303R
- log A
Answer: (B) −Eₐ/2.303R
From log k = log A − Eₐ/(2.303RT), plotting log k vs 1/T gives slope = −Eₐ/2.303R and intercept = log A.
- Q3A substance A decomposes by a first order reaction starting initially with [A] = 2.00 M, and after 200 min [A] becomes 0.15 M. For this reaction t₁/₂ is:
- 53.49 min
- 46.45 min
- 50.49 min
- 48.45 min
Answer: (A) 53.49 min
k = (2.303/t)log([A]₀/[A]) = (2.303/200)log(2.00/0.15) = (2.303/200)log(13.33) = (2.303/200)(1.1249) = 0.01295 min⁻¹. t₁/₂ = 0.693/k = 0.693/0.01295 = 53.49 min.
- Q4What is the time required for 75 % completion of a first order reaction if rate constant is 23.03 minute⁻¹ ?
- 12.00 s
- 3.6 s
- 36 s
- 6.0 s
Answer: (B) 3.6 s
B. Tests: the first order time for a stated percentage completion, plus the minute to second unit change. Why B: 75% complete leaves one quarter, so t = (2.303)/(k)log 4 = (2.303)/(23.03) × 0.6021 = 0.1 × 0.6021 = 0.0602 minute. Converting, 0.0602 × 60 = 3.61 s, which is option B. Why not A: 12.00 s is 0.2 minute, and (2.303)/(23.03)log(100)/(x) = 0.2 needs log(100)/(x) = 2, that is 99% completion, not 75% [misconception A: “the 100 in a percentage is itself the concentration ratio to put in the log”]. Why not C: 36 s is 0.602 minute, exactly ten times too long, and it follows from using k = 2.303 minute⁻¹ in place of 23.03 minute⁻¹ so that the 2.303 appears to cancel [misconception C: “the 2.303 in the formula always cancels against k”]. Why not D: 6.0 s is 0.1 minute, which needs log(100)/(x) = 1, that is 90% completion rather than 75%. Remember: 75% complete means one quarter left, so t = (2.303log 4)/(k), and always check whether k is per minute or per second before quoting the answer.
- Q5For a first-order reaction, the half-life period is independent of:
- first power of final concentration
- cube root of initial concentration
- square root of final concentration
- initial concentration
Answer: (D) initial concentration
For a first order reaction t₁/₂ = 0.693/k contains no concentration term, so the half-life is independent of the initial concentration of the reactant.
- Q6The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.6021
- 38.04 kJ/mol
- 380.4 kJ/mol
- 3.80 kJ/mol
- 3804 kJ/mol
Answer: (A) 38.04 kJ/mol
A. Tests: the two-temperature form of the Arrhenius equation. Why A: log(k₂)/(k₁) = (Eₐ)/(2.303R) × (T₂-T₁)/(T₁T₂) with T₁ = 300 K, T₂ = 330 K and (k₂)/(k₁) = 4. So Eₐ = 2.303 × 8.314 × 0.6021 × (T₁T₂)/(T₂-T₁) = 19.147 × 0.6021 × (99000)/(30) = 19.147 × 0.6021 × 3300 = 38044 J mol⁻¹, that is 38.04 kJ mol⁻¹. Why not B: reading R as 83.14 instead of 8.314 gives 2.303 × 83.14 × 0.6021 × 3300 = 380470 J mol⁻¹, exactly 380.4 kJ mol⁻¹ [misconception B: “the decimal point in R does not matter, the answer is in kJ anyway”]. Why not C: dividing T₁T₂ by T₁ instead of by T₂-T₁ gives (99000)/(300) = 330, and 19.147 × 0.6021 × 330 = 3804 J mol⁻¹, that is 3.80 kJ mol⁻¹ [misconception C: “the bracket is T₁T₂/T₁“]. Why not D: it is the same 3804 J mol⁻¹ produced by that wrong bracket, but with the unit written as kJ mol⁻¹, so the number and the unit are both wrong at once. Remember: put kelvin temperatures into log(k₂)/(k₁) = (Eₐ)/(2.303R)·(T₂-T₁)/(T₁T₂) and the answer comes out in joules per mole.
