Thermodynamics (AQA A Level Chemistry): Flashcards

1. Enthalpy of atomisation of Na (s) → Na (g)

2. Enthalpy of atomisation of ½Cl₂ (g) → Cl (g)

3. First ionisation energy: Na (g) → Na⁺ (g) + e^–

4. First electron affinity: Cl (g) + e^– → Cl^- (g)

5. Lattice enthalpy: Na⁺ (g) + Cl^- (g) → NaCl (s)

The enthalpy of formation links the elements directly to NaCl (s).

For MgO, the Born-Haber cycle includes the first and second ionisation energies of Mg, and the first and second electron affinities of O.

False.

The direction of each arrow must be correct but the relative size of the steps does not matter. However, electrons must be shown in the ionisation step.

Route 1 (direct): formation of the compound from its elements — the standard enthalpy of formation.

Route 2 (indirect): atomisation + ionisation + electron affinity + lattice enthalpy (sum of all steps via gaseous ions).

Using Hess's Law on a Born-Haber cycle: ΔH_f^ꝋ = ΔH₁^ꝋ + ΔH_latt^ꝋ, so ΔH_latt^ꝋ = ΔH_f^ꝋ - ΔH₁^ꝋ.

The sum of all enthalpy changes required to convert the elements in their standard states to gaseous ions — including atomisation enthalpies, ionisation energies and electron affinities.

True.

There are two moles of Cl atoms in MgCl₂, so two moles of electrons are added to two moles of Cl (g) to form two moles of Cl^- (g). The electron affinity value must therefore be multiplied by two.

ΔH_latt^ꝋ = ΔH_f^ꝋ − [(+90) + (+122) + (+418) + (−349)]

= (−437) − (+281)

= −718 kJ mol^-1

In a Born-Haber cycle calculation, the stage you are asked to calculate is always the direct route, and the known stages form the indirect route.

False.

Any stage in the cycle can be calculated if all other values are known. For example, you could be given the lattice enthalpy and asked to find the enthalpy of formation or an electron affinity.

To avoid sign errors when substituting multiple positive and negative enthalpy values. Missing a bracket can change the sign of a term and produce an incorrect final answer.

A half-life is the time taken for the concentration of a reactant to fall to half its initial value. For a first-order reaction, the half-life is constant and independent of initial concentration.

A rate-determining step is the slowest step in a multi-step reaction mechanism. It determines the overall rate of the reaction and the form of the rate equation.

As ionic radius increases, the lattice enthalpy becomes less exothermic because the electrostatic forces of attraction between oppositely charged ions are weaker.

True.

Ca²⁺ and O^2- have greater ionic charges than K⁺ and Cl^-, resulting in stronger electrostatic attraction and a more exothermic lattice enthalpy. Their smaller ionic radii also contribute.

Zn²⁺ is small with a high charge, distorting the electron cloud of the larger S^2- ion. This creates covalent character that the purely ionic theoretical model does not account for, leading to a larger discrepancy.

On a graph of lattice enthalpy against ionic radius for Group 1 halides, as ionic radius increases the lattice enthalpy values become less exothermic, showing weaker electrostatic attraction.

True.

The theoretical model assumes electrostatic attraction only between point charges. Close agreement (as for NaCl) indicates the ionic model is valid. A large discrepancy suggests covalent character.

1. Ionic charge: greater charge → stronger electrostatic attraction → more exothermic lattice enthalpy.

2. Ionic radius: larger ions → ions further apart → weaker electrostatic attraction → less exothermic lattice enthalpy.

A Maxwell–Boltzmann distribution is a graph showing the distribution of kinetic energies (or speeds) of particles in a gas at a given temperature. The area under the curve to the right of the activation energy represents the fraction of molecules that can react.

The enthalpy of solution is related to other terms by: ΔH_sol^ꝋ = reverse lattice enthalpy + ΔH_hyd^ꝋ (sum of hydration enthalpies of all ions).

False.

The enthalpy of solution can be either exothermic (negative) or endothermic (positive), depending on the balance between the energy needed to break the lattice and the energy released by hydration.

Ion-dipole attractions form because water is a polar molecule. The δ− oxygen is attracted to positive ions; the δ+ hydrogen atoms are attracted to negative ions.

ΔH_hyd^ꝋ [KCl] = (−711) + (+26) = −685 kJ mol^-1

ΔH_hyd^ꝋ [Cl^-] = (−685) − (−322) = −363 kJ mol^-1.

