NEET 2027 · Chemistry · Chemical Kinetics · Topic 01 of 15
Tier 1 · highest priority. The single most-asked unit in the chapter — concepts from the ground up, four animations, a complete formula sheet, and 76 worked questions.
A first order reaction is a reaction where the speed depends on how much reactant is left — and on nothing else. Twice as much stuff left means twice the speed. Half as much left means half the speed. That single sentence is the whole chapter unit; everything below is just that sentence written in maths.
Imagine a room full of popcorn kernels on a hot pan. Every second, each kernel has the same chance of popping. It does not matter whether the kernel is at the edge or the middle, whether it is early in the cooking or late. Every unpopped kernel, every second, has the same small chance.
At the start there are 1000 kernels, so lots of pops per second — it sounds loud. Later only 100 kernels are left, so you hear about a tenth as many pops per second. The kernels did not get lazy. Each one still has exactly the same chance. There are simply fewer of them left to pop.
That is a first order reaction. The rate falls not because the molecules slow down, but because there are fewer molecules left to react.
Now here is the strange and beautiful part. Ask: how long until half the kernels have popped? Say it takes 30 seconds to go from 1000 to 500. How long from 500 to 250? Also 30 seconds. From 250 to 125? Again 30 seconds. The time to halve is always the same, no matter how much you start with. That fixed time has a name: the half-life, written t½.
This is why first order questions are so easy once you see it. You are not really doing chemistry. You are counting halvings.
Take R → P. First order means the rate is proportional to the first power of [R]:
Separate the variables — put everything with [R] on one side and everything with t on the other:
Integrate both sides. The left side integrates to ln[R], the right to −kt plus a constant:
Find I using the starting condition. At t = 0 the concentration is [R]₀, so ln[R]₀ = I. Put that back:
Rearrange into the three forms you will actually use in the exam:
The last one has the 2.303 because we switched from natural log to base-10 log: ln x = 2.303 log x. Indian exam papers overwhelmingly use the log₁₀ form, so make that your default.
Half-life means the time when [R] has fallen to [R]₀/2. Substitute that into the log form:
The [R]₀ cancels — this cancellation is the entire reason half-life is independent of starting concentration. With log 2 = 0.301:
Note 0.693 is just ln 2. So the cleaner way to remember it is t½ = ln2 / k.
Because each half-life removes half of whatever remains, after n half-lives:
| n | Time elapsed | Fraction left | % completed | Recognise it as |
|---|---|---|---|---|
| 1 | t½ | 1/2 | 50% | One halving |
| 2 | 2t½ | 1/4 | 75% | Very commonly asked |
| 3 | 3t½ | 1/8 | 87.5% | Commonly asked |
| 4 | 4t½ | 1/16 | 93.75% | NCERT Q 3.16 |
| 5 | 5t½ | 1/32 | 96.875% | Occasional |
| 10 | 10t½ | 1/1024 | ≈99.9% | NCERT Example 3.8 |
When the fraction is not a power of ½ you must use logs — but four values come up so often that learning them outright is worth it. Setting t = (2.303/k)·log(100/(100−x)):
| Completion | k·t equals | In half-lives | Where it appears |
|---|---|---|---|
| 50% | 0.693 | 1.00 | Definition of t½ |
| 75% | 1.386 | 2.00 | Exactly 2 half-lives |
| 90% | 2.303 | 3.32 | Very frequent |
| 99% | 4.606 | 6.64 | NCERT Q 3.18 |
| 99.9% | 6.909 | 9.97 ≈ 10 | NCERT Example 3.8 |
NCERT tells you that all natural and artificial radioactive decay follows first order kinetics, then moves on. It is worth one extra minute of thought, because NEET dresses kinetics questions in radioactive clothing regularly.
A radioactive nucleus has no memory and no age. A carbon-14 nucleus that has been sitting in a tree for 4000 years is not "more likely" to decay today than one made yesterday. Each nucleus, each second, has the same fixed chance of falling apart — exactly like the popcorn kernels. Nothing external influences it: heating it, freezing it, dissolving it or squeezing it changes nothing.
So radioactive decay is automatically first order, and every formula on this page applies unchanged. Just read "number of nuclei N" wherever the page says "concentration [R]".