- Q7The reaction A → B follows first order kinetics. The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 h. The time taken for the conversion of 0.9 mol of A to 0.675 mol of B is:
- 0.25 h
- 1 h
- 0.5 h
- 2 h
Answer: (B) 1 h
Case 1: A goes 0.8→0.2 (0.6 mol B formed), ratio [A]₀/[A] = 0.8/0.2 = 4 in 1 h. Case 2: A goes 0.9→0.225 (0.675 mol B formed), ratio 0.9/0.225 = 4. Same ratio ⇒ same number of half-life equivalents ⇒ same time = 1 h (k is independent of starting amount).
- Q8For a first order reaction A arrow Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min⁻¹ is
- 0.4606
- 0.2303
- 1.3818
- 0.9212
Answer: (D) 0.9212
B. Tests: the first order integrated rate law in its base-10 logarithm form. Why B: k = (2.303)/(t)log([A]₀)/([A]) = (2.303)/(5)log(0.1)/(0.001) = (2.303)/(5)log 100 = (2.303 × 2)/(5) = 0.9212 min⁻¹. Why not A: (2.303 × 3)/(5) = 1.3818, which is the answer when the ratio is taken as 1000, that is a final concentration of 0.0001 M [misconception A: “count the decimal zeros instead of forming the ratio”]. Why not C: (2.303 × 1)/(5) = 0.4606, the answer for a ratio of only 10, as if A had fallen just to 0.01 M. Why not D: back-solving, 0.2303 min⁻¹ needs log([A]₀)/([A]) = (0.2303 × 5)/(2.303) = 0.5, a fall of only about 3.2 times, and no consistent substitution of 0.1 M and 0.001 M gives this. Remember: log(0.1)/(0.001) = 2, so the whole calculation collapses to (2.303 × 2)/(t).
- Q9Activation energy of a chemical reaction can be determined by:
- evaluating rate constants at two different temperatures
- evaluating rate constant at standard temperature
- evaluating velocities of reaction at two different temperatures
- changing concentration of reactants
Answer: (A) evaluating rate constants at two different temperatures
From the Arrhenius equation log(k₂/k₁) = (Eₐ/2.303R)[(T₂−T₁)/(T₁T₂)], measuring the rate constants at two different temperatures lets you solve for Eₐ.
- Q10For the chemical reaction N₂(g) + 3H₂(g) arrow 2NH₃(g) the correct option is:
- -(d[N₂])/(dt) = (1)/(2)(d[NH₃])/(dt)
- -(d[N₂])/(dt) = 2(d[NH₃])/(dt)
- -(1)/(3)(d[H₂])/(dt) = -(1)/(2)(d[NH₃])/(dt)
- 3(d[H₂])/(dt) = 2(d[NH₃])/(dt)
Answer: (A) -(d[N₂])/(dt) = (1)/(2)(d[NH₃])/(dt)
B. Tests: writing one single rate of reaction from stoichiometric coefficients, with correct signs. Why B: the rate of reaction is -(d[N₂])/(dt) = -(1)/(3)(d[H₂])/(dt) = +(1)/(2)(d[NH₃])/(dt), each species divided by its coefficient, minus for reactants and plus for products. Equating the first and the last term gives exactly -(d[N₂])/(dt) = (1)/(2)(d[NH₃])/(dt). Why not A: it multiplies by the coefficient 2 instead of dividing by it, so nitrogen would vanish twice as fast as ammonia appears; since 1 mol N₂ makes 2 mol NH₃, nitrogen must disappear at half the ammonia rate [misconception: “multiply by the stoichiometric coefficient”]. Why not C: it uses +(d[H₂])/(dt) with no minus sign, equating a disappearing reactant to an appearing product, and it places 3 and 2 as multipliers rather than divisors [misconception: “signs do not matter in rate expressions”]. Why not D: it attaches a minus sign to the ammonia term, which would mean NH₃ is being consumed; the correct relation is -(1)/(3)(d[H₂])/(dt) = +(1)/(2)(d[NH₃])/(dt) [misconception: “apply the reactant minus sign to every term”]. Remember: divide every d[species]/dt by its own coefficient, minus for reactants, plus for products.
