Foundations
Fix my algebra
Simple swaps, inverse operations and equations with several factors.
a = b/c → c = b/a
Maths skills for chemistry · AQA and OCR A
Rearranging equations
Practise the algebra behind chemistry calculations, from a = b/c to Arrhenius and time of flight. Start with the bit you need.
What do you want to work on?
Pick any route. The harder sections are available straight away, so jump directly to the algebra you want.
A level chemistry
Use chemistry equations
Moles, concentration, calorimetry, rates and the ideal gas equation.
pV = nRT → T = pV/nR
AQA and OCR A
Logs and Arrhenius
Move between exponential and logarithmic forms, rearrange for Ea, A or T, and derive the two temperature equation.
k = Ae−Eₐ/RT → ln k = ln A − Eₐ/RT
AQA · optional for OCR A
TOF relationships
Derive the speed and flight time equations, then prove the familiar mass shortcut and extend it to different ionic charges.
t₂ = t₁√(m₂/m₁)
Not sure where to start?
The quick check samples the foundation skills and suggests what to practise. Arrhenius and TOF stay available whatever your result.
Rearrange without guessing
The reliable rule is simple: do the same operation to both sides. Once a pattern is familiar, you can use the shortcut because you know which operations it represents.
When can I just swap two symbols?
For a = b/c, multiply both sides by c to get ac = b, then divide by a. That gives c = b/a.
Once you know that route, a = b/c ↔ c = b/a is a useful one swap pattern. It works because the equation contains multiplication and division in that simple arrangement.
a = b + c needs subtraction, so c = a − b. Use the operation already in the equation and apply its inverse.
Chemistry equations
The same algebra appears everywhere in chemistry. From n = m/Mᵣ, you can write Mᵣ = m/n. From c = n/V, you get V = n/c.
For q = mcΔT, divide by the factors multiplying m: m = q/(cΔT). For pV = nRT, making T the subject gives T = pV/(nR).
Powers and square roots
If y = x² and x is a positive physical quantity, x = √y. With y = ax², divide by a first: x = √(y/a).
This becomes useful in time of flight because KE = ½mv² gives v = √(2KE/m).
From Arrhenius to the logarithmic form
Start with:
k = Ae−Eₐ/(RT)
Take ln of both sides:
ln k = ln[Ae−Eₐ/(RT)]
Use ln(xy) = ln x + ln y and ln(ex) = x:
ln k = ln A − Eₐ/(RT)
Now write the temperature term as 1/T:
ln k = (−Eₐ/R)(1/T) + ln A
That is y = mx + c. A plot of ln k against 1/T has gradient −Eₐ/R and intercept ln A.
Derive the two temperature Arrhenius equation
Write the logarithmic equation at T₁ and T₂:
ln k₁ = ln A − Eₐ/(RT₁)
ln k₂ = ln A − Eₐ/(RT₂)
Subtract the first equation from the second. The ln A terms cancel:
ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂)
You can then rearrange this for Eₐ or any of the other quantities. OCR A supplies both Arrhenius forms, but seeing where this one comes from is useful algebra.
Prove the TOF mass shortcut
For an ion with kinetic energy KE:
KE = ½mv² → v = √(2KE/m)
Flight time is t = d/v, so:
t = d√(m/(2KE))
For two ions with the same charge, accelerating voltage and flight path, d and KE are the same. Therefore t ∝ √m:
t₂/t₁ = √(m₂/m₁)
and hence:
t₂ = t₁√(m₂/m₁)
This is directly useful for AQA TOF work. OCR A students can use it as optional extension.
TOF with different ionic charges
If ions are accelerated through the same potential difference, KE = qV. Combining that with the flight time expression gives:
t ∝ √(m/q)
For two ions:
t₂/t₁ = √(m₂q₁/(m₁q₂))
The familiar √(m₂/m₁) shortcut is the equal charge case.
Check a rearrangement with numbers or units
If you are unsure about a result, choose sensible values in the original equation and check that the rearranged version gives the same relationship. A unit check can catch an inverted fraction too.
For example, E = hf gives f = E/h. J ÷ J s has units s⁻¹, which fits frequency.
Extra challenge
This section is optional. It includes general TOF charge relationships, power law linearisation and algebra with the same shape as Arrhenius. It can go beyond what A level Chemistry requires.
For example, y = Axⁿ becomes ln y = ln A + n ln x, so a plot of ln y against ln x has gradient n.