3.1.8 thermodynamics Flashcards
how does a giant ionic lattice form?
- ions have to be formed from atoms
- these ions have to get close enough for attractive forces to form between them (ionic bonding)
lattice enthalpy of formation
enthalpy change when 1 mole of a SOLID ionic compound is FORMED from its constituent IONS in the GAS phase
always exothermic because is the formation of ionic bonding between ions
-LEF=+LED
lattice enthalpy of dissociation
enthalpy change when one mole of a SOLID ionic compound is BROKEN UP into its constituent IONS in the GAS phase
always endothermic because requires input of heat energy to dissolve/melt (some are insoluble because lattice is too strong and water can’t overcome it. therefore to free the ions in lattice have to melt it) the ionic compound, breaking the ionic bonds.
-LEF=+LED
what are born-haber cycles and what are they for
the principle of the cycle is hess’ law; that the energy involved in a single step will equal the sum of energy involved in several steps that start and end in the same place (enthalpy change of a reaction is independent of route taken)
the BH cycle is a specific application of hess’ law for ionic compounds. it is a thermodynamic model used to calculate the lattice enthalpy of an ionic compound and analyse the formation of ionic compounds from their elements.
these data are hard to get experimentally, but the alternative route where the constituent elements become atoms become ions become cations and anions become lattice is all easier to get
what data do they use/involve
- enthalpy of formation of ionic compound
- enthalpy of atomisation and or bond dissociation enthalpy
- ionisation enthalpy (1st, 2nd, 3rd)
- electron affinity (1st, 2nd, 3rd)
- lattice enthalpy of formation/dissociation
it will be up to them obvs what they give you and expect you to find.
enthalpy of formation
enthalpy change when one mole of a substance is formed from its constituent elements standard conditions standard states
exothermic for most substances
enthalpy of atomisation
enthalpy change when one mole of gaseous atoms is produced from an element in its standard states
eg 0.5I2 (iodine) (s) –> I (g)
endothermic
bond dissociation enthalpy
enthalpy change when one mole of covalent bonds is broken in the gaseous state
eg I2 (iodine) (g) –> 2I (g)
endothermic
relationship between bond dissociation enthalpy and enthalpy of atomisation
BDE is when one mole of X2 becomes 2 moles of gaseous atoms because the bond has broken. the BDE is to do with 1 mole of bonds are broken
the atomisation is to do with forming one mole of atoms by def. therefore half mole of X2 will form 1 mole of X
this is for diatomic elements only, so oxygen, nitrogen, halogens
depending on the compound, this will be important. because if you need 2 Cl- atoms eg for the BH cycle of BaCl2, you need to use BDE. but if you’re ot given it. you need to use atomisation energy and MULTIPLY IT BY TWO.
likewise if you need one Cl- atom for the BH cycle of NaCl, so need atomisation energy, but are given dissociation; need to use BDE and DIVIDE IT BY 2
atom = 1/2 diss
diss = atom x2
ionisation enthalpy (1st, 2nd, 3rd)
- endothermic; requires energy to remove the e- and overcome the eFoA (recall; ionisation energy)
- applies to gas atom that will be the cation in the compound
FIRST
enthalpy change when each atom in one mole of GASEOUS ATOMS loses one e- to form one mole of gaseous 1+ ions
SECOND
enthalpy change when each ION in one mole of GASEOUS 1+ IONS loses one e- to form one mole of gaseous 2+ ions
because gas atoms, in BH cycle, has to come after they’ve been made into gaseous atoms from standard state elements
electron affinity (1st, 2nd, 3rd)
FIRST
- enthalpy change when one mole of GASEOUS atoms gains one e- to form one mole of gaseous 1- ions
- happens to gaseous atom that will be the anion in the compound
- is exothermic mostly because +ve charge of nucleus is attractive, and when e- gets close enough to e- cloud will be attracted into it
SECOND
- enthalpy change when one mole of GASEOUS 1- ions gains one e- to form one mole of gaseous 2- ions
- will happen for anions with 2-, 3-, 4- charges
- is ENDOTHERMIC because adding -ve electron to a -ve ion. they would repel
so in the BH cycle the arrow representing the 1st electron affinity will be pointing down, exo; the 2nd will point UP, ENDO.
