1 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: 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.

Four candidate graphs (t-half vs [R]0 and Rate vs Concentration) for identifying a first order reaction

Chemical Kinetics

Half-life of first order reactions

Options

(A)Graph of

t_{1/2}

(y-axis) versus

[R]_0

(x-axis): a straight line with positive slope through the origin

(B)Graph of

t_{1/2}

(y-axis) versus

[R]_0

(x-axis): a horizontal straight line (

t_{1/2}

independent of

[R]_0

)

(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

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).

first order reactionhalf-lifet½ = 0.693/kindependent of initial concentrationrate constantintegrated rate equationzero order comparison

Marking Scheme

11 mark: correctly selecting option (B).
2No partial marks for MCQ; justification ( $t_{1/2} = 0.693/k$ independent of $[R]_0$ ) expected in written form only if asked.

Hint

For first order, $t_{1/2} = 0.693/k$ — it does not contain $[R]_0$ .

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.

Analysis & Explanation

Concept:

The half-life is the time in which the concentration of a reactant falls to half its initial value.

First order: $t_{1/2} = 0.693/k$ → independent of [R]₀ → horizontal line (option B is correct).
Zero order: $t_{1/2} = [R]_0/2k$ → directly proportional to [R]₀ → straight line through origin.

Why the distractors are wrong:

(A) $t_{1/2} \propto [R]_0$ (straight line through origin) is the signature of a zero order reaction, not first order.
(C) Rate independent of concentration (horizontal Rate vs [conc.] line) again describes a zero order reaction ( $\text{Rate} = k$ ).
(D) A Rate that decreases as concentration increases is unphysical for a simple positive-order reaction; for first order the correct Rate vs [conc.] plot is a straight line through the origin with positive slope ( $\text{Rate} = k[R]$ ).

Exam trap: Students confuse the t½–[R]₀ plot for zero and first order. Remember: constant t½ ⇒ first order.

Common Mistakes

1Selecting (A) — confusing the zero-order $t_{1/2} \propto [R]_0$ line with the first-order case.
2Assuming t½ always increases with concentration; for first order it is constant.
3Mixing up a Rate vs concentration plot with a t½ vs concentration plot.

Interesting Facts

Radioactive decay is a classic first order process, which is why every radioisotope (e.g. Carbon-14, half-life ≈ 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 ≈ 5 hours) that applies whatever the dose.

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Frequently Asked Questions

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.

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.