Derive an expression to calculate time required for completion of zero order reaction.

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Q. Derive an expression to calculate time required for completion of zero order reaction.

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Concept understanding — Zero Order Kinetics

Zero Order Kinetics: The Drug That Doesn't Care How Much You Give It

Imagine you're filling a bathtub. You turn the tap to a fixed flow rate — say, 5 litres per minute. The amount of water in the tub increases by exactly 5 litres every minute, regardless of whether the tub is empty or already half full. That's the core intuition behind zero order kinetics: a constant amount disappears per unit time, no matter how much is left.

Now contrast this with what you probably expect. Most processes in nature follow first order kinetics: the rate depends on how much is present. If you have 100 molecules, 10 might react per second; if you have 10 molecules, only 1 reacts per second. The fraction lost is constant, but the amount lost per second shrinks as the quantity shrinks. Zero order is the opposite — the amount lost per second is fixed, so the fraction lost actually increases as the quantity drops.

The Precise Statement

-\frac{d[A]}{dt} = k_0

Where $[A]$ is the concentration of the substance (or amount, depending on context), $t$ is time, and $k_0$ is the zero order rate constant with units of concentration per time (e.g., mg/L per hour, or simply mg/hour if we're talking about total amount).

The negative sign indicates the substance is being removed. The key point: the rate does not depend on $[A]$ . It's a flat, constant rate.

The Integrated Form and Half-Life

If you integrate the differential equation, you get a straight line:

[A]_t = [A]_0 - k_0 t

This is the equation of a line with slope $-k_0$ and intercept $[A]_0$ . Plot concentration vs. time, and you get a straight line sloping downward until it hits zero.

The half-life — the time for the concentration to fall to half its initial value — is:

t_{1/2} = \frac{[A]_0}{2k_0}

Notice something crucial: the half-life depends on the initial concentration. Double the starting amount, and the half-life doubles. This is completely different from first order kinetics, where half-life is constant regardless of starting concentration.

Watch out

A common mistake: students assume half-life is always constant. For zero order, it is not. The half-life changes with the starting amount. If you start with 100 mg, half-life might be 5 hours; start with 200 mg, half-life becomes 10 hours.

Where Does Zero Order Kinetics Actually Happen?

In pharmacology, zero order kinetics is most famously seen with ethanol (alcohol) and aspirin at high doses. The reason is saturation of enzymes.

Your body metabolises alcohol using an enzyme called alcohol dehydrogenase. At low alcohol levels, the enzyme works efficiently and the rate depends on how much alcohol is present (first order). But at higher concentrations — say, after a few drinks — the enzyme becomes saturated. It's working at maximum speed, like a factory running at full capacity. Adding more raw material (alcohol) doesn't make it work faster. The rate becomes constant: a fixed amount of alcohol is metabolised per hour, regardless of how much is in your blood.

This is why alcohol elimination follows a straight line when you plot blood alcohol concentration vs. time. A typical person eliminates about 0.015 g/dL per hour — a fixed amount, not a fixed fraction.

Note

The same saturation principle applies to some drug transporters in the kidneys. When the transport proteins are working at maximum capacity, drug excretion becomes zero order. This is why high doses of certain drugs (like phenytoin) can lead to unexpectedly long elimination times — the system is overwhelmed.

A Quick Comparison Table

Property	Zero Order	First Order
Rate depends on	Nothing (constant)	Concentration
Rate equation	$-\frac{d[A]}{dt} = k_0$	$-\frac{d[A]}{dt} = k_1[A]$

Why this formula?

Zero Order Kinetics: Why the Formula Holds

Let's build this from the ground up — understanding the why before the what.

The Core Idea

Zero order kinetics describes a process where the rate is constant — it does not depend on the concentration of the reactant.

This is the definition, but why would that ever happen?

Why the Rate is Constant

Imagine a reaction happening on a solid surface (like a catalyst or a tablet dissolving). The reactant molecules must first adsorb onto the surface before reacting.

If the surface is saturated with reactant molecules, adding more reactant in solution doesn't help — the surface is already full.
The reaction proceeds at a fixed speed determined by how fast the surface can process the adsorbed molecules.

Key insight: The rate is limited by the surface, not by how much reactant is floating around.

