The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: | Experiment | [A]/$\text{mol L}^{-1}$ | [B]/$\text{mol L}^{-1}$ | Initial rate/$\text{mol L}^{-1}\text{min}^{-1}$ | |---|---|---|---| | I | 0.1 | 0.1 | $2.0\times10^{-2}$ | | II | – | 0.2 | $4.0\times10^{-2}$ | | III | 0.4 | 0.4 | – | | IV | – | 0.2 | $2.0\times10^{-2}$ |

> [!TLDR] > The rate law is $\text{Rate} = k[A]^1[B]^0 = k[A]$. Using the data from Experiment I, we find $k = 0.2\ \text{min}^{-1}$. Then, for Experiment II, $[A] = 0.2\ \text{mol L}^{-1}$; for Experiment III, the rate is $8.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$; and for Experiment IV, $[A] = 0.1\ \text{mol L}^{-1}$. The problem gives us a reaction that is **first order with respect to A** and **zero order with respect to B**. That means the rate depends only on the concentration of A, and not at all on the concentration of B. This is the key insight — B’s concentration is irrelevant to the speed of the reaction, so any change in B alone will not affect the rate. The general form of the rate law is: $$ \text{Rate} = k[A]^1[B]^0 = k[A] $$ where $k$ is the rate constant. Our job is to find the missing values in the table using this relationship. 1. **Find the rate constant $k$ from Experiment I.** We have $[A] = 0.1\ \text{mol L}^{-1}$ and $\text{Rate} = 2.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$. Substituting into the rate law: $$ 2.0 \times 10^{-2} = k \times 0.1 $$ Solving for $k$: $$ k = \frac{2.0 \times 10^{-2}}{0.1} = 0.2\ \text{min}^{-1} $$ This $k$ is constant for all experiments at the same temperature. 2. **Find $[A]$ in Experiment II.** Here $[B] = 0.2\ \text{mol L}^{-1}$ and $\text{Rate} = 4.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$. Since the rate depends only on $[A]$, we write: $$ 4.0 \times 10^{-2} = 0.2 \times [A] $$ So: $$ [A] = \frac{4.0 \times 10^{-2}}{0.2} = 0.2\ \text{mol L}^{-1} $$ Notice that $[B]$ doubled from Experiment I (0.1 to 0.2), but the rate doubled because $[A]$ doubled — not because of B. This confirms zero order in B. 3. **Find the rate in Experiment III.** We have $[A] = 0.4\ \text{mol L}^{-1}$ and $[B] = 0.4\ \text{mol L}^{-1}$. Using the rate law: $$ \text{Rate} = 0.2 \times 0.4 = 0.08\ \text{mol L}^{-1}\

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the bl

Exercises · 3.12

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Experiment	[A]/ $\text{mol L}^{-1}$	[B]/ $\text{mol L}^{-1}$	Initial rate/ $\text{mol L}^{-1}\text{min}^{-1}$
I	0.1	0.1	$2.0\times10^{-2}$
II	–	0.2	$4.0\times10^{-2}$
III	0.4	0.4	–
IV	–	0.2	$2.0\times10^{-2}$

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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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The rate law is $\text{Rate} = k[A]^1[B]^0 = k[A]$ . Using the data from Experiment I, we find $k = 0.2\ \text{min}^{-1}$ . Then, for Experiment II, $[A] = 0.2\ \text{mol L}^{-1}$ ; for Experiment III, the rate is $8.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$ ; and for Experiment IV, $[A] = 0.1\ \text{mol L}^{-1}$ .

The problem gives us a reaction that is first order with respect to A and zero order with respect to B. That means the rate depends only on the concentration of A, and not at all on the concentration of B. This is the key insight — B’s concentration is irrelevant to the speed of the reaction, so any change in B alone will not affect the rate.

The general form of the rate law is:

\text{Rate} = k[A]^1[B]^0 = k[A]

where $k$ is the rate constant. Our job is to find the missing values in the table using this relationship.

Find the rate constant $k$ from Experiment I. We have $[A] = 0.1\ \text{mol L}^{-1}$ and $\text{Rate} = 2.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$ . Substituting into the rate law:

2.0 \times 10^{-2} = k \times 0.1

Solving for $k$ :

k = \frac{2.0 \times 10^{-2}}{0.1} = 0.2\ \text{min}^{-1}

This $k$ is constant for all experiments at the same temperature.

Find $[A]$ in Experiment II. Here $[B] = 0.2\ \text{mol L}^{-1}$ and $\text{Rate} = 4.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$ . Since the rate depends only on $[A]$ , we write:

4.0 \times 10^{-2} = 0.2 \times [A]

So:

[A] = \frac{4.0 \times 10^{-2}}{0.2} = 0.2\ \text{mol L}^{-1}

Notice that $[B]$ doubled from Experiment I (0.1 to 0.2), but the rate doubled because $[A]$ doubled — not because of B. This confirms zero order in B.

