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Chemistry: CHEM 5 > Thermodynamics > Flashcards

Thermodynamics Flashcards

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1
Q

What is enthalpy change?

A

The heat change of a reaction at constant pressure.

ΔH.

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2
Q

What are standard condition?

A
  • Temperature of 298K (25°C)
  • Pressure of 100kPa
  • All reactants and products in their standard state.
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3
Q

What is the standard enthalpy of formation, ΔHᶱf?

A

The enthalpy change when one of a compound is formed from its constituent elements under standard conditions, all reactants and products in their standard state.

(The more exothermic the enthalpy of formation, the more thermodynamically stable a compound is.)

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4
Q

What is the standard enthalpy of formation of an element?

A

Zero, by definition

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5
Q

What is the standard enthalpy of atomisation, ΔHᶱat?

A

The enthalpy change which accompanies the formation of one mole of gaseous atoms from the element in its standard state, under standard conditions.
E.g. Mg(s) –> Mg(g) ΔHᶱat = +147.7

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6
Q

What is first ionisation energy, ΔHᶱi1?

A

The standard enthalpy change when one mole of gaseous atoms is converted to a mole of gaseous ions each with a single positive charge.
E.g. Na(g) –> Na+(g) + e- ΔHᶱi1 = +496

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7
Q

What is second ionisation energy, ΔHᶱi2?

A

The standard enthalpy when one mole of electrons are removed from one mole of gaseous ions each with a single positive charge to form a mole of gaseous ions each with two positive charges.
E.g. Na+(g) –> Na2+(g) + e- ΔHᶱi2 = +4563

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8
Q

Why is second ionisation energy much larger than first ionisation energy?

A

Second IE requires much more energy than the first because it is removing an electron from a +1 ion so there is a higher nuclear attraction.

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9
Q

What is first electron affinity, ΔHᶱea?

A

The standard enthalpy charge when a mole of gaseous atoms is converted to a mole of gaseous ions, each with a single negative charge.
E.g. O(g) + e- –> O-(g) ΔHᶱea = -141.1
First EA are always negative values, showing a release of energy.

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10
Q

What is the trend in first electron affinity going down a group?

A

As you go down a group, first electron affinities get less. This is due to shielding and distance from the nucleus, meaning there is a lower attraction between the incoming electron and the nucleus.

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11
Q

What is second electron affinity, ΔHᶱea?

A

The standard enthalpy change when a mole of electrons is added to a mole of gaseous ions each with a single negative charge to form ions each with two negative charges.
E.g. O-(g) + e- –> O2-(g) ΔHᶱea = +798
Second electron affinities have positive values, showing an intake of energy. This is because you are forcing an electron into a small, very electron-dense space.

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12
Q

What is lattice formation enthalpy, ΔHᶱL?

A

The standard enthalpy change when one mole of solid ionic compound is formed from its gaseous ions.
E.g. Na+(g) + Cl-(g) –> NaCl(s) ΔHᶱL = -788
ΔHᶱL is always negative because energy is given out when a lattice forms.

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13
Q

What is lattice dissociation enthalpy, ΔHᶱL?

A

The standard enthalpy change when one mole of solid ionic compound separated into its gaseous ions.
E.g. NaCl(s) –> Na+(g) + Cl-(g) ΔHᶱL = +788
ΔHᶱL has the same numerical value as lattice formation enthalpy but is always positive as it is the opposite process.

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14
Q

What are the factors affecting lattice enthalpy?

A
  • Ion radius

- Charge

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15
Q

How does ion radius affect lattice enthalpy?

A

Lattice enthalpy because less exothermic (less -ve) as the size of the ions increase. Attraction decreases due to increased distance between the centres of the oppositely charged ions, shielding is also increased as ion radius increases.

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16
Q

How does charge affect lattice enthalpy?

A

As charge increases (across a period), it produces a greater attraction between the positive and negative ions. The ionic radius decreases resulting in the ions in the lattice being closer together, producing more attraction. As charge increases, lattice enthalpies become more exothermic and more negative.

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17
Q

How can lattice enthalpy be calculated?

A

Lattice enthalpy cannot be calculated directly because it is impossible to calculate the lattice enthalpy charge when converting gaseous ions solid crystals. Therefore we use Hess’s law to construct a Born-Haber cycle to calculate lattice enthalpy from experimental data.

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18
Q

What is Hess’s Law?

A

Hess’s Law states that the enthalpy change for a chemical reaction is the same, whatever route is taken from reactants to products.

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19
Q

What is a Born-Haber cycle?

A

A thermochemical cycle that includes all the enthalpy changes involved in the formation of an ionic compound, based on Hess’s Law.

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20
Q

How can lattice enthalpy be calculated from a Born-Haber cycle?

A

ΣΔHf =Σclockwise enthalpy changes

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21
Q

Why might the theoretical value for lattice enthalpy be different from the experimental value?

A

Theoretical lattice enthalpies assumes a perfect ionic model where the ions are 100% ionic and spherical and the attractions are purely electrostatic.

  • There are some compounds where there is a large discrepancy between the two values. This may be because extra bonding is present due to the compound having some covalent character so the compound is not purely ionic.
  • The additional bonding means the forces in lattice are stronger than pure ionic attraction.
22
Q

What are the factors that increase polarisation meaning that ionic compounds may not be purely ionic?

A
  • Cation: small size, high charge (to pull electron density towards it).
  • Anion: Large size (so electrons are further from the nucleus), high charge (so there are lots of electrons to pull)
23
Q

What is enthalpy of solution, ΔHᶱsol?

