Decide whether a reaction removes a fixed fraction or a fixed amount in each time interval.
JEE Main · JEE Advanced · NEET UG38 min study guide9 indexed sections
BY THE END, YOU CAN
Derive and apply first-order and zero-order integrated disappearance laws.
Use half-lives, fractional conversion and logarithmic plots correctly.
Extract reactant concentration from pressure or instrument signals before fitting a law.
Identify pseudo-first-order conditions and the limits of apparent rate constants.
Check units, stoichiometric rate conventions and the physical range of a model.
BEFORE YOU START
Concentration, logarithms and exponentials.
Balanced reaction coefficients and ideal-gas partial pressures.
Elementary slope and integration ideas; derivations are explained here.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: NEET UG 2020. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: First-Order Kinetics.
NEET UG · 2020 · 13 September · Booklet E1 · Chemistry · Q132
Paraphrased exam question
Answer checked against the official final key. Worked solution written for this lesson.
A first-order process has k=4.606×10⁻³ s⁻¹. How long does a reactant amount take to fall from 2.0 g to 0.2 g?
Reveal answer & worked solution
Answer: 500 s (original option 3).
First-order decay gives t=ln(m₀/m)/k.
Here m₀/m=10, so ln10≈2.303.
Therefore t=2.303/(4.606×10⁻³)=500 s.
Source: NEET (UG) 2020, E1, Q132; official final key. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Choose the rate definition before choosing a formula
A differential rate law tells you the instantaneous speed of composition change. An integrated rate law tells you the composition after a time interval. The bridge between them is accumulation: add all the small concentration changes that occur as the rate itself changes. A reaction that slows as reactant disappears will not generally lose equal amounts in equal times.
Throughout this lesson, k is defined in the reactant-disappearance law −d[A]/dt = k[A]^n at fixed volume and temperature. That convention makes the formulas unambiguous. For a balanced reaction aA → products, the stoichiometrically normalized reaction rate is r = −(1/a)d[A]/dt. If a problem instead defines r = k_r[A]^n, then the disappearance-law constant is k = a k_r. Do not silently mix these two definitions.
The integrated expressions assume the stated law and constant k remain valid during the observation. Temperature changes, changing catalyst activity, light intensity, volume, product inhibition or substantial reverse reaction can invalidate a simple fit. A reaction may exhibit one order over one concentration range and a different order elsewhere; reaction order is an observed kinetic description, not a permanent label inferred from a formula.
Concentrations and amounts are proportional only when volume is constant. In a fixed-volume experiment, ratios of amounts can replace ratios of concentrations. A changing-volume flow process, expanding gas vessel or evaporating solution needs an appropriate balance. Time units must also agree: a k in min⁻¹ cannot be multiplied by seconds without conversion.
02 / UNDERSTAND & APPLY
2. Derive the exponential disappearance law
For first-order disappearance, −d[A]/dt = k[A]. The larger the population of A, the greater the total number disappearing per unit time. Rearrange as d[A]/[A] = −k dt, then integrate from [A]₀ at time zero to [A]t at time t. The left integral is a logarithm and the right is −kt, giving ln([A]t/[A]₀) = −kt.
Exponentiating gives [A]t = [A]₀e^(−kt). The exponent must be dimensionless, so k has units of inverse time. This equation predicts a constant fraction remaining after each equal time interval, rather than a constant concentration loss. Doubling the initial concentration doubles the concentration at every later time under the same k, while leaving the fractional decay pattern unchanged.
The ratio inside the logarithm is dimensionless and positive. This is the cleanest form for calculation. The familiar plot of ln[A] against time assumes a fixed reference concentration or fixed concentration unit for the numerical logarithm; that convention does not affect its slope. A negative computed k for a simple disappearing reactant usually signals an inverted ratio or incorrect sign.
If the initial amount is a and the amount reacted is x, the remaining amount is a−x, so k = (1/t)ln[a/(a−x)] at fixed volume. Do not substitute x itself in the denominator unless the question has defined x as the amount remaining. “80% reacted” means 20% remains, and this translation should occur before touching a logarithm.
