The units of the rate constant are a fingerprint. Show me the units and I will tell you
the order, without seeing the reaction, the data, or anything else. This is the cheapest mark in the
whole chapter — and it takes about twenty minutes to own permanently.
Story track
Think of a shop selling apples. The rate at which you spend money is rupees per minute — that
is fixed, whatever else changes. Now, the "rate constant" is whatever you must multiply your apple
count by to get that spending rate.
If your spending does not depend on how many apples you have at all, then the constant must
already be in rupees per minute by itself. If your spending is proportional to your apple count, the
constant must be rupees per minute per apple — the apples cancel. If it depends on apples
squared, the constant needs per apple squared.
So the constant carries whatever units are needed to fix up the difference between what you
multiply and the answer you want. The higher the order, the more concentration units the constant has
to cancel — and it cancels them by carrying them upside down.
The one rule, derived
Maths track
Start from the rate law and rearrange for k:
Rate = k[A]ⁿ ⟹ k = Rate / [A]ⁿ
Rate always has units of concentration per time — mol L⁻¹ s⁻¹. Concentration to the power n has
units (mol L⁻¹)ⁿ. Divide:
units of k = (mol L⁻¹ s⁻¹)/(mol L⁻¹)ⁿ = (mol L⁻¹)^(1−n) · s⁻¹
That is the whole unit. Everything below is just this formula with different values of n substituted.
Order n
(1 − n)
Units of k
Equivalent form
0
1
mol L⁻¹ s⁻¹
Same as the rate itself
1
0
s⁻¹
Time only — no concentration
2
−1
mol⁻¹ L s⁻¹
L mol⁻¹ s⁻¹
3
−2
mol⁻² L² s⁻¹
L² mol⁻² s⁻¹
1/2
1/2
mol1/2 L−1/2 s⁻¹
Half order
3/2
−1/2
mol−1/2 L1/2 s⁻¹
Acetaldehyde decomposition
The pattern that makes it memorable. Read down the mol
exponents: +1, 0, −1, −2. They just step down by one as the order steps up by one. The L exponents do
the opposite: −1, 0, +1, +2. And the s⁻¹ never changes. Once you see the staircase you do not need
the table.
Reading it backwards — the actual exam skill
NEET almost always gives you the units and asks for the order. Work it in reverse:
No concentration in the units at all (just s⁻¹, min⁻¹, yr⁻¹) → first order
Units identical to a rate (mol L⁻¹ s⁻¹) → zero order
mol to a negative power → order is above 1; the power of mol is (1 − n), so n = 1 − (power of mol)
n = 1 − (exponent of mol in the units of k)
Test it: for mol⁻¹ L s⁻¹ the exponent of mol is −1, so n = 1 − (−1) = 2. For mol⁻² L² s⁻¹ the
exponent is −2, so n = 1 − (−2) = 3. It works every time, including fractions.
Pressure units — the variant that catches people
For gas-phase reactions the concentration is often replaced by partial pressure. The rule survives
untouched; only the concentration unit changes:
units of k = (pressure)^(1−n) · time⁻¹
So for a 3/2 order gas reaction measured in bar and minutes, the units are bar^(1−3/2) min⁻¹ =
bar^(−1/2) min⁻¹. NCERT Q 3.4 asks exactly this for dimethyl ether decomposition, and it is
not in any standard table — you must derive it.
The reordering trap.L mol⁻¹ s⁻¹ and
mol⁻¹ L s⁻¹ are the same units written in a different sequence. Examiners
routinely put one in the question and the other in the options to see whether you are matching
patterns or actually reading. Convert everything to the mol-first form before comparing.
Beyond the textbook
Why first order constants are special. Only for n = 1 do the
units of k contain no concentration. This is why first order rate constants can be compared directly
between reactions, why half-life is independent of concentration, and why radioactive decay constants
are quoted in plain s⁻¹ or yr⁻¹. The three facts are the same fact.
A quick consistency check on any answer. Whatever k you
calculate, multiply it back by the concentration terms. If you do not land on mol L⁻¹ s⁻¹, you have
made an error somewhere. This check catches order mistakes for free.
