The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ an

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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	$-\frac{d[A]}{dt} = k_0$	$-\frac{d[A]}{dt} = k_1[A]$
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.

For a zero-order reaction, the rate is constant and equals the rate constant $k$ . Using the stoichiometry of $2NH_3 \rightarrow N_2 + 3H_2$ , the rate of production of $N_2$ is $2.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}$ and that of $H_2$ is $7.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}$ .

The key to this problem lies in understanding what "zero order" means — and then connecting that to the stoichiometric coefficients of the reaction.

In a zero-order reaction, the rate of the reaction does not depend on the concentration of the reactant. The rate is constant throughout the reaction, and the given rate constant $k$ is the rate of the reaction itself (not merely the raw rate of disappearance of one particular species).

The decomposition reaction is:

2NH_3(g) \xrightarrow{Pt} N_2(g) + 3H_2(g)

For every 2 molecules of $NH_3$ that disappear, 1 molecule of $N_2$ and 3 molecules of $H_2$ appear.

aA \rightarrow bB + cC

Let’s apply this step by step.

Write the rate of reaction in terms of $NH_3$ . Since the reaction is zero order, the rate of reaction is:

-\frac{1}{2}\frac{d[NH_3]}{dt} = k = 2.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

This means the actual rate of disappearance of $NH_3$ is $-\frac{d[NH_3]}{dt} = 2k = 5.0 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}$ .

Relate this to the rate of production of $N_2$ . From the balanced equation:

-\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt}

Both sides equal $k$ , so:

\frac{d[N_2]}{dt} = k = 2.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

Relate to the rate of production of $H_2$ . From the balanced equation:

-\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt}

So:

\frac{d[H_2]}{dt} = 3k = 3 \times (2.5 \times 10^{-4}) = 7.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

Watch out

A common mistake is to treat the given $k$ as the raw rate of disappearance of $NH_3$ ( $-d[NH_3]/dt = k$ ) and then divide by the coefficient again when finding the rate of production of $N_2$ and $H_2$ — that double-counts the stoichiometric factor. By the standard definition, $k$ (the zero-order rate constant) already equals the overall rate of reaction $-\frac{1}{2}\frac{d[NH_3]}{dt}$ , so the rate of production of $N_2$ is simply $k$ , and of $H_2$ is $3k$ .

✓Final answer

N_2

Method: Stoichiometric Rate Analysis for Zero-Order Reactions

Why this method?

In a zero-order reaction, the rate is independent of concentration and equals the rate constant $k$ . For a balanced chemical equation, the rates of appearance/disappearance of species are linked by stoichiometric coefficients.

Step 1: Write the balanced equation

The decomposition of ammonia on platinum is:

2NH_3(g) \rightarrow N_2(g) + 3H_2(g)

Step 2: Write the rate expression in terms of $NH_3$

For a zero-order reaction:

\text{Rate} = -\frac{1}{2}\frac{d[NH_3]}{dt} = k

Given:

$k = 2.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}$

So:

-\frac{d[NH_3]}{dt} = 2k = 5.0 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

Step 3: Relate rates of production of $N_2$ and $H_2$

From stoichiometry:

-\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt}

Since each equals $k$ :

\frac{d[N_2]}{dt} = k = 2.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

\frac{d[H_2]}{dt} = 3k = 7.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

Final Answer

Rate of production of $N_2$ = $\mathbf{2.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}}$
Rate of production of $H_2$ = $\mathbf{7.5 \times 10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}}$

Key Exam Tip

In zero-order kinetics, the rate constant $k$ directly gives the rate of reaction. Multiply by the stoichiometric coefficient of the product (as written in the balanced equation) to get its rate of appearance.

Here are the common mistakes students make on this exact type of zero-order kinetics problem, and how to avoid each.

