Starting from the zero order integrated rate equation $[R] = [R]_0 - kt$, derive an expression for the half-life $t_{1/2}$ of a zero order reaction, and explain why this half-life, unlike a first order reaction's, is not a fixed characteristic of the reaction alone.

Question

Accepted Answer

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	$-\frac{d[A]}{dt} = k_0$	$-\frac{d[A]}{dt} = k_1[A]$
Units of rate constant	concentration/time	1/time
Plot of $[A]$ vs. $t$	Straight line	Curved (exponential decay)
Half-life	$[A]_0 / 2k_0$ (depends on starting amount)	$\ln 2 / k_1$ (constant)
Real example	Alcohol elimination at high doses	Most drug metabolism at therapeutic doses

The Intuition Check

If someone asks you "Is this process zero order?", ask yourself: does the rate stay the same even when the amount drops? If you have a machine that destroys exactly 5 units per hour, and you start with 100 units, after 10 hours you'll have 50 units left. After another 10 hours, you'll have 0. The machine doesn't slow down as the pile shrinks — that's zero order.

If instead the machine destroys 5% of what's left per hour, then in the first hour it destroys 5 units (5% of 100), but in the tenth hour it destroys only about 3 units (5% of 60). The amount destroyed per hour keeps dropping — that's first order.

Zero order is the exception, not the rule. It happens when something is saturated — enzymes, transporters, or any system that has a maximum processing capacity. When you meet it in an exam problem, the dead giveaway is a straight line on a concentration-time graph, or a half-life that changes with the starting dose.

Zero order kinetics is a distinctive case within the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘zero order reaction graph’ or ‘zero order kinetics examples’ are recurring important-question searches for board exams as well as JEE Main and NEET. Recognising a zero-order reaction from its straight-line concentration-time graph is a skill directly tested in competitive-exam MCQs.

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.

Real-World Examples (for context)

Example	Why it's zero order
Drug dissolution (e.g., a sustained-release tablet)	Surface area is constant; drug saturates the boundary layer
Enzyme-catalyzed reactions (at high substrate)	Enzyme active sites are fully occupied (saturation)
Photochemical reactions (with constant light)	Light intensity (not reactant concentration) limits the rate

Quick Summary for Exams

Property	Zero Order
Rate law	$-\frac{d[A]}{dt} = k$
Integrated form	$[A]_t = [A]_0 - kt$
Plot for straight line	$[A]_t$ vs. $t$
Slope	$-k$
Half-life	$t_{1/2} = \frac{[A]_0}{2k}$
Units of $k$	$\text{concentration} \cdot \text{time}^{-1}$

Remember: The reason for zero order is always a saturation or surface limitation — the rate can't go faster because something else (not the reactant) is the bottleneck.

Substituting $[R] = [R]_0/2$ and $t = t_{1/2}$ into $[R] = [R]_0 - kt$ gives $\dfrac{[R]_0}{2} = [R]_0 - k\,t_{1/2}$ , so $k\,t_{1/2} = [R]_0 - \dfrac{[R]_0}{2} = \dfrac{[R]_0}{2}$ , giving $t_{1/2} = \dfrac{[R]_0}{2k}$ . Because $t_{1/2}$ is directly proportional to $[R]_0$ , it is not a fixed number characteristic of the reaction alone (unlike $k$ itself) -- it explicitly depends on how much reactant is present at the start of the interval being considered. Consequently, for a zero order reaction, the SECOND half-life (starting from $[R]_0/2$ ) is only half as long as the first (since the 'initial' concentration for that second interval is itself only $[R]_0/2$ ), the third half-life is half of the second, and so on -- each successive half-life is half the one before it, rather than staying constant as it does for a first order reaction. [!ANSWER] $t_{1/2} = [R]_0/(2k)$ , proportional to the starting concentration; since the 'starting' concentration keeps halving as the reaction proceeds, each successive half-life for a zero order reaction is itself only half as long as the one before it -- it is not a fixed constant of the reaction.

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

CBSE 2025Set 56/6/11 markMCQ
$Q. The unit of rate and rate constant are same for a : (A) First order reaction (B) Second order reaction (C) Zero order reaction (D) Third order reaction$
›Reveal solutionSolution
For a zero-order reaction, the rate is independent of concentration, so the rate itself has units of concentration/time — which is exactly the same as the units of the rate constant $k$ . Hence, the correct option is (C).

