DAT General Chemistry: Zero-Order and Second-Order Kinetics

DAT High-Yield Concept

Zero-Order Reaction Rate: Independent of reactant concentration ( $\text{Rate} = k$ ). Plotting $[\ce{A}]$ vs. $t$ yields a straight line with slope $-k$ . Rate constant units are $\pu{M s^{-1}}$ .
Second-Order Reaction Rate: Proportional to the square of concentration ( $\text{Rate} = k[\ce{A}]^2$ ). Plotting $\frac{1}{[\ce{A}]}$ vs. $t$ yields a straight line with a POSITIVE slope $+k$ . Rate constant units are $\pu{M^{-1} s^{-1}}$ .
Half-Life Behavior: Zero-order half-life ( $t_{1/2} = \frac{[\ce{A}]_0}{2k}$ ) is directly proportional to initial concentration and gets shorter over time. Second-order half-life ( $t_{1/2} = \frac{1}{k[\ce{A}]_0}$ ) is inversely proportional to initial concentration and doubles each half-life.

Zero-Order Kinetics

A zero-order reaction has a rate that is completely independent of reactant concentration. This behavior occurs when the reaction rate is limited by a non-concentration factor, such as light intensity or saturated enzyme/catalyst surface sites.

Rate Law: $\text{Rate} = k[\ce{A}]^0 = k$

Integrated Rate Law: $[\ce{A}]_t = -kt + [\ce{A}]_0$

Rate Constant Units: $\pu{M s^{-1}}$ (or $\pu{mol L^{-1} s^{-1}}$ )

Half-Life Equation: $t_{1/2} = \frac{[\ce{A}]_0}{2k}$

Plotting reactant concentration $[\ce{A}]$ against time $t$ gives a straight line with a y-intercept of $[\ce{A}]_0$ and a negative slope equal to $-k$ . Because $t_{1/2}$ is directly proportional to concentration, each successive half-life becomes shorter as reactant is consumed.

Worked Example: Zero-Order Calculation

Problem: The surface-catalyzed decomposition $\ce{2NH3 -> N2 + 3H2}$ follows zero-order kinetics with $k = \pu{0.020 M s^{-1}}$ . If $[\ce{NH3}]_0 = \pu{0.50 M}$ , calculate $[\ce{NH3}]$ at $t = \pu{10.0 s}$ and the initial half-life.

Step 1 (Concentration): Apply the integrated rate law:

$[\ce{NH3}]_{10.0} = [\ce{NH3}]_0 - kt = \pu{0.50 M} - (\pu{0.020 M s^{-1}} \times \pu{10.0 s}) = \pu{0.50 M} - \pu{0.20 M} = \pu{0.30 M}$

Step 2 (Half-Life): Apply the zero-order half-life formula:

$t_{1/2} = \frac{[\ce{NH3}]_0}{2k} = \frac{\pu{0.50 M}}{2(\pu{0.020 M s^{-1}})} = \pu{12.5 s}$

Exam Trap: Assuming zero-order decay continues indefinitely like an exponential curve. Zero-order reactions reach complete reactant consumption ( $\pu{0 M}$ ) in finite time: $t_{\text{complete}} = \frac{[\ce{A}]_0}{k}$ .

Second-Order Kinetics

A second-order reaction has a rate proportional to the square of a single reactant's concentration or to the product of two reactant concentrations. It typically reflects a bimolecular collision mechanism.

Rate Law: $\text{Rate} = k[\ce{A}]^2$

Integrated Rate Law: $\frac{1}{[\ce{A}]_t} = kt + \frac{1}{[\ce{A}]_0}$

Rate Constant Units: $\pu{M^{-1} s^{-1}}$ (or $\pu{L mol^{-1} s^{-1}}$ )

Half-Life Equation: $t_{1/2} = \frac{1}{k[\ce{A}]_0}$

Plotting $\frac{1}{[\ce{A}]}$ against time $t$ yields a straight line with a y-intercept of $\frac{1}{[\ce{A}]_0}$ and a positive slope equal to $+k$ . Because $t_{1/2}$ is inversely proportional to concentration, each successive half-life doubles as concentration drops by half.

Worked Example: Second-Order Calculation

Problem: The dimerization reaction $\ce{2NO2 -> N2O4}$ is second-order in $\ce{NO2}$ with $k = \pu{0.50 M^{-1} s^{-1}}$ . For $[\ce{NO2}]_0 = \pu{0.20 M}$ , determine $[\ce{NO2}]$ at $t = \pu{30.0 s}$ and the initial half-life.