True.

MgCl₂ contains two moles of Cl^- ions per formula unit, so the hydration enthalpy contribution from Cl^- must be multiplied by two before substituting into the equation.

A rate constant is the proportionality constant k in the rate equation. Its value depends on temperature and the nature of the reaction, but not on concentration.

An ester is an organic compound formed by the condensation reaction between a carboxylic acid and an alcohol (or phenol), with the elimination of water and the formation of an ester linkage (–COO–).

When CaCO₃ (s) decomposes to CaO (s) and CO₂ (g), entropy increases because a gas is formed, which is more disordered than the solid reactant.

False.

Feasibility describes only whether a reaction is energetically favourable. It takes no account of rate. A feasible reaction may be very slow, such as the rusting of iron.

ΔS_system^ꝋ = ΣΔS_products^ꝋ − ΣΔS_reactants^ꝋ

Entropy values for elements are not zero (unlike enthalpies of formation).

ΔS_system^ꝋ = (2 × 38.20) − (2 × 32.60 + 205.0) = (76.40) − (270.2) = −193.8 J K^-1 mol^-1.

False.

Entropy changes are an order of magnitude smaller than enthalpy changes, so entropy is measured in J K^-1 mol^-1 (joules, not kilojoules).

The entropy of the universe is always increasing. Spontaneous (feasible) reactions occur because they lead to greater overall disorder in the universe.

ΔG^ꝋ = ΔH_reaction^ꝋ − TΔS_system^ꝋ

where ΔG is in kJ mol^-1, T is in kelvin, and ΔS must be converted from J K^-1 mol^-1 to kJ K^-1 mol^-1 before substituting.

Before substituting ΔS^ꝋ into the Gibbs equation, it must be converted from J K^-1 mol^-1 to kJ K^-1 mol^-1 by dividing by 1000.

True.

When ΔG^ꝋ < 0 the reaction is feasible and likely to occur. When ΔG^ꝋ > 0 the reaction is not feasible. The borderline is ΔG^ꝋ = 0.

ΔS^ꝋ = +344 ÷ 1000 = +0.344 kJ K^-1 mol^-1

ΔG^ꝋ = (+135) − (298 × 0.344) = +135 − 102.51 = +32.5 kJ mol^-1

The positive value shows the reaction is not feasible at 298 K.

ΔG^ꝋ can also be calculated using: ΔG^ꝋ = ΣΔG_products^ꝋ − ΣΔG_reactants^ꝋ.

True.

Like standard enthalpy of formation, the standard Gibbs free energy change of an element in its standard state is defined as zero.

Enthalpy change (ΔH^ꝋ) and entropy change (ΔS^ꝋ) are combined via the Gibbs equation: ΔG^ꝋ = ΔH_reaction^ꝋ − TΔS_system^ꝋ.

A negative ΔG indicates a feasible reaction.

A homogeneous catalyst is a catalyst that is in the same phase as the reactants. It provides an alternative reaction pathway with a lower activation energy, increasing the rate of reaction.

ΔG^ꝋ must be negative (or zero at the boundary). A positive ΔG^ꝋ means the reaction is not feasible.

An endothermic reaction with a positive ΔS becomes feasible only at high temperatures, when the −TΔS term becomes negative enough to overcome ΔH.

True.

Both ΔH (negative) and −TΔS (negative, because ΔS is positive) contribute negatively to ΔG, making it always negative.

Set ΔG = 0: T = ΔH^ꝋ / ΔS^ꝋ

Convert ΔS: 581 ÷ 1000 = 0.581 kJ K^-1 mol^-1

T = 1336 / 0.581 = 2299 K

The reaction becomes feasible above this temperature.

On a ΔG vs T graph, the gradient equals −ΔS^ꝋ and the y-intercept equals ΔH^ꝋ.

False.

When ΔH > 0 and ΔS < 0, ΔG is always positive at all temperatures, so the reaction is never feasible.

The temperature at which ΔG = 0 — the point at which the reaction changes from non-feasible to feasible (or vice versa). Above or below this temperature the feasibility changes according to the sign of ΔS.

A spontaneous reaction is one that proceeds without continuous external input of energy. It occurs when ΔG is negative, meaning the reaction is thermodynamically feasible under the given conditions.