Rate Law Calculator
Calculate reaction rates using the rate law expression. Enter the rate constant, reactant concentrations, and reaction orders to find the rate.
Rate Law Calculator
Calculator
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Formula: rate = k [A]^m [B]^n
Worked example โ Rate = 0.005 M/s, Half-life = 13.86 s
Formula
rate = k [A]^m [B]^n
The rate law relates reaction rate to the rate constant k and reactant concentrations [A] and [B] raised to their respective orders m and n. The overall reaction order is m + n. The rate constant k incorporates temperature dependence via the Arrhenius equation.
Worked Examples
Example 1: First Order Reaction
Problem:A first-order reaction has k = 0.05 per second and [A] = 0.1 M. Find the reaction rate.
Solution:rate = k[A]^1 = 0.05 * (0.1)^1 = 0.005 M/s Half-life = ln(2) / 0.05 = 13.86 s
Result:Rate = 0.005 M/s, Half-life = 13.86 s
Example 2: Second Order Two-Reactant Reaction
Problem:For rate = k[A]^1[B]^1, k = 2.5 L/(mol*s), [A] = 0.2 M, [B] = 0.3 M.
Solution:rate = 2.5 * (0.2)^1 * (0.3)^1 = 2.5 * 0.06 = 0.15 M/s Overall order = 1 + 1 = 2
Result:Rate = 0.15 M/s, Overall order = 2
Frequently Asked Questions
What is a rate law in chemistry?
A rate law is a mathematical expression that relates the rate of a chemical reaction to the concentrations of its reactants raised to specific powers called reaction orders. The general form is rate = k[A]^m[B]^n, where k is the rate constant, [A] and [B] are molar concentrations, and m and n are the reaction orders with respect to each reactant. Rate laws must be determined experimentally and cannot be predicted from the balanced equation alone. The overall reaction order is the sum of all individual orders (m + n).
How do you determine the rate constant k?
The rate constant k is determined experimentally by measuring how the reaction rate changes when reactant concentrations are varied. Using the method of initial rates, you run multiple experiments where only one reactant concentration changes at a time. By comparing the ratios of rates to the ratios of concentrations, you can solve for the exponents and then calculate k. The units of k depend on the overall reaction order: for zeroth order k has units of M/s, for first order 1/s, and for second order 1/(M*s).
What is the difference between reaction order and molecularity?
Reaction order is an experimentally determined value that describes how the rate depends on reactant concentrations and can be zero, fractional, or negative. Molecularity refers to the number of reactant molecules that participate in an elementary reaction step and is always a positive integer. For elementary reactions, the reaction order equals the molecularity, but for complex multi-step reactions, the overall rate law is determined by the rate-limiting step and the orders may differ from the stoichiometric coefficients.
How does temperature affect the rate law?
Temperature does not change the form of the rate law or the reaction orders, but it significantly affects the rate constant k. According to the Arrhenius equation, k = A * exp(-Ea/RT), where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is temperature in Kelvin. A 10-degree Celsius increase typically doubles or triples the rate constant. Higher temperatures provide more molecules with sufficient energy to overcome the activation energy barrier.
What is the difference between a differential rate law and an integrated rate law?
The differential rate law, rate = k[A]^m[B]^n, links the instantaneous rate to the concentrations present at that moment, which is what Rate Law Calculator evaluates. The integrated rate law is obtained by integrating that expression with respect to time and instead gives concentration as a function of time, such as ln[A] = ln[A]0 - kt for first order. Use the differential form when you have concentration data, and the integrated form when you have a concentration-versus-time curve.
Why can reaction orders not be read off the balanced equation?
The balanced equation records overall stoichiometry, while the rate law reflects only the steps up to and including the rate-determining step of the mechanism. Orders match coefficients only for a single elementary step. A standard counterexample is the reaction of nitrogen dioxide with fluorine to give nitryl fluoride, where two molecules of NO2 appear in the balanced equation but the experimental rate law is first order in NO2 and first order in F2, because the slow step involves one of each.
How does a catalyst appear in a rate law?
A catalyst is regenerated and so does not appear in the balanced equation, yet it can appear in the rate law with a positive order because it takes part in the rate-determining step. Acid-catalysed ester hydrolysis, for example, is typically first order in hydrogen ion. On a heterogeneous catalyst the order in a reactant often falls toward zero once the surface is saturated, because the rate is then limited by available active sites rather than by concentration.
What is a pseudo-order rate law and when is it used?
If one reactant is present in large excess, its concentration barely changes during the run, so it can be folded into the rate constant. For rate = k[A][B] with B in excess, the observed law becomes rate = k_obs[A] with k_obs = k[B]0, a pseudo-first-order form. This isolation technique is how orders are measured one reactant at a time, and it is why hydrolysis in bulk water is usually treated as first order even though water is a reactant.
How much does the rate change when concentrations are doubled?
Each reactant scales the rate by 2 raised to its own order. Doubling a reactant that is first order doubles the rate, second order quadruples it, and zero order leaves it unchanged. If a reaction is rate = k[A][B] squared, doubling both reactants multiplies the rate by 2 x 4 = 8, which is 2 raised to the overall order of 3. Predicting these factors is the quickest sanity check that the orders entered into the rate law are the ones the data support.
Why might a rate law fit early data but fail later in the reaction?
Rate laws are usually established from initial rates, when products are absent. As the run proceeds the reverse reaction becomes significant near equilibrium, a product may inhibit the catalyst or bind an active site, ionic strength changes in solution can alter the rate constant, and an autocatalytic product can accelerate the reaction. When late-stage data curve away from the fitted law, the mechanism has extra steps that the initial-rate expression does not capture.
References
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Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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