- Q11The activation energy for a simple chemical reaction A → B in the forward direction is Eₐ. The activation energy for the reverse reaction:
- is always double of Eₐ
- is negative of Eₐ
- can be less than or more than Eₐ
- is always less than Eₐ
Answer: (C) can be less than or more than Eₐ
Eₐ(reverse) = Eₐ(forward) − ΔH. If the forward reaction is exothermic (ΔH<0), Eₐ(reverse) > Eₐ(forward); if endothermic (ΔH>0), Eₐ(reverse) < Eₐ(forward). So it can be less than or more than the forward Eₐ depending on the reaction.
- Q12In collision theory the steric factor P is introduced to account for:
- the change in enthalpy
- the energy of activation
- the temperature dependence of k
- the orientation requirement of colliding molecules
Answer: (D) the orientation requirement of colliding molecules
Simple collision theory overestimates rates because it ignores orientation. The probability (steric) factor P corrects for the fact that only properly oriented collisions lead to reaction.
- Q13The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. It is because the
- rate is proportional to the surface coverage
- rate is independent of the surface coverage
- rate of decomposition is very slow
- rate is inversely proportional to the surface coverage
Answer: (A) rate is proportional to the surface coverage
A. Tests: linking surface coverage to the observed order of a heterogeneously catalysed decomposition. Why A: on a solid catalyst the reaction happens only on adsorbed molecules, so rate ∝ θ, the fraction of surface covered. At low pressure the Langmuir isotherm reduces to θ ≈ Kp, so coverage is directly proportional to the phosphine pressure. Therefore rate ∝ p, which is first order. At high pressure θ arrow 1, coverage saturates, and the same reaction turns zero order. Why not B: an inverse proportionality would make the rate fall as more phosphine adsorbs, which contradicts the observed rise in rate with pressure in the low pressure regime [misconception: “more adsorbed molecules block the surface”]. Why not C: a rate independent of coverage is the saturated, high pressure case, and that is exactly what produces zero order, not first order [misconception: “surface reactions are always zero order”]. Why not D: how slow a reaction is says nothing about its order; order is fixed by how the rate responds to concentration, not by the size of k [misconception: “slow reactions are first order”]. Remember: low pressure means coverage tracks pressure so the order is 1, high pressure means a saturated surface so the order is 0.
- Q14Observe the following reactions at T (K). I. A arrow products. II. 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) arrow 3Br₂(aq) + 3H₂O(l) Both the reactions are started at 10.00 am. The rates of these reactions at 10.10 am are same. The value of -(Δ[Br⁻])/(Δ t) at 10.10 am is 2 × 10⁻⁴ mol L⁻¹ min⁻¹. The concentration of A at 10.10 am is 10⁻² mol L⁻¹. What is the first order rate constant (in min⁻¹) of reaction I?
- 2 × 10⁻³
- 10⁻³
- 4 × 10⁻³
- 10⁻²
Answer: (C) 4 × 10⁻³
A. Tests: converting a species disappearance rate into the rate of reaction using its stoichiometric coefficient, then feeding that into a first order rate law. Why A: For reaction II the rate of reaction is -(1)/(5)(Δ[Br⁻])/(Δ t) = (2 × 10⁻⁴)/(5) = 4 × 10⁻⁵ mol L⁻¹ min⁻¹. Reaction I has the same rate at that instant, and it is first order, so k[A] = 4 × 10⁻⁵, giving k = (4 × 10⁻⁵)/(10⁻²) = 4 × 10⁻³ min⁻¹. Why not B: no consistent substitution gives this; 2 × 10⁻³ is half the correct value and would need a reaction rate of 2 × 10⁻⁵, that is the Br⁻ rate divided by 10 rather than by its coefficient 5. Why not C: no consistent substitution gives this either; 10⁻³ would need a reaction rate of 10⁻⁵, which is the Br⁻ rate divided by 20. Why not D: 10⁻² is just the concentration of A copied back as the answer; ignoring the coefficient 5 altogether would give (2 × 10⁻⁴)/(10⁻²) = 2 × 10⁻², which is not this value either [misconception D: “the number in the stem must be the answer”]. Remember: divide a species rate by its coefficient before calling it the rate of reaction.
