Partial Fractions — Maths for Thermodynamics & Kinetics

Splitting a fraction to conquer it

The key that unlocks second-order rate laws

A fraction with a factored denominator can be rewritten as a sum of simpler fractions, each over one factor:

\frac{1}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}

Find A and B with the cover-up method: multiply through by the full denominator, then substitute the value of x that makes one factor zero — the other term vanishes, leaving only one unknown to solve. Each simple piece then integrates to a logarithm.

In chemistry the method appears whenever you need to integrate a rate equation that contains a product of two linear concentration terms. The second-order rate law for $A + B \to products$ with unequal initial concentrations requires integrating $1/ ([A] [B])$ — impossible without first splitting into partial fractions. Consecutive first-order reactions give the same type of integral.

Setup

\frac{P ( x )}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}

Cover-up for A

Set

x = a

; the (x−a) cancels, leaving

A = P (a) / (a - b)

Each piece integrates

\int \frac{A}{x - a} d x = A ln ∣ x - a ∣ + C

Chemist's use case

2nd-order integrated rate law:

\int \frac{d x}{( a _{0} - x ) ( b _{0} - x )}

Step zero: is the fraction proper? If the degree of the numerator is greater than or equal to the degree of the denominator, do a polynomial long division first to get a polynomial plus a proper fraction. Only the proper-fraction remainder is then decomposed. Skipping this step produces a completely wrong decomposition.

Repeated linear factors require an extra term for each repeat. If one factor is $(x - a)^{2}$ , the partial-fraction form must include both $A / (x - a)$ and $B / (x - a)^{2}$ . Setting $x = a$ finds B (cover-up still works for the highest power), but you need a second equation (e.g. x = 0 or comparing coefficients) to find A.

In chemistry the cover-up method is often presented as a shortcut, but the underlying principle is simply multiplying both sides by one factor at a time and evaluating at its root. The same procedure scales to any number of distinct linear factors.

Common pitfalls

Improper fractions: reduce first. If the numerator degree ≥ denominator degree, perform polynomial division first. Attempting partial fractions directly on an improper fraction gives wrong constants.
Repeated factors need two terms. $1/ (x - a)^{2}$ decomposes to $A / (x - a) + B / (x - a)^{2}$ , not just $A / (x - a)$ . Using only one term leaves the decomposition incomplete.
Sign errors in cover-up. When setting $x = - 1$ in a factor $(x + 1)$ , the value substituted everywhere is $- 1$ , including in the numerator. Substituting 1 instead is very common.
Forgetting to integrate each piece separately. After decomposing, each term $A / (x - a)$ integrates to $A ln ∣ x - a ∣$ , not to $A / (x - a)^{2}$ .

Why bother

The integrated second-order rate law needs ∫1/[(a−x)(b−x)] dx, and consecutive first-order reactions produce similar fractions. Partial fractions is the only way through. Try the decomposer — watch the split curve land exactly on the original.

Decomposer

Choose an example fraction — the widget shows its decomposition and integral, and overlays the two curves to confirm they match.

Example:

original sum of partials

Fraction

\frac{1}{x ( x + 1 )}

Decomposes to

\frac{1}{x} - \frac{1}{x + 1}

Integral

ln ∣ x ∣ - ln ∣ x + 1∣ + C

Cover-up method: A = 1, B = −1. The two curves sit exactly on top of each other — the decomposition is the same function, just written so each piece integrates to a logarithm.

Worked example 1Cover-up method step by step

Decompose $\frac{3 x + 1}{( x - 2 ) ( x + 1 )}$ into partial fractions and hence find the integral.

Worked example 2Second-order integrated rate law

For $A + B \to products$ with rate $= k [A] [B]$ , initial concentrations $a_{0}$ and $b_{0}$ ( $a_{0} \neq = b_{0}$ ), and $[A] = a_{0} - x, [B] = b_{0} - x$ , show that integrating the rate equation gives the log form of the second-order integrated rate law.

Worked examples

Consecutive first-order reactions and a repeated factor — the two cases beyond the basics.

Worked example 3Consecutive first-order reactions — concentration of intermediate

For the scheme $A k_{1} B k_{2} C$ , the Laplace-transform solution for [B](t) contains a factor $1/ (k_{1} - k_{2})$ from a partial-fraction split. Write the integral $\int_{0}^{t} k_{1} e^{- k_{1} t^{'}} \frac{1}{k _{2} - k _{1}} (e^{- k_{1} t^{'}} - e^{- k_{2} t^{'}}) d t^{'}$ in terms of its partial-fraction result and state [B](t).

Worked example 4Repeated factor — the equal-rate limiting case

Decompose $\frac{1}{( s + k ) ^{2}}$ (a repeated linear factor, appearing when $k_{1} = k_{2} = k$ in consecutive reactions) and integrate (or inverse-Laplace) to find [B](t).

Worked example 5Second-order rate constant from integrated law

For the reaction $N O_{2} + CO \to NO + C O_{2}$ , second-order kinetics are observed with $a_{0} = [N O_{2}]_{0} = 0.0400 M$ and $b_{0} = [CO]_{0} = 0.0600 M$ .

After 120 s, $[N O_{2}] = 0.0310 M$ (so x = 0.0090 M). Use the integrated second-order rate law to find k.

Worked example 6Fresh decomposition then integrate — second-order kinetics 1/([A][B])

Decompose $\frac{1}{( 0.0500 - x ) ( 0.0800 - x )}$ into partial fractions (where x represents the extent of reaction) and hence write out the integral that gives the integrated rate law.

Worked example 7Repeated linear factor — decompose then integrate

Decompose $\frac{2 x + 1}{( x + 1 ) ^{2} ( x - 1 )}$ into partial fractions and hence find $\int \frac{2 x + 1}{( x + 1 ) ^{2} ( x - 1 )} d x$ .

ChallengeChallenge — Bateman equation: set up and verify [B] maximum

For the consecutive scheme $A k_{1} B k_{2} C$ with $[A]_{0} = 1.00 M$ , $k_{1} = 0.100 s^{- 1}$ , and $k_{2} = 0.300 s^{- 1}$ :

(a) Write down the Bateman equation for [B](t) (derived from the partial-fraction approach in example 3 above).

(b) Find the time $t_{max}$ at which [B] is a maximum, and evaluate $[B]_{max}$ .

Check yourself

Four questions on partial-fraction decomposition and its applications.

Question 1 of 4 · Score 0

Partial-fraction decomposition rewrites one complicated fraction as:

Choose an answer.