Carbon dating works because a living tree keeps swapping carbon with the air, holding its ¹⁴C level fixed. The moment it dies the swapping stops and the ¹⁴C clock starts running down. Measure how much is left, count the halvings, get the age.
Each one isolates a single idea. Work through them in order; the fourth is the one that stops the zero-order/first-order half-life confusion for good.
Press the step button and one half-life passes. Green dots are unreacted molecules, grey are reacted. Notice that the number removed shrinks each step (128, then 64, then 32…) but the fraction removed is always exactly one half. That is the definition of first order, shown rather than stated.
The left panel is what nature does — a curve that flattens out and never quite touches zero. The right panel is the same data with ln[R] on the vertical axis, and it is dead straight with slope −k. This is the whole reason chemists plot logs: a straight line is easy to read a rate constant off, a curve is not. Slide k and watch both panels respond together. The green dashed steps on the left mark successive half-lives — they are evenly spaced, always.
Drag the slider to move along a timeline measured in half-lives. The brass ticks mark the percentages NEET asks about. Two things to absorb: 75% sits exactly on 2 half-lives (a clean number worth memorising), while 90% sits at 3.32 — not 3 — which is exactly why 90% questions need logs and 75% questions do not.
Slide the starting concentration. On the zero-order side the half-life marker slides with it, because t½ = [R]₀/2k is proportional to where you started. On the first-order side the marker does not budge, no matter how high or low you start. If you understand this one picture, you will never confuse the two half-life formulas again.
Everything you need for this unit. The five bolded working forms — log form, half-life, halving ladder, percentage completion and the standard kt values — cover well over ninety percent of questions.
| Quantity / situation | Formula | When you use it |
|---|---|---|
| Differential rate law | Rate = −d[R]/dt = k[R] | Definition of first order; starting point for the derivation |
| Integrated law (ln form) | ln[R] = ln[R]₀ − kt | Straight-line form; use when a graph is involved |
| Integrated law (exponential) | [R] = [R]₀·e^(−kt) | When asked for concentration left after a given time |
| Integrated law (log₁₀ form) | k = (2.303/t)·log([R]₀/[R]) | The default working form for NEET numericals |
| Two-time form | k = 2.303/(t₂−t₁) · log([R]₁/[R]₂) | When neither reading is the initial concentration |
| Half-life | t½ = 0.693/k = ln2/k | The most-used single line in the chapter |
| After n half-lives | [R] = [R]₀/2ⁿ ; n = t/t½ | Whenever the fraction is 1/2, 1/4, 1/8, 1/16… |
| Fraction remaining at time t | [R]/[R]₀ = e^(−kt) = (1/2)^(t/t½) | Either form works; pick whichever the data suits |
| Time for x% completion | t = (2.303/k)·log(100/(100−x)) | For 90%, 99%, 99.9% and other non-power-of-2 values |
| Standard completion times | kt₅₀=0.693, kt₇₅=1.386, kt₉₀=2.303, kt₉₉=4.606, kt₉₉.₉=6.909 | Memorise; converts most questions to one division |
| Useful ratios | t₉₉ = 2t₉₀ ; t₉₉.₉ = 3t₉₀ = 10t½ ; t₇₅ = 2t½ | Asked directly as 'show that' or as one-liners |
| Units of k | time⁻¹ (s⁻¹, min⁻¹, h⁻¹, yr⁻¹) | Seeing these units identifies first order instantly |
| Graph: ln[R] vs t | slope = −k, intercept = ln[R]₀ | Straight, falling |
| Graph: log([R]₀/[R]) vs t | slope = k/2.303, passes through origin | Straight, rising |
| Graph: t½ vs [R]₀ | horizontal line | The signature test for first order |
| Radioactive decay | N = N₀e^(−λt) ; λ = 0.693/t½ ; A/A₀ = N/N₀ | Same maths, different symbols |
| Mean (average) life — gap content | τ = 1/k = 1.44 t½ | Not in NCERT; appears in decay-flavoured questions |
| General-order half-life — gap content | t½ ∝ 1/[R]₀^(n−1) | For 'find the order from how t½ changes' questions |
Four graph questions and three assertion–reason questions are included, marked with coloured left borders. Questions tagged PYQ pattern follow forms that have appeared in NEET/AIPMT papers or are lifted from NCERT exercises — exact year attributions are deliberately omitted rather than guessed. Attempt each question first, then open the solution and check not only your answer but your route: in this unit the fast route and the slow route give the same mark but not the same time.
