Rearranging Equations — Maths for Chemistry

Solving for the symbol you actually want

The most-used skill in all of chemistry

An equation is a balance. To “make x the subject” you peel away everything attached to x by doing the inverse operation to both sides — add undoes subtract, multiply undoes divide, exponentiate undoes a log. Work from the outside in: the last operation performed on x is the first one you undo.

P V = n RT \div n R T = \frac{P V}{n R}

Check your answer by substituting back, and always carry the units through the algebra — they must be consistent in the rearranged form. If pressure has units of Pa and volume of m³, then nRT also has units of Pa m³ = J, which is correct for energy.

Logarithmic and exponential functions introduce a special step: to free a variable trapped inside $ln (\cdot)$ you apply $e^{(\cdot)}$ to both sides, and vice versa. This is what converts $Δ G = - RT ln K$ into a formula for the equilibrium constant K.

Key formulas you will rearrange in chemistry

Ideal gas law

P V = n RT \Rightarrow P, V, n, T all isolable

Concentration

c = \frac{n}{V} \Rightarrow n = c V, V = \frac{n}{c}

Thermodynamics

Δ G = - RT ln K \Rightarrow K = e^{- Δ G / RT}

Arrhenius equation

k = A e^{- E_{a} / RT} \Rightarrow E_{a} = - RT ln (k / A)

Equations with logarithms or exponentials demand an extra move. When you isolate $ln (something)$ you exponentiate both sides; when the unknown sits in an exponent you take the natural log of both sides. Sign discipline is critical — the minus sign in $Δ G = - RT ln K$ means dividing by $- RT$ flips the sign of the right-hand side. The combined gas law links two states of the same sample:

\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}

Here any one of the six variables can be made the subject by straightforward cross-multiplication. The Henderson–Hasselbalch equation $pH = p K_{a} + lo g ([A^{-}] / [HA])$ is rearranged for the acid/base ratio by subtracting $p K_{a}$ and then applying the inverse log (antilog): $[A^{-}] / [HA] = 1 0^{pH - p K_{a}}$ .

Nernst equation

E = E^{\circ} - \frac{RT}{n F} ln Q \Rightarrow Q = e^{n F (E^{\circ} - E) / RT}

Henderson–Hasselbalch

\frac{[ A ^{-} ]}{[ HA ]} = 1 0^{pH - p K_{a}}

Arrhenius for T

T = \frac{- E _{a}}{R l n ( k / A )}

van der Waals

p = \frac{n RT}{V - nb} - \frac{a n ^{2}}{V ^{2}}

Common pitfalls

Dividing vs subtracting to isolate. If the target is multiplied by something, divide both sides — do not subtract. E.g. in $n RT = P V$ , to get $n$ you divide by $RT$ , not subtract $RT$ .
Sign when taking logs. In $Δ G = - RT ln K$ , dividing by $- RT$ gives $ln K = - Δ G / RT$ (not $+ Δ G / RT$ ). Missing the minus sign is the most common error.
Units after rearranging. If you change subjects, the units change too. $P = n RT / V$ gives pressure in Pa only when $R = 8.314 J mo l^{- 1} K^{- 1}$ and $V$ is in m³.
Log of a ratio, not a difference. $ln (k_{2} / k_{1})$ is not $ln k_{2} - ln k_{1}$ as an arithmetic operation on numbers — they are equal algebraically, but students sometimes compute $ln k_{2}$ then subtract rather than taking $ln$ of the ratio.

Everywhere

Ideal gas law, dilution, rate laws, the Nernst and Arrhenius equations, ΔG = −RT ln K — you will rearrange every one of these. Try the ideal-gas solver and switch the subject.

Make the subject

Choose which variable to solve for and drag the other three — the rearranged form updates live.

Solve the ideal gas law for:

P V = n RT \Rightarrow P = \frac{n RT}{V}

V volume2 L

n amount0.5 mol

T temperature298 K

P =6.11 atm

Pick a different subject and the same equation rearranges itself. Master this one move — isolate the symbol you want by undoing operations on both sides — and every formula in chemistry opens up.

Worked example 1PV = nRT solved for pressure

A 2.50 L flask at 310 K contains 0.120 mol of gas. Using PV = nRT with $R = 8.314 J mo l^{- 1} K^{- 1}$ , find the pressure in kPa.

Worked example 2Moles from concentration and volume

A student dissolves a solute to make 250.0 mL of a $0.400 mol d m^{- 3}$ solution. How many moles of solute are present?

Worked examples

More complex rearrangements — try each before revealing the solution.

Worked example 3Equilibrium constant from ΔG°

At 298 K the standard Gibbs energy change for a reaction is $Δ G^{\circ} = - 17.1 kJ mo l^{- 1}$ . Use $Δ G^{\circ} = - RT ln K$ to find K.

Worked example 4Activation energy from Arrhenius

The rate constant for a reaction is $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ at 300 K and $k_{2} = 8.0 \times 1 0^{- 3} s^{- 1}$ at 320 K.

Using the two-temperature Arrhenius form $ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ , find $E_{a}$ .

Worked example 5Dilution: finding the initial concentration

A student dilutes a stock solution: 25.0 cm³ of stock is made up to 250 cm³ with water. The final concentration is $0.0400 mol d m^{- 3}$ . Find the stock concentration using $c_{1} V_{1} = c_{2} V_{2}$ .

Worked example 6Arrhenius — rearrange for temperature T

A reaction has $A = 1.0 \times 1 0^{13} s^{- 1}$ and $E_{a} = 80.0 kJ mo l^{- 1}$ . At what temperature will $k = 1.0 \times 1 0^{- 3} s^{- 1}$ ?

Worked example 7Nernst equation — rearrange for the reaction quotient Q

At 298 K the cell potential for the half-reaction Cu²⁺/Cu is measured as $E = 0.290 V$ , with $E^{\circ} = 0.337 V$ and $n = 2$ electrons transferred. Find the reaction quotient Q. ( $F = 96485 C mo l^{- 1}$ )

Worked example 8Henderson–Hasselbalch — rearrange for the acid/base ratio

A phosphate buffer at pH 7.0 uses the $H_{2} P O_{4}^{-} /HP O_{4}^{2 -}$ couple ( $p K_{a} = 7.20$ ). Calculate the molar ratio $[HP O_{4}^{2 -}] / [H_{2} P O_{4}^{-}]$ required.

ChallengeChallenge — van der Waals equation solved for pressure

The van der Waals equation for a real gas is:

(p + \frac{a n ^{2}}{V ^{2}}) (V - nb) = n RT

(a) Rearrange for p. (b) For N₂ ( $a = 1.390 L^{2} atm mo l^{- 2}$ , $b = 0.03913 L mo l^{- 1}$ ) find the pressure of 2.00 mol N₂ in a 1.00 L vessel at 300 K, and compare with the ideal-gas value. ( $R = 0.08206 L atm mo l^{- 1} K^{- 1}$ )

Check yourself

Four questions on making a symbol the subject.

Question 1 of 4 · Score 0

Rearrange PV = nRT to make T the subject.

P V = n RT

Choose an answer.