The single most reliable way to lose marks on "factors affecting rate" questions is to confuse which half-life formula belongs to which reaction order. Zero-order half-life depends on initial concentration; first-order half-life does not. Mixing them up hands NTA a free negative mark.
What factors affect reaction rate? NCERT Class 12 Chemistry Chapter 3 (page 66) lists concentration, temperature, catalyst, and nature of reactants. Each factor changes the rate constant k or the effective collision frequency.
Temperature dependence — the Arrhenius equation. The rate constant increases exponentially with temperature: k = A·e^(−Eₐ/RT). A higher activation energy Eₐ means the rate is more sensitive to temperature changes. To compare rate constants at two temperatures, use the two-temperature form: ln(k₂/k₁) = (Eₐ/R)·(1/T₁ − 1/T₂). A common sign error: students swap T₁ and T₂ in the subtraction. Check: if T₂ > T₁, then k₂ > k₁, so ln(k₂/k₁) must be positive. Since 1/T₁ > 1/T₂ when T₂ > T₁, the right-hand side is positive. If your answer comes out negative, you swapped the temperatures.
Concentration dependence — order matters. For a first-order reaction, ln([A]₀/[A]) = kt, and the half-life is t₁/₂ = 0.693/k — constant regardless of how much reactant you start with. For a zero-order reaction, [A] = [A]₀ − kt, and the half-life is t₁/₂ = [A]₀/(2k) — directly proportional to initial concentration.
Watch-out: When a problem says "half-life doubles when initial concentration doubles," that signature points to zero-order, not first-order. First-order half-life would stay the same. NEET distractors exploit exactly this distinction.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
A student plots ln k versus 1/T for a reaction (as in the Arrhenius equation, ln k = −Eₐ/RT + ln A) and obtains a straight line with a POSITIVE slope, instead of the usual negative slope. Since slope = −Eₐ/R, what does this indicate about the reaction?
Answer: A. The Arrhenius equation gives ln k = −Eₐ/RT + ln A, so the slope of ln k vs 1/T is −Eₐ/R (NCERT Class 12 Chemistry Chapter 3, page 80). If the observed slope is positive, then −Eₐ/R > 0, and since R > 0, this forces Eₐ < 0. A negative Eₐ means k decreases as T increases — the opposite of the usual increase-with-temperature behaviour the equation normally describes.
Why B is wrong: B is wrong because it treats a typical pattern (Eₐ > 0 for ordinary reactions, giving a negative slope) as a fixed restriction of the equation itself. Slope = −Eₐ/R is positive whenever Eₐ is negative — nothing in the derivation of the equation forbids that. (trap: confusing what is usually observed with what the algebra permits)
Why C is wrong: C is wrong because the slope −Eₐ/R depends only on Eₐ, not on A — A appears only in the intercept (ln A), so a sign change in the slope says nothing about A's sign. (trap: mixing up the two constants the Arrhenius plot separately gives)
Why D is wrong: D is wrong because zero activation energy would make the slope exactly zero (a horizontal line, since −0/R = 0), not positive. A nonzero positive slope specifically requires a nonzero, negative Eₐ. (trap: confusing 'zero slope' with 'nonzero positive slope')
For a first-order reaction, the half-life is:
Answer: A. For a first-order reaction, t₁/₂ = 0.693/k. The initial concentration [A]₀ does not appear in the formula, so the half-life is independent of it. NCERT Class 12 Chemistry Chapter 3.
Why B is wrong: B is wrong because t₁/₂ = [A]₀/(2k) is the zero-order half-life formula, not first-order. Students confuse the two orders. (trap: zero-order vs first-order half-life — trap: half life zero vs first)
Why C is wrong: C is wrong because t₁/₂ = 1/(k[A]₀) is the second-order half-life formula. First-order has no [A]₀ dependence. (trap: applying wrong-order formula)
Why D is wrong: D is wrong because no standard kinetic order has half-life proportional to [A]₀². This is a fabricated relationship. (trap: guessing a mathematical pattern without checking the integrated rate law)
For a zero-order reaction, the half-life is given by:
Answer: B. The zero-order integrated rate law [A] = [A]₀ − kt gives t₁/₂ = [A]₀/(2k) when [A] = [A]₀/2. NCERT Class 12 Chemistry Chapter 3.