See it move — 3 animations
Three animations here rather than four — this unit is small enough that a fourth would be padding. Spend your time on the fingerprint matcher instead.
ANIM 1
The unit builder — watch the exponents cancel
Slide the order and watch the algebra happen line by line. The key moment is at n = 1, where the concentration exponent hits zero and mol and L disappear together, leaving nothing but s⁻¹. That vanishing act is the whole reason first order rate constants look different from every other kind.
ANIM 2
The fingerprint matcher — tap a unit, reveal the order
Tap any card above to reveal its order.
Six unit sets, including the two awkward ones (a fractional order and a pressure-based one) that no standard table contains. Cover the reveals and test yourself first; the aim is to reach the point where you recognise these on sight rather than deriving them under exam pressure.
ANIM 3
The exponent staircase
Read the rows rather than memorising the table. As the order climbs 0, 1, 2, 3, the mol exponent steps down +1, 0, −1, −2 and the L exponent steps up in exact mirror image. The s exponent never moves at all. Two staircases and a constant — that is the whole system, and it extends to fractional orders without any extra work.
Formula sheet
Learn the master rule and the reverse rule; the table below them is only there for checking.
Quantity / situation
Formula
When you use it
Master rule ★
units of k = (mol L⁻¹)^(1−n) · time⁻¹
Derive everything from this one line
Reverse rule ★
n = 1 − (exponent of mol in units of k)
When given units and asked for order
Zero order
mol L⁻¹ s⁻¹
Identical to the units of rate
First order
s⁻¹ (or min⁻¹, h⁻¹, yr⁻¹)
No concentration term at all
Second order
mol⁻¹ L s⁻¹ ≡ L mol⁻¹ s⁻¹
Same units, two spellings
Third order
mol⁻² L² s⁻¹
NCERT Q 3.2
Half order
mol^{1/2} L^{−1/2} s⁻¹
Derive; not in the standard table
Three-halves order
mol^{−1/2} L^{1/2} s⁻¹
Acetaldehyde decomposition
Pressure-based form
units of k = (pressure)^(1−n) · time⁻¹
Gas reactions; e.g. bar^{−1/2} min⁻¹
Units of rate
mol L⁻¹ s⁻¹, or atm s⁻¹ for gases
Never contains a bare concentration
Exponent staircase
mol: +1, 0, −1, −2 as n goes 0, 1, 2, 3
L takes the opposite signs; s⁻¹ never changes
Time-unit conversion
min⁻¹ → s⁻¹: divide by 60
Only the time part changes
Consistency check
k × (concentration)ⁿ must equal mol L⁻¹ s⁻¹
Free self-check on any answer
Units → half-life behaviour
s⁻¹ ⇒ t½ fixed ; mol L⁻¹ s⁻¹ ⇒ t½ ∝ [R]₀
Chain the units to the order to the behaviour
39 NEET-type questions with worked solutions
Four graph questions and three assertion–reason questions are included, marked by their coloured left borders. Questions tagged PYQ pattern follow forms that have appeared in NEET/AIPMT papers or come directly from NCERT exercises — exact year attributions are deliberately omitted rather than guessed.
Q01PYQ pattern
Identify the reaction order from the rate constant k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹.
(a) second order
(b) first order
(c) zero order
(d) third order
Given
k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹
Asked
Order
Concept
Match the units against the general rule.
Formula
units of k = (mol L⁻¹)^(1−n)·s⁻¹
Baby steps
Rewrite L mol⁻¹ s⁻¹ in mol-first form: mol⁻¹ L s⁻¹.
The exponent of mol is −1.
Using n = 1 − (exponent of mol): n = 1 − (−1) = 2.
So this is a second order reaction.
Answer · (a) second order
Shortcut · mol⁻¹ → second order. mol⁻² → third order. Just add one to the magnitude.
Q02PYQ pattern
Identify the reaction order from k = 3 × 10⁻⁴ s⁻¹.
