WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Mathematical Skills in Chemistry
Symbols, rearranging equations and logarithms: substituting into equations for Kc, enthalpy and pH
- 3 things to know
- All units
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
Answer each one before you reveal it.
Two acids have pKa values of 3.75 and 4.76. Which is the stronger, and by what factor do their Ka values differ?
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
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
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
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
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
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
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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