[A]t = [A]₀e^(−kt)
First-order disappearance at constant k and fixed volume.
k = ln([A]₀/[A]t)/t = (2.303/t) log₁₀([A]₀/[A]t)
The factor 2.303 converts a base-ten logarithm to a natural logarithm.
ORIGINAL WORKED EXAMPLE
Calculate k from a remaining fraction
A first-order reactant falls from 0.600 M to 0.150 M in 20.0 min. Find k and the concentration at 30.0 min.
The remaining fraction after 20.0 min is 0.150/0.600 = 1/4.
k = ln(4)/20.0 = 0.0693 min⁻¹.
At 30.0 min, kt = 2.07944 and [A] = 0.600e^(−2.07944) = 0.0750 M.
The extra ten minutes constitutes another half-life in this example, consistent with halving 0.150 M.
Answer k = 0.0693 min⁻¹; [A] at 30.0 min = 0.0750 M.
03 / UNDERSTAND & APPLY
3. Equal half-lives are a first-order fingerprint
A half-life is the time required for the currently considered reactant amount to fall by half. Substituting [A]t/[A]₀ = 1/2 into the first-order law gives t½ = ln2/k. Because [A]₀ cancels, every successive halving takes the same interval under unchanged conditions. Starting with 80 units, the sequence after one, two and three half-lives is 40, 20 and 10.
After n half-lives, the remaining fraction is (1/2)^n. The number n need not be an integer: n = t/t½. To find the time for a specified fractional conversion f, the remaining fraction is 1−f and t = −ln(1−f)/k. Complete disappearance is approached asymptotically in the ideal continuous model; setting [A] exactly to zero would require infinite time.
Equal increments in conversion percentage do not take equal times. Going from 0% to 50% conversion takes one half-life, while going from 50% to 75% takes another, and from 75% to 87.5% takes a third. The amounts consumed shrink while the time intervals stay equal. This observation helps distinguish first order from zero order without plotting.
An experimental detection threshold is finite. “Reaction complete” in an analytical report may mean below the instrument’s threshold rather than mathematically zero concentration. If the threshold is specified, use it as the positive final concentration. Do not criticize an experimental completion time merely because the ideal exponential never reaches exact zero.
ORIGINAL WORKED EXAMPLE
Conversion versus remainder
A first-order reaction has half-life 12.0 min. How long does 93.75% conversion take?
93.75% conversion leaves 6.25%, or 1/16, of the initial reactant.
1/16 = (1/2)^4, so four half-lives are required.
t = 4 × 12.0 = 48.0 min.
Using the converted fraction 0.9375 inside ln([A]₀/[A]t) would answer a different question.
Answer 48.0 min.
04 / UNDERSTAND & APPLY
4. Zero order consumes a fixed amount per unit time
For zero-order disappearance, −d[A]/dt = k. The observed rate is independent of [A] over the range in which the model applies. Integrating gives [A]t = [A]₀−kt, a straight declining concentration line. The units of k are concentration per time, such as mol L⁻¹ s⁻¹, rather than inverse time.
A zero-order regime can arise when a catalytic surface is saturated or another process supplies a constant rate-limiting flux. It does not mean reactant concentration is physically irrelevant at all possible values. As the reactant is depleted, a surface may cease to be saturated and the kinetic law may change. The linear model cannot be extrapolated to negative concentrations.
Setting [A]t = [A]₀/2 gives t½ = [A]₀/(2k). Unlike first order, the half-life depends on the starting concentration. Successive halving intervals become shorter: the same fixed disappearance rate removes half of a progressively smaller amount. If [A]₀ doubles while k remains unchanged, the initial half-life doubles.
The formal depletion time [A]₀/k is the boundary of the simple linear model, not a license to continue the line below zero. A realistic process can slow or switch kinetic regime before that point. When asked about concentration at a later time, explain the model limit and use the stated physical assumptions rather than report a negative amount.
[A]t = [A]₀ − kt; t½ = [A]₀/(2k)
Zero-order disappearance law, valid only over its specified concentration regime.
