Enthalpy Definitions and equations

Question 1

what is the definition of enthalpy change

Answer 1

Heat energy transferred in a reaction at constant pressure

Question 2

what are the “standard conditions”

Answer 2

298K 100kpa

Question 3

how many definitions do I need to remember LOLLLL

Answer 3

12 :0 HAHAHAHAHAHHAHAH

Question 4

Enthalpy change of formation defintion

Answer 4

Is the enthalpy change where 1 mole of a compound is formed from its elements in their standard states under standard conditions

Question 5

Bond dissociation enthalpy defintion

Answer 5

The enthalpy change when all the bonds of the same type in 1 molecule of gaseous molecules are broken

Question 6

Enthalpy of atomisation of an element def

Answer 6

Enthalpy change where 1 mole of gaseous atoms is formed from an element in its standard state

Question 7

Enthalpy of atomisation of a compound def

Answer 7

Enthalpy change when 1 mole of a compound in its standard state is converted to gaseous atoms

Question 8

First ionisation energy

Answer 8

Enthalpy change where 1 mole of gaseous 1+ ions is formed from 1 mole of gaseous atoms

Question 9

second ionisation energy

Answer 9

Enthalpy change when 1 mole of gaseous 2+ ions is formed from 1 mole of gaseous 1+ ions

Question 10

First electron affinity def

Answer 10

Enthalpy change when 1 mole of gaseous 1- ions is formed from 1 mole of gaseous atoms

Question 11

second electron affinity

Answer 11

Enthalpy change when 1 mole of gaseous 2- ions is formed from 1 mole of gaseous 1- ions

Question 12

Enthalpy change of hydration def

Answer 12

Enthalpy change when 1 mole of aqueous ions is formed from 1 mole of gaseous ions

Question 13

Enthalpy change of solution def

Answer 13

This is the enthalpy change where 1 mole of solute is dissolved in enough solvent, that no further enthalpy change occurs on further dilution.

Question 14

Lattice enthalpy of formation def

Answer 14

Enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions under standard conditions

Question 15

Lattice enthalpy of dissociation def

Answer 15

Enthalpy change when 1 mole of a solid ionic compound, is completely dissociated into gaseous ions under standard conditions

Question 16

Soluble substances tend to have (exothermic/endothermic) enthalpies of solution

Answer 16

Exothermic

Question 17

entropy change equation

Answer 17

triangle s= sum of products - sum of reactants

Question 18

what actually is entropy ( definition)

Answer 18

Entropy is the number of ways that particles can be arranged and the number of ways that energy can be shared between particles

Question 19

The more particles you have the higher the….

Answer 19

entropy

more disorder= more entropy

Big boi positive value of entropy= WHOLE LOTTA DISORDER

Question 20

Why do particles try to become disordered

Answer 20

To increase entropy

- As at higher entropy particles are more energetically stable

Question 21

Gibbs equation ( free energy change)

Answer 21

delta g= delta H - T(delta s)

Question 22

what happens if free energy change ( delta G) is negative or equal to zero?

Answer 22

Then a reaction could happen by itself

• Free energy change is used to predict whether a reaction is feasible or not.

Question 23

Whats the best combination of values for the gibbs equation?

Answer 23

• negative delta H
• positive delta s
• As this will mean the reaction is feasible at any temperature.

Question 24

What happens if delta H and delta S are both positive?

Answer 24

The reaction is not feasible at SOME TEMPERATURES(eg low temps)

• But may still be feasible at higher temperatures.

Question 25

What happens if delta H is positive and delta S is negative

Answer 25

REACTION IS NOT FEASIBLE AT ANY TEMP

Question 26

What happens is delta H and delta S are both negative

Answer 26

reaction may not be feasible at high temperatures | - But are instead feasible at low temps.

Question 27

temperature and Delta H and delta S equation:

Answer 27

T= delta H/ delta S - you can only use this equation when delta G is zero

Question 28

Applying y= mx + c to gibbs

Answer 28

delta G= - delta s*t + delta H delta g=y negative delta s = gradient (m) t= x value delta H = y intercept (c)