- Q15Assertion (A): The reactions 2NO + O₂ arrow 2NO₂ and 2CO + O₂ arrow 2CO₂ proceed at the same rate because they are similar. Reason (R): Both the reactions have same activation energy.
- Both (A) and (R) are false.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- Both (A) and (R) are true, but (R) is not the correct explanation of (A).
- (A) is true and (R) is false.
Answer: (A) Both (A) and (R) are false.
D. Tests: whether a similar looking balanced equation says anything about rate, which is set by activation energy and orientation. Why D: Assertion is false. NO is oxidised by O₂ readily at room temperature, which is why NO fumes brown in air, while CO and O₂ sit together for a very long time without a spark or a catalyst. Reason is also false, since the two reactions have very different activation energies, that of CO oxidation being far larger. Rate follows k = Ae^(-Eₐ/RT), not the look of the equation, so both statements fail. Why not A: this needs both statements true and causally linked, but neither is true. Why not B: this needs both statements true; the two reactions neither share an activation energy nor a rate [misconception B: “same stoichiometry implies same energetics”]. Why not C: this needs the assertion to be true, yet NO oxidation is fast at room temperature while CO oxidation is immeasurably slow. Remember: similar equations can hide wildly different barriers, so never infer rate from stoichiometry.
- Q16Which of the following statements is not correct about order of a reaction?
- The order of a reaction can be a fractional number.
- The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression.
- Order of a reaction is an experimentally determined quantity.
- The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.
Answer: (D) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.
C. Tests: separating order, which comes from experiment, from molecularity and stoichiometry, which come from the equation. Why C: Order is fixed by the experimentally found rate law, not by the balanced equation. For 2N₂O₅ arrow 4NO₂ + O₂ the coefficients of the reactant add to 2, yet the reaction is first order, rate = k[N₂O₅]. Only for a single step elementary reaction do the two numbers happen to agree, so the word “always” makes the statement false. Why not A: this is correct, fractional orders are real; the thermal decomposition of acetaldehyde follows rate = k[CH₃CHO]^(3/2), order 1.5. Why not B: this is correct, order is obtained by measuring how rate responds to concentration, for example by the initial rate method. Why not D: this is correct, it is the definition of order, the sum of the exponents of the concentration terms in the rate law. Remember: order is measured in the lab, molecularity is counted from the equation of a single step.
- Q17Correct statements regarding Arrhenius equation among the following are : A. Factor e^(-Eₐ/RT) corresponds to fraction of molecules having kinetic energy less than Eₐ. B. At a given temperature, lower the Eₐ, faster is the reaction. C. Increase in temperature by about 10°C doubles the rate of reaction. D. Plot of log k vs (1)/(T) gives a straight line with slope =-(Eₐ)/(R). Choose the correct answer from the options given below :
- B and D Only
- A and C Only
- A and B Only
- B and C Only
Answer: (D) B and C Only
C. Tests: statement-by-statement checking of the Boltzmann factor, the barrier-rate link, the ten degree rule and the slope of a log k plot. Why C: In k = Ae^(-Eₐ/RT) the exponential gives the fraction of molecules with energy equal to or greater than Eₐ, so statement A is wrong. Lowering Eₐ at fixed T raises e^(-Eₐ/RT) and hence raises k, so statement B is right. The standard rule that the rate roughly doubles for a 10°C rise makes statement C right. Taking log₁₀ of the Arrhenius equation gives log k = log A – (Eₐ)/(2.303RT), so the slope of log k vs (1)/(T) is -(Eₐ)/(2.303R), not -(Eₐ)/(R), and statement D is wrong. Only B and C survive. Why not A: it keeps statement A, but the factor e^(-Eₐ/RT) counts the molecules at or above the barrier, not the ones below it [misconception A: “the Boltzmann factor counts the slow molecules”]. Why not B: it keeps statement D, which drops the 2.303 that appears when the natural logarithm is converted to base 10 [misconception B: “a ln k plot and a log k plot have the same slope”]. Why not D: it keeps the false statement A and rejects B, even though a smaller barrier always gives a larger rate constant at a fixed temperature. Remember: log k vs 1/T has slope -Eₐ/2.303R, only ln k vs 1/T has slope -Eₐ/R.
