4.1 Zero-order integrated rate law

Zero-Order Integrated Rate Law

Zero-order reactions are unique in chemical kinetics because the reaction rate stays constant no matter how much reactant is present. The zero-order integrated rate law, $[A] = [A]_0 - kt$, lets you predict reactant concentration at any point in time. This behavior shows up most often in enzyme-catalyzed reactions (where the enzyme is saturated) and surface reactions (where the catalyst surface is fully covered).

Derivation of the Zero-Order Rate Law

The derivation starts from the differential rate law for a zero-order reaction:

$\frac{d[A]}{dt} = -k$

Notice that the right side is just $-k$, with no concentration term. That's what makes it zero-order: the rate doesn't depend on $[A]$ at all.

From here, you can derive the integrated form in a few steps:

1. Rearrange to isolate $d[A]$: $d[A] = -k \cdot dt$

2. Integrate both sides, using $[A]_0$ (initial concentration) and $[A]$ (concentration at time $t$) as your limits: $\int_{[A]_0}^{[A]} d[A] = -k \int_{0}^{t} dt$

3. Evaluate both integrals: $[A] - [A]_0 = -kt$

4. Rearrange to get the zero-order integrated rate law: $[A] = [A]_0 - kt$

This equation has the form $y = b + mx$, which is why a plot of $[A]$ vs. $t$ gives a straight line for zero-order reactions.

Applying the Zero-Order Rate Law

To find the concentration of reactant A at any time $t$:

1. Identify your known values: initial concentration $[A]_0$, rate constant $k$, and the time $t$ you're interested in.
2. Plug them into $[A] = [A]_0 - kt$ and solve.

Example: If $[A]_0 = 1.0 \text{ M}$, $k = 0.2 \text{ M/s}$, and $t = 3 \text{ s}$:

$[A] = 1.0 \text{ M} - (0.2 \text{ M/s})(3 \text{ s}) = 0.4 \text{ M}$

One thing to watch for: the equation can give you a negative concentration if you plug in a large enough $t$. Physically, concentration can't go below zero. The reaction simply stops when $[A] = 0$, so the linear model only applies up to that point.

Characteristics of Zero-Order Reactions

A zero-order reaction has several distinctive features you should recognize:

• A plot of $[A]$ vs. $t$ is a straight line with a negative slope.
• The slope of that line equals $-k$ (the negative of the rate constant).
• The y-intercept equals $[A]_0$, the initial concentration.
• Concentration decreases at a steady rate over time, unlike first- or second-order reactions where the rate changes as reactant is consumed.

The rate itself is constant and independent of $[A]$. This typically happens when something other than reactant availability controls the rate. For example, in an enzyme-catalyzed reaction at high substrate concentration, the enzyme is fully saturated, so adding more substrate doesn't speed things up. Similarly, in a surface-catalyzed reaction, if every active site on the catalyst is occupied, the rate depends only on how fast the surface processes molecules.

Calculating the Rate Constant

You can rearrange the integrated rate law to solve for $k$:

$k = \frac{[A]_0 - [A]}{t}$

To calculate $k$ from experimental data:

1. Measure the initial concentration $[A]_0$ and the concentration $[A]$ at a known time $t$.
2. Substitute into the equation above and solve.

Example: If $[A]_0 = 1.0 \text{ M}$, $[A] = 0.4 \text{ M}$, and $t = 3 \text{ s}$:

$k = \frac{1.0 \text{ M} - 0.4 \text{ M}}{3 \text{ s}} = 0.2 \text{ M/s}$

Note the units of $k$ for a zero-order reaction: M/s (or mol·L⁻¹·s⁻¹). This is different from first-order (s⁻¹) and second-order (M⁻¹·s⁻¹) rate constants. Checking your units is a quick way to verify you've identified the correct reaction order.