Integrated Rate Law (2nd Order) Equation, Derivation & Rearrangements

Core idea

Overview

The integrated rate law for a second-order reaction describes the concentration of a reactant over time when the reaction rate is proportional to the square of its concentration. It is characterized by a linear relationship between the reciprocal of the reactant concentration and time, where the slope represents the rate constant.

When to use: Use this equation when kinetic experiments show that a plot of 1/[A] versus time produces a straight line. It is applicable to elementary bimolecular reactions where two identical molecules collide, or situations where two different reactants have equal initial concentrations.

Why it matters: This law is essential for modeling industrial dimerization processes and environmental pollutant degradation. Understanding second-order kinetics allows chemical engineers to predict how effectively concentration changes can accelerate or slow down a reaction compared to first-order systems.

Symbols

Variables

1/[A] = 1 / Concentration, k = Rate Constant, t = Time, 1/[A]_0 = Initial 1/[A]0

1/[A]

1 / Concentration

M^{-} 1

k

Rate Constant

M^{-} 1 s^{-} 1

t

Time

s

1/ [A]_{0}

Initial 1/[A]0

M^{-} 1

Walkthrough

Derivation

Derivation of Second-Order Integrated Rate Law

Gives concentration as a function of time for a reaction second order in a single reactant A.

Rate law is $Rate = k [A]^{2}$ .

1

Start with the Differential Rate Law:

Rate of disappearance is proportional to [A]^2.

- \frac{d [ A ]}{d t} = k [A]^{2}

2

Separate Variables and Integrate:

Integrate with appropriate limits.

\int_{[A]_{0}}^{[A]_{t}} - \frac{1}{[ A ] ^{2}} d [A] = \int_{0}^{t} k d t

3

State the Integrated Form:

A plot of 1/[A] against t is linear with gradient k.

\frac{1}{[ A ] _{t}} = k t + \frac{1}{[ A ] _{0}}

Result

\frac{1}{[ A ] _{t}} = k t + \frac{1}{[ A ] _{0}}

Source: Standard curriculum — A-Level Chemistry (Kinetics extension)

Free formulas

Rearrangements

Solve for 1/[A]

Make invA the subject

\frac{1}{[ A ]} = k t + \frac{1}{[ A ] _{0}}

Exact symbolic rearrangement generated deterministically from calculator baseLaTeX for invA.

Difficulty: 2/5

Solve for $k$

Make k the subject

k = \frac{\frac{1}{[ A ]} - \frac{1}{[ A ] _{0}}}{t}

Exact symbolic rearrangement generated deterministically from calculator baseLaTeX for k.

Difficulty: 2/5

Solve for $t$

Make t the subject

t = \frac{\frac{1}{[ A ]} - \frac{1}{[ A ] _{0}}}{k}

Exact symbolic rearrangement generated deterministically from calculator baseLaTeX for t.

Difficulty: 2/5

Solve for $1/ [A]_{0}$

Make c the subject

\frac{1}{[ A ] _{0}} = \frac{1}{[ A ]} - k t

Exact symbolic rearrangement generated deterministically from calculator baseLaTeX for c.

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph displays a straight line where the slope represents the rate constant k and the y-intercept corresponds to the initial value of 1/[A]0. For a chemistry student, this linear relationship indicates that as time increases, the inverse of the concentration increases at a constant rate. The most important feature of this plot is that the positive slope confirms a second order reaction, meaning that the time required for the concentration to change is directly proportional to the difference in the inverse concentration values.

Graph type: linear

Why it behaves this way

Intuition

A straight line on a coordinate plane where the vertical axis represents the increasing 'rarity' or 'spacing' of molecules as they are consumed over time.

[A]

Instantaneous molar concentration of the reactant

Represents the current density of reactant molecules available to collide and react.

k

Second-order rate constant

Quantifies the frequency and efficiency of collisions; in second-order reactions, it scales the rate by the square of the concentration.

1/[A]

Reciprocal of concentration

Can be thought of as the average volume of space 'occupied' by or available to a single molecule; this value grows as the reaction consumes the reactant.

t

Elapsed reaction time

The independent variable representing the duration over which collisions have been occurring.

Signs and relationships

kt: The positive sign indicates that the reciprocal concentration increases over time as the actual concentration of the reactant decreases.
1/[A]: The reciprocal transform is used because the rate of change of concentration is proportional to the square of the concentration, which linearizes into an inverse relationship.