(a) first order
(b) zero order
(c) second order
(d) half order
Given
k = 3 × 10⁻⁴ s⁻¹
Asked
Order
Concept
Absence of any concentration unit is the first order signature.
Formula
For n = 1, (mol L⁻¹)^0 = 1, leaving s⁻¹
Baby steps
The units contain time only, with no mol or L anywhere.
This requires the exponent (1 − n) to be zero.
1 − n = 0 gives n = 1.
First order.
Answer · (a) first order
Q03PYQ pattern
For 3NO(g) → N₂O(g) with Rate = k[NO]², the dimensions of the rate constant are:
(a) mol⁻¹ L s⁻¹
(b) s⁻¹
(c) mol L⁻¹ s⁻¹
(d) mol⁻² L² s⁻¹
Given
Rate = k[NO]², so order = 2
Asked
Units of k
Concept
Read the order from the exponent, then apply the unit rule.
Formula
units = (mol L⁻¹)^(1−n)·s⁻¹
Baby steps
The exponent in the rate law is 2, so n = 2 (ignore the coefficient 3).
1 − n = −1.
(mol L⁻¹)⁻¹ = mol⁻¹ L.
Units = mol⁻¹ L s⁻¹.
Answer · (a) mol⁻¹ L s⁻¹
Q04PYQ pattern
For H₂O₂(aq) + 3I⁻(aq) + 2H⁺ → 2H₂O(l) + I₃⁻ with Rate = k[H₂O₂][I⁻], the units of k are:
(a) mol⁻¹ L s⁻¹
(b) s⁻¹
(c) mol⁻² L² s⁻¹
(d) mol L⁻¹ s⁻¹
Given
Rate = k[H₂O₂][I⁻]
Asked
Units of k
Concept
Overall order is the sum of exponents, here 1 + 1 = 2.
Formula
units = (mol L⁻¹)^(1−n)·s⁻¹
Baby steps
Order = 1 + 1 = 2. H⁺ does not appear in the rate law, so it contributes nothing.
1 − n = 1 − 2 = −1.
Units = mol⁻¹ L s⁻¹.
Answer · (a) mol⁻¹ L s⁻¹
Shortcut · Ignore every species missing from the rate law, no matter how large its coefficient in the equation.
Q05PYQ pattern
For CH₃CHO(g) → CH₄(g) + CO(g) with Rate = k[CH₃CHO]^{3/2}, the units of k are:
(a) mol^{−1/2} L^{1/2} s⁻¹
(b) mol^{1/2} L^{−1/2} s⁻¹
(c) mol⁻¹ L s⁻¹
(d) s⁻¹
Given
Order = 3/2
Asked
Units of k
Concept
The rule handles fractional orders without modification.
Formula
units = (mol L⁻¹)^(1−n)·s⁻¹
Baby steps
1 − n = 1 − 3/2 = −1/2.
(mol L⁻¹)^{−1/2} = mol^{−1/2} L^{1/2}.
Units = mol^{−1/2} L^{1/2} s⁻¹.
Answer · (a) mol^{−1/2} L^{1/2} s⁻¹
Q06PYQ pattern
The decomposition of dimethyl ether has Rate = k(p_{CH₃OCH₃})^{3/2}. If pressure is in bar and time in minutes, the units of rate and rate constant respectively are:
(a) bar min⁻¹ and bar^{−1/2} min⁻¹
(b) bar min⁻¹ and bar^{3/2} min⁻¹
(c) bar⁻¹ min and bar^{1/2} min⁻¹
(d) mol L⁻¹ min⁻¹ and bar^{−1/2} min⁻¹
Given
Order 3/2, pressure in bar, time in minutes
Asked
Units of rate and of k
Concept
Rate is always change of the measured quantity per unit time; the unit rule uses that same quantity.
Formula
units of k = (pressure)^(1−n)·time⁻¹
Baby steps
Rate is a change in pressure per unit time, so its units are bar min⁻¹.
k = Rate/(pressure)^{3/2}, so units = bar min⁻¹ / bar^{3/2}.
