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Mathematical Skills in Chemistry

Symbols, rearranging equations and logarithms: substituting into equations for Kc, enthalpy and pH

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Overview

Substituting into and rearranging the Arrhenius, dilution, Kc, titration and enthalpy equations, Hess's law, lattice and bond enthalpy calculations, and using logarithms for pH and pKa.

What you need to know

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Numbers

25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide is exactly neutralised by 12.50 cm³ of sulfuric acid. Calculate the concentration of the acid.

Numbers

Two acids have pKa values of 3.75 and 4.76. Which is the stronger, and by what factor do their Ka values differ?

Numbers

Calculate the enthalpy change for CaCO₃(s) → CaO(s) + CO₂(g) from the enthalpy changes of formation: CaCO₃ −1207, CaO −635 and CO₂ −394 kJ mol⁻¹.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Using Ea = RT ln(A/k), calculate the activation energy for a reaction with k = 8.43 × 10⁻³ s⁻¹ and A = 1.20 × 10¹¹ s⁻¹ at 298 K. (R = 8.31 J K⁻¹ mol⁻¹)

    Model answer

    A/k = (1.20 × 10¹¹) ÷ (8.43 × 10⁻³) = 1.42 × 10¹³, and ln(1.42 × 10¹³) = 30.29. Ea = 8.31 × 298 × 30.29 = 7.50 × 10⁴ J mol⁻¹, which is 75.0 kJ mol⁻¹.

  2. 2

    Use the two-point form of the Arrhenius equation to find the activation energy when k = 2.00 × 10⁻⁴ at 300 K and 3.60 × 10⁻³ at 320 K. (R = 8.31 J K⁻¹ mol⁻¹)

    Model answer

    ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Here ln(3.60 × 10⁻³ ÷ (2.00 × 10⁻⁴)) = ln 18.0 = 2.890, and 1/300 − 1/320 = 2.083 × 10⁻⁴ K⁻¹. So Ea = (8.31 × 2.890) ÷ (2.083 × 10⁻⁴) = 1.15 × 10⁵ J mol⁻¹, that is 115 kJ mol⁻¹.

  3. 3

    Rearrange the dilution relationship c₁V₁ = c₂V₂ and calculate the concentration obtained when 25.0 cm³ of 1.00 mol dm⁻³ solution is made up to 250 cm³.

    Model answer

    c₂ = c₁V₁ ÷ V₂ = (1.00 × 25.0) ÷ 250 = 0.100 mol dm⁻³. Because the same volume unit appears on both sides, cm³ can be used throughout without conversion.

  4. 4

    Substitute into the expression for Kc to find its value for H₂(g) + I₂(g) ⇌ 2HI(g), given an equilibrium mixture in 1.00 dm³ of 0.107 mol H₂, 0.107 mol I₂ and 0.786 mol HI.

    Model answer

    Since the volume is 1.00 dm³ the amounts are numerically the concentrations. Kc = [HI]² ÷ ([H₂][I₂]) = (0.786)² ÷ (0.107 × 0.107) = 0.6178 ÷ 0.01145 = 54.0. The units cancel, so Kc has none.

  5. 5

    A 2.00 dm³ vessel at equilibrium contains 0.0400 mol N₂O₄ and 0.0800 mol NO₂. Calculate Kc for N₂O₄(g) ⇌ 2NO₂(g), including units.

    Model answer

    Concentrations must be found first: [N₂O₄] = 0.0400 ÷ 2.00 = 0.0200 mol dm⁻³ and [NO₂] = 0.0800 ÷ 2.00 = 0.0400 mol dm⁻³. Kc = [NO₂]² ÷ [N₂O₄] = (0.0400)² ÷ 0.0200 = 0.00160 ÷ 0.0200 = 0.0800 mol dm⁻³.

  6. 6

    An equilibrium mixture of ethanoic acid, ethanol, ethyl ethanoate and water contains 0.33 mol of each reactant and 0.67 mol of each product in a fixed volume. Calculate Kc.

    Model answer

    Kc = ([ester][H₂O]) ÷ ([acid][alcohol]) = (0.67 × 0.67) ÷ (0.33 × 0.33) = 0.4489 ÷ 0.1089 = 4.1, to two significant figures. The volume cancels because there are equal numbers of terms above and below, so amounts in moles may be used directly, and Kc has no units.

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