Why A is wrong: A is wrong because t₁/₂ = 0.693/k is the first-order half-life formula, not zero-order. (trap: zero-order vs first-order half-life mix-up — mistake: half life zero first)
Why C is wrong: C is wrong because t₁/₂ = 1/(k[A]₀) is the second-order half-life formula. Zero-order half-life has [A]₀ in the numerator, not the denominator. (trap: confusing order-specific formulas)
Why D is wrong: D is wrong because the factor is 1/2, not 2. From [A]₀ − kt₁/₂ = [A]₀/2, solving gives t₁/₂ = [A]₀/(2k), not 2[A]₀/k. (trap: algebraic inversion error)
A first-order reaction has a rate constant k = 1.386 × 10⁻² min⁻¹. What is the half-life of this reaction?
Answer: A. t₁/₂ = 0.693/k = 0.693/(1.386 × 10⁻² min⁻¹) = 50 min. NCERT Class 12 Chemistry Chapter 3.
Why B is wrong: B is wrong because 25 min results from using t₁/₂ = 0.693/(2k), incorrectly doubling the denominator. (trap: inserting a spurious factor of 2 from the zero-order formula — mistake: half life zero vs first)
Why C is wrong: C is wrong because 100 min is two half-lives, the time for 75% completion; the half-life itself is 0.693/k = 50 min.
Why D is wrong: D is wrong because 36.1 min results from using t₁/₂ = 0.500/k, substituting 0.500 instead of 0.693 (ln 2). (trap: confusing ln 2 ≈ 0.693 with 0.500)
For a zero-order reaction with k = 2.0 × 10⁻³ mol L⁻¹ s⁻¹ and [A]₀ = 0.20 mol L⁻¹, the half-life is:
Answer: D. t₁/₂ = [A]₀/(2k) = 0.20/(2 × 2.0 × 10⁻³) = 0.20/4.0 × 10⁻³ = 50 s. NCERT Class 12 Chemistry Chapter 3.
Why A is wrong: A is wrong because 200 s puts the 2 in the numerator instead of the denominator: 2[A]₀/k = 0.40/(2.0 × 10⁻³) = 200 s.
Why B is wrong: B is wrong because 100 s results from using t₁/₂ = [A]₀/k, forgetting the factor of 2 in the denominator. (trap: incomplete recall of zero-order half-life formula — mistake: half life zero first)
Why C is wrong: C is wrong because 346.5 s results from using the first-order formula t₁/₂ = 0.693/k = 0.693/(2.0 × 10⁻³) = 346.5 s, ignoring the zero-order specification. (trap: applying first-order formula to zero-order reaction — trap: half life zero vs first)
The rate constant of a reaction doubles when temperature is raised from 300 K to 310 K. Using the Arrhenius two-temperature equation and R = 8.314 J mol⁻¹ K⁻¹, the approximate activation energy is:
Answer: C. ln(k₂/k₁) = ln 2 = 0.693. (Eₐ/R)·(1/T₁ − 1/T₂) = (Eₐ/8.314)·(1/300 − 1/310) = (Eₐ/8.314)·(10/93000). So Eₐ = 0.693 × 8.314 × 93000/10 = 0.693 × 8.314 × 9300 ≈ 53,598 J mol⁻¹ ≈ 53.6 kJ mol⁻¹. NCERT Class 12 Chemistry Chapter 3.
Why A is wrong: A is wrong because 536 kJ mol⁻¹ results from multiplying by 10 instead of dividing correctly — a units-handling error where J is not converted to kJ or the temperature difference is miscomputed. (trap: order-of-magnitude error)
Why B is wrong: B is wrong because 5.36 kJ mol⁻¹ results from a decimal-place error — dividing the correct answer by 10 (e.g., using 930 instead of 9300 in the calculation). (trap: arithmetic slip in temperature reciprocal subtraction)
Why D is wrong: D is wrong because 26.8 kJ mol⁻¹ results from using ln(k₂/k₁) = 0.693/2 = 0.3465 instead of 0.693 — incorrectly halving the logarithm. (trap: confusing 'rate doubles' with a half-value somewhere in the formula)
A catalyst increases the rate of a reaction by:
Answer: D. D is correct. With a lower activation energy, a larger fraction of molecules has enough energy to react at the same temperature, so the rate rises. The catalyst is regenerated at the end, and it speeds up the forward and backward reactions equally.
Why A is wrong: A is wrong because raising the average kinetic energy is what an increase in TEMPERATURE does. A catalyst leaves the temperature, and so the energy distribution, unchanged.
Why B is wrong: B is wrong because a catalyst speeds up the forward and reverse reactions equally, so it helps equilibrium to be reached sooner but does not change where it lies.
