Write the rate equation for the reaction $2A + B \rightarrow C$ if the order of the reaction is zero.

> [!TLDR] > In zero-order kinetics, the rate is independent of reactant concentrations, so the rate equation is simply $\text{Rate} = k$, where $k$ is the rate constant with units of concentration/time. ## The Concept: What Zero Order Really Means When we say a reaction is **zero order**, we mean the rate does **not** depend on how much reactant is present. Think of it like a conveyor belt moving at a fixed speed — it doesn't matter if the belt is full or nearly empty, it moves at the same pace. For the reaction $2A + B \rightarrow C$, zero order means changing the concentrations of $A$ or $B$ has no effect on the rate. This is different from first or second order, where the rate changes with concentration. Zero order often occurs when a catalyst surface is saturated (like in many enzyme reactions) or when a reaction is driven by light absorption. > [!FORMULA] > For any zero-order reaction: > $$ \text{Rate} = k $$ > where $k$ is the rate constant (units: $\text{mol L}^{-1} \text{s}^{-1}$ or similar concentration/time). ## Step-by-Step Derivation 1. **Recall the definition of reaction order.** The order of a reaction is the sum of the exponents to which concentration terms are raised in the rate law. For a general reaction $aA + bB \rightarrow \text{products}$, the rate law is: $$ \text{Rate} = k[A]^m[B]^n $$ where $m$ is the order with respect to $A$, $n$ is the order with respect to $B$, and the overall order is $m + n$. 2. **Apply the zero-order condition.** If the overall order is zero, then $m + n = 0$. Since orders are typically non-negative integers (or fractions, but here it's given as zero), the only possibility is $m = 0$ and $n = 0$. This means the rate depends on neither $[A]$ nor $[B]$. 3. **Write the rate equation.** Substituting $m = 0$ and $n = 0$ into the general form: $$ \text{Rate} = k[A]^0[B]^0 $$ Any quantity raised to the power zero equals 1, so: $$ \text{Rate} = k \times 1 \times 1 = k $$ 4. **Interpret

Write the rate equation for the reaction 2A + B arrow C if the order of the reaction is zero.

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2A + B \rightarrow C

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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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In zero-order kinetics, the rate is independent of reactant concentrations, so the rate equation is simply $\text{Rate} = k$ , where $k$ is the rate constant with units of concentration/time.

The Concept: What Zero Order Really Means

When we say a reaction is zero order, we mean the rate does not depend on how much reactant is present. Think of it like a conveyor belt moving at a fixed speed — it doesn't matter if the belt is full or nearly empty, it moves at the same pace. For the reaction $2A + B \rightarrow C$ , zero order means changing the concentrations of $A$ or $B$ has no effect on the rate.

This is different from first or second order, where the rate changes with concentration. Zero order often occurs when a catalyst surface is saturated (like in many enzyme reactions) or when a reaction is driven by light absorption.

\text{Rate} = k

Step-by-Step Derivation

Recall the definition of reaction order. The order of a reaction is the sum of the exponents to which concentration terms are raised in the rate law. For a general reaction $aA + bB \rightarrow \text{products}$ , the rate law is:

\text{Rate} = k[A]^m[B]^n

where $m$ is the order with respect to $A$ , $n$ is the order with respect to $B$ , and the overall order is $m + n$ .

Apply the zero-order condition.

If the overall order is zero, then $m + n = 0$ . Since orders are typically non-negative integers (or fractions, but here it's given as zero), the only possibility is $m = 0$ and $n = 0$ . This means the rate depends on neither $[A]$ nor $[B]$ .
Write the rate equation.

Substituting $m = 0$ and $n = 0$ into the general form:

\text{Rate} = k[A]^0[B]^0

Any quantity raised to the power zero equals 1, so:

\text{Rate} = k \times 1 \times 1 = k

Interpret the result. …

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Here are the most common mistakes students make on this question, along with how to avoid each one.

Mistake 1: Writing a Rate Law Based on Stoichiometric Coefficients

The Mistake:

Students see the coefficients (2 for A, 1 for B) and write:

\text{Rate} = k[A]^2[B]^1

This is the most frequent error. It confuses the molecularity (how many molecules collide in the elementary step) with the order (which is determined experimentally).

Why it happens:

In school, many first examples (like $2HI \rightarrow H_2 + I_2$ ) happen to have the rate law match the coefficients. Students memorize "rate = k[reactant]^coefficient" without understanding that this is only true for elementary reactions — and even then, the problem explicitly tells you the order here.

How to Avoid:

Always separate "stoichiometry" from "order". The coefficients tell you how much of each reactant is consumed, not how the rate depends on concentration.
Read the question carefully. The phrase "order of the reaction is zero" is the only instruction you need. Ignore the coefficients entirely when writing the rate law.

Mistake 2: Including the Reactants in the Rate Equation at All

The Mistake:

Students write something like:

\text{Rate} = k[A]^0 \quad \text{or} \quad \text{Rate} = k[A]^0[B]^0

Why it happens:

They know zero order means exponent = 0, but they still feel compelled to list the reactants. They think the rate law must include every reactant.

How to Avoid:

Remember the definition: For a zero-order reaction, the rate is independent of the concentration of any reactant.
Simplify: $[A]^0 = 1$ and $[B]^0 = 1$ . So the correct equation is simply:

\text{Rate} = k

Think of it as a constant. The rate does not change as the reaction proceeds (until the reactant runs out). There is no variable concentration in the rate law.

Mistake 3: Confusing the Rate Equation with the Integrated Rate Law

The Mistake:

Students write the integrated form (how concentration changes with time) instead of the differential rate equation:

[A] = [A]_0 - kt

Why it happens: …

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Q.Write the rate equation for the reaction $2A + B \rightarrow C$ if the order of the reaction is zero.

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: What Zero Order Really Means

Step-by-Step Derivation

Method: Direct Application of the Zero-Order Rate Law

Why this method works

Steps

Mistake 1: Writing a Rate Law Based on Stoichiometric Coefficients

Mistake 2: Including the Reactants in the Rate Equation at All

Mistake 3: Confusing the Rate Equation with the Integrated Rate Law

Q.Write the rate equation for the reaction 2A+B→C2A + B \rightarrow C2A+B→C if the order of the reaction is zero.

The Concept: What Zero Order Really Means

Step-by-Step Derivation

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Q.Write the rate equation for the reaction $2A + B \rightarrow C$ if the order of the reaction is zero.

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