H2 Chemistry Formula List

H2 Chemistry 9476 equations with units, assumptions, sign conventions and a worked calculation.

  • GCE A-Level H2 Chemistry 9476-2027
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A topic-organised calculation reference for H2 Chemistry (9476). Choose the model and check the units before substituting. Conditions in the final column matter as much as the equation.

Course hub · Definitions

Moles, composition and reacting quantities

RelationshipSymbols and unitsWhen to use it
Mᵣ = ∑ AᵣMᵣ and Aᵣ have no unitInclude every subscript and bracket multiplier in the formula.
n = m/M; m = nMn mol, m g, M g/molMatch the mass unit to the molar-mass unit; this does not require a reaction ratio.
mean Aᵣ = ∑(A_(r,i) fᵢ)Fractional isotope abundance fᵢ has no unit; ∑ fᵢ = 1For percentage abundances, divide each percentage by 100.
total positive charge + total negative charge = 0Ionic charges in units of elementary chargeChoose the smallest whole-number ion ratio; never change an ion’s own formula.
CₙH₂ₙ₊₂; CₙH₂ₙn is the number of carbon atomsAcyclic alkanes / acyclic alkenes with one double bond; not all hydrocarbons.
x = Mᵣ(compound)/Mᵣ(empirical formula)Whole-number multiplier x; relative masses have no unitMultiply every empirical-formula subscript by x to obtain the molecular formula; first find the simplest mole ratio.
N = nN_AParticle count N; N_A in mol⁻¹Specify the counted entity; N_A ≈ 6.02 × 10²³ mol⁻¹ for calculations.
n(A)/a = n(B)/bMoles; equation coefficients a and bFor aA → bB, use a balanced equation and the limiting reactant.
n = V/VₘGas volume V and molar gas volume Vₘ in matching unitsAt r.t.p. the school reference is Vₘ = 24 dm³/mol; use the supplied value.
c = n/V; n = cVc mol/dm³; solution volume V dm³Convert cm³ to dm³ before substitution: divide by 1000.
cₘₐₛₛ = m/V = cMMass concentration g/dm³; M g/molDistinguish mass concentration from molar concentration.
c₁V₁ = c₂V₂Matching concentration and volume unitsDilution of the same solute with no reaction; not the general titration equation.
c_AV_A/a = c_BV_B/bConcentrations mol/dm³ and volumes dm³Titration at the reacting ratio a:b; equal volumes or equal concentrations need not result.

Enthalpy and calorimetry

RelationshipSymbols and unitsWhen to use it
q = mcΔ T; Δ H = -q/nq J; mass g; c J/(g K); Δ T K; n molFor a reaction at constant pressure, q is heat gained by the measured surroundings. Negate for reaction heat, convert J to kJ and use the stated molar basis; neglect losses only if justified.
Δ H°ᵣ = ∑νΔ H_f°(products)-∑νΔ H_f°(reactants)Enthalpy changes kJ/mol; coefficients νSame states and standard reference conditions; formation enthalpies of elements in their standard states are zero.
Δ H ≈ ∑ E_broken-∑ E_formedAverage bond energies kJ/molApproximate gas-phase estimate; count all bonds and use the reaction’s coefficients.
Δ H_(route 1) = Δ H_(route 2)Enthalpy changes for the same initial and final statesHess’s law; reverse signs and scale enthalpies when reversing or scaling equations.

Rates and equilibrium

RelationshipSymbols and unitsWhen to use it
r = k[A]^m[B]ⁿ; overall order = m + nRate commonly mol/(dm³ s); concentration mol/dm³Orders are determined from evidence; hold temperature and other concentrations fixed when comparing initial rates.
r₂/r₁ = ([A]₂/[A]₁)^mDimensionless ratiosOnly A changes; factor 2 in concentration and factor 4 in rate implies m = 2.
[k] = [r]/[c]^(m + n)For overall order s: (mol/dm³)^(1-s)s⁻¹State the concentration and time units; zero-, first- and second-order constants have different units.
K_c = [C]^c[D]^d/([A]^a[B]^b)Equilibrium concentrations; powers from the balanced equationFor aA + bB ⇌ cC + dD; omit pure solids and pure liquids. Fixed temperature.
K_(c,reverse) = 1/K_c; K_(c,scaled) = K_c^sScaling factor s multiplies every coefficientA constant belongs to the exact equation; concentration-based units can change when it is rescaled.