A first order reaction has a rate constant k = 5.5 × 10⁻¹⁴ s⁻¹. Its half-life is:
k = 5.5 × 10⁻¹⁴ s⁻¹, first order
t½
For first order the half-life is fixed by k alone.
t½ = 0.693/k
Answer · (a) 1.26 × 10¹³ s
The time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is first order, the rate constant is:
t½ = 60 min, first order
k in s⁻¹
Rearrange the half-life relation; then convert the time unit.
k = 0.693/t½
Answer · (a) 1.925 × 10⁻⁴ s⁻¹
For a first order reaction the half-life is 6.93 minutes. The rate constant, in min⁻¹, is:
t½ = 6.93 min
k
Direct substitution.
k = 0.693/t½
Answer · (a) 0.1 min⁻¹
The half-life of a first order reaction is 20 minutes. The rate constant, expressed in s⁻¹, is closest to:
t½ = 20 min = 1200 s
k in s⁻¹
Convert to seconds before dividing, so no second conversion is needed.
k = 0.693/t½
Answer · (a) 5.8 × 10⁻⁴ s⁻¹
A first order reaction is 50% complete in 30 minutes. The time taken for it to be 50% complete when the initial concentration is doubled will be:
First order, t½ = 30 min at some [R]₀; then [R]₀ is doubled
new t½
The defining property of first order: half-life is independent of initial concentration.
t½ = 0.693/k — no [R]₀ term appears
Answer · (a) 30 minutes
If the rate constant of a first order reaction is 2.31 × 10⁻³ s⁻¹, the half-life in minutes is about:
k = 2.31 × 10⁻³ s⁻¹
t½ in minutes
Compute in seconds, then convert.
t½ = 0.693/k
Answer · (a) 5 minutes
For a first order reaction, if k = 0.0693 min⁻¹, the time in which the reaction is 50% complete is:
k = 0.0693 min⁻¹
time for 50% completion
50% completion IS the half-life — no separate calculation needed.
t½ = 0.693/k
Answer · (a) 10 min
The rate constant of a first order reaction is 3 × 10⁻⁴ s⁻¹. The reaction is carried out at a higher temperature where k doubles. The half-life then:
k doubles
effect on t½
t½ and k are inversely proportional.
t½ = 0.693/k
Answer · (a) becomes half
The rate constant for a first order reaction is 60 s⁻¹. The time taken to reduce the initial concentration of the reactant to its 1/16th value is:
k = 60 s⁻¹; final concentration = [R]₀/16
t
1/16 is a power of one half, so count halvings rather than using logs.
[R] = [R]₀/2ⁿ ; t = n·t½ ; t½ = 0.693/k
Answer · (a) 4.62 × 10⁻² s
A first order reaction has a half-life of 10 minutes. What fraction of the reactant remains after 40 minutes?
t½ = 10 min, t = 40 min
fraction remaining
Count how many half-lives fit into the elapsed time.
n = t/t½ ; fraction = (1/2)ⁿ
Answer · (a) 1/16
1 g of a radioactive substance with half-life 5 years is stored. The mass remaining after 20 years is:
N₀ = 1 g, t½ = 5 yr, t = 20 yr
mass remaining
Radioactive decay is first order; the halving ladder applies unchanged.
N = N₀/2ⁿ, n = t/t½
Answer · (a) 0.0625 g
A first order reaction is 75% complete in 60 minutes. Its half-life is:
75% complete in 60 min
t½
75% complete means 25% (= 1/4) remains, and 1/4 is exactly two half-lives.
t₇₅ = 2 t½
Answer · (a) 30 min
A sample decays to 1/8th of its original amount in 30 minutes by first order kinetics. The rate constant is:
Fraction left = 1/8 after 30 min
k
Convert the fraction to half-lives, get t½, then get k.