(this energy is basc how easy it is for an e- to be gained by this atom/ion.)
how do you construct a BH cycle
- arrows
- how to start
-features to have
- each arrow represents 1 enthalpy change. need to do separate arrows for each ionisation enthalpy and electron affinity
- process that is endothermic, using energy, will point upwards
- exothermic processes, releasing energy, will point downwards
- always start with the elements in their standard states
- be mindful of direction of arrows, use defs of what they represent to inform you
- label each arrow and need eq
- ?for ionisation enthalpy eqs is the e- accumulative or one lost at each stage shown as one? Accumulative I think
how do you construct a BH cycle: what goes where
- what trying to make? start with the standard state elements. eg NaCl BH cycle, start with line upon which is Na (s) and Cl2 (g)
- how many ions of each thing do i need? informs whether use atom/diss for X2 elements to get the gaseous atoms. now can go from elements to gaseous atoms. go one species at a time
- gaseous atoms to ions, one species at a time. do Na first, so the will be cation. use ionisation enthalpies and make sure get to correct ionic state. for Na, easy, 1+, for others need separate arrows to get to 2+ ad 3+. need to show the Cl (will be anion) in gas atom form and the lost e- at each stage
- at top have the cation, the final lost e- and the will be anion gas atom/s
- e affinity, pointing down for 1st one as exothermic, where the e- is passed to Cl, forming Cl-. if was something that must get 2- or 3- etc charge, the subsequent e- affinities are ENDO, arrow goes up
- when have the gaseous ions then there is exothermic arrow down to the solid ionic compound, representing the LEF
- the solid ionic compound line is below the line of standard state elements from before. an arrow connects them pointing down from standard state elements to the solid ionic compound; the enthalpy of formation of the ionic compound is exothermic. don’t get confused with LEF and enthalpy of formation
what are lattice enthalpies an indication of
the strength of ionic bonding
the higher the LE, the stronger the ionic bonding
what determines strength of ionic bonding
depends on the charge of the ions and the ionic radius
smaller ion -> higher charge density and closer to oppositely charged ion (can get closer) so stronger eFoA between them
stronger charges -> stronger eFoA
stronger ionic bonding means higher lattice enthalpies
if the LED is larger what does that mean
more endothermic
so requires higher input of energy
so harder to break lattice, overcome ionic bonds
larger LEF means what
more exothermic
more energy release when solid ionic compound is formed
so more stable
how can lattice enthalpies be calculated
experimentally via a BH cycle. even though calculating via arithmetic not analytically, the values used to determine it are gotten experimentally
theoretically via the perfect ionic model, similar to ideal gas. uses an eq, and assumes that the structure is perfectly ionic, the ions being perfect points of positive or negative charge
takes into account
- size of ions
- charges of ions
- arrangement in the lattice; which differs due to unequal charges eg NaCl and MgCl2
why is the experimental value for lattice enthalpies the true value, not the PIM value
bc of the assumption that structures are always perfectly ionic
there is changes to the value (so how strong the bonding is) due to polarisation. polarisation effects the experimental values, making the BH cycle value more
polarisation
the distortion of the electron cloud around an (atom or) ion by another nearby (atom or) ion.
smaller cations with a high charge have a high charge density. so they are very attractive to the e- cloud of the anion.
this is what gives Al ions its covalent character. it so strongly attracts the e- cloud of the anion that it is like a cov bond rather than the perfect points of +/- charge like in the PIM
what will give stronger polarisation
the smaller and more highly charged the cation is
and
the larger the anion. why? because the outer e- is further away and having more shielding so less eFoA. so the outer e- would be more easily attracted to high +ve charge density of cations. big anions are polarised more easily
how can you compare the experimental (BH) value with the theoretical (PIM) value, and why?
finding the % difference
exp - theoretical
———————– x 100
experimental
the % difference between them shows
- big: then there is more covalent character in the ionic lattice that the PIM can’t account for
- small: then the closer to having a perfect ionic structure, so no distortion/polarisation, ions are more like perfect points of +/- charge
why does more polarisation make the LEs stronger
if more cov character, stronger LE bc there is more attraction to overcome. need to overcome polarisation and ionic bonding
why are some ionic compounds insoluble in water
water molecules can’t provide enough energy to overcome ionic bonding as the solvent. lattice enthalpy is too high, ionic lattice is too strong
water doesn’t provide enough energy to overcome eFoA because eFoA is too strong. too strong lattice enthalpy
LE is high and significant cov character therefore water molecules can’t provide the energy
enthalpy of solution
enthalpy change when one mole of an ionic solid dissolves in an amount of water large enough so that the dissolved ions are well separated and do not interact with each other
solid ionic lattice —»> free ions in solution
VARIES - will explain
what happens when a solid ionic lattice is dissolved in solution
solid ionic lattice
↓
↓[[endothermic; energy needed to break ionic bonds]]
↓
ionic bonds between oppositely charged ions are broken
↓
↓[[exothermic; forming FoA between h2o and ions]]
↓
ions become surrounded by water molecules (polar solvent)
↓
free ions in solution
water is a polar solvent. how does it dissolve the ions
delta -ve O is attracted to cation. so the water surrounds them, Os facing the cation.