- Q18According to the Arrhenius equation k = A·e^(−Eₐ/RT), the rate constant increases when:
- Eₐ increases
- temperature increases
- A decreases
- temperature decreases
Answer: (B) temperature increases
As T increases, the exponent −Eₐ/RT becomes less negative, so e^(−Eₐ/RT) increases and k increases. Higher Eₐ or lower A would decrease k.
- Q19In a first order reaction 20 millimole of reactant is reduced to 10 millimole in 1.151 minute. Find the rate constant.
- 2.010 minute⁻¹
- 6.120 minute⁻¹
- 0.3010 minute⁻¹
- 0.6023 minute⁻¹
Answer: (D) 0.6023 minute⁻¹
A. Tests: identifying a half-life from a halving of amount and converting it to a first order rate constant. Why A: halving 20 millimole to 10 millimole takes exactly one half-life, so t_(1/2) = 1.151 min. Then k = (0.693)/(t_(1/2)) = (0.693)/(1.151) = 0.6023 min⁻¹. Why not B: 6.120 min⁻¹ matches no substitution of the given numbers; a lost power of ten would give 6.023, and back-substituting 6.120 gives t_(1/2) = 0.693/6.120 = 0.113 min, not the stated 1.151 min [misconception: “accepting a familiar looking number instead of dividing 0.693 by the half-life”]. Why not C: 0.3010 is log 2 itself, used as though the first order law read k = log 2; the law actually needs ln 2 = 0.693 divided by the half-life [misconception: “use log 2 where ln 2 belongs”]. Why not D: 2.010 is essentially (2.303)/(1.151) = 2.001, that is the prefactor 2.303 divided by the time with the log([A]₀)/([A]) factor forgotten entirely [misconception: “drop the log term from k = (2.303)/(t)log([A]₀)/([A])“]. Remember: one halving equals one half-life, and k = 0.693/t_(1/2) holds for first order only.
- Q20Two reactions have activation energies Eₐ(1) = 40 kJ/mol and Eₐ(2) = 80 kJ/mol with equal pre-exponential factors at the same temperature. Which is faster and why?
- Reaction 2, higher Eₐ means more energy
- Both equal, A is the same
- Reaction 1, lower Eₐ means larger k
- Cannot be determined
Answer: (C) Reaction 1, lower Eₐ means larger k
k = A·e^(−Eₐ/RT). A lower Eₐ makes the exponential term larger, giving a larger k and a faster reaction. Reaction 1 with Eₐ = 40 kJ/mol is faster.
- Q21Assertion (A): A catalyst does not change the equilibrium constant of a reversible reaction. Reason (R): A catalyst lowers the activation energy of both forward and backward reactions equally. Choose the correct option.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer: (A) Both A and R are true and R is the correct explanation of A
A catalyst lowers Eₐ for both forward and reverse reactions by the same amount, so both rates increase equally and equilibrium is reached faster but Kc is unchanged. R correctly explains A.