The key here is to understand what "unit of rate" and "unit of rate constant" actually mean, and why they coincide only for one specific order.

For any reaction, the rate is defined as the change in concentration per unit time. So its unit is always:
$\text{mol L}^{-1} \text{s}^{-1} \quad \text{(or M s}^{-1}\text{)}$
That never changes — it's the definition of rate.

The rate constant $k$ , however, gets its units from the rate law. The rate law for an $n$ th-order reaction is:
$\text{Rate} = k [\text{reactant}]^n$
If you plug in the units:
$[\text{Rate}] = [k] \times ([\text{concentration}])^n$
So:
$\text{M s}^{-1} = [k] \times \text{M}^n$
Therefore:
$[k] = \text{M}^{1-n} \text{s}^{-1}$
Now, the question asks: when are the units of rate and rate constant the same? That means:
$\text{M s}^{-1} = \text{M}^{1-n} \text{s}^{-1}$
For this to hold, the powers of M must match:
$1 = 1 - n \quad \Rightarrow \quad n = 0$
So only for zero-order reactions does $k$ have the same units as the rate itself.

Let's check each option:

First order ( $n=1$ ): $[k] = \text{M}^{1-1} \text{s}^{-1} = \text{s}^{-1}$ . That's per time, not concentration per time — different from rate.

Second order ( $n=2$ ): $[k] = \text{M}^{1-2} \text{s}^{-1} = \text{M}^{-1} \text{s}^{-1}$ . Different again.

Zero order ( $n=0$ ): $[k] = \text{M}^{1-0} \text{s}^{-1} = \text{M s}^{-1}$ . Exactly the same as rate.

Third order ( $n=3$ ): $[k] = \text{M}^{1-3} \text{s}^{-1} = \text{M}^{-2} \text{s}^{-1}$ . Different.

Watch out

A common mistake is to think that "unit of rate and rate constant are same" means the numerical values are equal — they're not. It's about the dimensions (units) being identical. For zero order, both are concentration/time.

Tip

You can also think physically: in zero-order kinetics, the rate doesn't depend on how much reactant is left — it's constant. So the rate constant is the rate itself (just a fixed number), hence they share units. For any other order, $k$ has to "adjust" its units to cancel the concentration dependence.

✓Final answer
$The correct option is (C) Zero order reaction .$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Example of a zero order reaction is:$
(a) CH3CHO → CH4 + CO
(b) 2NH3 --Pt--> N2 + 3H2
(c) NO + O3 → NO2 + O2
(d) N2O5 → 2NO2 + ½ O2.
›Reveal solutionSolution

A zero-order reaction has a rate independent of reactant concentration; the decomposition of ammonia on a hot Pt surface is the classic textbook example, because the metal surface (not the gas concentration) controls the rate once it is saturated.

In a zero-order reaction, rate $= k[A]^0 = k$ — the rate does not depend on the concentration of the reactant at all. This typically happens in heterogeneous catalytic reactions where the reaction occurs only on the catalyst's surface, and once the surface is fully covered (saturated) with reactant molecules, adding more gas-phase reactant cannot speed up the reaction any further — the rate is then limited only by the fixed number of active surface sites.

The decomposition of ammonia on a hot platinum surface,

$2NH_3 \xrightarrow{Pt \text{ surface}} N_2 + 3H_2$

is the standard NCERT example of a zero-order reaction, because at ordinary pressures the Pt surface is saturated with $NH_3$ molecules, so the rate stays constant regardless of $[NH_3]$ .

(The other options — CH₃CHO decomposition, NO + O₃, and N₂O₅ decomposition — are first or second order gas-phase reactions.)

✓Final answer
$2NH_3 \xrightarrow{Pt} N_2 + 3H_2$
CBSE 2024Set ANNUAL1 markMCQ
$Q. The units of reaction rate constant are mol L-1 s-1, what is the order of this reaction?$
(a) 2
(b) 0
(c) 1
(d) 4
›Reveal solutionSolution

The units of a rate constant reveal the reaction order; mol L⁻¹ s⁻¹ (same units as rate itself) means order = 0.

