Thermodynamics Flashcards

(27 cards)

1
Q

define enthalpy change of formation

A

enthalpy change when 1 mole of substance/compound is formed from its constituent elements in their standard states under standard conditions

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2
Q

define lattice enthalpy of formation

A

enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions under standard conditions

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3
Q

define lattice enthalpy of dissociation

A

the enthalpy change when 1 mole of a solid ionic compound is dissociated into its gaseous ions under standard conditions

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4
Q

define enthalpy change of dissociation

A

enthalpy change when 1 mole of bonds of the same type of molecule in the gaseous state is broken

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5
Q

define enthalpy change of 1st ionisation

A

enthalpy change when 1 mole of gaseous 1+ ions are made from 1 mole of gaseous atoms

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6
Q

define enthalpy change of atomisation

A

enthalpy change when 1 mole of gaseous atoms is made from an element in its standard state

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7
Q

define enthalpy change of 2nd ionisation

A

enthalpy change when 1 mole of gaseous 2+ ions are made from 1 mole of gaseous 1+ ions

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8
Q

define 1st electron affinity

A

enthalpy change when 1 mole of gaseous 1- ions are made from 1 mole of gaseous atoms

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9
Q

define 2nd electron affinity

A

enthalpy change when 1 mole gaseous 2- ions are made from 1 mole of gaseous 1- ions

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10
Q

What are born-haber cycles?

A

useful to calculate lattice enthalpies

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11
Q

why can theoretical and experiment lattice enthalpies be different?

A

different depending on how ‘purely ionic’ the compound is
Perfectly ionic model:
- ions are perfectly spherical
- the charge is evenely distributed in this sphere

if experimental value is differently it shows the compound does not follow the perfectly ionic model and has some covalent characteristics

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12
Q

why does the experimental value sometimes have some covalent characteristics?

A

positive ion distorts the charge distribution in the negative ion, therefore the positive ion polarises the negative ion

the more polarisation, the more covalent character there will be

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13
Q

define enthalpy of solution

A

enthalpy change when 1 mole of an ionic substance is dissolved in the minimum amount of solvent to ensure no further enthalpy change is observed upon further dilution

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14
Q

what is needed for a substance to dissolve?

A

substance bonds must break (endo)
new bonds formed between the solvent and the substance (exo)

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15
Q

what occurs when a substance is dissolved in a solvent?

A

substance bonds are broken to create free moving ions
bonds are formed between ions and water, the ions become hydrated

most ionic compounds are dissolved in solvents such as H2O, the delta positive H is attracted to negative ions, delta negative O is attracted to positive ions

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16
Q

what must happen for hydration to occur?

A

the new bonds formed must be the same strength or greater than those broken, if not then the substance is unlikely to dissolve

17
Q

how can the enthalpy of solution be calculated?

A

from knowing the lattice dissociation enthalpy and the enthalpy of hydration

18
Q

how do you calculate the enthalpy of a solution?

A

enthalpy of lattice dissociation + enthalpy of hydration

19
Q

what is entropy?

A

a measure of disorder, the more disordered the substance, the greater its entropy value. entropy will increase as temperature increases as the particles gain energy and move further apart, becoming more disordered

20
Q

which state has the greatest entropy?

21
Q

how do you calculate the overall entropy change?

A

S (total) = S products - S reactants

22
Q

what is gibbs free energy?

A

allows entropy change to be found without needing to measure the effects on the surroundings

23
Q

what is the equation for gibbs free energy? (G)

24
Q

what value must the Gibbs free energy be in order for a reaction to be feasible?

A

negative or zero

25
how does temperature effect the gibbs energy value and feasibility?
H and S both positive = will make T.S greater therefore will be feasible at certain temperatures H negative, S positive = makes G become more negative, therefore always feasible H positive, S negative = no effect G will always be positive, so never feasible H and S both negative, usually feasible at lower temperatures
26
how would you rearrange the equation to find the temperature at which gibbs free energy is zero?
T = H/S
27
what must be done in order to get the correct gibbs value?
enthalpy change must be converted from kJmol-1 to Jmol-1, to do this x1000