1. Mistake: Forgetting the Stoichiometric Ratios

The error:

Students often assume the rate of disappearance of $NH_3$ equals the rate of appearance of $N_2$ or $H_2$ . They write:

-\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{d[H_2]}{dt}

This is wrong because the balanced equation shows different coefficients.

How to avoid:

Always write the balanced chemical equation first:

2NH_3(g) \xrightarrow{Pt} N_2(g) + 3H_2(g)

Then use the stoichiometric relationship:

-\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt}

2. Mistake: Treating the Given $k$ as the Rate of Disappearance of $NH_3$ Instead of the Rate of Reaction

The error:

For a zero-order reaction, the rate constant $k$ is the overall rate of reaction:

\text{Rate} = -\frac{1}{2}\frac{d[NH_3]}{dt} = k

Some students instead assume $k = -\frac{d[NH_3]}{dt}$ directly (i.e., that $k$ is the raw rate of disappearance of $NH_3$ , ignoring its own coefficient of 2). This introduces an extra, incorrect factor of $\tfrac12$ into every downstream answer.

How to avoid:

By the standard definition used throughout NCERT and Indian board exams, for a reaction $aA \rightarrow bB + cC$ , the rate constant of a zero-order reaction is the common value:

k = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

So here, $k = -\frac12\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac13\frac{d[H_2]}{dt} = 2.5\times10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$ .

3. Mistake: Confusing Rate of Reaction with Rate of Appearance/Disappearance

The error:

Students write:

\text{Rate} = -\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{d[H_2]}{dt}

This ignores coefficients.

How to avoid:

Remember the definition:

\text{Rate of reaction} = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}

For $2NH_3 \rightarrow N_2 + 3H_2$ :

\text{Rate} = -\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt} = k

So:

$\frac{d[N_2]}{dt} = k$
$\frac{d[H_2]}{dt} = 3k$

4. Mistake: Using Integrated Rate Laws Instead of Differential Rate

The error:

Students try to use $[A] = [A]_0 - kt$ to find rates, but the question asks for instantaneous rates at any time (since zero order, rate is constant).

How to avoid:

For zero order, the rate is constant and equal to $k$ (the rate constant for the reaction). You do not need initial concentration or time. Just use the differential rate law:

\text{Rate} = k

Then apply stoichiometry.

5. Mistake: Not Checking the Units of the Final Answer

The error:

Students give answers without units, or with wrong units (e.g., $\text{s}^{-1}$ instead of $\text{mol L}^{-1} \text{s}^{-1}$ ).

How to avoid:

Always include units. For zero-order, rates of appearance/disappearance have units of concentration per time:

\text{mol L}^{-1} \text{s}^{-1}

✓ Final Correct Answer (Summary)

Given $k = 2.5\times10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}$ , which is the rate of the reaction ( $-\frac12\frac{d[NH_3]}{dt}$ ):

Rate of production of $N_2$ :

\frac{d[N_2]}{dt} = k = 2.5\times10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

Rate of production of $H_2$ :

\frac{d[H_2]}{dt} = 3k = 7.5\times10^{-4}\ \text{mol L}^{-1} \text{s}^{-1}

Key takeaway: Always start with the balanced equation, define the rate of reaction, and use stoichiometric coefficients to convert between species — and remember that the given rate constant $k$ for a zero-order reaction already IS the rate of reaction, not the raw disappearance rate of one specific reactant.

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Q.The decomposition of NH3NH_3NH3​ on platinum surface is zero order reaction. What are the rates of production of N2N_2N2​ and H2H_2H2​ if k=2.5×10−4 mol−1 L s−1k = 2.5\times10^{-4}\ \text{mol}^{-1}\,\text{L}\,\text{s}^{-1}k=2.5×10−4 mol−1Ls−1?

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Q.The decomposition of $NH_3$ on platinum surface is zero order reaction. What are the rates of production of $N_2$ and $H_2$ if $k = 2.5\times10^{-4}\ \text{mol}^{-1}\,\text{L}\,\text{s}^{-1}$ ?

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