Chemical Kinetics Mind Maps, Part 2: Rate Laws, Half-Life & Arrhenius

In part 1 we set up what reaction rate even means. Now for the part that earns marks: rate laws, half-life, and the Arrhenius equation. These three show up in JEE every year. Get them straight here and kinetics stops feeling like guesswork.

Rate laws

So what is a rate law? Remember from part 1, we wrote rate as a change in concentration over time:

$\text{Rate} = \frac{\text{Change in concentration}}{\text{Time}}$

That relationship — tying rate to concentration — is the heart of a rate law. Written properly, it looks like this:

$\text{Rate} = k[A]^m[B]^n$

Here $k$ is the rate constant, and the exponents $m$ and $n$ are the orders of the reaction with respect to each reactant. Add them up and you get the overall order. That's it. No mystery.

Now the one thing students trip on: a rate law comes in two forms.

• Differential form — relates rate to concentration. Example: $\text{Rate} = k[A]$.
• Integrated form — relates concentration to time. For a first-order reaction:

$\ln[A] = \ln[A]_0 - kt$

Both describe the same reaction. You just use them for different questions.

When do you reach for which? Simple rule:

• The problem links rate and concentration → use the differential form.
• The problem links concentration and time → use the integrated form.

Keep both in your head. Confusing the two is the single most common way to lose marks here.

Your turn. A reaction is first order in $A$. Its concentration drops from $0.80\,\text{M}$ to $0.40\,\text{M}$ over some time. Which form do you use to find the rate constant?

Check: You're given concentration and time, so use the integrated form, $\ln[A] = \ln[A]_0 - kt$. Plug in the two concentrations and the time, solve for $k$.

What is half-life?

In any reaction, reactants get used up. Their concentration falls with time. At some point the concentration drops to half of where it started.

The time that takes is the half-life, written $t_{1/2}$.

At one half-life, 50% of the reactant is gone.

Say "used up," not "remaining." It sounds like a small thing, but it saves you from a classic trap. If you train yourself to think used up, you can read any of these cleanly. What is $t_{75\%}$? It's the time when 75% of the reactant is used up — not the time when 75% is left. JEE asks it exactly this way to catch the half-asleep student.

For a first-order reaction, half-life has a clean property — it doesn't depend on how much you started with:

$t_{1/2} = \frac{0.693}{k}$

The same $k$, the same half-life, every time. That constant half-life is itself a signature of first-order kinetics.

Your turn. A first-order reaction has $k = 0.0231\,\text{s}^{-1}$. What is its half-life?

Check: $t_{1/2} = \dfrac{0.693}{0.0231} = 30\,\text{s}$. Notice we never needed the starting concentration — for first order, we never do.

The Arrhenius equation

Picture two scientists running the same reaction with the same reactants — one at the equator, one at the North Pole. The rates come out different. Why?

Temperature, of course.

The story is made up, but the physics is real: temperature changes how fast a reaction goes. Arrhenius worked out why.

A reactant can't just slide straight into product, even when it carries plenty of energy. First it has to climb an energy hill — the activation energy, $E_a$. Only molecules that make it over the top become product.

Think of it as a barrier sitting between reactants and products:

$\text{Reactants} \xrightarrow{\text{must climb } E_a} \text{Products}$

The energy profile rises from the reactants up to a peak (the activated complex), then drops down to the products. The height of that peak above the reactants is $E_a$.

Arrhenius tied it all together in one equation:

$k = A\,e^{-E_a / RT}$

where $A$ is the frequency factor, $R$ the gas constant, and $T$ the absolute temperature. Take logs and it turns into a straight line — the form JEE actually tests:

$\ln k = \ln A - \frac{E_a}{RT}$

Plot $\ln k$ against $1/T$ and you get a line with slope $-E_a/R$. That's how activation energy gets measured.

Now read the equation. Raise $T$ and the term $-E_a/RT$ gets less negative, so $k$ grows — the reaction speeds up. The physical picture: hotter molecules carry more energy, collide harder and more often, and far more of them clear the activation barrier. More successful collisions, faster reaction.

Your turn. Two reactions run at the same temperature. Reaction X has a higher activation energy than reaction Y. Which is faster?

Check: Reaction Y. A higher $E_a$ means a taller hill, so fewer molecules make it over — a bigger $E_a$ drags the rate down.

The three you must own

• Rate law links rate to concentration: $\text{Rate} = k[A]^m[B]^n$. Differential form for rate-vs-concentration, integrated form ($\ln[A] = \ln[A]_0 - kt$) for concentration-vs-time.
• Half-life is the time for a reactant to fall to half. Think "used up," not "remaining." First order: $t_{1/2} = 0.693/k$, independent of starting amount.
• Arrhenius: $k = A\,e^{-E_a/RT}$. Higher temperature or lower activation energy means a faster reaction.

Lock these three down and you can handle most kinetics questions JEE throws at you. Next time you hit a kinetics problem, your first move is to ask: rate and concentration, or concentration and time? That one question tells you which tool to grab.