Reaction Rate and Kinetics, Explained
After reading this you can predict how much reactant is left after a given time, find a half-life for any of the three simple reaction orders, and use the Arrhenius equation to see how a rate constant climbs with temperature.
What chemical kinetics measures
Thermodynamics tells you whether a reaction can happen. Kinetics tells you how fast. Those are different questions. A mixture of hydrogen and oxygen sits stable in a bottle for years even though burning it releases a lot of energy, because at room temperature almost no collisions carry enough energy to start the reaction.
The central quantity is the rate constant k. It links the speed of a reaction to the concentrations of the reactants. Once you know k and the reaction order, you can integrate the rate law and get a clean formula for concentration as a function of time.
Here is the hook. A first-order reaction with k = 0.05\ \text{s}^{-1} starting at 1\ \text{mol/L} falls to about 0.607 mol/L after 10 seconds, and its half-life is about 13.9 s no matter what concentration you start from. That concentration independence is the signature of first order, and it is worth understanding why.
When to use these models, and when not
The integrated rate laws here assume a single reactant governs the rate, with a fixed order of 0, 1 or 2. That covers a great many textbook reactions and a fair number of real ones: radioactive decay (first order), many enzyme-saturated processes (zero order), and simple bimolecular collisions (second order).
They break down when the order is fractional, when the mechanism has several slow steps, when the reaction reverses appreciably, or when temperature drifts during the run. If your data plot does not straighten out under any of the three linear forms below, the reaction is not one of these simple cases, and forcing a fit will mislead you.
Order is an experimental fact, not something you read off the balanced equation. A reaction written as 2A → products can turn out to be first order in A. You determine the order by measuring how rate changes with concentration, then pick the matching formula.
The three integrated rate laws
Each order comes from integrating a differential rate law. For a reactant A with concentration [A], the three forms are as follows.
Zero order. The rate does not depend on concentration at all, so the concentration falls in a straight line with slope -k. Here [A]_0 is the starting concentration, t is time, and k has units of \text{mol}\,\text{L}^{-1}\,\text{s}^{-1}.
First order. The rate is proportional to [A], so concentration decays exponentially. Here k has units of \text{s}^{-1}. A plot of \ln[A] against time is a straight line of slope -k.
Second order. The rate is proportional to [A]^2. Now the reciprocal of concentration rises linearly with time, and k has units of \text{L}\,\text{mol}^{-1}\,\text{s}^{-1}.
The half-life, the time for [A] to fall to half its start, follows directly from each. For zero order it is t_{1/2} = [A]_0 / (2k). For first order it is t_{1/2} = \ln 2 / k, with no concentration in it. For second order it is t_{1/2} = 1 / (k[A]_0). Only first order gives a constant half-life, which is why radioactive dating works: the fraction remaining depends on elapsed half-lives and nothing else.
A worked example matching the demo
First order, k = 0.05, starting at 1 mol/L, after 10 s
These are the tool's default values. Order is 1, so use the exponential form.
- Compute the exponent: -kt = -0.05 \times 10 = -0.5.
- Take the exponential: e^{-0.5} \approx 0.6065.
- Multiply by the start: [A] = 1 \times 0.6065 = 0.6065\ \text{mol/L}.
- Half-life: t_{1/2} = \ln 2 / k = 0.6931 / 0.05 = 13.86\ \text{s}.
So after 10 seconds about 60.7 percent of A remains, and it takes 13.86 seconds to reach the halfway point. Check the half-life against the decay: after 13.86 s the fraction is e^{-0.05 \times 13.86} = e^{-0.693} = 0.5, exactly half. The numbers agree.
The Arrhenius equation and temperature
The rate constant itself depends on temperature. The Arrhenius equation captures that dependence.
Here A is the pre-exponential factor (roughly the collision frequency with correct orientation), E_a is the activation energy, R is the gas constant 8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}, and T is absolute temperature in kelvin. The factor e^{-E_a/(RT)} is the fraction of collisions with enough energy to react.
Take the tool defaults: A = 10^{13}, E_a = 50\ \text{kJ/mol} = 50000\ \text{J/mol}, and T = 298\ \text{K}. Then RT = 8.314 \times 298 = 2478\ \text{J/mol}, so the exponent is -50000 / 2478 = -20.18. The fraction is e^{-20.18} \approx 1.72 \times 10^{-9}, and k = 10^{13} \times 1.72 \times 10^{-9} \approx 1.72 \times 10^{4}\ \text{s}^{-1}.
