Units for k: Rate Constant Explained

This calculator works out the units of the rate constant k for a chemical reaction from its overall order, and it also converts between common unit sets (mol/L, atm, and pressure-based). You enter the overall reaction order and choose your concentration and time units, and you get back the exact units of k as a single fraction, ready to use in your rate law. The result is shown both as a symbolic expression (like L mol�¹ s�¹) and as a plain-language reading of the units.

Formula

The rate constant k appears in the rate law for a reaction:

rate = k [A]^m [B]ⁿ …

where rate is the reaction rate (usually in mol L�¹ s�¹ or mol L�¹ min�¹), [A] and [B] are concentrations (mol/L or pressure units), and m, n, … are the orders with respect to each reactant. The overall order is the sum of the exponents: order = m + n + ….

To find the units of k, you solve the rate law for k:

k = rate / ([A]^m [B]ⁿ …)

Because the denominator has the units of concentration raised to the overall order, the units of k are:

units of k = (concentration units)^{1 − order} / (time units)

For example, with concentration in mol/L and time in seconds, and an overall order of 2:

units of k = (mol/L)¹⁻² / s = (mol/L)⁻¹ / s = L mol�¹ s�¹

If you use pressure units (like atm) for concentration, the same formula works: for a second-order reaction, k would be atm�¹ s�¹.

Worked Example

Using the calculator defaults: overall order = 2, concentration unit = mol/L, time unit = seconds.

1. Start with the formula: units of k = (concentration)^{1 − order} / time.
2. Plug in the order: (mol/L)^{1 − 2} / s = (mol/L)⁻¹ / s.
3. Simplify the negative exponent: (mol/L)⁻¹ = L/mol.
4. So units of k = (L/mol) / s = L mol�¹ s�¹.

The calculator shows L mol�¹ s�¹ for these inputs, matching the step-by-step result.

Why Units Matter

The units of k are not arbitrary—they are forced by the rate law. The rate always has units of concentration per time (e.g., mol L�¹ s�¹). When you divide that by the concentration terms raised to the reaction order, the exponents must cancel to leave the correct units for k. If you write a rate law with the wrong units for k, the numerical value of k will be wrong, and any calculation using it (like half-life or concentration at a given time) will be off.

For a zero‑order reaction, the rate does not depend on concentration, so k has the same units as the rate: mol L�¹ s�¹ (or M/s). For a first‑order reaction, k has units of s�¹ (or min�¹, h�¹). For a second‑order reaction, k has units of L mol�¹ s�¹ (or M�¹ s�¹). For a third‑order reaction, k has units of L² mol�² s�¹. The pattern continues: each increase in order adds one concentration unit to the denominator (and thus a negative exponent on the concentration).

How to Choose the Right Units

When you are given a rate constant, the units tell you the overall order. For example, if k = 0.034 L mol�¹ min�¹, you know the reaction is second order (because the concentration exponent is −1, so 1 − order = −1, giving order = 2). Conversely, if you know the order, you can predict the units.

To use this calculator, simply enter the overall order (a positive integer, often 0, 1, 2, or 3) and select the concentration and time units you are working with. The calculator will output the units of k in that system. If you are converting between unit sets, you can run the calculator twice—once with the original units and once with the desired units—to see the difference in the symbolic form. Note that the calculator does not convert the numerical value of k; it only shows the units. To convert a numerical k value, you would need to apply conversion factors for concentration and time (e.g., 1 M = 1 mol/L, 1 atm = 101.325 kPa, 1 min = 60 s).

Common Mistakes

• Forgetting the exponent on concentration: The formula uses (concentration)^1−order. For order 2, that is concentration⁻¹, not concentration⁰.
• Mixing time units: If the rate is given in mol L�¹ s�¹ but k is quoted in min�¹, you must convert time before using k in calculations.
• Using pressure and molarity interchangeably: For gases, concentration can be expressed as pressure (e.g., atm). The units of k will differ (e.g., atm�¹ s�¹ vs. L mol�¹ s�¹). Always check which concentration measure the rate law uses.
• Assuming k is unitless: k always has units unless the reaction is zero order and the rate is dimensionless (rare).
• Rounding too early: When you calculate a numerical k from experimental data, keep extra digits until the final answer to avoid rounding errors.

Limits of the Formula

The formula units of k = (concentration)^1−order / time assumes that the rate law is of the form rate = k [A]^m[B]ⁿ… with constant orders. It does not apply to reactions with more complex rate laws, such as those with fractional orders or those that involve a pre‑equilibrium. For elementary reactions, the order is usually a small integer, but for complex mechanisms, the overall order can be fractional. The formula still works for fractional orders—just plug in the fractional order and the calculator will give the corresponding units (e.g., concentration^0.5).

Also, the formula assumes that the concentration units are consistent across all species. If you use mol/L for some and atm for others, you must convert to a single set before applying the formula.

Frequently Asked Questions

What are the units of k for a first-order reaction?

For a first-order reaction, the units of k are simply 1/time, such as s�¹, min�¹, or h�¹. This is because the rate (mol L�¹ s�¹) divided by concentration (mol L�¹) leaves only time�¹.

Why does k have different units for different orders?

Because the rate law must be dimensionally consistent. The rate always has units of concentration/time. To get that, k must carry whatever units are needed to cancel the concentration terms raised to the reaction order. Higher orders require more concentration units in the denominator, so k gets a higher negative exponent on concentration.

Can k be unitless?

Yes, in one special case: if the reaction is zero order and the rate is defined as a dimensionless quantity (e.g., a fractional conversion per time), then k can be unitless. In most chemical kinetics, however, rate has units of concentration/time, so k for a zero-order reaction has units of concentration/time (e.g., mol L�¹ s�¹).

How do I convert k from L mol�¹ s�¹ to L mol�¹ min�¹?

Since the time unit changes from seconds to minutes, you multiply by 60 (because 1 min = 60 s). For example, k = 0.02 L mol�¹ s�¹ becomes 0.02 × 60 = 1.2 L mol�¹ min�¹. Note that the concentration part stays the same.

Does the units of k depend on temperature?

No, the units of k depend only on the reaction order and the units you choose for concentration and time. The numerical value of k changes with temperature (according to the Arrhenius equation), but its units remain the same for a given order and unit system.

What if my reaction order is not an integer?

The formula still works. For example, if the overall order is 1.5, then the units of k would be (concentration)^−0.5 / time, such as L^0.5 mol^−0.5 s�¹. The calculator handles fractional orders just fine.

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in n | Overall reaction order | - | 2 | 0
sel conc | Concentration unit | mol/L=1, atm=1, kPa=1, mmHg=1
sel time | Time unit | s=1, min=1, h=1
out units | Units of k | - | conc^(1-n) / time | 0
out reading | Read as | - | conc^(1-n) per time | 0