thermodynamics Flashcards

1
Q

ENTHALPY CHANGE OF FORMATION (ΔfH)

A

THE enthalpy change when 1 mole of a substance is formed from its elements in their standard states under standard conditions. 2C (s) + 2H2 (g) —–> c2h6 (g)

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2
Q

ΔlatticeH - lattice enthalpy of formation

A

enthalpy when 1 mole of a solid ionic compound is formed from its gaseous ions under standard conditions
ca2 (g) + + 2cl- (g) —–>cacl2 (s)

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3
Q

Δlattice H - lattice enthalpy of dissociation

A

enthalpy change when 1 mole of a solid ionic compound is dissociated into its gaseous ions under standard conditions. cacl2 (s) ——-> ca2+ (g) + 2cl- (g)

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4
Q

ΔdissH - enthalpy change of dissociation

A

then enthalpy change when 1 mole of bonds of the same type of molecule in the gaseous state is broken.
f2 (g) —–> 2F (g)

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5
Q

Δie1H - enthalpy change of 1st ionisation

A

enthalpy change when 1 mole of gaseous 1+ ions are made from 1 mole of gaseous atoms
Na(g) —-> Na+ (g)

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6
Q

ΔatH - enthalpy change of atomisation

A

enthalpy change when 1 mol of gaseous atoms is made from an element in its standard state
1/2F2(g)—-> F(g)

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7
Q

Δie2H - enthalpy change of 2nd ionisation

A

the enthalpy change when 1 mole of gaseous 2+ ions are made from 1 mol of gaseous 1+ ions
ca+(g)—–> ca2+(g)

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8
Q

Δea1H - 1st electron affinity

A

enthalpy change when 1 mole of gaseous 1- ions are made from 1 mole of gaseous atoms.
O(g) —> O- (g)

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9
Q

Δea1H - 2nd electron affinity

A

enthalpy change when 1 mole of gaseous 2- ions are made from 1 mol of gaseous 1- ions
O- (g)—–> o2- (g)

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10
Q

like in hess’s cycle - in born haber cycles the total enthalpy change of a reaction is…..

A

the same no matter what route is taken

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11
Q

theoretical and experimental values of lattice enthalpies can be different depending on what?

A

how purely ionic the compound is

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12
Q

how can theoretical lattice enthalpies be calculated?

A

from data assuming a perfectly ionic model

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13
Q

what is a perfectly ionic model

A

model where:
1) ions are perfectly spherical
2) the charge is evenly distributed in the spheres (point charges)

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14
Q

what does this tell us about the fact that the experimental lattice enthalpy value worked out is different from the theoretical value?

A

it tells us the compound being experimented on doesnt follow the perfectly ionic model (dont assume the perfectly ionic modeL) and instead has some covalent character. this is due to:
most of the time the positive ion disorts the charge distribution in the negative ion. therefore the positive ion polarises the negative ion. the more the polarisation, the more covalent character there will be (sharing of electrons).

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15
Q

lattice enthalpy values tells us how much a substance is …

A

purely ionic

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16
Q

covalent character is caused by what?

A

larger distortions in the negative ion

17
Q

the bigger the difference in lattice enthalpy values between experimental and theoretical, the more what?

A

the more polarisation present and hence the greater the covalent character

18
Q

ΔsolutionH - enthalpy change 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.

19
Q

for a substance to dissolve what must happen, in terms of bonding?

A

1) substance bonds must break (endothermic)
2) new bonds formed between the solvent and substance (exothermic)

20
Q

shiw how hydration (of ionic lattice in solid form/ salt) takes place and state what must happen in terms of bond formation strength compared with bond broken and explain why soluble substances tend to have exothermic enthalpies of solution as a result?

A

https://merchanttaylorsschools-my.sharepoint.com/:w:/g/personal/tofoma_merchanttaylors_com/EdTowe9OV9JNm7hylVtX2PwBZ9BovTgXncw9TdIR-nTENw?e=Dn8XaC

21
Q

ENTHALPY OF HYDRATION

A

ENTHALPY CHANGE WHEN 1 MOLE OF AQUEOUS IONS IS MADE FROM 1 MOL OF GASEOUS IONS.

22
Q

GIVE THE CYCLE FOR CALCULATING ENTHALPY OF SOLUTION USING YNTHALPY OF HYDRATION AND LATTICE DISSOSCIATION ENTHALPY, AND STATE WHAT WE ASSUME WHEN DISSOLVING SALT/IONIC LATTICES IN SOLUTION.

A

https://merchanttaylorsschools-my.sharepoint.com/:w:/g/personal/tofoma_merchanttaylors_com/EUyH8uEO55BOrUnI6tW7_-MBJ3faJpay42O6lv8IwMyZ0g?e=bc9kaA

23
Q

ENTROPY IS A MEASURE OF ….

A

MEASURE OF DISORDER

24
Q

WHAT IS ENTROPY (S)

A

THE Number of ways energy can be shared between particles

25
the more disorder there is the higher the level of ...
entropy
26
GOING FROM SOLID TO GAS, WHAT HAPPENS TO THE DISORDER AND HENCE THE ENTROPY
https://merchanttaylorsschools-my.sharepoint.com/:w:/g/personal/tofoma_merchanttaylors_com/ERK3NKho-4NFvilEUuf2Ro4BccCGG4OEOhQJLdZtnPLFIw?e=TkkdDB
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APART FROM THE ARRANGEMENT OF PARTICLES (whether its solid. liquid or gas), what also affects entropy change
the number of particles also (i.e the number of moles in a reaction in the same state). more moles=entropy increases because this means there are more ways energy can be distributed. eg, n2o4(g) ----> 2no2 (g)
27
reaction can be spontaneous (feasible) even if it is what, and state why? and give an eg of a spontaneuos enthalpically not favourable endothermic reaction
enthalpically unfavourable (i.e endothermic). a reaction will tend towards more disorder and hence increase entropy. increasing entropy is energetically favourable and some reactions that are enthalpically unfavourable (endothermic) can still spontaneosuly react if changes in entropy overcome changes in enthalpy. https://merchanttaylorsschools-my.sharepoint.com/:w:/g/personal/tofoma_merchanttaylors_com/EcNsJuNl3mtOkCvm81eNoHEBKJb8rftMmnIZwZAb5S1fow?e=HITwva
27
a reaction is entropically feasible when ΔS is....
positive
28
ΔS formula and standard entropy conditions
https://merchanttaylorsschools-my.sharepoint.com/:w:/g/personal/tofoma_merchanttaylors_com/ESyUhYDRXRlNpZ6J-nGY8dgBWhGESwLR84dynqkhv8jagQ?e=QHXcEN
29
what does ΔG tell us about a reaction.
whether its feasible or not
30
for a reaction to be feasible ΔG must be...
negative or 0
31
if a reaction is calculated to be feasible, (as ΔG is negative or 0), u may not observe a reaction occuring. why?
due to the activation energy being too high or the rate of reaction bering very slow.
32
exothermic reactions with a positive entropy would always be...
feasible whatever the temp is
33
endothermic reactions with a negative entropy would never be....
feasible whatever the temp is
34
different scenarios when ΔH and Δs is changed and hence ΔG value is changed, state wether the reaction is feasible at each change, and at which temps.
https://merchanttaylorsschools-my.sharepoint.com/:w:/g/personal/tofoma_merchanttaylors_com/ERpnxY43Fw5PkDoCuCPVvf0BQZ_ff1Y80pElm1Lw5oegkw?e=LM8htx
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