Find the rate in Experiment III. We have $[A] = 0.4\ \text{mol L}^{-1}$ and $[B] = 0.4\ \text{mol L}^{-1}$ . Using the rate law:

\text{Rate} = 0.2 \times 0.4 = 0.08\ \text{mol L}^{-1}\text{min}^{-1}

In scientific notation: $8.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$ .

Find $[A]$ in Experiment IV. Here $[B] = 0.2\ \text{mol L}^{-1}$ and $\text{Rate} = 2.0 \times 10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$ . Again:

2.0 \times 10^{-2} = 0.2 \times [A]

So:

$[A] = \frac{2.0 \times 10^{-2}}{0.2} = 0.1\ \text{mol L}^{-1}$ …

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Method: Rate Law Analysis Using the Method of Initial Rates

Step 1: Write the rate law from the given orders

The reaction is:

First order with respect to A
Zero order with respect to B

So the rate law is:

\text{Rate} = k[\text{A}]^{1}[\text{B}]^{0} = k[\text{A}]

This means the rate depends only on $[\text{A}]$ , not on $[\text{B}]$ .

Step 2: Find the rate constant $k$ using Experiment I

From Experiment I:

[\text{A}] = 0.1\ \text{mol L}^{-1}, \quad \text{Rate} = 2.0 \times 10^{-2}\ \text{mol L}^{-1} \text{min}^{-1}

Substitute into the rate law:

2.0 \times 10^{-2} = k \times 0.1

k = \frac{2.0 \times 10^{-2}}{0.1} = 0.2\ \text{min}^{-1}

Step 3: Fill Experiment II — find $[\text{A}]$

Given: $[\text{B}] = 0.2$ , Rate $= 4.0 \times 10^{-2}$

Since rate depends only on $[\text{A}]$ :

4.0 \times 10^{-2} = 0.2 \times [\text{A}]

[\text{A}] = \frac{4.0 \times 10^{-2}}{0.2} = 0.2\ \text{mol L}^{-1}

Experiment II: $[\text{A}] = \mathbf{0.2}$

Step 4: Fill Experiment III — find the rate

Given: $[\text{A}] = 0.4$ , $[\text{B}] = 0.4$

Rate depends only on $[\text{A}]$ :

\text{Rate} = 0.2 \times 0.4 = 8.0 \times 10^{-2}\ \text{mol L}^{-1} \text{min}^{-1}

Experiment III: Rate $= \mathbf{8.0 \times 10^{-2}}$

Step 5: Fill Experiment IV — find $[\text{A}]$

Given: $[\text{B}] = 0.2$ , Rate $= 2.0 \times 10^{-2}$

$2.0 \times 10^{-2} = 0.2 \times [\text{A}]$ …

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Mistake 1: Forgetting that zero order with respect to B means B does not affect the rate

What students do wrong:

They write the rate law as

\text{Rate} = k[A]^1[B]^0

but then still try to use the concentration of B to compare rates between experiments.

Example of error:

In Experiment II, they see [B] doubled (0.1 → 0.2) and think the rate should double — but it doesn’t.

How to avoid:

Remember: zero order means the exponent is 0, so $[B]^0 = 1$ .
The rate depends only on [A] (since it’s first order in A).
So the rate law is simply:

\text{Rate} = k[A]

Mistake 2: Using the wrong experiment to find the order or the constant

What students do wrong:

They try to compare Experiment I and II to find the order with respect to B, but [A] is unknown in II.

How to avoid:

To find the order with respect to a reactant, you must compare experiments where only that reactant’s concentration changes and all others are constant.
Here, you cannot find order from I and II because both [A] and [B] change.
Instead, use the given orders to write the rate law directly.

Mistake 3: Incorrectly calculating the missing [A] in Experiment II

What students do wrong:

They assume [A] in II is the same as in I (0.1), then get confused when the rate doubles.

Correct approach:

From I:

2.0 \times 10^{-2} = k (0.1) \quad \Rightarrow \quad k = 0.2 \, \text{min}^{-1}

From II:

4.0 \times 10^{-2} = k [A]_{\text{II}} = 0.2 [A]_{\text{II}}

[A]_{\text{II}} = \frac{4.0 \times 10^{-2}}{0.2} = 0.2 \, \text{mol L}^{-1}

How to avoid:

Always calculate $k$ from a complete row first.
Then use that $k$ to find unknowns in other rows.

Mistake 4: Forgetting to check consistency — Experiment IV is a trap

What students do wrong:

They see [B] = 0.2 in IV and [B] = 0.2 in II, and assume [A] must also be the same. But the rates are different.

Correct reasoning:

In IV: rate = $2.0 \times 10^{-2}$ , so

2.0 \times 10^{-2} = 0.2 [A]_{\text{IV}} \quad \Rightarrow \quad [A]_{\text{IV}} = 0.1 \, \text{mol L}^{-1}

So [A] in IV is 0.1, not 0.2.
This shows that even with same [B], rate changes with [A] — confirming B has no effect.

How to avoid: …

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