- Q22Integrated rate law equation for a first order gas phase reaction is given by (where Pᵢ is initial pressure and Pₜ is total pressure at time t)
- k = (2.303)/(t) × log(2Pᵢ)/((2Pᵢ – Pₜ))
- k = (2.303)/(t) × log(Pᵢ)/((2Pᵢ – Pₜ))
- k = (2.303)/(t) × (Pᵢ)/((2Pᵢ – Pₜ))
- k = (2.303)/(t) × log((2Pᵢ – Pₜ))/(Pᵢ)
Answer: (B) k = (2.303)/(t) × log(Pᵢ)/((2Pᵢ – Pₜ))
A. Tests: recasting the first order integrated law in total pressure variables for a decomposition A(g) arrow B(g) + C(g). Why A: let x be the pressure of A that has reacted. Then the pressure of A left is Pᵢ – x and the total is Pₜ = (Pᵢ – x) + x + x = Pᵢ + x, so x = Pₜ – Pᵢ and the unreacted A has pressure Pᵢ – (Pₜ – Pᵢ) = 2Pᵢ – Pₜ. Substituting pressures for concentrations in k = (2.303)/(t)log([A]₀)/([A]) gives k = (2.303)/(t)log(Pᵢ)/(2Pᵢ – Pₜ). Why not B: it inverts the ratio inside the logarithm. Since Pₜ grows above Pᵢ, the quantity 2Pᵢ – Pₜ is smaller than Pᵢ, so log(2Pᵢ – Pₜ)/(Pᵢ) is negative and the formula would return a negative rate constant [misconception: “the log ratio can be written either way up”]. Why not C: it drops the logarithm and keeps the bare ratio. At t close to zero, Pₜ ≈ Pᵢ so the ratio tends to 1, and the correct expression needs log 1 = 0; this form instead returns a non-zero k from no reaction at all [misconception: “the factor 2.303 replaces the logarithm”]. Why not D: it doubles the numerator to 2Pᵢ. The pressure of unreacted A at time t is indeed 2Pᵢ – Pₜ, but at time zero the pressure of A is Pᵢ, not 2Pᵢ [misconception: “apply the factor 2 to both pressures in the ratio”]. Remember: for A(g) decomposing into two gaseous products, unreacted A is 2Pᵢ – Pₜ, and Pᵢ stays on top of the log.
- Q23The unit of rate of reaction is:
- L mol⁻¹ s⁻¹
- s⁻¹
- mol L⁻¹
- mol L⁻¹ s⁻¹
Answer: (D) mol L⁻¹ s⁻¹
Rate = change in concentration / time = (mol L⁻¹)/s = mol L⁻¹ s⁻¹. This is true regardless of order; only the rate constant’s units depend on order.
- Q24The bromination of acetone in acid, CH₃COCH₃ + Br₂ → CH₃COCH₂Br + H⁺ + Br⁻, gave the data: [CH₃COCH₃],[Br₂],[H⁺] then rate of disappearance of Br₂: (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. The rate equation is:
- rate = k[CH₃COCH₃][Br₂][H⁺]
- rate = k[CH₃COCH₃][Br₂]
- rate = k[CH₃COCH₃][H⁺]
- rate = k[CH₃COCH₃][Br₂][H⁺]²
Answer: (C) rate = k[CH₃COCH₃][H⁺]
Runs 1 and 2: [Br₂] doubles (0.05→0.10), rate unchanged ⇒ order in Br₂ = 0. Runs 2 and 3: [H⁺] doubles (0.05→0.10), rate doubles (5.7→11.4×10⁻⁵≈1.2×10⁻⁴) ⇒ order in H⁺ = 1. Order in acetone = 1. So rate = k[CH₃COCH₃]¹[Br₂]⁰[H⁺]¹ = k[CH₃COCH₃][H⁺].
- Q25For the reaction BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺ → 3Br₂(l) + 3H₂O(l), the rate of appearance of bromine (Br₂) is related to the rate of disappearance of bromide ion as:
- d[Br₂]/dt = −(3/5)d[Br⁻]/dt
- d[Br₂]/dt = +(3/5)d[Br⁻]/dt
- d[Br₂]/dt = −(5/3)d[Br⁻]/dt
- d[Br₂]/dt = +(5/3)d[Br⁻]/dt
Answer: (A) d[Br₂]/dt = −(3/5)d[Br⁻]/dt
Unified rate = −(1/5)d[Br⁻]/dt = +(1/3)d[Br₂]/dt. Therefore d[Br₂]/dt = (3/5)(−d[Br⁻]/dt) = −(3/5)d[Br⁻]/dt (the minus keeps Br₂ formation positive since d[Br⁻]/dt is negative).