Chemical Kinetics — Mind Map (Part 1)

Chemical Kinetics shows up again and again in JEE. The good news: once your basics click, most of its questions feel easy. So let me give you the whole chapter as a mind map you can hold in your head — the kind you can redraw on the back of a page before the exam.

This is Part 1. We'll nail down what rate means and what order tells you.

What Chemical Kinetics actually asks

Pick any reaction. You can ask three different questions about it, and a different branch of chemistry answers each one:

• Will it happen at all? → that's thermodynamics.
• How much product do I get? → that's equilibrium.
• How long will it take? → that's Chemical Kinetics.

Kinetics is the speed branch. We call that speed the rate of reaction. That's it:

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

Which speeds even matter?

Not every speed is worth your attention. Think about it:

• A job you finish in 1 second. Shave off 10% and you save 0.1 second. Who cares?
• A job that takes 100 years — you start it for your grandson. Cut 10% and it still takes 90 years. Far too slow to matter.
• A job that takes 10 days. Cut 10% and you save a whole day. That matters.

So we care about reactions that are neither too fast nor too slow. Those are the ones where nudging the rate actually buys us something.

Defining the rate properly

By convention, rate is always positive. But reactants disappear — their concentration drops — so we stick a minus sign in front to keep the number positive. Take a general reaction:

$aA + bB \longrightarrow cC + dD$

Reactants vanish, products build up:

• Rate of disappearance of A $= -\dfrac{d[A]}{dt}$
• Rate of disappearance of B $= -\dfrac{d[B]}{dt}$
• Rate of appearance of C $= +\dfrac{d[C]}{dt}$
• Rate of appearance of D $= +\dfrac{d[D]}{dt}$

Now the trap. Which of these is the rate of reaction? None of them on its own.

A reaction has only one rate at a given temperature. To get it, divide each term by its stoichiometric coefficient:

$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$

Miss the coefficient and your answer is off by a factor. JEE loves that slip. And remember the unit — rate is always $\text{mol}\,\text{L}^{-1}\text{s}^{-1}$.

Your turn. For $N_2 + 3H_2 \longrightarrow 2NH_3$, the rate of disappearance of $H_2$ is $6 \times 10^{-3}\,\text{mol}\,\text{L}^{-1}\text{s}^{-1}$. What is the rate of the reaction?

Check: Rate $= -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{3}\times 6\times 10^{-3} = 2\times 10^{-3}\,\text{mol}\,\text{L}^{-1}\text{s}^{-1}$. Divide by the coefficient of $H_2$, which is 3.

Order of a reaction

Order tells you how concentration affects the rate. Three things to lock in:

• It is experimentally determined. You cannot read it off the balanced equation — you have to look at the data.
• It can be zero, positive, or negative (and even a fraction).
• It is the sum of the powers of the concentration terms in the rate law.

For a rate law like $\text{Rate} = k[A]^x[B]^y$, the order is $x + y$.

The whole chapter, as one mind map

Here's Part 1 collapsed into a nested outline. Copy it onto a card and you've got your revision sheet:

• Chemical Kinetics — the speed branch of chemistry
  • Rate of reaction $= \dfrac{\Delta[\,]}{\Delta t}$
    • always positive (minus sign for reactants)
    • per unit stoichiometric coefficient → only one rate
    • $\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}$
    • unit: $\text{mol}\,\text{L}^{-1}\text{s}^{-1}$
  • Order of reaction
    • found by experiment, not from the equation
    • can be zero / positive / negative / fractional
    • = sum of powers in the rate law
    • shows how concentration controls rate
  • Rate law $\text{Rate} = k[A]^x[B]^y$
    • $k$ = rate constant
    • $x, y$ = orders w.r.t. each reactant

Quick recap

• Kinetics answers how fast — not whether or how much.
• One reaction, one rate: always divide by the stoichiometric coefficient.
• Rate's unit is $\text{mol}\,\text{L}^{-1}\text{s}^{-1}$.
• Order comes from experiment and is the sum of the concentration powers.

In Part 2 we'll map the rate laws themselves — zero, first, and second order — plus their half-lives. For now, redraw the outline above from memory. If you can, the basics are yours.