Chemical Kinetics — Rate Law, Order & Activation Energy

Chemical kinetics is the study of reaction rates: how fast a reaction happens, what changes that speed, and how to describe it with a rate law. Reach for kinetics whenever speed matters: combustion, atmospheric chemistry, battery materials, corrosion, enzyme behavior, drug stability, and reactor design. In most intro problems you work with three pieces, in this order: the rate law (how rate depends on concentration), the reaction order (how strong that dependence is), and the activation energy (why temperature and catalysts change the rate constant).

Step 1: Read The Rate Law

A rate law connects reaction rate to concentration for a specific reaction under specific conditions. A common form is

where $k$ is the rate constant, $[A]$ and $[B]$ are concentrations, and $m$ and $n$ are the reaction orders with respect to each reactant. The exponents tell you how sensitive the rate is to concentration, and $k$ sets the scale of the rate for those conditions. Do not take $m$ and $n$ from the overall balanced equation unless the step is explicitly elementary; for an overall reaction, the rate law usually comes from experiment.

Step 2: Interpret The Reaction Order

Reaction order tells you how the rate changes when concentration changes.

• Zero order in $A$: changing $[A]$ does not change the rate in that range.
• First order in $A$: doubling $[A]$ doubles the rate.
• Second order in $A$: doubling $[A]$ makes the rate four times larger.

The overall order is the sum of the exponents. In $rate = k[A]^2[B]$ the reaction is second order in $A$, first order in $B$, and third order overall.

Step 3: Account For Temperature And Activation Energy

Even when molecules collide, not every collision reacts; they need enough energy to reach a higher-energy transition state. That energy barrier is the activation energy $E_a$, and a higher $E_a$ usually means a smaller fraction of collisions can react. The Arrhenius equation links it to the rate constant:

The practical takeaways: higher temperature usually increases $k$, larger activation energy makes the rate more temperature-sensitive, and a catalyst can raise the rate by providing a pathway with lower effective activation energy. That last point has a condition: a catalyst changes the pathway, not the overall stoichiometric equation. Note also that reaction order and the rate constant are different. Order comes from the exponents and describes the concentration response; $k$ is the proportionality constant under given conditions. When temperature changes, $k$ usually changes, while the order usually stays the same for the same mechanism.

Worked Example: A Second-Order Case, Start To Finish

Suppose experiments give

Compare two experiments at the same temperature. In experiment 1, $[A] = 0.10\ \mathrm{M}$ and $[B] = 0.20\ \mathrm{M}$. In experiment 2, $[A]$ is doubled to $0.20\ \mathrm{M}$ while $[B]$ stays the same. Because the rate depends on $[A]^2$, doubling $[A]$ multiplies the rate by

So experiment 2 runs four times as fast as experiment 1, with everything else held constant. If instead you kept $[A]$ fixed and doubled $[B]$, the rate would only double, because $[B]$ appears to the first power. The skill is always the same: change one variable at a time, read its exponent, and convert that exponent into a rate factor.

Where The Procedure Trips People Up

• Step 1, taking reaction order from the balanced equation. That works only for an elementary step; for an overall reaction the order comes from experimental data.
• Step 2, confusing reaction rate with equilibrium. A fast reaction reaches its result quickly, but that does not mean it produces more product at equilibrium.
• Step 3, forgetting the temperature condition. Arrhenius reasoning uses absolute temperature, so use kelvin, not degrees Celsius.
• Step 3, assuming a catalyst changes the final outcome. A catalyst changes the rate by changing the pathway; it does not change the equilibrium expression or the overall balanced equation.

In practice kinetics answers questions like: Will a reaction be useful at room temperature? How much faster will it run when heated? Will a catalyst make the process practical?

Practice The Full Procedure

Take $rate = k[A]^2[B]$ and test two new cases. First, double both $[A]$ and $[B]$ together: the rate multiplies by $2^2 \times 2 = 8$. Second, halve $[A]$ while doubling $[B]$: the rate multiplies by $(1/2)^2 \times 2 = 1/2$. Working both confirms the rate law and reaction order really click. To go further, compare this topic with activation energy or reaction engineering, which connect the rate law on paper to what changes in a real process.