= bar^(1 − 3/2) min⁻¹ = bar^{−1/2} min⁻¹.
Answer · (a) bar min⁻¹ and bar^{−1/2} min⁻¹
Shortcut · Swap 'mol L⁻¹' for 'bar' throughout and the rule works unchanged. Nothing else needs adjusting.
Q07
A rate constant has units mol⁻² L² s⁻¹. The order of the reaction is:
(a) 3
(b) 2
(c) 1
(d) 0
Given
k in mol⁻² L² s⁻¹
Asked
Order
Concept
Use the reverse rule.
Formula
n = 1 − (exponent of mol)
Baby steps
The exponent of mol is −2.
n = 1 − (−2) = 3.
Third order.
Answer · (a) 3
Q08PYQ pattern
For a zero order reaction, the units of the rate constant are:
(a) mol L⁻¹ s⁻¹
(b) s⁻¹
(c) mol⁻¹ L s⁻¹
(d) dimensionless
Given
n = 0
Asked
Units of k
Concept
For zero order the rate equals k, so their units must match.
Formula
Rate = k[A]⁰ = k
Baby steps
1 − n = 1 − 0 = 1.
(mol L⁻¹)¹ = mol L⁻¹.
Units = mol L⁻¹ s⁻¹ — identical to the units of the rate itself, exactly as expected when Rate = k.
Answer · (a) mol L⁻¹ s⁻¹
Q09
If k = 0.5 mol⁻¹ L min⁻¹, the half-life of the reaction depends on the initial concentration as:
(a) t½ ∝ 1/[R]₀
(b) t½ is independent of [R]₀
(c) t½ ∝ [R]₀
(d) t½ ∝ [R]₀²
Given
k in mol⁻¹ L min⁻¹
Asked
Dependence of half-life on initial concentration
Concept
Get the order from the units, then apply the general half-life proportionality.
Formula
n = 1 − (exponent of mol) ; t½ ∝ 1/[R]₀^(n−1)
Baby steps
Exponent of mol is −1, so n = 1 − (−1) = 2. The reaction is second order.
General relation: t½ ∝ 1/[R]₀^(n−1).
With n = 2, the exponent (n − 1) = 1.
So t½ ∝ 1/[R]₀.
Answer · (a) t½ ∝ 1/[R]₀
Shortcut · Units → order → half-life behaviour. This three-step chain answers several questions at once.
Q10
Which of the following sets of units is NOT a valid set for any rate constant?
(a) mol L s⁻¹
(b) s⁻¹
(c) mol L⁻¹ s⁻¹
(d) mol⁻¹ L s⁻¹
Given
Four candidate unit sets
Asked
The invalid one
Concept
The mol and L exponents must always be equal and opposite.
Formula
units = (mol L⁻¹)^(1−n)·s⁻¹ — so mol and L carry opposite exponents
Baby steps
In (mol L⁻¹)^p the exponent of mol is p and of L is −p, always opposite in sign.
Option (a) has mol +1 and L +1, both positive. That can never arise, so it is invalid.
Answer · (a) mol L s⁻¹
Shortcut · Check the signs of the mol and L exponents. If they match rather than oppose, the unit set is impossible.
Q11
A first order rate constant is given as 0.0693 min⁻¹. Expressed in s⁻¹ it is:
(a) 1.155 × 10⁻³ s⁻¹
(b) 4.158 s⁻¹
(c) 0.0693 s⁻¹
(d) 1.155 × 10⁻² s⁻¹
Given
k = 0.0693 min⁻¹
Asked
k in s⁻¹
Concept
A per-minute quantity is converted to per-second by dividing by 60.
Formula
k(s⁻¹) = k(min⁻¹)/60
Baby steps
One minute contains 60 seconds, so a quantity per minute is 60 times a quantity per second.
k = 0.0693/60.
= 1.155 × 10⁻³ s⁻¹.
Answer · (a) 1.155 × 10⁻³ s⁻¹
Shortcut · Per-time constants get smaller when converted from minutes to seconds. Multiplying instead of dividing is the classic slip.
