The half life period of a first order reaction is 10 minutes. Then its rate constant is : (a) 6.93×10² min⁻¹ (

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Q. The half life period of a first order reaction is 10 minutes. Then its rate constant is :

(a)

6.93\times10^{2}

min

^{-1}

(b)

0.693\times10^{-2}

min

^{-1}

(c)

6.93\times10^{-2}

min

^{-1}

(d)

69.3\times10^{-1}

min

^{-1}

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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:

$t_{1/2} = \frac{[A]_0}{2k}$ …

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Using the standard first-order half-life relation $k=0.693/t_{1/2}$ with $t_{1/2}=10$ min gives $k=6.93\times10^{-2}\text{ min}^{-1}$ .

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Q.The half life period of a first order reaction is 10 minutes. Then its rate constant is :

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