Chemistry

Second-Order Integrated Rate Law

Track concentration for a single-reactant quadratic disappearance law.

\frac{1}{C _{t}} = \frac{1}{C _{0}} + k t

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Source & review

Difficulty

Intermediate

Read time

3 min

Prerequisites

Algebra, logarithms, and molar quantities

Review statusNeeds review

Source

OpenStax, Chemistry 2e, §12.4

Adapted from Teng-Jui Lin’s CHEM 155 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

FormuLab adapted notation and chemical symbols, added variable meanings, units, applicability conditions, and an original worked example. Restricted the integrated law to single-reactant disappearance, with any stoichiometric normalization included in k.

Definition

Understanding the core concept

When disappearance rate is proportional to the square of one reactant concentration, inverse concentration increases linearly with time. This form applies to −dC/dt = kC², rather than every reaction with overall order two.

Variables & Units

Understanding each component

Symbol	Meaning	Units
$C_{t}$	Reactant concentration at elapsed time t	$mol L^{- 1}$
$C_{0}$	Positive initial reactant concentration	$mol L^{- 1}$
$k$	Second-order disappearance-rate constant	$L mo l^{- 1} s^{- 1}$
$t$	Elapsed time	$s$

Real-World Applications

Where this formula is used in practice

Second-order decay

Estimate concentration for a quadratic disappearance law.

Kinetic plots

Relate a linear inverse-concentration plot to a second-order rate constant.

Worked Example

Step-by-step calculation with real numbers

Problem

A reactant follows −dC/dt = kC² with C_0 = 0.100 mol/L and k = 0.500 L/(mol·s). Find its concentration after 20.0 s.

Given

C_{0} = 0.100 mol L^{- 1}

k = 0.500 L mo l^{- 1} s^{- 1}

t = 20.0 s

Solution

\frac{1}{C _{t}} = \frac{1}{C _{0}} + k t

C_{t} = \frac{1}{C _{0}^{- 1} + k t}

C_{t} = \frac{1}{\frac{1}{0.100} + ( 0.500 ) ( 20.0 )} mol L^{- 1}

C_{t} \approx 0.05 mol L^{- 1}

Final Answer

C_{t} \approx 0.05 mol L^{- 1}

Formula Guide

Info

DifficultyIntermediate

Read Time3 min

PrerequisitesAlgebra, logarithms, and molar quantities

Tools

Source & review

StatusNeeds review

Last reviewedPending

Source

OpenStax, Chemistry 2e, §12.4

Adapted from Teng-Jui Lin’s CHEM 155 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

Assumptions

• Use a constant-volume system at fixed temperature with the stated disappearance-rate law and constant rate coefficient.
• Use consistent concentration and time units; apply the model only while the reactant remains present.
• Use the single-reactant law −dC/dt = kC² with positive C_0 and nonnegative k and elapsed time.

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