Q12
The rate constant for a reaction has units mol^{1/2} L^{−1/2} s⁻¹. The order is:
(a) 1/2
(b) 3/2
(c) 2
(d) 1
Given
k in mol^{1/2} L^{−1/2} s⁻¹
Asked
Order
Concept
Apply the reverse rule with a fractional exponent.
Formula
n = 1 − (exponent of mol)
Baby steps
The exponent of mol is +1/2.
n = 1 − 1/2 = 1/2.
The reaction is half order.
Answer · (a) 1/2
Q13
Given Rate = k[A]²[B], the units of k are:
(a) mol⁻² L² s⁻¹
(b) mol⁻¹ L s⁻¹
(c) s⁻¹
(d) mol L⁻¹ s⁻¹
Given
Rate = k[A]²[B]
Asked
Units of k
Concept
Add the exponents to get the overall order first.
Formula
n = 2 + 1 = 3
Baby steps
Overall order = 2 + 1 = 3.
1 − n = −2.
(mol L⁻¹)⁻² = mol⁻² L².
Units = mol⁻² L² s⁻¹.
Answer · (a) mol⁻² L² s⁻¹
Q14
For a reaction with k measured in atm⁻¹ s⁻¹, the order is:
(a) 2
(b) 1
(c) 0
(d) 3
Given
k in atm⁻¹ s⁻¹
Asked
Order
Concept
Pressure replaces concentration but the exponent rule is identical.
Formula
units = (pressure)^(1−n)·time⁻¹
Baby steps
The exponent of the pressure unit is −1.
So 1 − n = −1.
n = 2, second order.
Answer · (a) 2
Q15
The rate constant of a reaction is 5 × 10⁻² mol L⁻¹ s⁻¹. Which statement follows?
(a) The rate is constant and equal to 5 × 10⁻² mol L⁻¹ s⁻¹ while reactant remains
(b) The reaction is first order
(c) t½ is independent of initial concentration
(d) The rate doubles when the concentration doubles
Given
k in mol L⁻¹ s⁻¹
Asked
The valid conclusion
Concept
Those units identify zero order, where the rate equals k.
Formula
Rate = k[A]⁰ = k
Baby steps
Units matching the rate itself means n = 0.
For zero order, Rate = k, a constant independent of concentration.
So the rate stays at 5 × 10⁻² mol L⁻¹ s⁻¹ until the reactant runs out.
Options (b), (c) and (d) all describe first order behaviour and are wrong here — for zero order t½ = [R]₀/2k does depend on [R]₀.
Answer · (a) The rate is constant and equal to 5 × 10⁻² mol L⁻¹ s⁻¹ while reactant remains
Q16Graph
A plot of rate against concentration gives a horizontal straight line. The units of the rate constant for this reaction are:
(a) mol L⁻¹ s⁻¹
(b) s⁻¹
(c) mol⁻¹ L s⁻¹
(d) mol⁻² L² s⁻¹
Given
Rate versus concentration is horizontal
Asked
Units of k
Concept
Read the order off the graph shape, then apply the unit rule.
Formula
flat plot ⇒ n = 0 ; units = (mol L⁻¹)^1·s⁻¹
Baby steps
A horizontal line means the rate does not change with concentration, so n = 0.
1 − n = 1.
Units = mol L⁻¹ s⁻¹, the same as the rate itself.
Answer · (a) mol L⁻¹ s⁻¹
Q17Graph
A plot of log(rate) against log[A] is a straight line of slope 1. The units of the rate constant are:
(a) s⁻¹
(b) mol L⁻¹ s⁻¹
(c) mol⁻¹ L s⁻¹
(d) mol⁻² L² s⁻¹
Given
log–log plot of slope 1
Asked
Units of k
Concept
On a log–log plot the slope is the order.
Formula
log Rate = log k + n log[A] ⇒ slope = n = 1
Baby steps
The slope of a log(rate) vs log[A] plot equals the order, so n = 1.
1 − n = 0.
(mol L⁻¹)⁰ = 1, leaving units of s⁻¹.
