Indices, Powers & Roots — Maths for General Chemistry

The rules of powers

Indices keep rate laws and big/small numbers under control

An index (exponent) counts repeated multiplication. A handful of rules let you combine them without ever writing the multiplication out in full. The same three laws reappear everywhere in chemistry: in the exponents of rate laws $rate = k [A]^{m} [B]^{n}$ , in equilibrium expressions where stoichiometric coefficients become powers, and in the conversion between standard-form numbers.

Negative exponents are reciprocals ( $x^{- n} = 1/ x^{n}$ ), which is why dm⁻³ means “per cubic decimetre”. Fractional exponents are roots, so the root-mean-square speed $u_{rms} = (3 RT / M)^{1/2}$ is just a square root in disguise.

Multiply → add

x^{a} x^{b} = x^{a + b}

Divide → subtract

x^{a} / x^{b} = x^{a - b}

Power of a power

(x^{a})^{b} = x^{ab}

Special cases

x^{0} = 1, x^{- n} = \frac{1}{x ^{n}}, x^{1/2} = x

Fractional exponents as roots appear in many physical-chemistry formulas. The root-mean-square speed $u_{rms} = (3 RT / M)^{1/2}$ is a ½ power (square root). The most-probable speed involves the same structure. In general, $x^{p / q} = (x^{p})^{1/ q} = q x^{p}$ — the denominator is the root order, the numerator is the power. This matters whenever you need to solve $K_{s p} = s^{3}$ (for a 1 : 2 salt) by taking a cube root: $s = (K_{s p})^{1/3}$ .

Standard form and index laws often combine. Multiplying $(3.0 \times 1 0^{- 4}) \times (2.0 \times 1 0^{- 11})$ means multiplying the coefficients and adding the exponents of 10 (not multiplying them): $6.0 \times 1 0^{- 15}$ . A very common slip is to multiply exponents instead — that would be the rule for a power of a power, not for a product.

Fractional power as root

x^{1/ n} = n x, x^{p / q} = q x^{p}

Zero power

x^{0} = 1

for any

x \neq = 0

. In

K_{c}

expressions, pure solids and solvents are assigned activity 1 for the same reason.

Negative power in rate law

A negative order means the rate decreases with concentration:

rate = k [A]^{- 1}

→

rate = k / [A]

. Rare but real (e.g. some catalytic cycles).

Adding vs multiplying exponents

1 0^{a} \cdot 1 0^{b} = 1 0^{a + b}

(not

1 0^{ab}

). The latter is only for

(1 0^{a})^{b}

Common pitfalls

$1 0^{a} \cdot 1 0^{b}$ adds exponents, never multiplies. Students sometimes write $1 0^{- 4} \times 1 0^{- 11} = 1 0^{44}$ . Correct: $1 0^{- 4 + (- 11)} = 1 0^{- 15}$ .
$x^{0} = 1$ , not 0. Any non-zero number raised to the power zero equals 1. This is why the activity of a pure solid is 1 in equilibrium expressions.
$(x^{a})^{b} = x^{ab}$ , but $x^{a} \cdot x^{b} = x^{a + b}$ . Mixing these two rules is the source of many wrong answers. Check: is it a product of two powers, or a power raised to a power?
Fractional exponent in the wrong order. $x^{2/3}$ means cube-root first, then square — or equivalently square first, then cube-root. Either way, remember the denominator is the root order.

In chemistry

Rate = k[A]ᵐ[B]ⁿ multiplies powers of concentration; equilibrium constants stack powers from the stoichiometry; the root-mean-square speed uses a ½ power. Open the playground to see the rules in action.

Index-law playground

Adjust the base and exponents — the chosen law and its numerical result update instantly.

Operation:

x^{a} \cdot x^{b} = x^{a + b}

2^{3} \times 2^{2} = 2^{3 + 2} = 2^{5}

= 32

base x2

exponent a3

exponent b2

Same base, simple rule: multiply → add exponents, divide → subtract, power of a power → multiply. Negative exponents are reciprocals; fractional exponents are roots.

Worked example 1Rate law with two reactants

The rate law for a reaction is $rate = k [NO]^{2} [O_{2}]^{1}$ . If $[NO]$ is tripled and $[O_{2}]$ is doubled, by what factor does the rate change?

Worked example 2Equilibrium constant and stoichiometry

For the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , write the expression for $K_{p}$ in terms of partial pressures, and explain why the exponents match the stoichiometric coefficients.

Worked examples

Standard-form calculations and roots that appear in real chemistry problems.

Worked example 3Root-mean-square speed

The rms speed of a gas molecule is $u_{rms} = (\frac{3 RT}{M})^{1/2}$ . Calculate the rms speed of N₂ (M = 0.02802 kg mol⁻¹) at 298 K.

Worked example 4Standard-form product using index laws

In a very dilute solution, $[H^{+}] = 3.2 \times 1 0^{- 4} M$ and $[O H^{-}] = 5.0 \times 1 0^{- 11} M$ . Verify whether the product equals the water autoionisation constant $K_{w} = 1.0 \times 1 0^{- 14}$ .

Worked example 5Negative exponents in unit analysis

A rate constant has the value $k = 4.5 \times 1 0^{- 3} L mo l^{- 1} s^{- 1}$ . Rewrite it in SI base units (m³ mol⁻¹ s⁻¹) given that 1 L = 10⁻³ m³.

Worked example 6Molar solubility of a 1:2 electrolyte — fractional exponent

For CaF₂ dissolving as $Ca F_{2} (s) ⇌ C a^{2 +} + 2 F^{-}$ , $K_{sp} = 3.45 \times 1 0^{- 11}$ . If the molar solubility is s, then $[C a^{2 +}] = s$ and $[F^{-}] = 2 s$ . Find s using index laws.

Worked example 7Reaction order from the rate-concentration power

In three experiments the initial rate was measured with [A] varied and [B] held constant. When [A] doubled (all else equal) the rate increased by a factor of 8. When [A] tripled the rate increased by a factor of 27. Determine the order with respect to A.

ChallengeChallenge — root-mean-square speed: derive and evaluate

The Maxwell–Boltzmann root-mean-square speed is given by:

u_{rms} = (\frac{3 RT}{M})^{1/2}

(a) Show that the units work out to m s⁻¹ when R is in J mol⁻¹ K⁻¹ and M is in kg mol⁻¹. (b) Calculate $u_{rms}$ for H₂ (M = 0.002016 kg mol⁻¹) and O₂ (M = 0.03200 kg mol⁻¹) at 298 K. (c) Without recalculating, by what factor is $u_{rms}$ (H₂) larger than $u_{rms}$ (O₂)?

Check yourself

Four questions on index laws in chemical contexts.

Question 1 of 4 · Score 0

Simplify x³ × x⁴.

x^{3} \times x^{4}

Choose an answer.