1/8 = (1/2)³ ; t½ = t/3 ; k = 0.693/t½
Answer · (a) 0.0693 min⁻¹
For a first order reaction, 87.5% of the reactant is consumed in 45 minutes. The rate constant is nearest to:
87.5% consumed in 45 min
k
87.5% gone leaves 12.5% = 1/8, which is three half-lives.
t½ = t/3 ; k = 0.693/t½
Answer · (a) 0.0462 min⁻¹
Consider A → products with k = 2.0 × 10⁻² s⁻¹. If [A]₀ = 1.0 mol L⁻¹, the concentration of A remaining after 100 s is:
k = 2.0 × 10⁻² s⁻¹, [A]₀ = 1.0 M, t = 100 s
[A] at 100 s
The exponential form is fastest when time and k are both given.
[A] = [A]₀ e^(−kt)
Answer · (a) 0.135 mol L⁻¹
Sucrose decomposes in acid solution into glucose and fructose by first order kinetics with t½ = 3.00 hours. The fraction of sucrose remaining after 8 hours is about:
t½ = 3.00 h, t = 8 h
fraction remaining
Non-integer number of half-lives — use the power form directly.
fraction = (1/2)^(t/t½)
Answer · (a) 0.158
After 3 half-lives, the percentage of a first order reactant that has reacted is:
n = 3 half-lives
percentage reacted
Distinguish 'remaining' from 'reacted' — the question asks for reacted.
remaining = (1/2)ⁿ ; reacted = 1 − (1/2)ⁿ
Answer · (a) 87.5%
A first order reaction has a rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of the reactant take to reduce to 3 g?
k = 1.15 × 10⁻³ s⁻¹, [R]₀ = 5 g, [R] = 3 g
t
Mass may be used in place of concentration because only their ratio matters.
t = (2.303/k)·log([R]₀/[R])
Answer · (a) 444 s
A first order reaction takes 40 minutes for 30% decomposition. Its t½ is nearest to:
30% decomposed in 40 min
t½
Find k from the log form first, then convert to half-life.
k = (2.303/t)·log(100/(100−x)) ; t½ = 0.693/k
Answer · (a) 77.7 min
For a first order reaction, the time required for 90% completion is:
90% completion, first order
t in terms of k
Substitute x = 90 into the completion formula and simplify.
t = (2.303/k)·log(100/(100−x))
Answer · (a) 2.303/k
Show-that type: for a first order reaction, the time required for 99% completion compared with that for 90% completion is:
First order
ratio t₉₉ : t₉₀
Both reduce to logs of powers of ten, so the ratio is clean.
t = (2.303/k)·log([R]₀/[R])
Answer · (a) twice
For a first order reaction, the time required for 99.9% completion is how many times the half-life?
99.9% completion
t₉₉.₉ / t½
Compare two expressions in k; k cancels.
t = (2.303/k)·log([R]₀/[R]) ; t½ = 0.693/k
Answer · (a) about 10 times
A first order reaction is 20% complete in 10 minutes. The time taken for it to be 60% complete is nearest to:
20% complete in 10 min
time for 60% completion
Get k from the first piece of data, then use it for the second.
k = (2.303/t)·log(100/(100−x))
Answer · (a) about 41 min
If a first order reaction is 10% complete in 20 minutes, the time for 19% completion is approximately:
10% complete in 20 min
time for 19% completion
Remaining fractions multiply. 0.90 × 0.90 = 0.81, so 19% completion is exactly two intervals.
fraction remaining after two equal intervals = (0.9)² = 0.81
Answer · (a) 40 min
The concentration of a reactant falls from 0.8 M to 0.2 M in 20 minutes by first order kinetics. The rate constant is:
[R]₀ = 0.8 M, [R] = 0.2 M, t = 20 min
k
0.8 → 0.2 is a factor of 4, which is two half-lives.