Deriving the Zero Order Rate Law

Step 1: Write the rate definition

For a reaction $A \rightarrow \text{products}$ , the rate of disappearance of $A$ is:

-\frac{d[A]}{dt} = k

where $k$ is the zero order rate constant (units: concentration/time, e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ).

Notice: No $[A]$ term on the right side — that's the signature of zero order.

Step 2: Separate variables and integrate

-d[A] = k \, dt

Integrate from initial time $t=0$ (concentration $[A]_0$ ) to time $t$ (concentration $[A]_t$ ):

-\int_{[A]_0}^{[A]_t} d[A] = k \int_0^t dt

-[A]_t + [A]_0 = kt

Step 3: Rearrange to the familiar form

\boxed{[A]_t = [A]_0 - kt}

This is the integrated rate law for zero order kinetics.

What This Formula Tells Us

Linear decrease: Concentration falls linearly with time (not exponentially like first order).
Slope = $-k$ : A plot of $[A]_t$ vs. $t$ gives a straight line with slope $-k$ .
Half-life depends on initial concentration:

Set $[A]_t = \frac{[A]_0}{2}$ :

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

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

Critical exam point: Unlike first order (where $t_{1/2}$ is constant), zero order half-life increases with higher initial concentration.

--- …

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For a zero-order reaction, the rate is constant and independent of concentration. The time for completion is simply the initial concentration divided by the rate constant: $t_{complete} = \frac{[A]_0}{k}$ .

The Concept: Why Zero-Order Reactions Are Different

Most reactions slow down as reactants get used up — that's first-order or second-order behaviour. But a zero-order reaction proceeds at a constant rate, regardless of how much reactant remains. This happens when the reaction rate is limited by something other than concentration — for example, a saturated enzyme surface in a biochemical reaction, or a metal catalyst surface in a heterogeneous catalytic reaction.

The key insight: if the rate doesn't depend on $[A]$ , then the concentration drops linearly with time. That straight-line decay makes the "time for completion" calculation trivial — it's just how long it takes to consume all the reactant at a fixed speed.

A \rightarrow \text{Products}

Deriving the Expression Step by Step

1. Start with the rate law.

For a zero-order reaction, the rate of disappearance of reactant $A$ is constant:

-\frac{d[A]}{dt} = k

The negative sign indicates $[A]$ is decreasing. The rate constant $k$ is positive.

2. Separate variables and integrate.

Rearrange to get all $[A]$ terms on one side and $dt$ on the other:

d[A] = -k \, dt

Integrate from initial time $t = 0$ (when $[A] = [A]_0$ ) to any later time $t$ (when $[A] = [A]_t$ ):

\int_{[A]_0}^{[A]_t} d[A] = -k \int_{0}^{t} dt

The left side integrates to $[A]_t - [A]_0$ , and the right side integrates to $-kt$ :

[A]_t - [A]_0 = -kt

3. Rearrange to the familiar integrated form.

[A]_t = [A]_0 - kt

This is a straight line with slope $-k$ and intercept $[A]_0$ . If you plot $[A]_t$ vs. $t$ , you get a line that falls steadily.

Tip

The linearity of $[A]_t$ vs. $t$ is the quickest way to identify a zero-order reaction from experimental data. If your concentration-time graph is a straight line with a negative slope, the reaction is zero-order.

4. Define "completion" of the reaction. …

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Common Mistakes: Time for Completion of Zero Order Reaction

Students often confuse zero order kinetics with first order — here are the most frequent errors and how to avoid each.

✗ Mistake 1: Using the First Order Formula

The error:

Plugging $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$ into a zero order problem.

Why it happens:

Memorising formulas without understanding the rate law behind them.

How to avoid:

Always start from the integrated rate equation for zero order:

[A]_t = [A]_0 - kt

For completion, $[A]_t = 0$ , so:

0 = [A]_0 - kt \quad \Rightarrow \quad t = \frac{[A]_0}{k}

Key point: Zero order completion time depends on initial concentration, not on a log term.

✗ Mistake 2: Forgetting Units of $k$

The error:

Writing $t = \frac{[A]_0}{k}$ but using $k$ in $\text{s}^{-1}$ (first order units).

Why it happens:

Not checking dimensional consistency.

How to avoid:

For zero order, $k$ has units of concentration/time (e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ).