ORIGINAL WORKED EXAMPLE
A finite-range linear model
A zero-order process has [A]₀ = 0.240 M and k = 0.00600 M min⁻¹. Find [A] after 15.0 min, the first half-life, and the model’s depletion time.
[A]15 = 0.240 − 0.00600(15.0) = 0.150 M.
t½ = 0.240/[2(0.00600)] = 20.0 min.
Formal depletion occurs at 0.240/0.00600 = 40.0 min.
An extrapolation to 50 min would give a negative concentration and must be rejected; the stated law cannot remain valid there.
Answer 0.150 M after 15.0 min; first half-life 20.0 min; formal model boundary 40.0 min.
05 / UNDERSTAND & APPLY
5. Use the right plot and inspect its residual pattern
For zero order, plotting [A] against t gives a line with slope −k and intercept [A]₀. For first order, plotting ln([A]/c°) against t gives slope −k; plotting log₁₀([A]/c°) gives slope −k/2.303. Confusing the two logarithm bases produces a factor-of-2.303 error even when the plotted points form a perfect line.
Equal fractional losses over equal time intervals support first order; equal absolute concentration losses support zero order. A handful of noisy points over a narrow range can appear nearly linear in several transformations, so a straight-looking graph alone is not proof of a mechanism. Look at a sufficiently wide range, compare residuals and repeat at other initial concentrations when possible.
For contrast, the simple second-order disappearance law −d[A]/dt = k[A]² gives 1/[A]t = 1/[A]₀ + kt and t½ = 1/(k[A]₀). That inverse-concentration plot has positive slope. This comparison is useful for model selection, but it applies to the stated single-reactant squared law, not automatically to every reaction with total order two such as A+B with unequal starting amounts.
Rate-constant units provide a useful independent check: zero order gives M time⁻¹, first order time⁻¹ and the simple second-order law M⁻¹ time⁻¹. A graph’s slope units should match the chosen equation. If a fit gives a slope in M min⁻¹, calling it a first-order rate constant is dimensionally inconsistent.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Original ideal-model curves, not experimental measurements. Zero-order law is valid only until reactant depletion at10seconds and is not extended to negative concentration. Logarithms use a dimensionless concentration ratio. Straight-line appearance over a tiny noisy interval alone does not establish mechanism.
ORIGINAL WORKED EXAMPLE
Distinguish two data series
At 0, 5 and 10 min, series I has concentrations 0.800, 0.400 and 0.200 M; series II has 0.800, 0.600 and 0.400 M. Classify each within the ideal candidate models.
Series I halves in each five-minute interval, so it matches first order with k = ln2/5 = 0.1386 min⁻¹.
Series II loses 0.200 M in each five-minute interval, so it matches zero order with k = 0.0400 M min⁻¹.
Their concentration at ten minutes differs even though they started with the same amount.
Three ideal constructed points support the models here; real experiments require checking measurement uncertainty and a wider range.
Answer Series I: first order. Series II: zero order.
06 / UNDERSTAND & APPLY
6. Fit the reactant signal, not every measured number
An instrument may measure absorbance, optical rotation, conductivity, titration volume or pressure rather than concentration directly. If a signal S is linearly related to reaction extent and approaches a known endpoint S∞, the remaining-reactant signal is often proportional to S_t−S∞, not S_t itself. The first-order ratio can then be formed from the endpoint-corrected differences.
For an ideal linear calibration, [A]t/[A]₀ = (S_t−S∞)/(S₀−S∞), provided the same proportionality holds throughout. The differences may both be negative for a growing signal; their ratio is still positive. Derive the relation from the actual response rather than inserting absolute values to hide a sign error. Multiple absorbing species or changing background can invalidate a one-signal linear assumption.
In gas reactions at fixed temperature and volume, total pressure is proportional to total gas moles, not necessarily to the remaining reactant. For A(g) → 2B(g) starting with pure A at pressure P₀, let x be the pressure-equivalent amount of A consumed. Then p_A = P₀−x, p_B = 2x and P_total = P₀+x. Therefore p_A = 2P₀−P_total. Inserting P_total into the decay formula would wrongly treat an increasing quantity as disappearing reactant.