1/4 = (1/2)² ; t½ = t/2 ; k = 0.693/t½
Answer · (a) 0.0693 min⁻¹
For a first order reaction, if 0.6 of the reactant remains after time t, then kt equals:
Fraction remaining = 0.6
kt
kt is just the natural log of the inverse fraction.
kt = ln([R]₀/[R]) = 2.303·log([R]₀/[R])
Answer · (a) 0.511
A first order reaction requires 100 minutes for 50% completion. The time required for 90% completion under identical conditions is nearest to:
t½ = 100 min
t₉₀
90% completion always corresponds to 3.32 half-lives.
t₉₀ = 2.303/k and t½ = 0.693/k, so t₉₀ = 3.32 t½
Answer · (a) 332 min
The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. The age of the sample is about:
t½ = 5730 yr, N/N₀ = 0.80
age t
Radioactive decay is first order; find λ from t½ then use the log form.
λ = 0.693/t½ ; t = (2.303/λ)·log(N₀/N)
Answer · (a) about 1845 years
During nuclear explosion, ⁹⁰Sr with half-life 28.1 years is produced. If 1 μg of ⁹⁰Sr was absorbed in the bones of a newborn, the amount remaining after 10 years (if not lost metabolically) is:
N₀ = 1 μg, t½ = 28.1 yr, t = 10 yr
N
Exponential decay with a non-integer number of half-lives.
N = N₀ e^(−λt), λ = 0.693/t½
Answer · (a) 0.781 μg
Continuing the previous data: the amount of ⁹⁰Sr remaining after 60 years is:
N₀ = 1 μg, t½ = 28.1 yr, t = 60 yr
N
Same exponential relation with a larger exponent.
N = N₀ e^(−λt)
Answer · (a) 0.228 μg
A radioactive isotope has a half-life of 20 days. What fraction of the original activity remains after 60 days?
t½ = 20 d, t = 60 d
fraction of activity remaining
Activity is proportional to the number of nuclei, so the same halving ladder applies.
A/A₀ = N/N₀ = (1/2)ⁿ
Answer · (a) 1/8
The activity of a radioactive sample falls to 1/32 of its original value in 25 minutes. Its half-life is:
A/A₀ = 1/32 in 25 min
t½
Read the exponent off the power of two.
1/32 = (1/2)⁵
Answer · (a) 5 min
A radioactive nucleus decays with rate constant λ. The mean (average) life of the nucleus is:
Decay constant λ
mean life τ
Gap content — mean life is the reciprocal of the decay constant, longer than the half-life.
τ = 1/λ = t½/0.693 = 1.44 t½
Answer · (a) 1/λ
Two radioactive samples X and Y have half-lives of 10 and 20 minutes respectively. If they start with equal numbers of nuclei, after 20 minutes the ratio N(X) : N(Y) will be:
t½(X) = 10 min, t½(Y) = 20 min, equal N₀, t = 20 min
N(X):N(Y)
Count half-lives separately for each sample.
N = N₀(1/2)^(t/t½)
Answer · (a) 1 : 2
For a first order reaction, which plot gives a straight line with slope equal to −k?
First order reaction, straight line of slope −k required
Which variables to plot
Put the integrated law into y = mx + c form and read off which quantity is y.
ln[R] = −kt + ln[R]₀
Answer · (a) ln[R] versus t
The graph shown plots half-life against initial concentration for a certain reaction. The order of the reaction is:
t½ versus [R]₀ is a horizontal line
order of reaction
A horizontal line means the half-life does not change with initial concentration — the signature of first order.
t½ ∝ 1/[R]₀^(n−1)
Answer · (a) first order
The plot of log([R]₀/[R]) against time for a first order reaction gives a straight line through the origin. Its slope equals:
Straight line of log([R]₀/[R]) versus t through the origin
slope
Rearrange the base-10 integrated form into y = mx.
k = (2.303/t)·log([R]₀/[R])
Answer · (a) k/2.303
Two first order reactions A and B are plotted as ln[R] against t on the same axes. Line A is steeper than line B. Which statement is correct?
Both first order; |slope of A| > |slope of B|
Comparison of k and t½
Slope magnitude is k; half-life is inversely proportional to k.
slope = −k ; t½ = 0.693/k
Answer · (a) A has the larger k and the shorter half-life
Assertion (A): The half-life of a first order reaction does not depend on the initial concentration of the reactant.
Reason (R): In the derivation of t½ for a first order reaction, the initial concentration term cancels out.
