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.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$T(K)$
›Reveal solutionSolution
The units of $k$ ( $mol\,L^{-1}\,min^{-1}$ ) identify this as a zero-order reaction; applying $[A]=[A]_0-kt$ gives $[A]=0.6\ mol\,L^{-1}$ after 20 minutes.

Concept and Intuition

The units of a rate constant reveal the reaction order without needing any other data:

Zero order: $k$ has units of concentration/time ( $mol\,L^{-1}\,s^{-1}$ or similar).

First order: $k$ has units of (time) $^{-1}$ only.

Second order: $k$ has units of $(concentration)^{-1}(time)^{-1}$ . Here $k = 0.02\ mol\,L^{-1}\,min^{-1}$ has units of concentration/time, so this is unambiguously a zero-order reaction, and the integrated rate law to use is the simple linear one, $[A] = [A]_0 - kt$ .

Step-by-Step Solution

Identify order from units of $k$ : $mol\,L^{-1}\,min^{-1} \Rightarrow$ zero order.

Zero-order integrated rate law: $[A] = [A]_0 - kt$ .

Substitute $[A]_0 = 1.0\ mol\,L^{-1}$ , $k=0.02\ mol\,L^{-1}\,min^{-1}$ , $t = 20$ min: …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$t = 10$
›Reveal solutionSolution
For a zero-order reaction the rate is constant with time, so the rate at $t=20$ min equals the rate at $t=10$ min, i.e. $x$ .

Concept and Intuition

For a zero-order reaction, rate $= k[A]^0 = k$ , a constant that does not depend on the concentration of reactant present. Physically, this happens for the catalytic decomposition of $NH_3$ on a hot metal (Pt) surface at high pressure, where the metal surface is fully saturated with adsorbed $NH_3$ molecules — the reaction rate is then limited only by the fixed number of active catalytic sites, not by how much $NH_3$ is left in the gas phase. As a result the rate does not change as the reaction proceeds (until the surface is no longer saturated).

Step-by-Step Solution

Zero-order kinetics: rate $= k$ (a constant), independent of reactant concentration and therefore independent of the time elapsed (as long as the surface stays saturated).

At $t = 10$ min, rate $= x$ (given). …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$[\mathrm{R}]_0$
›Reveal solutionSolution
Zero-order kinetics gives a linear concentration-vs-time relation; setting the remaining concentration to zero directly gives $t=[\mathrm{R}]_0/k$ .

Concept and Intuition

A zero-order reaction has rate independent of concentration: $\text{rate}=k$ , a constant. Integrating $-\dfrac{d[\mathrm{R}]}{dt}=k$ gives a straight-line decay of concentration with time, unlike first-order's exponential decay — so, unusually, a zero-order reaction can reach exactly zero concentration in finite time.

Step-by-Step Solution

Integrated zero-order rate law: $[\mathrm{R}]=[\mathrm{R}]_0-kt$ .

"100% completion" means all reactant is consumed: $[\mathrm{R}]=0$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\mathrm{AB_3(g)}$
›Reveal solutionSolution
A zero-order reaction's rate equals its rate constant at all times, unaffected by reactant concentration — so the rate here is simply the given $k$ .

Concept and Intuition

For a zero-order reaction, $\text{rate}=k[A]^0=k$ . This means the rate does not change as the reaction proceeds or as concentration varies — a defining and easily-testable feature of zero-order kinetics.

Step-by-Step Solution

Rate law for zero order: $\text{rate}=k$ .

Given $k=2.5\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}$ at 300 K. …
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AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$A \rightarrow P$
›Reveal solutionSolution
Zero-order kinetics means concentration decreases linearly with time; plugging in gives $[A] = 0.09$ mol/L after 10 s.

Concept and Intuition

For a zero-order reaction $A \to P$ , the rate is independent of concentration: $\text{rate}=k$ (constant). Integrating $-\dfrac{d[A]}{dt}=k$ gives the linear law $[A]_t = [A]_0 - kt$ — concentration drops at a constant rate over time, unlike first-order kinetics where it decays exponentially.

Step-by-Step Solution

Zero-order integrated rate law: $[A]_t = [A]_0 - kt$ .

Given: $[A]_0 = 0.1$ mol L $^{-1}$ , $k = 1\times10^{-3}$ mol L $^{-1}$ s $^{-1}$ , $t = 10$ s.

$kt = 1\times10^{-3} \times 10 = 0.01$ mol L $^{-1}$ .

$[A]_{10} = 0.1 - 0.01 = 0.09$ mol L $^{-1}$ . …
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AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$\mathrm{2NH_3(g) \xrightarrow[1130\,K]{Pt} N_2(g) + 3H_2(g)}$
›Reveal solutionSolution
For a zero-order reaction the "rate" equals $k$ directly, and each species' rate of formation/consumption is scaled by its stoichiometric coefficient — giving $3y\times10^{-4}$ for H $_2$ .

Concept and Intuition

For $2NH_3\to N_2+3H_2$ , the reaction rate is defined as $\text{Rate}=-\dfrac{1}{2}\dfrac{d[NH_3]}{dt}=\dfrac{d[N_2]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}$ . For a zero-order reaction this common rate equals the rate constant $k$ itself (units mol L $^{-1}$ s $^{-1}$ match).

Step-by-Step Solution

Zero order $\Rightarrow$ Rate $=k=y\times10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$ .

