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A question gives you k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹ and asks for the order of the reaction. There is no data table and no rate law, only units. That is enough. This post shows the single formula behind all rate-constant units, how to run it backwards, and how to handle fractional orders, pressure units and unit conversions.
Where does this appear in the syllabus?
The rationalised NCERT Class 12 chapter on Chemical Kinetics (Reprint 2026-27) derives the units of k in Section 3.2.3 and lists them in Table 3.3. The JEE (Main) 2026 syllabus names "rate constant and its units" in Unit 8. Check the syllabus for your exam year.
Where do the units of k come from?
For a rate law rate = k[A]ˣ[B]ʸ with overall order n = x + y, rearrange for k:
k = rate / [A]ˣ[B]ʸ
The rate is always concentration per time, mol L⁻¹ s⁻¹. The denominator has units (mol L⁻¹)ⁿ. So:
units of k = (mol L⁻¹) / (mol L⁻¹)ⁿ × s⁻¹ = (mol L⁻¹)¹⁻ⁿ s⁻¹
OpenStax Chemistry 2e writes the same result as L⁽ⁿ⁻¹⁾ mol⁽¹⁻ⁿ⁾ s⁻¹. Put in the common orders:
n = 0: mol L⁻¹ s⁻¹
n = 1: s⁻¹
n = 2: L mol⁻¹ s⁻¹
n = 3: L² mol⁻² s⁻¹
Worked example 1: reading the order from units
Find the order from each rate constant.
k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹. Write L mol⁻¹ as (mol L⁻¹)⁻¹. So 1 − n = −1, and n = 2. This is NCERT's Example 3.4, which gives second order.
k = 3 × 10⁻⁴ s⁻¹. No concentration term, so 1 − n = 0 and n = 1. First order, also from Example 3.4.
k = 4.0 × 10⁻³ mol L⁻¹ min⁻¹ (a sample value). 1 − n = 1, so n = 0. Zero order.
k = 1.2 × 10⁻² mol⁻¹ᐟ² L¹ᐟ² s⁻¹ (a sample value). Write it as (mol L⁻¹)⁻¹ᐟ². So 1 − n = −½ and n = 3/2.
The numerical value of k plays no part. Only the units carry the order.
What if the rate is measured in pressure?
For gas reactions, NCERT notes that rates can be written with partial pressures, giving units of atm s⁻¹. The same logic holds with atm in place of mol L⁻¹:
units of k = atm¹⁻ⁿ s⁻¹
So k in atm⁻¹ s⁻¹ means second order, and k in atm s⁻¹ means zero order.
To convert between the two kinds of constant, use p = cRT for an ideal gas. Substituting into −dp/dt = kₚpⁿ gives k_c = kₚ(RT)ⁿ⁻¹.
Worked example 2 (sample numbers). A second-order gas reaction has kₚ = 1.0 × 10⁻⁴ atm⁻¹ s⁻¹ at 500 K. With R = 0.0821 L atm mol⁻¹ K⁻¹:
k_c = kₚ × RT = 1.0 × 10⁻⁴ × 0.0821 × 500 = 4.1 × 10⁻³ L mol⁻¹ s⁻¹
Check the units: atm⁻¹ s⁻¹ × L atm mol⁻¹ = L mol⁻¹ s⁻¹, which is second order as expected. For a first-order reaction, n − 1 = 0 and the two constants are equal.
What happens to k when you change units?
Worked example 3. Convert each to seconds.
A first-order k = 0.36 min⁻¹ becomes 0.36 / 60 = 6.0 × 10⁻³ s⁻¹.
A second-order k = 1.2 L mol⁻¹ min⁻¹ becomes 1.2 / 60 = 0.020 L mol⁻¹ s⁻¹.
Now change the concentration unit instead. A first-order k does not change at all, because it has no concentration in its units. This fits the fact that a first-order half-life, t½ = 0.693/k, does not depend on the starting concentration. For any other order, changing from mol L⁻¹ to mol m⁻³ changes the number: since 1 L = 10⁻³ m³, a second-order k of 0.020 L mol⁻¹ s⁻¹ becomes 2.0 × 10⁻⁵ m³ mol⁻¹ s⁻¹.
Which mistakes cost marks?
Reading the power of L instead of mol. In L mol⁻¹ s⁻¹ the power of (mol L⁻¹) is −1, giving n = 2. Using the +1 on L gives n = 0, which is wrong.
Forgetting that "s⁻¹" alone means first order. It is not "no units".
Mixing min and s. Convert time units before using the value in an integrated equation.
Assuming order from the balanced equation. NCERT stresses that order is found by experiment; units are one experimental clue.
Treating pseudo first-order constants as first order. A k′ in s⁻¹ may hide a second-order reaction; see our post on pseudo first-order reactions.
A practice routine
Day 1: Derive the units for n = 0, ½, 1, 3/2, 2 and 3 from the formula, without the table.
Day 2: Ten rate constants with mixed units. Write the order for each in under 20 seconds.
Day 3: Three pressure-unit questions, converting kₚ to k_c and checking units.
Day 4: Combine with initial-rate data, as in our post on finding order from initial rates.
If you keep getting the exponent wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so each unit step is shown.
Tonight, write the units of k for a reaction of order 5/2 and check that you get mol⁻³ᐟ² L³ᐟ² s⁻¹.
