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Description
Flashcards by Andrei De Leon, updated more than 1 year ago
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Created by Andrei De Leon
over 5 years ago
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| Question | Answer |
| Enthalpy Change (kJmol⁻¹) ΔH | Heat energy change at constant pressure |
| Standard Enthalpy Change | Enthalpy change under standard conditions, all reactants and products in their standard states |
| Hess' Law | Enthalpy change is independent of route |
| Enthalpy Change of Formation ΔH(f) | Enthalpy change when 1 mole of a compound is formed from its elements under standard conditions, all reactants and products in their standard states |
| Explain why an experimental ΔH(f) will differ from the value given by a data book | Data books use the average ΔH(f) for a given bond in different compounds |
| Enthalpy Change of Combustion ΔH(c) | Enthalpy change when 1 mole of a substance is burned completely in the presence of oxygen, all reactants and products in their standard states |
| Bond Dissociation Enthalpy ΔH(diss) | Enthalpy change when 1 mole of covalent bonds are completely separated to form gaseous atoms |
| Explain why ionic bond enthalpies are not found in any data book | Bond enthalpy only applies to covalent substances |
| Enthalpy of Atomisation ΔH(at) | Enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state under standard conditions |
| Explain why ΔH(at) is half of ΔH(diss) for diatomic molecules | a1 mole of atoms are formed in atomisation 2 moles of chlorine are formed in bond dissociation |
| 1st Ionisation Enthalpy ΔH(i1) | Enthalpy change when 1 mole of gaseous 1+ ions is formed from 1 mole of gaseous atoms For compounds: enthalpy change when 1 mole of a compound in its standard state is converted to gaseous atoms under standard conditions |
| 2nd Ionisation Enthalpy ΔH(i2) | Enthalpy change when 1 mole of gaseous 2+ ions is formed from 1 mole of gaseous 1+ions |
| 1st Electron Affinity ΔH(ea1) | Enthalpy change when 1 mole, of gaseous 1- ions is formed from 1 mole of gaseous atoms |
| 2nd Electron Affinity ΔH(ea2) | Enthalpy change when 1 mole of gaseous 2- ion is formed from 1 mole of gaseous 1- ions |
| Lattice Enthalpy of Formation ΔH(L) | Enthalpy change when 1 mole of solid ionic compound is formed from its gaseous ions |
| Trends in Lattice Enthalpy | Decreases down a group- With larger ions the opposite charges are further apart so lattice enthalpies are smaller |
| The Perfect Ionic Model | 1.Ions are perfect spheres with charge evenly distributed 2. Bonding in the lattice is ionic with no covalent character |
| Explain partial covalent bonding in ionic compounds | Positive ions attract the outer electrons of negative ions. Electron clouds are polarised. |
| Trends in polarisation | Cations: Become more polarising as charge increases and size decreases (higher charge density) Anions: Become more polarisable as size increases |
| Enthalpy of Solution ΔH(soln) | Enthalpy change when 1 mole of ionic compound dissolves in water to give a solution of infinite dilution |
| Enthalpy of Hydration ΔH(hyd) | Enthalpy change when 1 mole of aqueous ions is formed from gaseous ions |
| How do you find enthalpy of solution? ΔH (soln) | ΔH (soln) = ΔH (L diss) + Σ ΔH (hydration) |
| Factors affecting solubility | 1. Higher charge (greater electrostatic forces of attraction) 2. Larger ion (less charge density) |
| What processes does dissolving involve? | 1. Lattice dissociation 2. Hydration of gaseous ions to produce aqueous ions |
| Entropy (Jmol⁻¹K⁻¹) ΔS | Measure of disorder in a system |
| Factors affecting entropy | 1. Change in physical state (Temperature) 2. Dissolving 3. Change in moles |
| Formula to find entropy change | ΔS = ΔS (products) - ΔS (reactants) ΔS= Entropy change |
| When is the entropy of an element zero? | 1. 0 Kelvin 2. Particles have no energy to vibrate/ distribute 3. Substance in a state of perfect order |
| What causes differences in entropy between substances in the same physical state? | 1. Larger Mr 2. More electrons, more energy levels |
| Why is ΔS for boiling/condensing > ΔS for melting/freezing? | Gases are more disordered than liquids |
| What is the relationship between ΔS and ΔH? | 1. ΔS ∝ ΔH 2. ΔS ∝ 1/T 3. ΔS ∝ ΔH/T 4. ΔS (surrounding) = -ΔH/T 5. ΔS(total) = ΔS(system) + ΔS(surroundings) 6. ΔS(total) = ΔS(system) -ΔH/T |
| Gibbs free energy change (kJmol⁻¹) ΔG | -TΔS(total) 1. The only thermodynamic term that can predict whether a reaction is feasible 2. Free energy refers to remaining ΔH, after sufficient energy is transferred to make ΔG ≤ 0 |
| Formula to find ΔG | ΔG = ΔH - TΔS ΔG = kJmol⁻¹ ΔH = kJmol⁻¹ T = K ΔS = JK⁻¹mol⁻¹ |
| Conditions required for a reaction to be feasible | ΔG ≤ 0 |
| Reasons why a reaction with ΔG ≤ 0 may not occur | 1. Activation energy is too high 2. Rate is too slow |
| How to find the minimum temperature at which a reaction will be feasible | 1. ΔG = 0 2. ΔH = TΔS 3. T = ΔH/ΔS |
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