Statements about first order half-life
Truth values and whether R explains A
Test each statement independently, then ask whether R is the actual cause of A.
k = (2.303/t½)·log([R]₀ / ([R]₀/2))
Answer · (a) Both A and R are true and R is the correct explanation of A
Assertion (A): A first order reaction can never be 100% complete in finite time.
Reason (R): The concentration in a first order reaction decays exponentially and approaches zero only asymptotically.
Statements about completion of a first order reaction
Truth values and explanation
An exponential function never reaches zero for any finite argument.
[R] = [R]₀ e^(−kt) ; t = (2.303/k)·log([R]₀/[R])
Answer · (a) Both A and R are true and R is the correct explanation of A
Assertion (A): For a first order reaction, doubling the initial concentration doubles the rate but leaves the rate constant unchanged.
Reason (R): The rate constant depends on temperature but not on the concentration of the reactants.
Statements distinguishing rate from rate constant
Truth values and explanation
Rate and rate constant are different quantities — the first depends on concentration, the second does not.
Rate = k[R] ; k = A e^(−Ea/RT)
Answer · (a) Both A and R are true and R is the correct explanation of A
Which of the following units unambiguously indicates a first order reaction?
Units of the rate constant
Which corresponds to first order
The general unit of k is (mol L⁻¹)^(1−n) time⁻¹; for n = 1 the concentration term vanishes.
units of k = (mol L⁻¹)^(1−n)·time⁻¹
Answer · (a) s⁻¹
For a first order reaction A → B, which of the following remains constant as the reaction proceeds?
A first order reaction proceeding with time
The constant quantity
Only k is fixed at a given temperature; everything else changes.
Rate = k[A]
Answer · (a) The rate constant k
The hydrogenation of ethene, C₂H₄(g) + H₂(g) → C₂H₆(g), has rate = k[C₂H₄]. This means the reaction is:
Rate = k[C₂H₄]
overall order
Overall order is the sum of the exponents in the experimental rate law.
order = sum of exponents in Rate = k[A]ˣ[B]ʸ
Answer · (a) first order overall
For a first order reaction, the plot of rate against concentration of the reactant is:
Rate = k[R], first order
Shape of rate versus concentration
Rate is directly proportional to concentration, which is the equation of a line through the origin.
Rate = k[R]
Answer · (a) a straight line through the origin
The decomposition of N₂O₅ at 318 K is first order with initial concentration 1.24 × 10⁻² mol L⁻¹, falling to 0.20 × 10⁻² mol L⁻¹ in 60 minutes. The rate constant is:
[R]₁ = 1.24 × 10⁻² M, [R]₂ = 0.20 × 10⁻² M, t = 60 min
k
Standard two-concentration log calculation.
k = (2.303/t)·log([R]₁/[R]₂)
Answer · (a) 0.0304 min⁻¹
A first order reaction has k = 0.1155 min⁻¹. The fraction of reactant left after 12 minutes is nearest to:
k = 0.1155 min⁻¹, t = 12 min
fraction remaining
Convert k to a half-life and count halvings — faster than exponentials.
t½ = 0.693/k ; fraction = (1/2)^(t/t½)
Answer · (a) 0.25
If a first order reaction has a rate constant of 10⁻³ s⁻¹, the time needed for the concentration to fall from 0.1 M to 0.025 M is:
k = 10⁻³ s⁻¹, 0.1 M → 0.025 M
t
0.1 → 0.025 is a factor of four, i.e. two half-lives.
t½ = 0.693/k ; t = n·t½
Answer · (a) 1386 s
For a first order reaction the rate becomes one-eighth of its initial value. The concentration of the reactant at that moment, relative to its initial value, is:
Rate falls to 1/8 of the initial rate
Relative concentration
For first order the rate is directly proportional to concentration, so the two fall in step.
