Half-life of a Reaction — Revision Notes

⚡ 30-Second Revision

Definition: — Time for reactant concentration to halve.
Zero-Order: — $t_{1/2} = \frac{[A]_0}{2k}$ (Directly proportional to $[A]_0$)
First-Order: — $t_{1/2} = \frac{0.693}{k}$ (Independent of $[A]_0$)
Second-Order: — $t_{1/2} = \frac{1}{k[A]_0}$ (Inversely proportional to $[A]_0$)
Radioactive Decay: — Always first-order.
Key: — Identify reaction order first!

2-Minute Revision

Half-life ($t_{1/2}$) is the time taken for a reactant's concentration to reduce to half its initial value. It's a quick measure of reaction speed. For zero-order reactions, $t_{1/2} = [A]_0 / 2k$, meaning it's directly proportional to the initial concentration.

If you double the starting amount, it takes twice as long to halve it. For first-order reactions, $t_{1/2} = 0.693 / k$, which is constant and independent of the initial concentration. This is a crucial characteristic, especially for radioactive decay.

For second-order reactions (type $2A \rightarrow P$), $t_{1/2} = 1 / (k[A]_0)$, showing an inverse proportionality to the initial concentration. Doubling the initial amount halves the half-life. Always remember to identify the reaction order before applying any half-life formula.

For problems involving multiple half-lives, especially for first-order, the amount remaining after 'n' half-lives is Initial Amount $imes (1/2)^n$.

5-Minute Revision

The half-life ($t_{1/2}$) is a fundamental concept in chemical kinetics, defining the time required for a reactant's concentration to decrease to half of its initial value. Its behavior is uniquely tied to the reaction's order.

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Zero-Order Reactions: — The rate is constant, independent of concentration. The half-life formula is $t_{1/2} = \frac{[A]_0}{2k}$. This shows a *direct proportionality* between $t_{1/2}$ and the initial concentration $[A]_0$. If you double $[A]_0$, the $t_{1/2}$ also doubles. The units of $k$ are $ext{mol L}^{-1}\text{s}^{-1}$.

*Example:* If $[A]_0 = 1.0,\text{M}$ and $k = 0.1,\text{M/s}$, then $t_{1/2} = 1.0 / (2 \times 0.1) = 5,\text{s}$. If $[A]_0 = 2.0,\text{M}$, $t_{1/2} = 2.0 / (2 \times 0.1) = 10,\text{s}$.

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First-Order Reactions: — The rate is directly proportional to the first power of concentration. The half-life formula is $t_{1/2} = \frac{0.693}{k}$. Crucially, $t_{1/2}$ is *independent* of the initial concentration. This means it takes the same amount of time to halve the concentration, regardless of how much you start with. Radioactive decay is a classic example. The units of $k$ are $ext{s}^{-1}$.

*Example:* If $k = 0.01,\text{s}^{-1}$, then $t_{1/2} = 0.693 / 0.01 = 69.3,\text{s}$. This value remains constant even if $[A]_0$ changes. For problems involving multiple half-lives, the amount remaining after $n$ half-lives is given by $N_t = N_0 (1/2)^n$, where $n = \text{Total time} / t_{1/2}$.

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Second-Order Reactions: — The rate is proportional to the square of the concentration (for $2A \rightarrow P$). The half-life formula is $t_{1/2} = \frac{1}{k[A]_0}$. This indicates an *inverse proportionality* between $t_{1/2}$ and the initial concentration $[A]_0$. If you double $[A]_0$, the $t_{1/2}$ is halved. The units of $k$ are $ext{L mol}^{-1}\text{s}^{-1}$.

*Example:* If $[A]_0 = 1.0,\text{M}$ and $k = 0.1,\text{L mol}^{-1}\text{s}^{-1}$, then $t_{1/2} = 1 / (0.1 \times 1.0) = 10,\text{s}$. If $[A]_0 = 2.0,\text{M}$, $t_{1/2} = 1 / (0.1 \times 2.0) = 5,\text{s}$.

Key Strategy: Always identify the reaction order first. Then, apply the correct formula. Pay attention to units and significant figures. For conceptual questions, understand the dependencies of $t_{1/2}$ on $[A]_0$ for each order.

Prelims Revision Notes

Half-life ($t_{1/2}$) - Key Facts for NEET

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Definition: — Time required for the concentration of a reactant to decrease to half of its initial value.

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Zero-Order Reaction:

* Rate Law: Rate $= k$ * Integrated Rate Law: $[A] = [A]_0 - kt$ * Half-life Formula: $t_{1/2} = \frac{[A]_0}{2k}$ * **Dependency on $[A]_0$:** Directly proportional to initial concentration ($t_{1/2} propto [A]_0$). * Characteristic: As $[A]_0$ increases, $t_{1/2}$ increases. * Units of k: $ext{mol L}^{-1}\text{s}^{-1}$

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First-Order Reaction:

* Rate Law: Rate $= k[A]$ * Integrated Rate Law: $ln[A] = ln[A]_0 - kt$ or lnleft(\frac{[A]_0}{[A]}\right) = kt * Half-life Formula: $t_{1/2} = \frac{ln 2}{k} = \frac{0.693}{k}$ * **Dependency on $[A]_0$:** Independent of initial concentration.

* Characteristic: $t_{1/2}$ is constant regardless of $[A]_0$. * Units of k: $ext{s}^{-1}$ * Applications: Radioactive decay, many decomposition reactions. * Amount Remaining: After $n$ half-lives, Amount remaining $= \text{Initial Amount} \times (1/2)^n$.

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**Second-Order Reaction (Type $2A \rightarrow P$ or $A+B \rightarrow P$ with $[A]_0 = [B]_0$):**

* Rate Law: Rate $= k[A]^2$ * Integrated Rate Law: $rac{1}{[A]} = \frac{1}{[A]_0} + kt$ * Half-life Formula: $t_{1/2} = \frac{1}{k[A]_0}$ * **Dependency on $[A]_0$:** Inversely proportional to initial concentration ($t_{1/2} propto \frac{1}{[A]_0}$). * Characteristic: As $[A]_0$ increases, $t_{1/2}$ decreases. * Units of k: $ext{L mol}^{-1}\text{s}^{-1}$

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General Relationship: — For all reaction orders, $t_{1/2}$ is inversely proportional to the rate constant $k$. A larger $k$ means a faster reaction and a shorter half-life.

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Key Problem-Solving Steps:

* Always identify the reaction order first. * Select the appropriate half-life formula. * Perform calculations carefully, especially with exponents and logarithms. * For multi-half-life problems, calculate 'n' first.

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Common Traps: — Confusing the dependencies of $t_{1/2}$ on $[A]_0$ for different orders. Miscalculating with scientific notation.

Vyyuha Quick Recall

To remember half-life dependencies: Zero-order: Zealous Always (t1/2 $propto$ [A]0) First-order: Fixed Independent (t1/2 is Independent of [A]0) Second-order: Shrinking Inverse (t1/2 $propto$ 1/[A]0)