Acids and bases

RelationshipSymbols and unitsWhen to use it
pH = - log ₁₀[H⁺]Use the numerical concentration in mol/dm³School dilute-solution concentration model; base-10 log, not natural log.
K_w = [H⁺][OH⁻]Concentrations mol/dm³; K_w commonly (mol/dm³)²K_w = 1.00 × 10⁻¹⁴ at 298 K in this model; neutral [H⁺] = [OH⁻].
[H⁺] = c; [OH⁻] = zcc mol/dm³; z hydroxide ions per formula unitCompletely ionised monobasic strong acid / completely dissociated soluble hydroxide; water’s contribution must be negligible.
Kₐ = [H⁺][A⁻]/[HA]; K_b = [BH⁺][OH⁻]/[B]Equilibrium concentrations mol/dm³Define the acid/base reaction with water; distinguish equilibrium concentration from prepared concentration.

Further H2 physical chemistry

RelationshipSymbols and unitsWhen to use it
pV = nRT; pᵢ = xᵢpₜₒₜₐₗp Pa; V m³; T K; R = 8.31 J mol⁻¹K⁻¹Ideal gas and ideal-gas mixture; xᵢ = nᵢ/nₜₒₜₐₗ.
Kₚ = p_C^cp_D^d/(p_A^ap_B^b)Equilibrium partial pressures in one consistent unitGas-phase species in the balanced reaction; do not insert total pressure for each gas.
Δ S° = ∑ν S°(products)-∑ν S°(reactants)Molar entropy J/(mol K)Use coefficients; standard molar entropies of elements are not zero.
Δ G° = Δ H°-TΔ S°T K; energy units consistentConvert entropy to kJ/(mol K) if enthalpy is in kJ/mol; negative standard Gibbs change favours reaction from standard-state reactants.
[A]ₜ/[A]₀ = (1/2)^(t/t_(1/2))Times in matching unitsRepeated half-lives for a first-order reactant with constant half-life; integrated rate equations are not required.
K_w = KₐK_b; pKₐ = - log ₁₀KₐSame concentration convention throughoutKₐ and K_b refer to a conjugate pair at the same temperature.
[H⁺] ≈ square root of Kₐc; [OH⁻] ≈ square root of K_bcc mol/dm³Weak monoprotic acid / weak monoacidic base alone in water; ionisation small relative to c and water contribution negligible. Check the approximation.
pH ≈ pKₐ + log ₁₀([A⁻]/[HA])Conjugate base and weak acid concentrationsBuffer after any neutralisation; both partners present in appreciable amounts. Near exhaustion the approximation fails.
Kₛₚ = [M^(a +)]^x[X^(b-)]^yEquilibrium dissolved-ion concentrationsFor solid MₓX_y in a saturated solution; ion ratio follows the dissolution equation.
E°_cell = E°_cathode-E°_anodeBoth tabulated reduction potentials in VStandard cell conditions; choose the reduction and oxidation half-cells correctly.
Δ G° = -nFE°_celln electron stoichiometric coefficient; F ≈ 9.65 × 10⁴ C/molΔ G° in J/mol for the cell equation; divide by 1000 to express in kJ/mol.
Q = It; nₑ = Q/F; n_product = Q/(zF)Charge C; current A; time s; z electrons per product entityConstant current and quantitative current efficiency; use the balanced electrode half-equation.

A quick application: calorimetry and signs

Suppose 50.0 g of solution warms by 6.00 K when 0.0200 mol of a reactant is consumed at constant pressure. Here the molar basis is one mole of that reactant. Taking c = 4.18 J g⁻¹K⁻¹ and neglecting heat losses, qₛₒₗᵤₜᵢₒₙ = 50.0 × 4.18 × 6.00 = 1254 J. The reaction released this heat. Converting J to kJ and dividing by the amount reacted gives Δ H = -1254/(1000 × 0.0200) = -62.7 kJ/mol.

The solution’s positive heat gain corresponds to a negative reaction enthalpy change. State which body gains the heat before assigning the sign.

Keep the equation and evidence connected

Use the balanced reaction to set the molar basis and stoichiometric powers. Quantitative thermodynamic feasibility does not establish reaction speed. Integrated rate equations are not required; half-life reasoning is used where the first-order model applies.

Return to the course hub for the lesson behind a term or relationship, then practise without this reference.