If an inert gas is present initially, or product stoichiometry differs, derive a new pressure relation. Pressure methods also require the stated fixed temperature and volume: a piston maintaining constant pressure removes the direct total-pressure progress signal. The experimental observable and the chemical concentration are separate layers of the calculation.
k = (1/t)ln[(S₀−S∞)/(S_t−S∞)]
First-order decay inferred from an endpoint-corrected signal linearly related to remaining reactant.
ORIGINAL WORKED EXAMPLE
A gas-pressure calculation
A(g) → 2B(g) is first order, starts with pure A at 100 kPa, and occurs at fixed temperature and volume. Total pressure is 150 kPa after 8.00 min. Find k.
For this stoichiometry, p_A = 2P₀−P_total = 200−150 = 50 kPa.
The remaining A fraction is 50/100 = 1/2.
k = ln(100/50)/8.00 = 0.0866 min⁻¹.
Total pressure has risen even as A disappears, confirming why total pressure itself was not the decay variable.
Answer k = 0.0866 min⁻¹.
07 / UNDERSTAND & APPLY
7. Apparent first order can hide another reactant
For a reaction with rate of A disappearance k₂[A][B], a large excess of B may keep its concentration nearly constant during the measurement. Then the law becomes approximately −d[A]/dt = k_obs[A], where k_obs = k₂[B]_approximately constant. The exponential A decay is called pseudo-first-order behavior. It does not prove the full rate law depends only on A.
The excess must be large relative to the stoichiometric consumption of B, not merely larger than A by an unspecified amount. If one mole B is consumed per mole A and B begins at one hundred times the A concentration, complete A consumption changes B by at most about 1%. If the coefficient of B is larger, include that coefficient in the estimate. The approximation should be checked at the largest conversion used.
Repeat the experiment at different excess B concentrations. If k_obs grows proportionally with [B], that supports first-order dependence on B in the underlying law. Dividing k_obs by [B] estimates k₂ with units M⁻¹ time⁻¹. A reported pseudo-first-order k is therefore tied to the chosen excess-reactant concentration as well as temperature.
Water as solvent is frequently present in huge excess, so its concentration can be absorbed into an observed rate constant over a useful range. Changes in solvent composition can nevertheless change both concentration and medium effects. Treat k_obs as conditional, not a universal constant transferable across arbitrary solvents or mixtures.
ORIGINAL WORKED EXAMPLE
Recover an underlying constant
A reaction follows −d[A]/dt = k₂[A][B]. With B held effectively at 0.500 M, exponential decay gives k_obs = 0.0200 s⁻¹. Find k₂ and predict k_obs at 0.800 M B if the same underlying law and medium apply.
k₂ = k_obs/[B] = 0.0200/0.500 = 0.0400 M⁻¹ s⁻¹.
At 0.800 M, k_obs = 0.0400 × 0.800 = 0.0320 s⁻¹.
The prediction assumes B remains effectively constant and the change does not alter the underlying k₂ through a medium effect.
8. Audit the assumptions before reporting a lifetime
An irreversible first-order exponential tends to zero reactant. A reversible A ⇌ B system generally tends to a nonzero equilibrium amount instead. Under suitable simple first-order forward and reverse laws, it is the deviation from equilibrium that decays exponentially, with a combined relaxation constant. Applying the irreversible formula directly to the total A concentration would give a misleading result.
Parallel first-order pathways A → B and A → C can make A disappear with k_total = k₁+k₂. Measuring only A establishes the sum, not each product-specific constant. A clean exponential thus does not uniquely identify the mechanism. Product analysis or independent experiments are required to distinguish these possibilities.
Do not infer a first-order molecular event from an overall stoichiometric coefficient. A balanced equation specifies atom conservation; the kinetic order is experimental. A zero-order law can involve several molecular steps, while a pseudo-first-order law can arise from a bimolecular reaction. “Order” and “molecularity of an elementary step” are different concepts.