Rate = k[R], so Rate/Rate₀ = [R]/[R]₀
Answer · (a) one-eighth
Two first order reactions have half-lives in the ratio 3 : 1. The ratio of their rate constants k₁ : k₂ is:
t½(1) : t½(2) = 3 : 1
k₁ : k₂
k and t½ are inversely proportional, so the ratio inverts.
k = 0.693/t½
Answer · (a) 1 : 3
The half-life of a first order reaction is 1.7 hours. The time required for 87.5% completion is:
t½ = 1.7 h, 87.5% completion required
t
87.5% gone leaves 12.5% = 1/8, which is exactly three half-lives.
remaining = (1/2)ⁿ
Answer · (a) 5.1 hours
In a first order reaction, the concentration of the reactant decreases from 800 mol dm⁻³ to 50 mol dm⁻³ in 2 × 10⁴ s. The rate constant is:
800 → 50 mol dm⁻³ in 2 × 10⁴ s
k
800/50 = 16, which is four halvings — no logs required.
t½ = t/n ; k = 0.693/t½
Answer · (a) 1.386 × 10⁻⁴ s⁻¹
A drug is eliminated from the body by first order kinetics with a half-life of 4 hours. If 200 mg is taken, the amount remaining after 12 hours is:
Initial 200 mg, t½ = 4 h, t = 12 h
amount remaining
Pharmacokinetics is first order; the halving ladder applies exactly as in chemistry.
amount = initial/2ⁿ, n = t/t½
Answer · (a) 25 mg
For the first order reaction 2N₂O₅ → 4NO₂ + O₂, the half-life at 318 K is 1440 s. The rate constant is:
t½ = 1440 s, first order
k
The half-life relation applies regardless of the stoichiometric coefficients, provided the reaction is first order.
k = 0.693/t½
Answer · (a) 4.81 × 10⁻⁴ s⁻¹
A first order reaction is 40% complete in 50 minutes. What percentage remains after 100 minutes?
40% complete in 50 min
percentage remaining at 100 min
Equal time intervals multiply the surviving fraction, so no logarithm is needed.
remaining after two equal intervals = (fraction per interval)²
Answer · (a) 36%
The rate constant of a first order reaction at 27 °C is 2.5 × 10⁻³ s⁻¹. The initial concentration is 0.4 M. The initial rate of the reaction is:
k = 2.5 × 10⁻³ s⁻¹, [R]₀ = 0.4 M
initial rate
Substitute directly into the differential rate law — no integration needed.
Rate = k[R]
Answer · (a) 1.0 × 10⁻³ mol L⁻¹ s⁻¹
A first order reaction has t½ = 30 s. The time in which the reaction is 99% complete is closest to:
t½ = 30 s, 99% completion
t
99% completion always takes 6.64 half-lives.
t₉₉ = 4.606/k and t½ = 0.693/k, so t₉₉ = 6.64 t½
Answer · (a) about 199 s
For a first order reaction, a plot of [R] against t is drawn. The instantaneous rate at any time equals:
Concentration versus time plot for first order
Meaning of instantaneous rate on the graph
Instantaneous rate is a derivative, and a derivative is the slope of the tangent.
r_inst = −d[R]/dt
Answer · (a) the magnitude of the slope of the tangent at that time
The half-life of a first order reaction doubles when the initial concentration is halved. The order of the reaction is actually:
t½ doubles when [R]₀ is halved
true order
Gap content — use the general-order half-life proportionality.
t½ ∝ 1/[R]₀^(n−1)
Answer · (c) second order
A first order reaction proceeds at 25 °C. Which of these changes will alter its half-life?
First order reaction; four proposed changes
Which changes t½
t½ depends only on k, and k depends only on temperature (and catalyst), not on concentration.
t½ = 0.693/k ; k = A e^(−Ea/RT)
Answer · (a) Raising the temperature
If the initial concentration of a first order reactant is 0.5 M and after 30 minutes it is 0.125 M, the rate constant is:
0.5 M → 0.125 M in 30 min
k
0.5 to 0.125 is a factor of four, which is two halvings.
t½ = t/n ; k = 0.693/t½
Answer · (a) 0.0462 min⁻¹
For a first order reaction, if 99.9% completion takes 30 minutes, the half-life is about:
t₉₉.₉ = 30 min
t½
99.9% completion is very nearly 10 half-lives.
t₉₉.₉ = 10 t½ (more precisely 9.97)
Answer · (a) about 3 min
Which of the following statements about a first order reaction is INCORRECT?
Four statements about first order kinetics
Identify the incorrect one
For first order the concentration falls exponentially, so the raw [R] versus t plot is curved.
