Thermodynamic- paper 1 Flashcards
Define Hess’ law
the enthalpy change for a chemical reaction is the same, regardless of the route taken from reactants to products
the standard enthalpy of formation △Høf
give an example for the formation of NaCl
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
The standard enthalpy of combustion △Høc
give an example using C2H6
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)
the standard enthalpy of atomisation, △Høat
use Na and Cl as an example
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)
Mean bond enthalpy △HøBE
use Cl2 as an example
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)
How are the enthalpy of atomisation and bond enthalpy related?
Bond enthalpy is exactly double of atomization
First Ionisation Enthalpy 1st△Høi
give an example using Mg
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-
Second ionisation enthalpy
use Mg as an example
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-
First electron affinity, 1st △Høea
use chlorine as an example
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)
Second electron affinity, 2nd△Høea
use Cl as an example
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
Lattice formation enthalpy △HLFO
use NaCl as an example
the standard enthalpy change when one mole of solid ionic compound is formed from its gaseous ions
Na+ (g) + Cl- (g) –> NaCl (s)
Lattice dissociation enthalpy △HLDO
Use NaCl as an example
The standard enthalpy change when one mole of solid ionic compound dissociates into its gaseous ions
NaCl (s) –> Na+ (g) + Cl-(g)
standard enthalpy of hydration △HhydO
Use Na+ and Cl-
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)
standard enthalpy of solution △HsolO
use NaCl as an example
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).
state the steps to follow when constructing a Born-Haber cycle
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
why is electron affinity/ any reaction endothermic
need to overcome repulsion so energy needs to be put in, negative electrons added to an already negative ion
how do you compare atoms
C: charge of the ion
R: radius/size of the ion
A:attraction between the ions
M:more exothermic or endothermic
What does the perfect ionic model feature
Ions are point charges and are perfect spheres
Theoretical vs experimental lattice enthalpy’s
Theorectical
-Perfect Ionic model
-Ions are point charges or perfect spheres
-Ionic bonding
Experimental
-Born haber
-ions are polarisable
- and covalent bonding
what can be determined if a ionic compound has no covalent character
it is a purely ionic bond and ions are not polarisable
which model predicts stronger bonding
Born Haber- it allows for covalent character so predicts stronger bonding.
When an ionic solid is dissolved in solution, what is happening in terms of bonds?
Strong ionic bonds must be broken, water (solvent) is polar
+ ions are attracted to Od- or negative ions attracted to Hd+
why are hydration enthalpies always negative
due to the electrostatic attraction between metal ion and Od- of water
define what a polar bond is
difference in electronegativity of 2 atoms in a covalent bond means that one atom attracts electron density more stronger making this side more negative
key facts for entropy
entropy- the amount of disorder within a system. entropy is given the symbol S and is always a positive number.
what are the units for entropy
jK -1 mol-1 although will sometimes needs to be converted to kJK-1 mol-1- so divide by 1000
what temperature in Kelvin is considered to be perfect temperature
0- perfect order has particles have no energy and therefore no entropy
how to determine whether a reaction is feasible
if the value is less than 0
Mg(s) +2HCl(aq)–> MgCl2(aq) + H2(g)
is the entropy positive or negative
fewer particles on thw product side but a gas has been produced so entropy is positive as it has increased.
how to calculate delta s
sum of products-sum of reactants
equation for gibbs free energy change, △G
△G= △H-T△S
give the units for △G
△H
T
△S
△G= KJmol-1
△H=KJmol-1
T=Kelvin
△S= KJK-1Mol-1
rearange the equation to calculate temperature
T= △S/△G
describe the feasibility of a reaction that has a positive △H and △S
the equation: -△H-(T△S)
△G= always negative
feasibility= reaction is feasible at any temperature
describe the feasibility of a reaction when △H is positive and △S is negative
the equation: △H + T △S
△G= always positive
feasibility= reaction is not feasible at any temperature
describe the feasibility of the reaction when △H is negative and △S is negative
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
describe the feasibility when △H and △S are both positive
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
kinetic factors:
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.
△G and graphs
△H= y intercept
△S= the gradient
y=mx + c alongside gibbs
y=mx+c
△G= -△ST + △H