Check:

\frac{[A]_0}{k} \rightarrow \frac{\text{mol L}^{-1}}{\text{mol L}^{-1} \text{s}^{-1}} = \text{s} \quad \checkmark

✗ Mistake 3: Confusing "Completion" with "Half-Life"

The error:

Using $t_{1/2} = \frac{[A]_0}{2k}$ and then doubling it to get completion time.

Why it happens:

Assuming half-life repeats identically — but for zero order, each successive half-life is shorter.

How to avoid:

Half-life: $t_{1/2} = \frac{[A]_0}{2k}$
Completion time: $t_{\text{complete}} = \frac{[A]_0}{k} = 2 \times t_{1/2}$

This works only for zero order (because it's exactly two half-lives). For first order, completion is never reached.

✗ Mistake 4: Writing the Derivation Backwards

The error:

Starting with $t = \frac{[A]_0}{k}$ and then "deriving" the integrated equation.

Why it happens:

Memorising the final answer instead of the logical flow.

How to avoid:

Always derive step-by-step: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$Q. When the initial concentration of a zero order reaction is doubled, the half-life of the reaction is: (A) doubled (B) not changed (C) tripled (D) halved$
›Reveal solutionSolution
For a zero‑order reaction, the half‑life is directly proportional to the initial concentration. Doubling the initial concentration therefore doubles the half‑life. The correct option is (A).

Concept and Intuition

In chemical kinetics, the half‑life $t_{1/2}$ is the time required for the concentration of a reactant to fall to half its initial value. For a zero‑order reaction, the rate is constant — it does not depend on the concentration of the reactant. That means the reactant disappears at a steady, unchanging speed.

Think of it like draining a tank of water at a constant rate: if you start with twice as much water, it will take twice as long to drain half of it. Similarly, for a zero‑order reaction, doubling the starting amount doubles the time needed to consume half of it. This is fundamentally different from first‑order reactions (where half‑life is constant) or second‑order reactions (where half‑life is inversely proportional to initial concentration).

Step‑by‑Step Derivation

Write the integrated rate law for a zero‑order reaction. For a reaction $A \rightarrow \text{products}$ with rate law $\text{rate} = k$ (where $k$ is the rate constant), the concentration of $A$ at time $t$ is:

$[A]_t = [A]_0 - kt$
This is a straight line with slope $-k$ .

Define the half‑life condition. At $t = t_{1/2}$ , the concentration is half the initial:

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

Substitute into the integrated law.

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

Solve for $t_{1/2}$ . Rearranging:

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

Interpret the result. …
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}$
›Reveal solutionSolution
For a zero‑order reaction, the concentration decreases linearly with time: $[A]_t = [A]_0 - kt$ .

Using the given values, the time taken is $t = \frac{0.2 - 0.024}{6.0 \times 10^{-3}} = 29.33\ \text{s}$ , so the correct option is (D).

Concept & Intuition

In a zero‑order reaction, the rate does not depend on the concentration of the reactant. That means the reactant disappears at a constant speed — like water draining from a tank at a fixed rate. The concentration vs. time graph is a straight line with slope $-k$ . So if you know how much concentration has dropped and the constant rate, you can directly find the time by dividing the change in concentration by the rate constant.

Step‑by‑Step Solution

Recall the zero‑order integrated rate law For a reaction $A \rightarrow \text{products}$ that is zero order in $A$ :

$[A]_t = [A]_0 - k t$
where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$ , and $k$ is the rate constant (with units $\text{mol L}^{-1} \text{s}^{-1}$ ).

Identify the given quantities

Initial concentration: $[A]_0 = 0.2\ \text{M}$

Final concentration: $[A]_t = 0.024\ \text{M}$

Rate constant: $k = 6.0 \times 10^{-3}\ \text{mol L}^{-1} \text{s}^{-1}$

Rearrange the equation to solve for time

From $[A]_t = [A]_0 - k t$ , we get:

$k t = [A]_0 - [A]_t \quad \Rightarrow \quad t = \frac{[A]_0 - [A]_t}{k}$

Plug in the numbers
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COMEDK 2024Set 2024-A1 markMCQ
$Q. The half-life for a zero order reaction is (A) Inversely proportional to the initial concentration and directly proportional to the rate constant (B) Directly proportional to the initial concentration and inversely proportional to the rate constant (C) Independent of rate constant, but depends on the initial concentration (D) Independent of initial concentration$
›Reveal solutionSolution
For a zero‑order reaction, the half‑life is directly proportional to the initial concentration and inversely proportional to the rate constant. The correct option is (B).

