Thermodynamic- paper 1 Flashcards

1
Q

Define Hess’ law

A

the enthalpy change for a chemical reaction is the same, regardless of the route taken from reactants to products

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2
Q

the standard enthalpy of formation △Høf
give an example for the formation of NaCl

A

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

Na(s) + 1/2Cl2 –> NaCl

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3
Q

The standard enthalpy of combustion △Høc
give an example using C2H6

A

The enthalpy change when one mole of a compound is completely burned in oxygen under standard conditions, all reactants and products in their standard state

C2H6(g) + 1/2O2(g) –> 2CO2(g) + 3H2O (g)

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4
Q

the standard enthalpy of atomisation, △Høat
use Na and Cl as an example

A

the enthalpy change when one mole of gaseous atoms if formed from an element in its standard state

1/2Cl2(g) –> Cl(g)

Na(s) –> Na(g)

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5
Q

Mean bond enthalpy △HøBE
use Cl2 as an example

A

The enthalpy change when one mole of gaseous molecules each break a covalent bond to form two free radicals, averaged over a range of compounds

Cl2(g) –> 2Cl(g)

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6
Q

How are the enthalpy of atomisation and bond enthalpy related?

A

Bond enthalpy is exactly double of atomization

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7
Q

First Ionisation Enthalpy 1st△Høi
give an example using Mg

A

the standard enthalpy change when one mole of electrons is removed from one mole of gaseous atoms to give one mole of gaseous ions each with a single positive charge

Mg(g) –> Mg+(g) + e-

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8
Q

Second ionisation enthalpy
use Mg as an example

A

The standard enthalpy change when one mole of electrons is removed from one mole of gaseous 1+ ions to give one mole of gaseous ions each with a 2+ charge

Mg +(g) –> Mg2+ (g) + e-

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9
Q

First electron affinity, 1st △Høea
use chlorine as an example

A

The standard enthalpy change when one mole of gaseous atoms is converted into a mole of gaseous ions, each with a single negative charge under standard conditions

Cl(g) + e- –> Cl- (g)

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10
Q

Second electron affinity, 2nd△Høea
use Cl as an example

A

the standard enthalpy change when one mole of electrons is added to a mole of gaseous ions, each with a single negative charge to form a mole of ions each with a two negative charge

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11
Q

Lattice formation enthalpy △HLFO
use NaCl as an example

A

the standard enthalpy change when one mole of solid ionic compound is formed from its gaseous ions

Na+ (g) + Cl- (g) –> NaCl (s)

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12
Q

Lattice dissociation enthalpy △HLDO
Use NaCl as an example

A

The standard enthalpy change when one mole of solid ionic compound dissociates into its gaseous ions
NaCl (s) –> Na+ (g) + Cl-(g)

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13
Q

standard enthalpy of hydration △HhydO
Use Na+ and Cl-

A

The standard enthalpy change when one mole of gaseous ions is converted into one mole of aqueous ions

Na+(g) + Cl-(g) –> Na+(aq) + Cl-(aq)

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14
Q

standard enthalpy of solution △HsolO
use NaCl as an example

A

the standard enthalpy change when one mole of solute dissolves in enough solvent to form a solution in which the ions are far enough apart to not interact with each other

NaCl (s) + aq –> Na+(aq) + Cl- (aq).

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15
Q

state the steps to follow when constructing a Born-Haber cycle

A

Start with the elements in their standard state
atomise the metal element
atomise the non-metal element
ionise the metal ion
electron affinity for non-metal
lasstice formation for whole ionic compound
enthalpy of formation

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16
Q

why is electron affinity/ any reaction endothermic

A

need to overcome repulsion so energy needs to be put in, negative electrons added to an already negative ion

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17
Q

how do you compare atoms

A

C: charge of the ion
R: radius/size of the ion
A:attraction between the ions
M:more exothermic or endothermic

18
Q

What does the perfect ionic model feature

A

Ions are point charges and are perfect spheres

19
Q

Theoretical vs experimental lattice enthalpy’s

A

Theorectical
-Perfect Ionic model
-Ions are point charges or perfect spheres
-Ionic bonding

Experimental
-Born haber
-ions are polarisable
- and covalent bonding

20
Q

what can be determined if a ionic compound has no covalent character

A

it is a purely ionic bond and ions are not polarisable

21
Q

which model predicts stronger bonding

A

Born Haber- it allows for covalent character so predicts stronger bonding.

22
Q

When an ionic solid is dissolved in solution, what is happening in terms of bonds?

A

Strong ionic bonds must be broken, water (solvent) is polar
+ ions are attracted to Od- or negative ions attracted to Hd+

23
Q

why are hydration enthalpies always negative

A

due to the electrostatic attraction between metal ion and Od- of water

24
Q

define what a polar bond is

A

difference in electronegativity of 2 atoms in a covalent bond means that one atom attracts electron density more stronger making this side more negative

25
Q

key facts for entropy

A

entropy- the amount of disorder within a system. entropy is given the symbol S and is always a positive number.

26
Q

what are the units for entropy

A

jK -1 mol-1 although will sometimes needs to be converted to kJK-1 mol-1- so divide by 1000

27
Q

what temperature in Kelvin is considered to be perfect temperature

A

0- perfect order has particles have no energy and therefore no entropy

28
Q

how to determine whether a reaction is feasible

A

if the value is less than 0

29
Q

Mg(s) +2HCl(aq)–> MgCl2(aq) + H2(g)
is the entropy positive or negative

A

fewer particles on thw product side but a gas has been produced so entropy is positive as it has increased.

30
Q

how to calculate delta s

A

sum of products-sum of reactants

31
Q

equation for gibbs free energy change, △G

A

△G= △H-T△S

32
Q

give the units for △G
△H
T
△S

A

△G= KJmol-1
△H=KJmol-1
T=Kelvin
△S= KJK-1Mol-1

33
Q

rearange the equation to calculate temperature

A

T= △S/△G

34
Q

describe the feasibility of a reaction that has a positive △H and △S

A

the equation: -△H-(T△S)
△G= always negative
feasibility= reaction is feasible at any temperature

35
Q

describe the feasibility of a reaction when △H is positive and △S is negative

A

the equation: △H + T △S
△G= always positive
feasibility= reaction is not feasible at any temperature

36
Q

describe the feasibility of the reaction when △H is negative and △S is negative

A

the equation: -△H + T △S
△G= △G becomes more negative at a lower temperature as △H > T △S
△gets more positive at higher temp
-T △S> △H
feasibility- reaction gets more feasible at lower temperatures

37
Q

describe the feasibility when △H and △S are both positive

A

equation= △H-(T △S)
△G= gets more negative at higher temperatures
T △S> △H
more positive at lower temps
△H>T △S
feasibility= reaction gets more feasible at higher temperatures

38
Q

kinetic factors:

A

the reaction may still have a very high activation energy such that very few particles have sufficient energy to react or it may occur at a very slow rate.

39
Q

△G and graphs

A

△H= y intercept
△S= the gradient

40
Q

y=mx + c alongside gibbs

A

y=mx+c
△G= -△ST + △H