Finally, report what the requested time actually means: half-life, time to a stated conversion, time to a detection threshold or a model-depletion boundary. Include units and temperature if supplied. A precise-looking numerical lifetime is not meaningful if it came from the wrong remaining fraction, an uncorrected instrument background or an invalid kinetic regime.
09 / UNDERSTAND & APPLY
9. Original practice with solved reasoning
These original problems use fixed volume, constant temperature and the stated disappearance-law convention unless another condition is given. Keep logarithm bases consistent and retain guard digits until the final value. None is labeled as a past-examination item.
When a question offers only a percentage, use a unit starting amount or concentration so it cancels in the ratio. When a graph or signal is supplied, identify its vertical variable before interpreting the slope. These habits turn many apparently different kinetics problems into the same conserved-quantity calculation.
ORIGINAL PRACTICE · 1
A first-order reaction has k = 0.0347 min⁻¹. Estimate its half-life.
Reveal answer & reasoning
20.0 min.
t½ = ln2/k = 0.693147/0.0347 ≈ 19.98 min. No starting concentration is required for this law.
ORIGINAL PRACTICE · 2
After three first-order half-lives, what percentage has reacted?
Reveal answer & reasoning
87.5%.
The remaining fraction is (1/2)³=1/8=12.5%. Reacted fraction is 100%−12.5%=87.5%.
ORIGINAL PRACTICE · 3
A first-order reactant is 90.0% consumed in 50.0 s. Find k.
Reveal answer & reasoning
0.0461 s⁻¹.
10.0% remains, so k=ln(1/0.100)/50.0=ln10/50.0. Using 0.900 would confuse converted and remaining fractions.
ORIGINAL PRACTICE · 4
A zero-order law has k=0.00200 M s⁻¹ and [A]₀=0.100 M. Find the first and second successive halving intervals.
Reveal answer & reasoning
25.0 s, then 12.5 s.
The first interval removes 0.0500 M and the next removes 0.0250 M at the same rate. Divide each amount by 0.00200 M s⁻¹.
ORIGINAL PRACTICE · 5
A log₁₀[A] versus time plot has slope −0.0150 min⁻¹. Find the first-order k.
Reveal answer & reasoning
0.0345 min⁻¹.
The base-ten slope is −k/2.303, so k=2.303×0.0150=0.0345 min⁻¹. A natural-log plot would have slope −k directly.
ORIGINAL PRACTICE · 6
A linear signal decreases from S₀=100 to S_t=40 in 10.0 min and has endpoint S∞=20. Find the first-order k.
Reveal answer & reasoning
0.139 min⁻¹.
The remaining-reactant ratio is (40−20)/(100−20)=1/4. Therefore k=ln4/10.0=0.1386 min⁻¹. The uncorrected ratio 40/100 is not the reactant fraction.
ORIGINAL PRACTICE · 7
For 2A → products, the normalized rate is r=k_r[A] with k_r=0.0100 s⁻¹. Find the disappearance-law half-life.
Reveal answer & reasoning
34.7 s.
r=−(1/2)d[A]/dt, so −d[A]/dt=2k_r[A] and k=0.0200 s⁻¹. Thus t½=ln2/0.0200=34.7 s.
ORIGINAL PRACTICE · 8
A zero-order model predicts [A]=−0.020 M at a requested late time. What should be reported?
Reveal answer & reasoning
The negative concentration is unphysical; the model has been extrapolated beyond its valid depletion range.
Locate the boundary [A]₀/k and state that the law must stop or change no later than depletion. Do not present a negative value as a real concentration or claim a new rate law without evidence.
KEEP THE BIG IDEAS
Quick revision
First order removes a constant fraction: [A]t=[A]₀e^(−kt), with t½=ln2/k.
Zero order removes a constant concentration amount: [A]t=[A]₀−kt, only within its valid range.
Converting reaction rate to disappearance rate may require a stoichiometric coefficient.
The slope of a base-ten logarithm plot differs from the natural-log slope by 2.303.
An instrument signal often needs an endpoint correction or stoichiometric conversion.
Pseudo-first-order k depends on the approximately constant excess reactant.
A successful simple fit does not uniquely establish a reaction mechanism.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.