Rate $=\dfrac13\dfrac{d[H_2]}{dt}\Rightarrow \dfrac{d[H_2]}{dt}=3\times\text{Rate}=3k$ .

$=3y\times10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$ . …
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AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$\mathrm{A \rightarrow products}$
›Reveal solutionSolution
Zero-order integrated rate law directly gives $t = \Delta[A]/k = 8.33\ s$ .

Concept and Intuition

In a zero-order reaction, the rate is independent of concentration -- the concentration falls linearly with time, unlike first/second order reactions where it falls exponentially or hyperbolically. The integrated rate law is simply $[A]_t = [A]_0 - kt$ , a straight line of slope $-k$ .

Step-by-Step Solution

Zero-order integrated rate law: $[A]_t = [A]_0 - kt$ .

Given $[A]_0 = 0.10\ M$ , $[A]_t = 0.075\ M$ , $k = 0.0030\ mol\,L^{-1}\,s^{-1}$ .

Rearranging: $t = \dfrac{[A]_0 - [A]_t}{k}$ .

Substitute: $t = \dfrac{0.10 - 0.075}{0.0030} = \dfrac{0.025}{0.0030} = 8.33\ s$ . …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$^{-1}$
›Reveal solutionSolution
A zero-order reaction's reactant concentration decreases linearly with time; using the given data to find $k$ , then applying stoichiometry ( $A\to3B$ ) gives the product concentration formed at $t=20$ s.

Concept and Intuition

For a zero-order reaction, rate $=k$ is constant (independent of concentration), so $[A]_t = [A]_0 - kt$ — a straight-line decay. The rate of formation of product is scaled by the stoichiometric coefficient: since 3 mol of B appear for every 1 mol of A consumed, $\dfrac{d[B]}{dt} = 3\times\left(-\dfrac{d[A]}{dt}\right) = 3k$ .

Step-by-Step Solution

From the data at $t=0$ ( $[A]=0.1$ ) and $t=15$ s ( $[A]=0.05$ ): $k = \dfrac{0.1-0.05}{15} \approx 0.0033\ \text{mol L}^{-1}\text{s}^{-1}$ .

Amount of A reacted by $t=20$ s: $x = k\times20 \approx 0.0033\times20 = 0.066\ \text{mol/L}$ (this is the " $x$ " in $[A]=0.1-x$ ). …
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AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
$P \rightarrow Q$
›Reveal solutionSolution
For a zero order reaction, concentration falls linearly with time; using the given data to find $k$ , and applying $[P] = [P]_0 - kt$ for $t=15\,\mathrm{s}$ gives $[P] = 0.025$ M.

Concept and Intuition

For a zero order reaction, the rate is independent of reactant concentration, so the integrated rate law is linear in time: $[P] = [P]_0 - kt$ , where $k$ is the (constant) rate. This means equal time intervals always remove the same amount (not the same fraction) of reactant — unlike first order kinetics where equal time intervals remove the same fraction.

Step-by-Step Solution

Use the given data (0.1 M → 0.05 M in 10 s) to find $k$ : $k = \dfrac{[P]_0 - [P]}{t} = \dfrac{0.1-0.05}{10} = \dfrac{0.05}{10} = 0.005\ \mathrm{mol\,L^{-1}\,s^{-1}}$ .

Apply the zero-order integrated law at $t = 15\,\mathrm{s}$ (measuring from the same $t=0$ , $[P]_0=0.1$ M): $[P] = [P]_0 - kt = 0.1 - (0.005)(15)$ . …
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AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$A \rightarrow$
›Reveal solutionSolution
If doubling a reactant's concentration leaves the rate unchanged, the reaction is zero order in that reactant. Answer: 0.

Concept and Intuition

The order of reaction with respect to a species tells you how sensitively the rate depends on that species' concentration: rate $\propto [A]^n$ . If increasing $[A]$ has no effect on the rate, the exponent $n$ must be zero, because any nonzero power of 2 (the doubling factor) would change the rate. Zero-order behaviour typically arises when the rate-determining step doesn't actually involve free $A$ in solution — e.g., a heterogeneous catalytic reaction where the catalyst surface is already saturated with A, so adding more A in solution can't speed anything up.

Step-by-Step Solution

Write the general rate law: rate $= k[A]^n$ .

Let the initial concentration be $[A]$ , so initial rate $r_1 = k[A]^n$ . …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$t_{1/2}$
›Reveal solutionSolution
Zero-order half-life is directly proportional to initial concentration: $t_{1/2}=\dfrac{[A]_0}{2k}$ , so the graph is a straight line through the origin with slope $\frac{1}{2k}$ .

Concept and Intuition

For a zero-order reaction the rate is constant ( $\text{rate}=k$ , independent of concentration), so the integrated rate law is linear in time:
$[A]_t = [A]_0 - kt$
Half-life is the time at which $[A]_t = \dfrac{[A]_0}{2}$ .

Step-by-Step Solution

Set $[A]_t = \dfrac{[A]_0}{2}$ in the integrated law: $\dfrac{[A]_0}{2} = [A]_0 - k t_{1/2}$ .

Solve: $k t_{1/2} = [A]_0 - \dfrac{[A]_0}{2} = \dfrac{[A]_0}{2}$ , so $t_{1/2} = \dfrac{[A]_0}{2k}$ . …
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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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