Step 1 (Concentration): Apply the integrated rate law:

$\frac{1}{[\ce{NO2}]_{30.0}} = kt + \frac{1}{[\ce{NO2}]_0} = (\pu{0.50 M^{-1} s^{-1}} \times \pu{30.0 s}) + \frac{1}{\pu{0.20 M}} = \pu{15.0 M^{-1}} + \pu{5.0 M^{-1}} = \pu{20.0 M^{-1}}$

$[\ce{NO2}]_{30.0} = \frac{1}{\pu{20.0 M^{-1}}} = \pu{0.050 M}$

Step 2 (Half-Life): Apply the second-order half-life formula:

$t_{1/2} = \frac{1}{k[\ce{NO2}]_0} = \frac{1}{\pu{0.50 M^{-1} s^{-1}} \times \pu{0.20 M}} = \pu{10.0 s}$

Exam Trap: Searching for a negative slope on a second-order linear plot. Zero-order and first-order linear plots have negative slopes ( $-k$ ), but the second-order plot of $\frac{1}{[\ce{A}]}$ vs. $t$ has a positive slope ( $+k$ ).

Comparing Concentration Fall in Zero-Order and Second-Order Reactions

The pace at which concentration drops over time differs significantly between zero-order and second-order reactions:

Zero-Order Decay: Concentration drops linearly at a fixed absolute rate ( $\pu{M s^{-1}}$ ). The speed does not slow down as reactant is consumed.
Second-Order Decay: Concentration drops extremely rapidly initially when $[\ce{A}]$ is high, but as reactant is consumed, collision frequency drops precipitously, causing the decay rate to flatten into a long tail.
Half-Life Progression: In zero-order reactions, successive half-lives get 50% shorter (e.g., $\pu{10 s} \rightarrow \pu{5 s} \rightarrow \pu{2.5 s}$ ). In second-order reactions, successive half-lives double (e.g., $\pu{10 s} \rightarrow \pu{20 s} \rightarrow \pu{40 s}$ ). (First-order half-life remains constant, covered in depth in its own lesson).

DAT Mnemonic for Reaction Plot Slopes:
"Zero Down, Two Up"
- Zero-order: Plot

[\ce{A}]

vs.

t

goes Down (slope =

-k

).
- Second-order: Inverse plot

\frac{1}{[\ce{A}]}

vs.

t

goes Up (slope =

+k

Property	Zero-Order Kinetics	Second-Order Kinetics
Rate Law	$\text{Rate} = k$	$\text{Rate} = k[\ce{A}]^2$
Integrated Rate Law	$[\ce{A}]_t = -kt + [\ce{A}]_0$	$\frac{1}{[\ce{A}]_t} = kt + \frac{1}{[\ce{A}]_0}$
Linear Plot (y vs x)	$[\ce{A}]$ vs. $t$	$\frac{1}{[\ce{A}]}$ vs. $t$
Slope of Line	$-k$	$+k$
Units of $k$	$\pu{M s^{-1}}$	$\pu{M^{-1} s^{-1}}$
Half-Life ( $t_{1/2}$ )	$\frac{[\ce{A}]_0}{2k}$	$\frac{1}{k[\ce{A}]_0}$
Successive $t_{1/2}$ Trend	Decreases by half	Doubles

Critical Thinking Questions

If the initial concentration of reactant is reduced by half, how does the initial half-life change for (a) a zero-order reaction and (b) a second-order reaction?
A plot of $\frac{1}{[\ce{A}]}$ versus time yields a straight line with a slope of $\pu{0.040 M^{-1} s^{-1}}$ and a y-intercept of $\pu{2.0 M^{-1}}$ . What is the initial reactant concentration, and what is the reactant concentration at $t = \pu{50.0 s}$ ?
Why do zero-order reactions reach a concentration of absolute zero in finite time, whereas second-order reactions asymptotically approach zero concentration?

References

Sources returned by Google Search grounding during lesson verification.

libretexts.org — vertexaisearch.cloud.google.com
purdue.edu — vertexaisearch.cloud.google.com
chemistrysteps.com — vertexaisearch.cloud.google.com
ucalgary.ca — vertexaisearch.cloud.google.com
12.5: Integrated Rate Laws - Chemistry LibreTexts — chem.libretexts.org
2.10: Zero-Order Reactions - Chemistry LibreTexts — chem.libretexts.org

Zero-Order and Second-Order Kinetics

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