Free study cues

Insight

Canonical usage

Ensuring dimensional consistency across all terms, particularly for concentration, time, and the rate constant.

Common confusion

A common mistake is using inconsistent units for time (e.g., seconds for k and minutes for t) or incorrectly assigning units to the rate constant k, especially confusing it with first-order rate constant units (time^-1).

Unit systems

[A]mol L^-1 · Commonly expressed as M (molar). Must be consistent with [A]_0.

$[A]_{0}$ mol L^-1 · Initial concentration, must be consistent with [A].

$k$ L mol^-1 s^-1 · The second-order rate constant. Its units depend on the chosen units for concentration and time. For mol L^-1 and s, it is L mol^-1 s^-1 (or M^-1 s^-1).

$t$ s · Time elapsed. Must be consistent with the time unit used in the rate constant k.

One free problem

Practice Problem

A decomposition reaction follows second-order kinetics with a rate constant of 0.250 M⁻¹s⁻¹. If the initial concentration of the reactant is 0.500 M, what will the concentration be after 10.0 seconds?

Rate Constant0.25 M^-1 s^-1

Initial 1/[A]02 M^-1

Time10 s

Solve for: invA

Hint: Calculate the reciprocal of the initial concentration first, then add the product of k and t.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

In dimerization reactions where two molecules combine, Integrated Rate Law (2nd Order) is used to calculate 1 / Concentration from Rate Constant, Time, and Initial 1/[A]0. The result matters because it helps connect measured amounts to reaction yield, concentration, energy change, rate, or equilibrium.

Study smarter

Tips

Confirm that the rate constant units are in M⁻¹ time⁻¹, such as M⁻¹s⁻¹ or M⁻¹min⁻¹.
A positive slope in the reciprocal plot (1/[A] vs t) is a unique signature of second-order reactions.
Unlike first-order reactions, the half-life of a second-order reaction increases as the initial concentration decreases.

Avoid these traps

Common Mistakes

Using ln[A] instead of 1/[A] for 2nd order.
Convert units and scales before substituting, especially when the inputs mix M^-1, M^-1 s^-1, s.
Interpret the answer with its unit and context; a percentage, rate, ratio, and physical quantity do not mean the same thing.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

Gives concentration as a function of time for a reaction second order in a single reactant A.

Use this equation when kinetic experiments show that a plot of 1/[A] versus time produces a straight line. It is applicable to elementary bimolecular reactions where two identical molecules collide, or situations where two different reactants have equal initial concentrations.

This law is essential for modeling industrial dimerization processes and environmental pollutant degradation. Understanding second-order kinetics allows chemical engineers to predict how effectively concentration changes can accelerate or slow down a reaction compared to first-order systems.

Using ln[A] instead of 1/[A] for 2nd order. Convert units and scales before substituting, especially when the inputs mix M^-1, M^-1 s^-1, s. Interpret the answer with its unit and context; a percentage, rate, ratio, and physical quantity do not mean the same thing.

In dimerization reactions where two molecules combine, Integrated Rate Law (2nd Order) is used to calculate 1 / Concentration from Rate Constant, Time, and Initial 1/[A]0. The result matters because it helps connect measured amounts to reaction yield, concentration, energy change, rate, or equilibrium.

Confirm that the rate constant units are in M⁻¹ time⁻¹, such as M⁻¹s⁻¹ or M⁻¹min⁻¹. A positive slope in the reciprocal plot (1/[A] vs t) is a unique signature of second-order reactions. Unlike first-order reactions, the half-life of a second-order reaction increases as the initial concentration decreases.

References

Sources

Atkins Physical Chemistry
McQuarrie & Simon, Physical Chemistry: A Molecular Approach
Wikipedia: Rate equation
Atkins' Physical Chemistry
McQuarrie, Donald A. 'Physical Chemistry: A Molecular Approach'
Atkins, P. W., & de Paula, J. (2014). Atkins' Physical Chemistry (10th ed.). Oxford University Press.
Chang, R. (2010). Chemistry (10th ed.). McGraw-Hill.
Standard curriculum — A-Level Chemistry (Kinetics extension)

Integrated Rate Law (2nd Order)

Overview

Variables

Derivation

Start with the Differential Rate Law:

Separate Variables and Integrate:

State the Integrated Form:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Integrated Rate Law (1st Order)

Frequently Asked Questions

Sources