Why C is wrong: C is wrong because ΔH depends only on the initial and final states, which a catalyst does not change. It lowers the energy of the transition state, not of the reactants or products.
A student uses the Arrhenius two-temperature equation to calculate Eₐ. Given k₁ = 2.0 × 10⁻³ s⁻¹ at T₁ = 300 K and k₂ = 8.0 × 10⁻³ s⁻¹ at T₂ = 350 K, the student writes: ln(k₂/k₁) = (Eₐ/R)·(1/T₂ − 1/T₁) and obtains a negative Eₐ. What is the student's error?
Answer: C. The correct formula is ln(k₂/k₁) = (Eₐ/R)·(1/T₁ − 1/T₂). The student wrote (1/T₂ − 1/T₁), which reverses the sign. Since k₂ > k₁, ln(k₂/k₁) > 0, but (1/T₂ − 1/T₁) < 0 when T₂ > T₁, forcing Eₐ to be negative — a physically impossible result for a standard reaction. NCERT Class 12 Chemistry Chapter 3.
Why A is wrong: A is wrong because using log₁₀ instead of ln would change the magnitude (by a factor of 2.303) but would not produce a negative Eₐ. Both ln and log₁₀ of k₂/k₁ are positive when k₂ > k₁. (trap: mistake: arrhenius t subtraction — the actual error is the sign in the temperature term, not the logarithm base)
Why B is wrong: B is wrong because the temperatures are already given in Kelvin (300 K and 350 K). No conversion is needed, and even if Celsius were mistakenly used, it would change the numerical value of Eₐ but not necessarily make it negative. (trap: assuming a unit-conversion error when the sign error is the real cause)
Why D is wrong: D is wrong because the student correctly used k₂/k₁ (as stated in the stem). Writing k₁/k₂ would make ln negative, but the stated error is in the temperature subtraction order, not the rate-constant ratio. (trap: misreading the problem — the stem specifies k₂/k₁ was used correctly)
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Pattern: First-order half-life application (NEET pattern: first order half life)
Given
A first-order reaction has a half-life of 20 minutes. A sample starts with [A]₀ = 0.80 mol L⁻¹.
Required
Find the concentration remaining after 60 minutes.
Concept
For a first-order reaction, the half-life is constant and independent of initial concentration. After each half-life, the concentration halves. Alternatively, use the integrated rate law ln([A]₀/[A]) = kt with k derived from the half-life. NCERT Class 12 Chemistry Chapter 3.
Formula
t₁/₂ = 0.693/k → k = 0.693/t₁/₂
ln([A]₀/[A]) = kt
Substitution
k = 0.693/20 = 0.03465 min⁻¹
ln(0.80/[A]) = 0.03465 × 60
Calculation
ln(0.80/[A]) = 2.079
0.80/[A] = e^(2.079) = 7.997 ≈ 8
[A] = 0.80/8 = 0.10 mol L⁻¹
Quick-check method: 60 min = 3 half-lives. After 1 half-life: 0.40. After 2: 0.20. After 3: 0.10 mol L⁻¹. ✓
Note: The number 3 (half-lives) and 0.693 (= ln 2, a mathematical constant) are exact values and do not limit significant figures. The answer is reported to 2 significant figures, matching the precision of the given data (0.80 mol L⁻¹ and 20 min).
Final answer
[A] = 0.10 mol L⁻¹ after 60 minutes.
Common trap
A student might use the zero-order formula [A] = [A]₀ − kt = 0.80 − 0.03465 × 60 = 0.80 − 2.079 = −1.279 mol L⁻¹, which is physically impossible (negative concentration). This is the hallmark error of applying the wrong-order integrated rate law. The negative result should be an immediate red flag that the wrong formula was used.
Similar NEET-style question
A first-order reaction has k = 0.0231 min⁻¹. What fraction of reactant remains after 100 minutes? (Answer: t₁/₂ = 0.693/0.0231 = 30 min. 100 min ≈ 3.33 half-lives. [A]/[A]₀ = e^(−0.0231 × 100) = e^(−2.31) ≈ 0.099, so about 10% remains.)
(1) Concentration of reactants (rate law). (2) Temperature (Arrhenius). (3) Catalyst (lowers Ea). (4) Surface area (heterogeneous reactions). (5) Pressure (gases). (6) Nature of reactants.
-- NCERT Class 12 Chemistry, Ch. 3, p. 66More in Chemical Kinetics: 5 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
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