Now raise the temperature to 308 K, just 10 degrees. The exponent becomes -50000 / (8.314 \times 308) = -19.53, the fraction is e^{-19.53} \approx 3.29 \times 10^{-9}, and k nearly doubles to about 3.29 \times 10^{4}\ \text{s}^{-1}. That "rate doubles per 10 degrees" rule of thumb comes straight from this exponential, and it holds only near this activation energy and temperature.
To compare two temperatures without recomputing A, use the two-point form: \ln(k_2/k_1) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right). With E_a = 50\ \text{kJ/mol}, moving from 298 to 308 K gives \ln(k_2/k_1) = 0.654, so k_2/k_1 = 1.92, matching the numbers above.
Reading and interpreting your results
Match the output to what you fed in. For the integrated modes you get a concentration left after the time and a half-life. If the concentration comes back negative in zero order, the reaction has already finished: [A] = [A]_0 - kt can only be trusted down to zero, at time [A]_0/k. Beyond that the model does not apply.
For Arrhenius the output is a single rate constant at your chosen temperature. Its units depend on the order of the reaction it belongs to, which Arrhenius alone does not tell you. Feed that k back into the matching integrated law to predict concentrations.
| Order | Half-life formula | Value at [A]₀ = 1 | Depends on [A]₀? |
|---|---|---|---|
| Zero | [A]₀ / (2k) | 10 s | Yes, grows with [A]₀ |
| First | ln2 / k | 13.86 s | No |
| Second | 1 / (k[A]₀) | 20 s | Yes, shrinks with [A]₀ |
All three use k = 0.05 in the matching units. Notice the half-lives differ even from the same start, so identifying the correct order matters before you quote a number.
Common mistakes
The first mistake is using the wrong order. The formulas give different answers, and the units of k differ, so a value fitted as first order cannot be dropped into the second-order law. Confirm the order from a linear plot: [A] versus t for zero, \ln[A] versus t for first, 1/[A] versus t for second. Whichever is straight is your order.
Activation energy is almost always tabulated in kJ/mol, but R is in J/(mol·K). Forgetting to multiply E_a by 1000 makes the exponent 1000 times too small and inflates k by an enormous factor. In the worked example, using 50 instead of 50000 would give an exponent of -0.0202 instead of -20.18.
The second mistake is putting temperature in Celsius. Arrhenius needs kelvin: 25 degrees Celsius is 298 K, not 25. The third is extrapolating a zero-order line past the point where reactant runs out. Check that kt \lt [A]_0 before trusting the result.
Related tools
Kinetics sits inside a larger workflow. Before you track how fast a reaction runs, you usually balance it with the Chemical Equation Balancer and set starting amounts with the Stoichiometry Calculator and the Mole Calculator. Prepare a solution of known concentration using the Dilution Calculator or the Concentration Converter. Once a reaction reaches balance rather than running to completion, the Equilibrium (ICE Table) Solver takes over. For the energy side of a reaction, see the Calorimetry Calculator.
Frequently asked questions
Why does first-order half-life not depend on concentration?
Because the decay is exponential. Halving takes the same number of "e-foldings" no matter where you start: from 1 to 0.5 or from 0.5 to 0.25, each step is \ln 2 / k seconds. With k = 0.05 that is 13.86 s every time.
How do I find the reaction order from data?
Plot the concentration data three ways: raw, natural log, and reciprocal. The plot that comes out as a straight line names the order, and its slope gives k.
What does the pre-exponential factor A represent?
Roughly, how often reactant molecules collide with the right orientation, ignoring energy. The exponential term then keeps only the collisions with enough energy. For gas-phase reactions A is often near 10^{13}\ \text{s}^{-1}.
Why does a 10 degree rise sometimes double the rate?
The exponent -E_a/(RT) changes with temperature. At E_a = 50\ \text{kJ/mol} near 300 K, a 10 K rise multiplies k by about 1.9. Larger E_a gives a bigger multiplier; smaller E_a gives less than a doubling.
Can I use these formulas for radioactive decay?
Yes. Radioactive decay is strictly first order, so N = N_0 e^{-kt} and t_{1/2} = \ln 2 / k apply exactly, with k the decay constant.
Related tools
Chemical Equation Balancer
Balance any chemical equation to whole numbers.
Ideal Gas Law Calculator (PV = nRT)
Solve for pressure, volume, moles or temperature.
Molar Mass Calculator
Molar mass and percent composition from a formula.
Mole Calculator (grams ↔ moles ↔ particles)
Convert between mass, moles and number of particles.
pH Calculator
pH, pOH and ion concentrations for acids and bases.
Boiling & Freezing Point Calculator
Boiling-point rise and freezing-point drop of solutions.