Question 11 markMCQSection A

$Which of the following curve represents the first order reaction?$

Q: Which of the following curve represents the first order reaction?

**Correct option: (B)** — a horizontal straight line of t½ versus [R]₀. **Key idea:** For a first order reaction, $t_{1/2} = 0.693/k$. - The half-life depends only on the rate constant k, **not** on the initial concentration [R]₀. - Hence the plot of t½ against [R]₀ is a horizontal line (constant value), which is exactly graph (B).

Q: Why is the half-life of a first order reaction independent of concentration?

Integrating the first order rate law gives $t_{1/2} = 0.693/k$, an expression that contains only the rate constant k. Since k depends on temperature (through the Arrhenius equation) and not on how much reactant you start with, the half-life stays the same whether you begin with a large or small concentration.

Q: How does the half-life of a zero order reaction differ?

For a zero order reaction $t_{1/2} = [R]_0/2k$, so the half-life is directly proportional to the initial concentration. A plot of t½ against [R]₀ is therefore a straight line through the origin — the opposite behaviour to first order, whose plot is a horizontal line.

Chemical Kinetics

Half-life of first order reactions

Options

(A)

t_{1/2}

(B)

t_{1/2}

(C)

Graph of Rate (y-axis) versus Concentration (x-axis): a horizontal straight line

(D)

Graph of Rate (y-axis) versus Concentration (x-axis): a decreasing curve

Official answer

t_{1/2} = 0.693/k

first order reaction

half-life

t½ = 0.693/k

independent of initial concentration

rate constant

integrated rate equation

zero order comparison

Marking scheme

1. $1 mark: correctly selecting option (B).$
2. $t_{1/2} = 0.693/k$

Hint

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

Quick oral answer

$For a first order reaction the half-life equals 0.693 divided by the rate constant, so it does not depend on the starting concentration — that gives a flat, horizontal t½ versus [R]₀ line.$

Common mistakes

1. $t_{1/2} \propto [R]_0$
2. $Assuming t½ always increases with concentration; for first order it is constant.$
3. $Mixing up a Rate vs concentration plot with a t½ vs concentration plot.$

Analysis and explanation

t_{1/2} = 0.693/k

Interesting facts

$Radioactive decay is a classic first order process, which is why every radioisotope (e.g. Carbon-14, half-life \approx 5730 years) has a fixed half-life regardless of sample size.$

$Because first-order half-life is concentration-independent, pharmacologists can quote a single 'drug half-life' (e.g. caffeine \approx 5 hours) that applies whatever the dose.$

Question details

Q1 of 37 questions

Board: CBSE
Class: Class 12
Subject: Chemistry
Year: 2026
Section: Section A
Marks: 1
Type: MCQ
Chapter: Chemical Kinetics
Topic: $Half-life of first order reactions$
Bloom's level: understand
Syllabus topic: $Unit 3: Chemical Kinetics — Integrated Rate Equations and Half-life of First Order Reactions$
Author: CBSE
Reviewed by: Padhantu Academic Team

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Frequently asked questions

$Why is the half-life of a first order reaction independent of concentration?$

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

$How does the half-life of a zero order reaction differ?$

$t_{1/2} = [R]_0/2k$