For a reaction of order n, the rate law is rate = k[conc]ⁿ, so k has units of mol^(1-n) L^(n-1) s⁻¹ (concentration in mol/L, time in s). Setting this general formula equal to the given units mol L⁻¹ s⁻¹ requires (1-n) = 1 and (n-1) = -1, both giving n = 0. Physically, k having the SAME units as the rate itself (mol L⁻¹ s⁻¹) is the signature of a zero-order reaction, where rate = k regardless of concentration.

✓Final answer
$The reaction is zero order (option b).$
CBSE 2023Set 56/3/11 markMCQ
$Q. The unit of the rate of reaction is the same as that of the rate constant for a: (A) first order reaction (B) second order reaction (C) zero order reaction (D) it cannot be same$
›Reveal solutionSolution
For a zero order reaction, the rate is independent of concentration and equals the rate constant itself, so both have the same units (e.g., mol L⁻¹ s⁻¹). The correct option is (C).

Why This Question Tests Your Understanding of Order

The trick here is not about memorising formulas — it's about what "order" actually means for the units. The rate of a reaction always has units of concentration per time (like $\text{mol L}^{-1} \text{s}^{-1}$ ). But the rate constant's units change depending on the order, because the rate law includes concentration terms raised to different powers.

For a general reaction of order $n$ , the rate law is:
$\text{Rate} = k [\text{reactant}]^n$
If you rearrange for $k$ :
$k = \frac{\text{Rate}}{[\text{reactant}]^n}$
So the units of $k$ are:
$\text{Units of } k = \frac{\text{concentration} \cdot \text{time}^{-1}}{(\text{concentration})^n} = (\text{concentration})^{1-n} \cdot \text{time}^{-1}$
Now, the question asks: when are the units of the rate and the rate constant the same? That happens when the rate law has no concentration dependence — meaning $n = 0$ .

Step-by-Step Reasoning

Write the general relationship.

For any reaction of order $n$ , the rate is $r = k [A]^n$ . The rate $r$ always has units of $\text{concentration} \cdot \text{time}^{-1}$ (e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ).

Find when $k$ has the same units as $r$ .

For $k$ to have the same units as $r$ , the factor $[A]^n$ must be dimensionless — i.e., $n = 0$ . Then $r = k [A]^0 = k$ , so the rate literally equals the rate constant.

Check each option.

(A) First order ( $n=1$ ): $k$ has units of $\text{time}^{-1}$ (e.g., $\text{s}^{-1}$ ), different from rate.

(B) Second order ( $n=2$ ): $k$ has units of $\text{concentration}^{-1} \cdot \text{time}^{-1}$ (e.g., $\text{L mol}^{-1} \text{s}^{-1}$ ), different.

(C) Zero order ( $n=0$ ): $k$ has units of $\text{concentration} \cdot \text{time}^{-1}$ , exactly the same as rate.

(D) It cannot be same: False, because zero order is a counterexample.

Watch out

A common mistake is to think that for a first order reaction, the rate constant's unit ( $\text{s}^{-1}$ ) somehow matches the rate's unit ( $\text{mol L}^{-1} \text{s}^{-1}$ ). They both have "per second", but the rate also has a concentration term — so they are not the same.

Tip

A quick way to remember: for zero order, the rate is constant (doesn't depend on concentration), so the rate constant is the rate. Hence, same units.

$\text{Units of } k = (\text{concentration})^{1-n} \cdot \text{time}^{-1}$
✓Final answer
$The correct option is (C) — zero order reaction.$
CBSE 2023Set TERM21 mark
$Q. The units of rate constant for zero order reaction are \dots\dots\dots\dots\dots$
›Reveal solutionSolution

For a zero order reaction, rate = k, so k carries the same units as rate: mol L⁻¹ time⁻¹.

For a general reaction of order $n$ , the rate law is $Rate = k[A]^n$ , and by dimensional analysis:

$k = \dfrac{Rate}{[A]^n} \;\Rightarrow\; \text{units of } k = (mol\,L^{-1})^{1-n} time^{-1}$

For a zero order reaction, $n = 0$ , so:

$Rate = k[A]^0 = k$

This means $k$ has the same units as the rate of reaction itself, i.e. $mol\,L^{-1}\,s^{-1}$ (or $mol\,L^{-1}\,time^{-1}$ , depending on the time unit used).

✓Final answer
$mol\,L^{-1}\,s^{-1}$

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