Units of k and Determining Order from Units

CHERate, Order & Molecularity

The units of the rate constant uniquely encode the overall order, so examiners give the units and ask for the order (or vice versa). One general formula covers every case and removes the need to memorise each separately.

General units of k (order n)

[k] = (mol\,L^{-1})^{1-n}\,s^{-1} = mol^{1-n}\,L^{n-1}\,s^{-1}

n

= overall order of reaction

Order-specific units

n{=}0:\ mol\,L^{-1}s^{-1} \quad n{=}1:\ s^{-1} \quad n{=}2:\ L\,mol^{-1}s^{-1} \quad n{=}3:\ L^2\,mol^{-2}s^{-1}

obtained by substituting

n

into the general rule

General rule for order $n$ : units of $k = (mol\,L^{-1})^{1-n}\,s^{-1} = mol^{1-n}\,L^{n-1}\,s^{-1}$ .
Zero order: $k$ in $mol\,L^{-1}s^{-1}$ (same units as rate).
First order: $k$ in $s^{-1}$ (concentration-independent units).
Second order: $k$ in $L\,mol^{-1}s^{-1}$ (equivalently $mol^{-1}L\,s^{-1}$ ).
Third order: $k$ in $L^2\,mol^{-2}s^{-1}$ .
Reverse-reading shortcut: count how the concentration unit appears — if $mol\,L^{-1}$ is to the power $(1-n)$ , solve for $n$ to get the order.
Because units pin down the order unambiguously, given units like $L\,mol^{-1}s^{-1}$ you can immediately state the reaction is second order.
The time unit ( $s^{-1}$ ) is always present; only the concentration part changes with order, so focus on the power of $mol\,L^{-1}$ .
Derivation: from $\text{rate}=k[\text{conc}]^n$ , $[k]=\dfrac{[\text{rate}]}{[\text{conc}]^n}=\dfrac{mol\,L^{-1}s^{-1}}{(mol\,L^{-1})^n}$ , giving the general rule.
If $k$ has no concentration units at all (pure $s^{-1}$ ), the reaction is first order; if its units equal those of rate, it is zero order.
Fractional orders give fractional powers of $mol\,L^{-1}$ in $k$ ; e.g. order $1.5$ gives $k$ in $mol^{-1/2}\,L^{1/2}\,s^{-1}$ .
Units can be quoted with any time base (min, h) — the inverse-time factor follows whatever unit was used for the rate measurement.

Where the marks go

Misremembering the exponent: the general rule has $(1-n)$ for the concentration power, not $(n-1)$ — invert carefully.
Confusing zero-order units ( $mol\,L^{-1}s^{-1}$ ) with the rate's units and then misidentifying the order.
Forgetting that first-order $k$ is just $s^{-1}$ with no concentration term — a common slip when matching units to order.
Dropping or mis-signing the power of $L$ versus $mol$ when rewriting $(mol\,L^{-1})^{1-n}$ in expanded form.
Assuming the time unit must be seconds — $min^{-1}$ or $h^{-1}$ are equally valid depending on how rate was measured.

How the board asks it

Define / stateorder-specific units of k
State the units of the rate constant $k$ for (i) a zero order reaction and (ii) a first order reaction.
Give reasonsunits pin down the order unambiguously
The rate constant of a reaction has units $L\,mol^{-1}\,s^{-1}$ . State the order of the reaction and give reasons for your answer.
Derive / prove $[k]=[\text{rate}]/[\text{conc}]^n$
Starting from $\text{rate}=k[A]^n$ , derive the general expression for the units of the rate constant $k$ for a reaction of order $n$ .
Identify / classifypure $s^{-1}$ implies first order
The rate constant of a gaseous decomposition is expressed in $s^{-1}$ . Identify the order of the reaction and write one example of such a reaction.
Assertion–Reasonzero-order units equal rate units
Assertion: For a zero order reaction the rate constant has the same units as the rate, $mol\,L^{-1}\,s^{-1}$ . Reason: The rate of a zero order reaction is independent of the concentration of the reactant. Select the correct option regarding these two statements.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.