delta +ve H is attracted to anion. so the water surrounds them, Hs facing the anion
forms hydration shells around the ions - NOT a complex, water molecules orienting themselves based on eFoA
hydration enthalpy
enthalpy change when one mole of gaseous ions become hydrated (dissolved in water)
exothermic
gas ions -> dissolved ions
how can you form a (hess) cycle using
- lattice enthalpies
- hydration enthalpies
- enthalpy of solution
gas ions
↓(1) ↓(2)
ionic solid →(3)→ dissolved ions
1- lattice enthalpy of formation
2- hydration enthalpies
3- enthalpy of solution
how can you form a (BH) cycle using
- lattice enthalpies
- hydration enthalpies
- enthalpy of solution
- solid ionic compound
- arrowing going up (endo) showing lattice enthalpy of dissociation
- goes to the ions (eg Na+ and Cl- (g/g))
- enthalpy of hydration going down for one of them, changes sign eg Na+ (g) to Na+ (aq)
- enthalpy of hydration for the other, going down again (so now its Cl-(aq) and Cl- (aq))
- connected by arrow going up from solid ionic compound to the aq ions, endothermic, representing the enthalpy of solution
using the cycles, why does the sign of the enthalpy of solution change
depending on how endothermic the LED is (if LEF remember must switch sign)
and how exothermic the sum of the hydration enthalpies
the enthalpy of solution could be +ve or -ve, endo or exo
if LED > sum of hydr enthalpies, then endo (+)
if LED < sum of hydr enthalpies, then exo (-)
spontaneous reaction/process
requires no energy input
- most exo reactions; a reaction that is exothermic will result in products that are more thermodynamically stable than the reactants. this causes reactions to be spontaneous (occur without any external influence).
- some exo reactions require input of heat energy eg combustion, thermal decomposition
- endo reactions can also be spontaneous.
what two thermodynamic factors driving reactions
what does this explain, in terms of endo reactions
enthalpy change of a reaction is insufficient to explain feasibility of reactions:
enthalpy - system’s internal energy
entropy - measure of a system’s disorder
this explains why some endo reactions are spontaneous, because if the reaction increases entropy, will be spontaneous (probs)
units of entropy
J K^-1 mol^-1
2nd law of thermodynamics
systems tend towards higher entropy
this means that, reactions with increase in entropy are favourable
calculation of change in entropy
sum of absolute entropy of products - sum of absolute entropy of reactants
entropy values will always be positive (except for crystals at 0K; 0). but the CHANGE IN entropy values can be negative, ie a decrease in entropy
if its +ve then disorder has increased
if its -ve then disorder has decreased
((where else do we see the sum of))
bond enthalpy calcs
∆H = sum of energy in bonds broken - sum of energy in bonds formed
bonds broken: R bond enthalpies
bonds formed: P bond enthalpies
why do gases have higher entropies
entropy and states
much more disordered
solids < liquids < gases in terms of entropy
because of this the same substance has different values of entropy depending on its state. the temperature at which there is a change in state is the temperature at which the reaction (of melting, of boiling) is feasible because change in gibbs free energy becomes 0
what happens as a solid substance is heated in terms of energy conversion
SOLID
heat energy from the source causing increase in temp causes the particles to have more kinetic energy. they are vibrating more, so have more entropy. when the vibrations are enough to break bonds/forces to become liquid,
LIQUID
it happens. there is a jump in entropy. the period of changing state, there is no increase in temp because heat energy –> overcoming bonds, not to kinetic energy. upon continued heating when forces/bonds(?) are overcome again,
GAS
conversion to gaseous state. much more highly disordered.
what are the different graphs depicting changes in state with temperature
temp on y, time on x
- at the points of melting and boiling, the graph is horizontal. because over time (as its reached the points) energy from the heat energy input is being used to overcome bonds/forces. so temp doesnt increase over time.
