Physical Unit 1.8: Thermodynamics Flashcards
define enthalpy of formation
the enthalpy change when 1 mole of a substance is formed from its constituent elements will all reactants & products in their standard states under standard conditions
e.e. 2Na(s) + 1/2O2(g) –> Na2O(s)
exothermic (-ve)
define standard enthalpy of combustion
enthalpy change when 1 mole of a substance is completely burned in oxygen with all reactants & products in standard states under standard conditions
e.g. H2 + 1/2O2(g) –> H2O(g)
exothermic (-ve)
define standard enthalpy of neutralisation
enthalpy change when 1 mole of water is formed in a reaction b/w an acid & an alkali under standard conditions
e.g.1/2H2SO4(aq) + NaOH(aq) –> 1/2NaSO4(aq) + H2O(aq)
exothermic (-ve)
define enthalpy of ionisation
first ionisation energy: enthalpy change when each atom in 1 mole of gaseous atoms loses 1 e- to form 1 mole of gaseous 1+ ions
e.g. Mg(g) –> Mg+(g) + e-
endothermic (+ve)
2nd: enthalpy change when each ion in 1 mole of gaseous 1+ ions loses 1 e- to form 1 mole of gaseous 2+ ions
e.g. Mg+(g) –> Mg2+(g) + e-
endothermic (+ve)
define electron affinity
1st electron affinity: enthalpy change when each atom in one mole of gaseous atoms gains one e- to form one mole of gaseous 1- ions
e.g. O(g) + e- –> O-(g)
exothermic (-ve)
2nd electron affinity: enthalpy change when each ion in one mole of gaseous 1- ions gains one e- to form one mole of gaseous 2- ions
e.g. O-(g) + e- –> O2-(g)
endothermic (+ve) bc adding -ve e- to -ve ion, which repel
define enthalpy of atomisation
enthalpy change when one mole of gaseous atoms is produced from an element in its standard state
e.g. 1/2I2(g) –> I(g)
endothermic (+ve)
define hydration enthalpy
enthalpy change when one mole of gaseous ions become hydrated/dissolved in water
e.g. Mg2+(g) + aq –> Mg2+(aq)
exothermic (-ve)
define enthalpy of solution
enthalpy change when one mole of an ionic solid dissolves in an amount of water large enough that the dissolved ions are well separated & do not interact with each other
e.g. MgCl2(s) + aq –> Mg2+(aq) + 2Cl-(aq)
varies endo or exo
more -ve ΔH =
increased enthalpy of solution =
more likely to dissolve
decreased solubility
define bond dissociation enthalpy
enthalpy change when one mole of covalent bonds is broken in the gaseous state
e.g. I2(g) –> 2I(g)
endothermic (+ve)
roughly in the hundreds
define lattice enthalpy of formation
enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gas phase
e.g. Mg2+(g) + 2Cl-(g) –> MgCL2(s)
exothermic (-ve)
define 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
e.g. MgCl2(s) –> Mg2+(g) + 2Cl-(g)
endothermic (+ve)
define enthalpy of vaporisation
enthalpy change when one mole of a liquid is turned into a gas
e.g. H2O(l) –> H2O(g)
endothermic (+ve)
define enthalpy of fusion
enthalpy change when one mole of a solid is turned into a liquid
e.g. Mg(s) –> Mg(l)
endothermic (+ve)
what is the enthalpy of formation of an element & why?
0
by definition
Hess’s Law
the enthalpy change for a reaction is independent of the route taken
calculations involving enthalpy of formation data
arrows from elements to compounds
calculations involving enthalpy of combustion data
arrows from compounds to oxides
define mean bond enthalpy
enthalpy change when one mole of covalent bonds is broken in the gas phase often averaged over a range of different compounds
why are enthalpies of reaction that have been calculated using mean bond enthalpy data not as accurate?
because the values used are averaged over a range of different compounds not specific values for the given compound
calculations involving bond enthalpy data
arrows from compounds to gas atoms
sometimes need to also use enthalpy of vaporisation to convert liquid to gas
what is one way to reduce heat loss in calorimetry?
measure the heat capacity of the calorimeter as a whole
calorimetry calculations
calculations involving enthalpy of solution
see booklet for cycle
do hydration enthalpy of each compound in separate steps
the greater the magnitude of lattice enthalpy,
the stronger the bonding
i.e. the more +ve the lattice enthalpy of dissociation or the more -ve the lattice enthalpy of formation
what factors cause lattice enthalpy to be greater?