Answer · (a) s⁻¹
Q18Graph
For a reaction, [R] plotted against t gives a falling straight line. The rate constant of this reaction has units:
(a) mol L⁻¹ s⁻¹
(b) s⁻¹
(c) mol⁻¹ L s⁻¹
(d) mol^{1/2} L^{−1/2} s⁻¹
Given
Concentration versus time is a falling straight line
Asked
Units of k
Concept
A linear [R] vs t plot is the signature of zero order.
Formula
[R] = [R]₀ − kt ⇒ n = 0
Baby steps
Only zero order gives a straight [R] vs t plot; first order gives a curve.
So n = 0, and the slope of this line is −k.
Units = (mol L⁻¹)^(1−0) s⁻¹ = mol L⁻¹ s⁻¹.
Consistency check: the slope of a concentration-versus-time graph must itself have units of concentration per time. It matches.
Answer · (a) mol L⁻¹ s⁻¹
Shortcut · The slope of any [R] vs t line has units mol L⁻¹ s⁻¹ by construction — which instantly confirms the zero order answer.
Q19Graph
A plot of ln[R] against t is a falling straight line of slope −0.05 s⁻¹. The units of the slope confirm the reaction is:
(a) first order
(b) zero order
(c) second order
(d) half order
Given
ln[R] vs t straight, slope = −0.05 s⁻¹
Asked
Order
Concept
Both the graph shape and the units of the slope point to the same conclusion.
Formula
ln[R] = ln[R]₀ − kt ; slope = −k
Baby steps
A straight ln[R] vs t plot is the first order signature.
The slope equals −k, so k = 0.05 s⁻¹.
Units of s⁻¹ with no concentration term confirm first order independently.
Two independent routes to the same answer — worth using as a self-check in the exam.
Answer · (a) first order
Q20Assertion–Reason
Assertion (A): The units of the rate constant of a first order reaction contain no concentration term. Reason (R): For a first order reaction the exponent (1 − n) in the general unit expression becomes zero.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about first order rate constant units
Asked
Truth values and explanation
Concept
The exponent vanishing is exactly what removes the concentration units.
Formula
units of k = (mol L⁻¹)^(1−n)·time⁻¹
Baby steps
Check A: first order constants are quoted as s⁻¹, min⁻¹ or yr⁻¹, with no mol or L. A is true.
Check R: putting n = 1 gives 1 − n = 0, and any quantity to the power zero equals 1. R is true.
Does R explain A? Yes — the concentration factor collapsing to 1 is precisely why no concentration unit survives.
This is also why first order rate constants can be compared directly across different reactions.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q21Assertion–Reason
Assertion (A): A rate constant with units mol L⁻¹ s⁻¹ belongs to a zero order reaction. Reason (R): For a zero order reaction the rate equals the rate constant, so the two must share the same units.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about zero order rate constant units
Asked
Truth values and explanation
Concept
Rate = k for zero order, forcing an exact unit match.
Formula
Rate = k[A]⁰ = k
Baby steps
Check A: substituting n = 0 into the unit rule gives mol L⁻¹ s⁻¹. A is true.
Check R: since [A]⁰ = 1, the rate law reduces to Rate = k, so they must have identical units. R is true.
Does R explain A? Yes — the equality of rate and rate constant is the direct cause of the unit match.
This is the quickest way to recall the zero order units without any algebra.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q22Assertion–Reason
Assertion (A): The units L mol⁻¹ s⁻¹ and mol⁻¹ L s⁻¹ describe rate constants of different orders. Reason (R): The order of a reaction depends on the sequence in which the units are written.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is false
Given
Statements about unit ordering
Asked
Truth values
Concept
Multiplication is commutative, so the written order of units carries no information.
Formula
L mol⁻¹ s⁻¹ ≡ mol⁻¹ L s⁻¹ — both correspond to n = 2
Baby steps
Check A: both expressions contain mol to the power −1, L to the power +1 and s to the power −1. They are identical, both second order. A is false.