- then line continues going up because temp is increasing (temp as measure of vibrations of particles, so heat energy converted to kinetic energy until enough to overcome bonds/forces)
——-
entropy on y, temp on x
- more important, more a level
- at the points of melting and boiling, graph line is vertical because entropy rapidly increases at the temp. the point at x of the vertical lines is the melting/boiling point
- the vertical line corresponding to boiling point is longer than melting, because the jump in entropy (to higher entropy; more disorder) is more, because gases are extremely highly disordered
how can you qualitatively determine the change in entropy given the equation for a reaction
- look at the states, and number of particles
- look at the number of particles in each state
- do these for each side of the equation
- remember aq means dissolves, dissociated, eg NaCl (aq) isnt one particle its two
- a decrease in the number of molecules reduces entropy
- a change in state will inc/dec entropy
- producing a gas will increase entropy, and vv
if a reaction produces more overall particles but less gas moles
and entropy decreases
what does that tell you
change in state > number of molecules for determining the entropy change. the decrease in disorder due to less gas moles is more then the increase in disorder from more particles formed
when can we say that change in s is negligible and unknown
if same number of molecules, same states
then the change in entropy will be negligible, but we dk if there is a decrease or increase without the absolute entropy values
to summarise, an increase in entropy is likely when
and a decrease in entropy is likely when
LIKELY INCREASE
- change in state from s –> l –> g
- increase in no. molecules
LIKELY DECREASE
- change in state from g –> l –> s
- decrease in no. molecules
how are these two thermodynamic factors linked
what is this
in the change in gibbs free energy
gibbs is the thermodynamic potential of a system to do work (?)
the change in gibbs free energy determines feasibility (can it occur) of a reaction
∆G = ∆H-T∆S
units of ∆G = ∆H-T∆S
- ∆G in kJmol^-1
- ∆H in kJmol^-1.
- T in Kelvin, add 273 to degrees celcius
- ∆S in J K^-1 mol^-1. to convert to kJ, divide by 1000
for G and H, they might give you in J, which you don’t need to change if they don’t specify what they want answer to be in. all either needs to be kJ or J
what is the change in gibbs free energy
driving force of all chemical processes. it must be equal to or less than 0 (so 0 or -ve value) for a reaction to be feasible
so for any spontaneous reaction, ∆G = 0 or a -ve value, which is why some endo reactions are spontaneous
why is temperature included in the eq ∆G = ∆H-T∆S
what is the implication of temp being in the equation
entropy value depends on state. for a given substance, entropy value changes based on state
it shows that temp can be calculated for when a reaction could take place, we can determine the temp at which a reaction would become feasible. reactions that arent feasible at a certain temp could/will be at another
what happens when a solid changes state in terms of ∆G
changes of state are controlled by this eq, as all reactions are.
SOLID. remains solid until ∆G=0, so when T is high enough (melting point) so that its 0. the reaction then becomes feasible
LIQUID. remains liquid until ∆G=0, so when T is high enough (boiling point) so that its 0. the reaction then becomes feasible
GAS
how can you use understanding of ∆G to calculate the melting or boiling point of a substance
if given (the means to calculate) enthalpy change of reaction and entropy change
then make ∆G=0
and rearrange for T to be subject
make sure units
calculate for T
feasibility usually = spontaneity of reactions for aqa, but when are they different
feasibility depends on temp, and is based on gibbs. it is whether it can occur
spontaneity is whether it does take place or not. a reaction that is thermodynamically feasible (or thermodynamically unstable, both of those to say that change in gibbs=0 or -ve) may be KINETICALLY stable, meaning the Ea is too high and reaction does not occur. even though it could
when will a reaction always be feasible
∆G = ∆H-T∆S
when change in gibbs will always be 0 or -ve
so if ∆H is negative AND ∆S is positive
because this means T∆S will always be a +ve number, and an already -ve number for ∆H - a positive, (eg -(+)3) will only make it more -ve. so ∆G will def be negative
when will a reaction never be feasible
∆G = ∆H-T∆S
when change in gibbs will always be +ve
so if ∆H is positive AND ∆S is negative
because this means T∆S will always be -ve (- x + = -). and a positive - -value, –> positive + value = positive number always
when will a reaction be feasible if it is exothermic and has a decrease in entropy
∆G = ∆H-T∆S
only if the -ve value of ∆H is MORE OR EQUAL TO T∆S
so the -ve value for change in entropy needs to be smaller than -ve ∆H, so that when its times by T,
this product, when added (because -(- = +) to the -ve value of ∆H, keeps it negative
THEREFORE the temperature must be LOW
when will a reaction be feasible if it is endothermic and has an increase in entropy
∆G = ∆H-T∆S
only if the product of T∆S is BIGGER OR EQUAL TO the +ve ∆H
because then the positive value of the product, when taken away from the less positive (smaller) value of ∆H,
will result in the ∆G being 0 or negative because its eg 2 - +5; a more positive number being taken away (T∆S) from a smaller positive number (∆H)
THEREFORE the temp must be HIGH