smaller ions
higher charge on ions
–> so stronger ionic attractions
Born-Haber cycles
see booklet
every line must balance for atoms & charges (i.e. no overall charge)
first line is elements in standard states
draw a separate step for each enthalpy change
2nd & 3rd electron affinities are endothermic so go up (not down)
write numerical values on each step
compare experimental lattice enthalpy (LE) to theoretical LE of a perfectly ionic model
experimental LE is calculated using a Born-Haber cycle with each ΔH value found by accurate measurement in experiments
theoretical LE is calculated by a theoretical mathematical calculation that considers the size, charge & arrangement of ions in the lattice
it is assumed that the structure is perfectly ionic
which is the real LE value?
the experimental LE value
describe covalent character
distortion of ions in ionic compounds (polarised) so ions are not perfectly spherical
= covalent character
+ve ions that are small &/or highly charged are very good at distorting
-ve ions that are large &/or highly charged are easier to distort
hoe does covalent character affect the properties of ionic compounds?
covalent character =
lower solubility in water
lower mp bc covalent character disrupts the ionic lattice
lower electrical conductivity
the bigger the % difference b/w the experimental & theoretical values of LE,
the greater the covalent character of an ionic compound
greater the distortion of ions, the greater the covalent character
solubility is not just about covalent character
if a compound has a large LE of formation (>1000), it is likely to be insoluble bc enthalpy of solution will be v large & +ve
diagram of covalent character
see booklet
define entropy
disorder
the more disordered something is, the greater the entropy
what are the units of entropy
Jmol-1K-1
describe the relative entropy of each state
gases have the most entropy as particles move rapidly & randomly
solids have the least entropy as particles vibrate about fixed positions
what is the tendency for entropy?
there is a tendency for entropy to increase = for things to become more disordered
2nd law of thermodynamics is that over time, entropy will naturally increase
how does the entropy of a substance vary with temp.?
3rd law of thermodynamics: entropy of a substance is 0 at absolute 0 (bc particles do not move so are in perfect order) & increases with temp.
the higher the temp., the faster the particles vibrate/move so the greater the entropy
there are big increases in entropy on state changes (melting & boiling)
entropy increase from liquid to gas is greater than increase from solid to liquid bc of the large amount of disorder in gases compared to l & s
graph of temp. (x) vs entropy (y)
see booklet
the more ordered the structure,
the lower the entropy
structures like diamond are v highly ordered so have v low entropy
smaller = more ordered
calculating entropy changes
see booklet
cycle
arrows from 0 Kelvin to compounds that are +ve
in reaction with an increase in entropy,
ΔS is +ve
with a decrease in entropy, ΔS is -ve
reactions with an increase in entropy are favourable
explanation for why entropy increases/decreases/does not change
state # moles of each state
increases: e.g. 1 mole of solid –> 3 moles of gas
so ΔS will be +ve & large bc large increase in disorder
decreases: e.g. 3 moles of gas –> 1 mole of solid
so ΔS will be -ve bc increase in order
no change: e.g. ΔS will be close to 0 bc no significant change in disorder
3 moles of gas –> 3 moles of gas
define Gibbs free energy change (ΔG)
combines enthalpy change (ΔH) & entropy change (ΔS)
what is the formula for ΔG & units?
ΔG = ΔH - TΔS
ΔH is in kJmol-1
T is in K
ΔS is in Jmol-1K-1
decrease in enthalpy (a -ve ΔH) is more favourable
increase in entropy (a +ve ΔS) is more favourable
when is a reaction feasible?
when ΔG ≤ 0
define feasible
a reaction can take place
depends on temp. - can be feasible at certain temps but not at others
the point at which a reaction switches from being feasible to not feasible is when ΔG = 0
why might a feasible reaction not take place?
bc it might have a v high activation energy - need to look at kinetics
spontaneous = feasible
watch units: must divide tΔS by 1000 to get kJmol-1
ΔG calculations
see booklet
describe changes of state in terms of ΔG
below the mp of a substance, melting is not feasible bc ΔG is +ve
at the mp, ΔG = 0 so melting becomes feasible (also freezing is feasible at the same T)
below the bp of a substance, boiling is not feasible bc ΔG is +ve
at the bp, ΔG = 0 so boiling becomes feasible (also condensing is feasible at the same T)
what is ΔG at the mp/bp?
0
calculation involving ΔG at a change of state
see booklet
what are the units of ΔG?
kJmol-1