A

The standard enthalpy change when one mole of solute dissolves in sufficient solvent to form a solution in which the molecules or ions are far enough apart not to interact with eachother.
E.g. NaCl(s) + aq –>Na+(aq) + Cl-(aq)

24
Q

What is enthalpy of hydration, ΔHᶱhyd?

A

The standard enthalpy change when water molecules surround one mole of gaseous ions.
E.g. Na+ + aq –> Na+(aq)
For ionic compounds the enthalpy of hydration has rather a small value and can be positive or negative.

25
Q

When is the enthalpy of hydration likely to be more exothermic?

A

The ion has a smaller radius, meaning it has a greater charge density. This means it has a stronger attraction to water.

26
Q

What is the process of dissolving an ionic compound in water?

A
  1. Breaking the lattice to give separate gaseous ions (ΔHᶱLE)
  2. Hydrating the cations (ΔHᶱhyd)
  3. Hydrating the anions (ΔHᶱhyd)
27
Q

How do you calculate enthalpy of solution, ΔHᶱsol?

A

ΔHᶱsol = ΔHᶱL(+ve) + ΔHᶱhyd(cation) + ΔHᶱhyd(anion)

28
Q

What is the relationship between ΔHᶱsol and the ability to dissolve?

A
  • Generally if ΔHᶱsol is negative (exothermic), the ionic compound is likely to dissolve.
  • If the reaction is endothermic, it is unlikely to dissolve, theoretically.
29
Q

How do you calculate enthalpy of solution, ΔHᶱsol, from experimental values?

A

q=mcΔT / mol of ionic compound

30
Q

What are the factors affecting the enthalpy of solution?

A
  • Ion radius: the smaller the ion, the higher the attraction so the higher ΔHᶱsol.
  • Charge: the higher charged the ion, the higher the attraction so the higher ΔHᶱsol.
31
Q

What is mean bond enthalpy, ΔHᶱdiss?

A

The enthalpy change when one mole of gaseous molecules each breaks a covalent bond to form two free radicals, averaged over a range of compounds.
E.g. CH4(g) –> C(g) + 4H(g) ΔHᶱdiss = 1664
Average for C-H bond = 1664/4 = +416

32
Q

How do you calculate mean bond enthalpy?

A

Σreactants - Σproducts

33
Q

What is using mean bond enthalpies less accurate than using experimental data?

A

Bond enthalpies are average values from a range of compounds, therefore they are not exact for that specific compound.

34
Q

What are mean bond enthalpies useful for?

A
  • Comparison of bond strength, to see which bond in a molecule is most likely to break (the lowest value)
  • Hess’s Cycles to calculate approximate enthalpies when this cannot be measured experimentally.
35
Q

What is entropy?

A

A numerical measure of disorder in a chemical system.

  • Entropy is a measure of the extent to which energy is dispersed. As system become energetically more stable when it becomes disordered.
  • The units are J K-1 mol-1.
36
Q

What is the relationship between disorder and feasibility?

A

A reaction is more feasible if it is more disordered.

  • All spontaneous changes produce an overall increase in total entropy.
  • In order to be considered spontaneous, the entropy must be positive.
37
Q

When does entropy increase?

A
  • A solid becomes and liquid or a gas
  • A liquid becomes a gas
  • The temperature rise (so more KE)
  • A solid dissolves in a liquid to form a solution
  • A reaction produces products with a greater degree of freedom of movement
  • Evolution of CO2 from hydrogencarbonates with acid
  • Less moles/reactants –> more moles/reactant.
38
Q

How is total entropy change, ΔS, calculated?

A

ΔStotal = ΔSsystem + ΔSsurroundings

39
Q

How is the entropy change of the system, ΔSᶱsy, calculated?

A

ΔSᶱsys = Σ(entropy of products) - Σ(entropy of reactants)

40
Q

Why do some reactions occur, even though ΔSᶱsys is negative?

A
  • Only ΔSᶱsys was calculated, the total entropy change must be calculated to decide if a reaction is spontaneous
  • If the reaction is exothermic, energy given out increases the entropy of surrounding, so ΔStotal may be +ve.
41
Q

What drives the reaction if ΔH = -ve and ΔS = +ve?

A

Both ΔH and ΔS drives the reaction.

42
Q

What drives the reaction if ΔH = +ve and ΔS = +ve?

A

Only ΔS drives the reaction.

43
Q

What drives the reaction if ΔH = -ve and ΔS = -ve?

A

Only ΔH drives the reaction.

44
Q

How is the feasibility of a reaction determined?

A

The balance between enthalpy and entropy (as well as temperature) determines the feasibility of a reaction. This is given by the relationship known as the Gibbs free energy change, ΔG.

45
Q

How is free energy change, ΔG, calculated?

A

ΔG = ΔH - T ΔS
ΔG=J mol-1 (/1000 = kJ mol-1)
ΔH=enthalpy changex1000, J mol-1, products-reactants
T=temp in kalvin ( +273)
ΔS=entropy of system, J K-1 mol-1, products-reactants.

46
Q

Is the reaction feasible if ΔG

A

ΔG

47
Q

Is the reaction feasible if ΔG > 0 (+ve)?

A

ΔG > 0, not feasible

48
Q

Is the reaction feasible if ΔG = 0?

A

ΔG = 0, reaction just about feasible, system at equilibrium (as ΔG is not changing).
This is only achieved when ΔH and ΔS are both either +ve or -ve.

49
Q

How can temperature be calculated from the ΔG formula?

A

T = ΔH/ΔS

50
Q

How could it be decided which compound is more stable from the ΔG value?

A

The compound with the lowest ΔG is the one that is the least feasible and therefore more stable.

Chemistry: CHEM 5 (6 decks)

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