Concept & Intuition

Half‑life ( $t_{1/2}$ ) is the time required for the concentration of a reactant to fall to half its initial value. For a zero‑order reaction, the rate is constant — it does not depend on concentration. That means the reactant is consumed at a steady pace, so the more you start with, the longer it takes to reach half that amount. Also, a larger rate constant means faster consumption, so the half‑life gets shorter. Hence $t_{1/2}$ should be proportional to $[A]_0$ and inversely proportional to $k$ .

Step‑by‑step derivation

Write the integrated rate law for a zero‑order reaction For a reaction $A \rightarrow \text{products}$ with rate $-\frac{d[A]}{dt} = k$ , integration gives:

$[A]_t = [A]_0 - kt$
where $[A]_0$ is the initial concentration and $[A]_t$ is the concentration at time $t$ .

Define the half‑life condition

At $t = t_{1/2}$ , the concentration is half the initial: $[A]_{t_{1/2}} = \frac{[A]_0}{2}$ .

Substitute into the integrated law

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

Solve for $t_{1/2}$ Rearranging:

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

Interpret the relationship …
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COMEDK 2024Set 2024-M1 markMCQ
$Q. Given below are 4 graphs [A], [B], [C] and [D] Identify the 2 graphs that represent a Zero order reaction? (A) [A] and [D] (B) [A] and [B] (C) [B] and [C] (D) [C] and [D]$
›Reveal solutionSolution
For a zero‑order reaction, concentration decreases linearly with time and the rate is independent of concentration. Graph [C] shows $[R]$ vs. time as a straight line with negative slope, and graph [D] shows rate constant vs. $[R]$ as a horizontal line. Thus the correct pair is [C] and [D].

The key idea is that a zero‑order reaction has a constant rate that does not depend on the concentration of the reactant. This leads to two characteristic plots:

Concentration vs. time: $[R] = [R]_0 - kt$ → a straight line with slope $-k$ .

Rate vs. concentration: $\text{Rate} = k$ → a horizontal line.

Let’s examine each graph in turn.

Graph [A] – vertical axis is $\ln[R]$ , horizontal is time.

For a first‑order reaction, $\ln[R] = \ln[R]_0 - kt$ , which gives a straight line with slope $-k$ .

This is not zero‑order; it’s first‑order decay.

Graph [B] – vertical axis is $[R]$ , horizontal is time.

The curve starts high, falls steeply, then flattens asymptotically.

This is the exponential decay of a first‑order reaction ( $[R] = [R]_0 e^{-kt}$ ).

Not zero‑order.

Graph [C] – vertical axis is $[R]$ , horizontal is time.

The plot is a straight line with constant negative slope, annotated $K = -\text{Slope}$ .

This matches $[R] = [R]_0 - kt$ exactly.

✓ Zero‑order.

Graph [D] – vertical axis is Rate, horizontal is $[R]$ .

The plot is a horizontal line – rate does not change as $[R]$ increases. …
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COMEDK 2023Set 2023-M1 markMCQ
$300 \mathrm{~K}$
›Reveal solutionSolution

$t_{1/2}$ is directly proportional to the initial pressure, which is the signature of a zero-order reaction.

For an $n^{\text{th}}$ -order reaction the half-life depends on the initial concentration (here pressure) as
$t_{1/2} \propto [A]_0^{\,1-n}.$
Given data:
$\frac{t_{1/2,1}}{t_{1/2,2}} = \frac{350}{175} = 2, \qquad \frac{P_1}{P_2} = \frac{40}{20} = 2.$
So …
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KCET 2022Set B-31 markMCQ
$\mathrm{CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH}$
›Reveal solutionSolution

The reaction is second order overall, so rearranging Rate $= k[A][B]$ for $k$ leaves units of $\mathrm{L\,mol^{-1}\,s^{-1}}$ .

Step 1 — Determine the overall order.

The rate law is given experimentally as
$\text{Rate} = k\,[\mathrm{CH_3COOC_2H_5}]^1[\mathrm{NaOH}]^1$
Order $=$ sum of the exponents $= 1 + 1 = \mathbf{2}$ . The reaction (saponification of ethyl acetate) is second order overall, first order in each reactant.

Step 2 — Do the algebra on the units.