Check R: the sequence in which units are written is purely cosmetic and cannot affect any physical quantity. R is false.
So both statements are false — option (d).
This is a deliberately constructed trap; always compare exponents rather than appearances.
Answer · (d) Both A and R are false
Shortcut · Rewrite every unit set in mol-first form before comparing anything. It removes this trap permanently.
Q23
For a reaction of order n, if concentration is measured in mol L⁻¹ and time in hours, the units of k are:
(a) (mol L⁻¹)^(1−n) h⁻¹
(b) (mol L⁻¹)^n h⁻¹
(c) (mol L⁻¹)^(n−1) h⁻¹
(d) (mol L⁻¹) h^(1−n)
Given
General order n, time in hours
Asked
General units of k
Concept
The time unit simply follows whatever the question uses; the concentration exponent is unchanged.
Formula
units of k = (concentration)^(1−n)·(time)⁻¹
Baby steps
k = Rate/[A]ⁿ, and Rate has units (mol L⁻¹) h⁻¹.
Dividing by (mol L⁻¹)ⁿ subtracts n from the concentration exponent.
Result: (mol L⁻¹)^(1−n) h⁻¹.
Note the time exponent is always −1 regardless of order.
Answer · (a) (mol L⁻¹)^(1−n) h⁻¹
Q24
A student writes the rate constant of a reaction as 4.2 mol L⁻¹ min⁻¹. The half-life of this reaction is:
(a) [R]₀/2k
(b) 0.693/k
(c) 1/k[R]₀
(d) independent of [R]₀
Given
k in mol L⁻¹ min⁻¹
Asked
Half-life expression
Concept
Identify the order from the units, then use the matching half-life formula.
Formula
units mol L⁻¹ time⁻¹ ⇒ n = 0 ⇒ t½ = [R]₀/2k
Baby steps
The units match those of a rate, so n = 0.
For zero order the integrated law is [R] = [R]₀ − kt.
t½ = [R]₀/2k, which does depend on the initial concentration.
Answer · (a) [R]₀/2k
Shortcut · Units first, then order, then the correct half-life formula. Skipping the middle step is how the wrong formula gets used.
Q25
Which order of reaction has a rate constant whose numerical value is unaffected by the choice of concentration unit (mol L⁻¹ versus mol dm⁻³ versus mol m⁻³)?
(a) first order
(b) zero order
(c) second order
(d) third order
Given
Comparison across concentration units
Asked
The order that is unaffected
Concept
Only when the units of k contain no concentration term does its value become unit-independent.
Formula
units of k = (concentration)^(1−n)·time⁻¹
Baby steps
For any order other than 1, the units of k contain concentration raised to a non-zero power.
Changing the concentration unit therefore rescales the numerical value of k.
For n = 1 the concentration exponent is zero, so no rescaling occurs.
Note mol L⁻¹ and mol dm⁻³ happen to be identical anyway, but mol m⁻³ differs by a factor of 1000 — and only a first order k survives that unchanged.
Answer · (a) first order
Shortcut · This is why first order constants (and radioactive decay constants) are quoted without any concentration reference.
Q26
The rate constant of a reaction is 2.0 × 10⁻³ mol⁻¹ L s⁻¹ and the concentration of the single reactant is 0.5 mol L⁻¹. The rate is:
(a) 5.0 × 10⁻⁴ mol L⁻¹ s⁻¹
(b) 1.0 × 10⁻³ mol L⁻¹ s⁻¹
(c) 4.0 × 10⁻³ mol L⁻¹ s⁻¹
(d) 2.0 × 10⁻³ mol L⁻¹ s⁻¹
Given
k = 2.0 × 10⁻³ mol⁻¹ L s⁻¹, [A] = 0.5 mol L⁻¹
Asked
Rate
Concept
The units of k reveal the order, which tells you what power to raise the concentration to.
Formula
units mol⁻¹ L s⁻¹ ⇒ n = 2 ⇒ Rate = k[A]²
Baby steps
Exponent of mol is −1, so n = 1 − (−1) = 2. Second order.