thermodyamics.

Question 1

and the trenWhat are born haber cycles?

Answer 1

= used to calculate lattice enthalpies
= cant be directly from experiments

Question 2

How is the cycle formed?

Answer 2

= bottom we have the ions
= enthalpy of formation- exothermic
= enthalpy of atomisation for both elements- endothermic
= 1st ionisation energy of one of the elements- endothermic
= 1st electron affininity, exothermi
= lattice enthalpy of formation

Question 3

What do we do for double ions?

Answer 3

= 2 ionisation energies
= 2 electron affininity

Question 4

What does theoretical and experimental lattice enthalpies depend on?

Answer 4

= depending on, how purley ionic the compound it is

Question 5

What is the perfect ionic model and how does this relate to theoretical lattice enthalpies?

Answer 5

= can be calculated assuming from data assuming perfect ionic model
= ions are spherical
= 100% ionic
= the attractions are purley electrostatic

Question 6

When experimental values of lattice enthalpies are calculated what is usually seen?

Answer 6

= number is different to the theoretical value
= compound being experimented on does not follow perfect ionoc model and has covalent characteristics
= positive ion distorts, the charge distribution in the negative ion, positive ion polarises negative ion
= the more polarisation the more covalent character and the more differrnce between the theoretical and experimental values

Question 7

what does lattice enthalpy values tell us?

Answer 7

= how much a substance is purley ionic
= experimental values are higher than the theroretical value
= some covalent chaarcter being displayed, larger distortion of negative ion
=

Question 8

How does size of the ion affect the lattice enthalpy?

Answer 8

= the larger the ions, the more positive lattice enthalpy, as the ions are larger the charges between them further apart, so weakr forces of attraction between them

Question 9

How does the charge on the ion affect lattice enthalpy?

Answer 9

= the bigger the charge on the ion, the greater attreaction between the ions so the stronger the lattice enthalpy

Question 10

What is needed for a asubstance to dissolve?

Answer 10

= substance bonds must break (endothermic)
= new bonds formed, between solvent and substance (exothermic)

Question 11

How is the ionic lattice dissolved?

Answer 11

= solid ionic compound
= substance bonds broken, to create free moving ions
= bonds form between ions, and water ions are hydrated

Question 12

chemmically how does it dissolve?

Answer 12

= = most ionic compounds dissolve in polar solvents, the partial positive hydrogen is attracted to the negative ions, and partial negative oxygen is attracted to the positive ions, the structure starts to break down
= water molecules surround the ions hydration
= for this to happen, the new bonds must be the same strength or greater than the bonds broken
= if not the then the substance is very unlikely to dissolve
= soluable substances tend to have exothermic enthalpies of solution for this reason

Question 13

How do we calculate enthalpy change of a solution?

Answer 13

= knowing
= lattice dissociation energy
= enthalpy of hydration
= on the left we have the ionic compound (s), on tje right we have the ions, (aq)
= at the bottom we havr the gaseous ions
= arrows go from left down
= downwatds left is the enthalpy lattice of dissociation
= upwatds is enthaly of hydration
= we assume break the solid lattice, up into it’s gaseous ions first, (lattice dissociation), dissolve the gaseous ions in water (enthalpy of hydration)

Question 14

What is entropy?

Answer 14

= measure of disorder in a system
= number of ways energy can be shared between particles
= the more disorder the higher entropy
=at t=0 thereis no entropy as the particles are stionary and not moving so there is balance in the sytsem, no entropy

Question 15

What is enropy like between phases of states?

Answer 15

= splids liquids and gases increased entropy
= solids have lowest level of disorder, particles are arranged neatly in rows
= liquids and gases have more disorder

Question 16

How does number of particles affect entropy?

Answer 16

= if the reaction is in the same state, but more moles are produced, then entropy increases
= there are more ways eneergy can be distributed

Question 17

How can a reaction be spontaneous?

Answer 17

= a reaction that is exothermic, will results in products that are more thermodyamically stable, than the ractants
= driving force for the reaction and causes them to be spontaneous
= when it is feasivle, so when gibs free energy is less than0
= a spontaneous reaction will proceed on its own without any further, external influence

Question 18

How does entropy ^S can be calculated between reactants and products?

Answer 18

= entropy of products - entropy of reactants
= Jk-1mol-1
= standard entropy,

Question 19

if entropy is positive what does this mean?

Answer 19

= entropically feasible

Question 20

WHat is gibbs free energy ^G?

Answer 20

= tells if a reaction is feasible or not
= ^G= ^H -T^S
= ^H is J mol
= ^g is Jmol-1
= the reaction is feasible if ^G is negative or 0
=need to covert from joules to kj

Question 21

Even if the reaction calculated is feasible why might we not observe a reaction occuring?

Answer 21

= due to activation energy being too high
= rate of reaction being too slow
=

Question 22

if enthalpy change is negative and entropy is positive what is feasabillity\?

Answer 22

= gibbs is always negative
= feasible at any temperature
=

Question 23

if the enthalpy change is positive and entropy is negative what is feasibility and gibbs value?

Answer 23

= gibbs is always positive
= feasible never at any temperature

Question 24

What happens if enthalpy change is negative and the entropy change is negative?

Answer 24

= the free givs is negative at lower temperatures
= feasible at lower tempertatures
=

Question 25

What happens if the enthalpy change is positive and and the entropy is positive?

Answer 25

= free gibs is negative at higher temperatures
= and feasible at higher temperatures

Question 26

How do we calculate the temperature of when the reaction just becomes feasible can be calculated?

Answer 26

= the reaction is just feasible if ^G is 0
= s we can rearrange the equation to when it is 0
= t= enthalpy change / entropy change
=

Question 27

How can we use the y=mx+c formula?

Answer 27

= y= ^G
= c: ^H
= the gradient is -^S

Question 28

What happens to the entropy if there is the same number of molecules?

Answer 28

= the sytsem is balanced.

Question 29

What is needed for an reaction to occur?

Answer 29

= ^G needs to be 0, feasible reaction less than 0,