Reaction Rate Law Calculator
Calculate the instantaneous reaction rate from the rate law v = k[A]^a[B]^b… Enter the rate constant, the concentration and order of each reactant, and get the rate, overall order and correct unit of k.
Calculate Reaction Rate from the Rate Law
Rate Law
Where:
- \(v\) — reaction rate, the change in concentration per unit time (mol/L·s or the chosen units)
- \(k\) — rate constant, a temperature-dependent experimental value
- \([A], [B], [C]\) — molar concentrations of the reactants
- \(a, b, c\) — reaction orders with respect to each reactant (determined experimentally)
The overall reaction order is the sum of the exponents: \(n = a + b + c + \cdots\). The rate constant carries units that depend on \(n\): \(k\) has units of \((\text{concentration})^{1-n}\,(\text{time})^{-1}\).
Rate Law – Calculation Example
Example: a reaction with rate law \(v = k[A][B]^2\), where \(k = 0.042\;\mathrm{M^{-2}\,s^{-1}}\), \([A] = 0.50\;\mathrm{M}\) and \([B] = 0.20\;\mathrm{M}\).
\[ v = 0.042 \times (0.50)^{1} \times (0.20)^{2} = 0.042 \times 0.50 \times 0.040 \]
\[ v = 8.4 \times 10^{-4}\;\mathrm{M/s} \]
Result: \(v = 8.4 \times 10^{-4}\) mol/(L·s). The overall order is \(n = 1 + 2 = 3\), so the unit of \(k\) is \(\mathrm{M^{-2}\,s^{-1}}\) (equivalently \(\mathrm{L^{2}\,mol^{-2}\,s^{-1}}\)).
How to use the calculator
- Enter the Rate constant (k) — the experimental value of \(k\) at the reaction temperature. Use the units that match your concentration and time choices; the correct unit of \(k\) is shown in the results.
- Choose the Concentration unit (M, mM or µM) and the Time unit (s, min or h) for your data.
- List the reactants in Concentration × Order, one per line or separated by semicolons — for example
0.50 × 1; 0.20 × 2for a rate law \(v = k[A][B]^2\). A plain concentration without an order counts as first order; zero, fractional and negative orders are all accepted. - Read the Reaction Rate in your units and converted to mol/(L·s), together with the substituted rate law, each reactant's contribution, the overall order and the unit of \(k\).
What the rate law describes
The rate law expresses how the instantaneous rate of a chemical reaction depends on the concentrations of its reactants. For a general reaction, the rate is proportional to each concentration raised to a power: \(v = k[A]^a[B]^b \cdots\). Importantly, the exponents (the orders) cannot be read off the balanced equation — except for elementary steps — they must be determined experimentally, classically from initial-rate measurements.
Each order tells you how sensitively the rate responds to that reactant. If \([A]\) doubles and the rate stays the same, the reaction is zero order in A. If the rate doubles, it is first order (rate \(\propto [A]\)). If the rate quadruples, it is second order (rate \(\propto [A]^2\)). The calculator's contribution table shows exactly this effect for each reactant you enter, including fractional and negative orders, where an inhibitor slows the reaction as its concentration rises.
The rate constant \(k\) absorbs everything else: temperature, catalyst and medium. Its numerical value is meaningless without its units, which follow from the overall order \(n\). For a first-order reaction \(k\) has units of s⁻¹; for second order, M⁻¹·s⁻¹; for third order, M⁻²·s⁻¹. This is why the calculator reports the unit of \(k\) alongside the rate — it lets you check that the \(k\) you entered is consistent with the orders you assumed.
Keep in mind that the result is the instantaneous rate at the concentrations entered. As the reaction proceeds, reactants are consumed and the rate falls. This is why chemists usually work with initial rates, measured before concentrations change appreciably.
Assumptions, limitations and tips
- Fixed temperature: \(k\) and the orders are valid only at the temperature at which they were measured. The rate constant rises strongly with temperature (Arrhenius behaviour); to compare \(k\) at two temperatures, use our Van 't Hoff Equation Calculator.
- Not for equilibrium mixtures: the rate law describes the forward kinetics, not the equilibrium composition. For gas-phase equilibria, see the Kp to Kc Calculator.
- First-order reactions only: simple half-life formulas apply only when the overall order is 1. For other orders the half-life depends on the starting concentration — the Reaction Half-Life Calculator handles those cases.
