Chemical Energetics I

Question 1

△H for endothermic reaction?

State a common endothermic process.

Answer 1

△H > 0

Bond breaking

Question 2

△H for exothermic reaction?

State a common exothermic process

Answer 2

△H < 0

Bond forming

Question 3

What are the 2 standard conditions for enthalpy changes?

Answer 3

298K and 1 bar

Question 4

What are 2 things to include in a thermochemical equation?

Answer 4

△H and state symbol

Question 5

Standard Enthalpy Change of Reaction, △Hr

Answer 5

Amount of heat absorbed or evolved when molar quantities of reactants as shown in chemical equation react together under standard conditions of 298K and 1 bar.

Question 6

Standard Enthalpy Change of Formation, △Hf

How can this be used to calculate enthalpy change of reaction?

Answer 6

Amount of heat absorbed or evolved when one mole of substance is formed from its constituent elements, all at their standard states at 298K and 1 bar.

△Hf(Products) - △Hf(Reactants)

Question 7

Two things to note about Standard Enthalpy Change of Formation, △Hf

Answer 7

1. △Hf(Elements) = 0

2. Positive △Hf –> Compound less stable than elements

Question 8

Standard Enthalpy Change of Combustion, △Hc

How can this be used to calculate enthalpy change of reaction?

Answer 8

Amount of heat evolved when one mole of substance in standard state completely burned in excess oxygen under standard conditions of 298K and 1 bar.
[Always exothermic (-ve)]

△Hc(Reactants) - △Hc(Products)

Question 9

Standard Enthalpy Change of Neutralisation, △Hneut

Answer 9

Amount of heat evolved when ONE mole of water is formed from neutralization between acid and base, under standard conditions of 298K and 1 bar.
[Always exothermic (-ve)] –> B.F.

Question 10

△Hneut of strong vs weak acid?

Answer 10

△Hneut of strong acid/base –> More exothermic (completely ionised)

△Hneut of weak acid/base –> Less exothermic (energy used to dissociate completely)

Question 11

Standard Enthalpy Change of Atomisation, △Hat

State one special case.

Answer 11

Amount of heat absorbed when one one mole of free gaseous atoms formed from its element under standard condition of 298K and 1 bar.
[Always endothermic (+ve)] –> B.B

△Hat = 0 for noble gases

Question 12

How to use bond energy to calculate enthalpy change of reaction?

State condition when doing so.

Answer 12

1. Draw structural formula
2. Identify type of B.B and B.F (Data Booklet)
3. BE (bonds broken) - BE (bonds formed)

Reactants and products all in gaseous state

Question 13

Experimental method to find enthalpy change.

Answer 13

1. Q = mc△T
2. Find moles of limiting reagent (n)
3. Endothermic: +Q/n
   Exothermic: -Q/n

Question 14

Assumptions made for Q=mc△T

One thing to note during calculation

Answer 14

1g cm^-3, c = 4.18J g-1 K-1 (Data booklet)

For m, only add mass if reactants both solution

Question 15

Ionisation Energy,IE

Answer 15

First IE is energy absorbed to remove one mole of electrons from one mole of gaseous atoms to form one mole of singly charged gaseous cations

[Always endothermic (+ve)]

e.g. Na (g) –> Na+ (g) + e-

Question 16

Electron Affinity, EA

Answer 16

1st: Amount of heat absorbed or evolved to add one mole of electrons to one mole of gaseous atoms to form one mole of gaseous singly negatively charged ions
e. g. O (g) + e- –> O- (g)

Question 17

1st EA vs 2nd EA

Answer 17

1st: Exothermic –> Attraction between nucleus and e-
2nd: Endothermic –> Overcome repulsion between 2 e-

Question 18

Lattice Energy, LE

Answer 18

Amount of heat evolved when one mole of solid ionic compound is formed from its constituent free gaseous ions under standard conditions of 298K and 1 bar.
[Always exothermic (-ve)]

X+ (g) + Y- (g) –> XY (s)

Question 19

Factors affecting LE

Answer 19

LE ∝ q+ x q- /r+ + r-

Radius of ions [D.B]

• Larger inter-ionic distance
• Weaker strength of ionic bond
• LE less exothermic

Charge of ions

• Charge of ions increase
• Electrostatic attraction between ions increase
• Stronger ionic bond
• LE more exothermic

Question 20

Limitation of ionic model [IE]

Answer 20

Ionic bond with covalent character [Chemical bonding]

- Bond form strong –> More exothermic

Question 21

Standard enthalpy change of hydration, △Hhyd

Answer 21

Amount of heat evolved when one mole of free gaseous ions dissolved in large amount of water forming solution at infinite dilution at 298K and 1 bar.
[Always endothermic (+ve)] –> Ion-dipole attraction

Ion (g) –> Ion (aq)

Question 22

Factors affecting △Hhyd

Answer 22

△Hhyd ∝ q/r

Increase charge density of ions, stronger ion-diole attraction between ions an H2O

Question 23

Standard enthalpy change of solution, △Hsol

Answer 23

Amount of heat absorbed or evolved when one mole of solvent is dissolved in an infinite volume of water under standard conditions of 298K and 1 bar.

XY (s) –> X+ (aq) + Y- (aq)

Question 24

△Hsol = ?

Soluble vs insoluble

Answer 24

△Hsol = △Hhyd(ions) - LE

Soluble: △Hsol < 0
Insoluble: △Hsol > 0

Question 25

Hess’s Law

Answer 25

Sum of △H CW = Sum of △H ACW

Question 26

Energy Cycle [Hess’s Law]

Answer 26

• Balanced eqn of △H of reaction to find OR Most complciation eqn
• Balanced eqn of other △H given
• Common species –> Alternate pathway

Question 27

3 Formula [Hess’s Law]

Answer 27

△H = △Hc (R) - △Hc (P)
-> △Hc (O2) = 0

△H = △Hf (P) - △Hf (R)
->△Hf (elements) = 0

△Hsol = △Hhyd (ions) - LE

Question 28

Algebraic [Hess’s Law]

Answer 28

Reverse eqn –> Reverse sign of △H

Try to make eqn suppose to find

Question 29

Born - Haber process

Answer 29

Refer to image 1

Question 30

What to take note when drawing energy level diagram?

Answer 30

• Start from △Hf (Elements –> 0]
• Completed should have 2 arrow to each line
• I.E (lose e-) before EA (Gain loss e-)