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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Q.Derive an expression to calculate time required for completion of zero order reaction.

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: Why Zero-Order Reactions Are Different

Deriving the Expression Step by Step

Method: Integrated Rate Law Approach for Zero-Order Reactions

Steps

Key Points for Exams …

Common Mistakes: Time for Completion of Zero Order Reaction

✗ Mistake 1: Using the First Order Formula

✗ Mistake 2: Forgetting Units of $k$

✗ Mistake 3: Confusing "Completion" with "Half-Life"

✗ Mistake 4: Writing the Derivation Backwards

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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