Thermodynamic- paper 1

Question 1

Define Hess’ law

Answer 1

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

Question 2

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

Answer 2

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

Question 3

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

Answer 3

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)

Question 4

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

Answer 4

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)

Question 5

Mean bond enthalpy △HøBE
use Cl2 as an example

Answer 5

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)

Question 6

How are the enthalpy of atomisation and bond enthalpy related?

Answer 6

Bond enthalpy is exactly double of atomization

Question 7

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

Answer 7

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-

Question 8

Second ionisation enthalpy
use Mg as an example

Answer 8

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-

Question 9

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

Answer 9

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)

Question 10

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

Answer 10

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

Question 11

Lattice formation enthalpy △HLFO
use NaCl as an example

Answer 11

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

Na+ (g) + Cl- (g) –> NaCl (s)

Question 12

Lattice dissociation enthalpy △HLDO
Use NaCl as an example

Answer 12

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

Question 13

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

Answer 13

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)

Question 14

standard enthalpy of solution △HsolO
use NaCl as an example

Answer 14

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).

Question 15

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

Answer 15

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

Question 16

why is electron affinity/ any reaction endothermic

Answer 16

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

Question 17

how do you compare atoms

Answer 17

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

Question 18

What does the perfect ionic model feature

Answer 18

Ions are point charges and are perfect spheres

Question 19

Theoretical vs experimental lattice enthalpy’s

Answer 19

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

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

Question 20

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

Answer 20

it is a purely ionic bond and ions are not polarisable

Question 21

which model predicts stronger bonding

Answer 21

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

Question 22

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

Answer 22

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

Question 23

why are hydration enthalpies always negative

Answer 23

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

Question 24

define what a polar bond is

Answer 24

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

Question 25

key facts for entropy

Answer 25

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

Question 26

what are the units for entropy

Answer 26

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

Question 27

what temperature in Kelvin is considered to be perfect temperature

Answer 27

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

Question 28

how to determine whether a reaction is feasible

Answer 28

if the value is less than 0

Question 29

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

Answer 29

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

Question 30

how to calculate delta s

Answer 30

sum of products-sum of reactants

Question 31

equation for gibbs free energy change, △G

Answer 31

△G= △H-T△S

Question 32

give the units for △G △H T △S

Answer 32

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

Question 33

rearange the equation to calculate temperature

Answer 33

T= △S/△G

Question 34

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

Answer 34

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

Question 35

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

Answer 35

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

Question 36

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

Answer 36

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

Question 37

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

Answer 37

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

Question 38

kinetic factors:

Answer 38

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.

Question 39

△G and graphs

Answer 39

△H= y intercept △S= the gradient

Question 40

y=mx + c alongside gibbs

Answer 40

y=mx+c △G= -△ST + △H