A zero order reaction $X \rightarrow \text{Product}$, with an initial concentration 0.02 M has a half life of 10 min. If one starts with concentration 0.04 M, then the half life is (a) 10 s (b) 5 min (c) 20 min (d) cannot be predicted using the given information

A zero order reaction X arrow Product, with an initial concentration 0.02 M has a half life of 10 min. If one

Choose the Best Answer · Q2

X \rightarrow \text{Product}

(a) 10 s

(b) 5 min

(d) cannot be predicted using the given information

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Concept understanding — Half Life of a Reaction

Half Life of a Reaction

Imagine you have a pile of 1000 coins, and every minute, exactly half of the coins that are still there vanish. After one minute, 500 coins remain. After another minute, 250 remain. Then 125, then about 62, and so on. The time it takes for the pile to shrink from 1000 to 500 is the same as the time it takes to shrink from 500 to 250, or from 250 to 125. That constant time interval is the half life.

In chemistry, a reaction's half life ( $t_{1/2}$ ) is the time required for the concentration of a reactant to fall to exactly one-half of its initial value. It is a simple, intuitive way to describe how fast a reaction proceeds — the shorter the half life, the faster the reaction.

The Precise Definition

For any reaction, if you start with an initial concentration $[A]_0$ , the half life $t_{1/2}$ is the time at which:

[A] = \frac{1}{2}[A]_0

That is all. But here is the crucial point: the half life is not a universal constant — it depends on the order of the reaction. For different reaction orders, the half life behaves very differently.

Half Life for a First Order Reaction

For a first order reaction, the rate law is:

\text{Rate} = k[A]

The integrated rate equation is:

[A] = [A]_0 e^{-kt}

Set $[A] = \frac{1}{2}[A]_0$ and solve for $t$ :

\frac{1}{2}[A]_0 = [A]_0 e^{-k t_{1/2}}

Cancel $[A]_0$ :

\frac{1}{2} = e^{-k t_{1/2}}

Take natural logarithms:

\ln\left(\frac{1}{2}\right) = -k t_{1/2}

-\ln 2 = -k t_{1/2}

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

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

This is the key result: for a first order reaction, the half life is independent of the initial concentration. Whether you start with 1 M or 0.001 M, the time to halve the concentration is exactly the same. This is a unique property of first order reactions — no other order behaves this way.

Why Does This Matter?

The constancy of $t_{1/2}$ for first order reactions is what makes radioactive decay predictable. Carbon-14 dating works because the half life of $^{14}\text{C}$ is always 5730 years, regardless of how much carbon is present. The same principle applies to many chemical reactions, especially decompositions and isomerisations.

Tip

For a first order reaction, if you know the half life, you can find the rate constant instantly: $k = 0.693 / t_{1/2}$ . This is often the fastest way to get $k$ from experimental data.

Contrast with Other Orders

For a zero order reaction, the half life depends on the initial concentration:

t_{1/2} = \frac{[A]_0}{2k}

Double the starting concentration, and the half life doubles.

For a second order reaction:

t_{1/2} = \frac{1}{k[A]_0}

Double the starting concentration, and the half life is cut in half.

Watch out

Never assume a half life is constant. Only for first order reactions is $t_{1/2}$ independent of concentration. If a problem gives you a half life and asks for the rate constant, first confirm the reaction order — otherwise you will use the wrong formula.

The Intuition in One Sentence

The half life of a first order reaction is the time it takes for any starting amount to drop by half — and it is always the same length of time, no matter how much you start with. That is why it is such a powerful and elegant concept.

Half Life of a Reaction is a numerical-heavy concept from the Chemical Kinetics chapter of NCERT/CBSE Class 12 Chemistry, and it shows up often in "Half Life of a Reaction numericals", "Half Life of a Reaction important questions", and "Half Life of a Reaction class 12 chemistry" searches because board exams, JEE Main, and NEET all test it through calculation-based problems.

Step 1. For a zero order reaction, $t_{1/2}=\dfrac{[A]_0}{2k}$ (Section 7.6) -- unlike first order, half life is DIRECTLY PROPORTIONAL to the initial concentration.

Step 2. From the given data, $10=\dfrac{0.02}{2k}\Rightarrow k=\dfrac{0.02}{20}=1\times10^{-3}\ \text{M min}^{-1}$ .

Step 3. For the new run, $[A]_0=0.04$ M: $t_{1/2}=\dfrac{0.04}{2\times1\times10^{-3}}=\dfrac{0.04}{0.002}=20$ min.

✓Final answer

(c) 20 min

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Q.A zero order reaction $X \rightarrow \text{Product}$ , with an initial concentration 0.02 M has a half life of 10 min. If one starts with concentration 0.04 M, then the half life is

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