[R] = [R]₀e^(−kt) — a curve ; ln[R] = ln[R]₀ − kt — a line
Answer · (a) A graph of [R] against t is a straight line
A first order gas-phase reaction has k = 2.303 × 10⁻³ s⁻¹. The time required for the pressure of the reactant to fall to one-tenth of its initial value is:
k = 2.303 × 10⁻³ s⁻¹, p/p₀ = 1/10
t
Pressure works exactly like concentration for a gas at constant temperature and volume.
t = (2.303/k)·log(p₀/p)
Answer · (a) 1000 s
For a first order reaction with rate constant k, the time at which the concentration falls to 1/e of its initial value is:
[R]/[R]₀ = 1/e
t
Taking natural logs makes this a one-line problem — and this time is the mean life.
[R] = [R]₀e^(−kt)
Answer · (a) 1/k
Which pairing of reaction and order is correct according to NCERT?
Four reaction–order pairings
The correct pairing
Recall the named examples given in the text.
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Answer · (a) Decomposition of N₂O₅ — first order
A first order reaction has k = 0.693 min⁻¹ at a certain temperature. Starting with 1 mol L⁻¹, the concentration after 3 minutes is:
k = 0.693 min⁻¹, [R]₀ = 1 M, t = 3 min
[R]
k = 0.693 makes t½ exactly 1 minute, so the elapsed minutes are the number of half-lives.
t½ = 0.693/k
Answer · (a) 0.125 mol L⁻¹
The fraction of a first order reactant remaining after a time equal to twice the half-life is:
t = 2 t½
fraction remaining
Two halvings.
fraction = (1/2)ⁿ
Answer · (a) 0.25
A reaction that is first order in A has its rate measured as 4 × 10⁻³ mol L⁻¹ s⁻¹ when [A] = 0.2 M. The rate constant is:
Rate = 4 × 10⁻³ mol L⁻¹ s⁻¹ at [A] = 0.2 M
k
Rearrange the differential rate law.
k = Rate/[A]
Answer · (a) 2 × 10⁻² s⁻¹
A first order reaction takes 100 minutes to go from 90% remaining to 81% remaining. How long would it take to go from 81% remaining to 72.9% remaining?
90% → 81% takes 100 min
time for 81% → 72.9%
For first order, equal fractional drops always take equal times, wherever they occur on the curve.
t depends only on the ratio [R]₁/[R]₂
Answer · (a) 100 minutes
If the half-life of a first order reaction is 6.93 × 10³ s, the time required for 75% of the reactant to be consumed is:
t½ = 6.93 × 10³ s, 75% consumed
t
75% consumed leaves 25% = 1/4, which is exactly two half-lives.
t₇₅ = 2 t½
Answer · (a) 1.386 × 10⁴ s
For a first order reaction, the ratio of the time taken for 99.9% completion to that for 90% completion is:
First order
t₉₉.₉ : t₉₀
Both times are proportional to the logarithm of the reciprocal remaining fraction.
t ∝ log([R]₀/[R])
Answer · (a) 3 : 1
The rate constant for a first order reaction is 1.54 × 10⁻³ s⁻¹. Calculate its half-life.
k = 1.54 × 10⁻³ s⁻¹
t½
Direct substitution.
t½ = 0.693/k
Answer · (a) 450 s
A first order reaction with rate constant k has an initial concentration [R]₀. The concentration at time t is best written as:
First order, initial concentration [R]₀
Expression for [R] at time t
Recognise the integrated forms of the three common orders.
[R] = [R]₀e^(−kt)
Answer · (a) [R]₀e^(−kt)
In a first order reaction, 60% of the reactant decomposes in 40 minutes. The rate constant is nearest to:
60% decomposed in 40 min
k
Standard log-form calculation with 40% remaining.
k = (2.303/t)·log(100/(100−x))
Answer · (a) 0.0229 min⁻¹
A first order reaction is carried out with a catalyst that doubles the rate constant. The time required for 75% completion will:
k doubles because of a catalyst
effect on t₇₅
Every characteristic time for a first order reaction is inversely proportional to k.
t₇₅ = 1.386/k
Answer · (a) be halved