Rearrange the rate law:
$k = \frac{\text{Rate}}{[\mathrm{CH_3COOC_2H_5}][\mathrm{NaOH}]}$
Now substitute the units. Rate is always a concentration change per unit time, $\mathrm{mol\,L^{-1}\,s^{-1}}$ , and each concentration is $\mathrm{mol\,L^{-1}}$ :
$[k] = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{(\mathrm{mol\,L^{-1}})(\mathrm{mol\,L^{-1}})} = \frac{\mathrm{mol\,L^{-1}\,s^{-1}}}{\mathrm{mol^2\,L^{-2}}}$
Step 3 — Simplify.
$[k] = \mathrm{mol}^{1-2}\,\mathrm{L}^{-1+2}\,\mathrm{s}^{-1} = \mathrm{mol^{-1}\,L\,s^{-1}}$ $\boxed{[k] = \mathrm{L\,mol^{-1}\,s^{-1}}}$
Step 4 — The general rule (learn this once, use it always). …
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KCET 2019Set A-11 markMCQ
$k =$
›Reveal solutionSolution
The key idea is to test each statement against the Arrhenius equation and the definition of activation energy. The wrong statement is (D), because a catalyst does alter the activation energy $E_a$ by providing an alternative path with a lower value.

The Relevant Concept

The Arrhenius equation is the backbone of chemical kinetics for temperature dependence:
$k = A e^{-E_a / RT}$
Here:

$k$ is the rate constant.

$A$ is the Arrhenius constant (or pre-exponential factor), related to collision frequency and orientation.

$E_a$ is the activation energy — the minimum energy reactant molecules must have for a reaction to occur.

$R$ is the gas constant.

$T$ is the absolute temperature.

The fraction of molecules with energy at least $E_a$ is given by the Boltzmann factor $e^{-E_a / RT}$ . Taking natural logs gives a linear form:
$\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$
A catalyst works by providing a different reaction pathway with a lower activation energy, which directly changes $E_a$ . Let's check each statement.

Step-by-Step Analysis

1. Statement (A): "Rate constant $k =$ Arrhenius constant $A$ : if $E_a = 0$ "

If $E_a = 0$ , then $e^{-E_a / RT} = e^0 = 1$ . The Arrhenius equation becomes $k = A \cdot 1 = A$ . This is mathematically correct. A reaction with zero activation energy would proceed at every collision, so the rate constant equals the collision-frequency factor. This statement is true.

2. Statement (B): " $\ln k$ vs $\frac{1}{T}$ plot is a straight line."

From $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$ , this is of the form $y = c + mx$ (with $y = \ln k$ , $x = 1/T$ , slope $m = -E_a/R$ , intercept $c = \ln A$ ). Over the temperature ranges where $E_a$ and $A$ are constant, this is indeed a straight line. This statement is true.

3. Statement (C): " $e^{-E_a/RT}$ gives the fraction of reactant molecules that are activated at the given temp" …
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Q.Derive an expression to calculate time required for completion of zero order reaction.

Concept understanding — Zero Order Kinetics

Zero Order Kinetics: The Drug That Doesn't Care How Much You Give It

The Precise Statement

The Integrated Form and Half-Life

Where Does Zero Order Kinetics Actually Happen?

A Quick Comparison Table

Why this formula?

Zero Order Kinetics: Why the Formula Holds

The Core Idea

Why the Rate is Constant

Deriving the Zero Order Rate Law

Step 1: Write the rate definition

Step 2: Separate variables and integrate

Step 3: Rearrange to the familiar form

What This Formula Tells Us

The Concept: Why Zero-Order Reactions Are Different

Deriving the Expression Step by Step

Method: Integrated Rate Law Approach for Zero-Order Reactions

Steps

Key Points for Exams …

Common Mistakes: Time for Completion of Zero Order Reaction

✗ Mistake 1: Using the First Order Formula

✗ Mistake 2: Forgetting Units of $k$

✗ Mistake 3: Confusing "Completion" with "Half-Life"

✗ Mistake 4: Writing the Derivation Backwards

Concept and Intuition

Step‑by‑Step Derivation

The Relevant Concept

Step-by-Step Analysis

Q.Derive an expression to calculate time required for completion of zero order reaction.

The Concept: Why Zero-Order Reactions Are Different

Deriving the Expression Step by Step

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