- Units of k must be consistent: if you enter concentrations in mM, the rate is returned in mM per time unit and \(k\) must correspondingly use mM. When in doubt, convert everything to M and s — the calculator also shows the rate in mol/(L·s).
- Orders need not be integers: complex or chain mechanisms often give fractional orders, and catalysts or inhibitors can give negative ones. Both are handled correctly here.
Quick reference: order and the effect of doubling a concentration
| Order in reactant X | Rate law term | Rate change if [X] doubles | Typical unit of k (n = this order only) |
|---|---|---|---|
| 0 | \([X]^0 = 1\) | no change | M·s⁻¹ |
| 1 | \([X]\) | × 2 | s⁻¹ |
| 2 | \([X]^2\) | × 4 | M⁻¹·s⁻¹ |
| 3 | \([X]^3\) | × 8 | M⁻²·s⁻¹ |
| 0.5 | \([X]^{0.5}\) | × 1.41 | M0.5·s⁻¹ |
| −1 (inhibitor) | \([X]^{-1}\) | × 0.5 (rate halves) | M²·s⁻¹ |
For a multi-reactant rate law, the effects multiply: in the example above, doubling both \([A]\) and \([B]\) increases the rate by \(2^1 \times 2^2 = 8\).
Frequently Asked Questions
How do I find the reaction orders experimentally?
Use the method of initial rates. Run several trials where you change one reactant's concentration at a time and measure the initial rate. If doubling \([A]\) multiplies the rate by 2, the order in A is 1; by 4, the order is 2; no change, the order is 0. In general, \(a = \log(r_2/r_1) / \log([A]_2/[A]_1)\). Once all orders are known, solve the rate law for \(k\).
Can I get the orders from the balanced equation?
No. Reaction orders must be determined experimentally and only coincidentally match the stoichiometric coefficients — matching is guaranteed only for elementary (single-step) reactions. For example, \(2N_2O_5 \to 4NO_2 + O_2\) is first order in \(N_2O_5\), not second. Never assume \(a\) and \(b\) from the coefficients of a multi-step overall equation.
How do I convert the rate constant k to different units?
Convert the concentration and time parts separately. Example: \(k = 0.5\;\mathrm{M^{-1}s^{-1}}\) to \(\mathrm{L\,mol^{-1}\,min^{-1}}\): since 1 min = 60 s, \(k = 0.5 \times 60 = 30\;\mathrm{L\,mol^{-1}\,min^{-1}}\). If the concentration unit changes (M to mM), the numeric value also changes because the exponent \(1-n\) applies. Easiest fix: convert everything to M and s before using the calculator.
What does a negative or fractional reaction order mean?
A negative order means the substance slows the reaction — it acts as an inhibitor, so raising its concentration decreases the rate. Fractional orders (e.g. 0.5 or 1.5) typically signal a complex multi-step mechanism; for instance the decomposition of acetaldehyde has an order of 3/2. Both are physically valid and handled correctly by this calculator — they are not input errors.
Why is my calculated rate different from my experimental measurement?
Check three things first: (1) the temperature — \(k\) is only valid where it was measured and changes strongly with temperature; (2) the units of \(k\) — a \(k\) in \(\mathrm{M^{-1}s^{-1}}\) used with concentrations in mM gives rates off by 1000×; (3) whether you measured an average rate over time, which is always smaller than the instantaneous initial rate because concentrations fall as the reaction proceeds.
Is the reaction rate always a positive number?
Yes — the rate reported here is the magnitude of the concentration change, \(v = |d[X]/dt|\) divided by the stoichiometric coefficient, so it is positive by convention. If your experiment shows a reactant concentration decreasing, the rate of disappearance \(-d[A]/dt\) is what equals \(k[A]^a[B]^b\). The calculator requires a positive \(k\) and positive concentrations; only the orders may be zero, fractional or negative.
How do temperature and catalysts affect the rate constant?
Both affect \(k\), not the orders in most simple cases. Temperature follows the Arrhenius relation \(k = Ae^{-E_a/RT}\), so \(k\) typically rises sharply — a 10 °C increase can double or triple it. A catalyst raises \(k\) by providing a lower-energy pathway. If your \(k\) was measured at 25 °C, do not use it at 40 °C without correcting it.