IB Chemistry on Energetics, Enthalpy Change and Thermodynamics

1. AverageKE same
Heat (q) transfer thermal energy
from hot to cool due to temp diff
2
..
2
1
vmKE 
Averagetranslationalenergy/KE
per particle
Heat Temperature
Heat vs Temperature
SymbolQ
Unit - Joule
Form of Energy SymbolT
Unit – C/K
Not Energy
At 80C
Distribution of molecular speed, Xe, Ar, Ne, He at same temp
2
.
2
1
vmKE 
80oC
Diff gases have same average KE per particle.
Click here Heat vs Temperature Click here specific heat capacity
He – mass low ↓ - speed v high ↑ Xe – mass high ↑ - speed v low ↓
Temp ᾳ AverageKE
Highertemp - HigheraverageKE
2
..
2
1
vmKE 
Movement of particle,KE.
Heat energy
(energy in transfer)
80oC 50oC
degreeof
hotness/coldness
Total KE/PE energy
of particlesin motion
1 liter water 80C2 liter water 80C
Same Temp
Same average kinetic energy per particle
Same average speed
Same temp
Diff amt heat

2. Specificheat capacity
Amountof heat neededto increase
temp of 1g of substanceby 1C
Q = Heat transfer
Click here specific heat capacityClick here specific heat capacity
80oC 50oC
Warmer body
higher amt average KE
Energy transfer as heat
Gold
0.126J/g/K
Silver
0.90J/g/K
Water
4.18J/g/K
Cold body
lower amt average KE
Q = mcθ
Heat
Total KE/PE energy
of particlesin motion
Symbol Q
Unit - Joule
Form of Energy
Amt heat energyQ, needto changetemp depend
m = mass c = specific
heat capacity
θ = Temp diff
Lowest Highest
specific heat capacity
0.126J 4.18J
↓ ↓
to raise 1g to 1 K to raise 1g to 1K
Click here themochemistry notes
Coffee-cup calorimeter
constant pressure – no gas
Calorimetry - techniques used to measure
enthalpy changes during chemical processes.
Bomb calorimeter
Constant vol – gas released
80C
50C

3. Heat capacitybomb
Heat released
∆Hc is calculated.
Combustion-exo- temp water increase.
Specificheat capacity
Amountof heat neededto increase
temp of 1 g of substanceby 1C
Q = Heat transfer
Q = mcθ
m = mass c = specific
heat capacity
θ = Temp diff
Coffee-cup calorimeter
constant pressure – no gas
Calorimetry- techniques used to measure
enthalpy changes during chemicalprocesses.
Bomb calorimeter
Constant vol – gas released
Cup calorimeter
Determinespecific heat capacityof X
m = 1000g
Heated 200 C
5000 ml water
m = 5000g
c = 4.18
Ti = 20 C
Tf = 21.8 C
Heat lost by X = Heat gain by water
mc∆T = mc∆T
X
1000 x c (200 – 21.8) = 5000 x 4.18 x (21.8 – 20)
c = 37620/178200
c = 0.211J/g/K
Benzoic acid – used std – combustion1g release 26.38 kJ
Combustion 0.579 g benzoic acid cause a 2.08°C increasein temp.
1.732g glucose combusted, temp increase of 3.64°C. Cal ΔHcomb X.
Bomb calorimeter(combustion)
Find heat capacityof bomb and ∆Hc X
Bomb sealed, fill with O2.
1g – 26.38kJ
0.579g – 26.38 x 0.579
Q = - 15.3kJ
∆Hc X = Qbomb
Find heat capacity bomb
Q bomb = c∆T
KkJc
T
Q
c
TcQ
/34.7
08.2
3.15


 Qbomb = c∆T
= 7.34 x 3.64
= 26.7 kJ
Insulated with water.
Combustion X
Find Q using
benzoic acid
1.732g – 26.7 kJ
180g – 2.78 x 103 kJmol-1
Click here bomb calorimeter
X
X
1. 2. 3.

4. System – rxn vessel
(rxn take place)
open system closed system isolated system
Enthalpy– Heat content/Amt heat energyin substance
- Energystored as chemicalbond+ intermolecular force as potentialenergy
Exchange
energy
Exchange
matter
Exchange
energy
NO Exchange
energy
NO Exchange
matter
Heat(q) transfer from system to surrounding
↓
Exothermic.∆H < 0
↓
HOT
Surrounding – rest of universe
Heat(q) transfer to system from surrounding
↓
Endothermic.∆H > 0
↓
COLD
H
Time
H
Time
Heat
energy
Heat
energy
∆H = + ve
∆H = - ve
∆H system = O
reactionsystem
surrounding
No heat loss from system
(isolated system)
∆Hrxn = Heat absorb water (mc∆θ)
∆Hrxn = mc∆θ
water
Enthalpy Change = Heat of reaction = -∆H
2Mg(s) + O2(g) →2MgO(s) ∆H = -1200kJ mol-1
Enthalpy/H
(heat content)
2Mg + O2
2MgO
∆H= -1200kJ mol-1
- Energyneithercreatednor destroyed - Converted from oneform into another.
- Amt heat lost by system equalto amt heat gain by its surrounding.
- Total energysystem plus its surrounding isconstant,if energyis conserved.

5. change
Energy Flow to/fromSystem
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Lose energy to surrounding as heat or work
E = sum kinetic energy/motion of molecule, and potential
energy represented by chemical bond bet atom
∆E = q + w
∆E = Change internal
energy
q = heat
transfer
w = work done
by/on system
Thermodynamics
Study of work, heat and energy on a system
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Gain energy from surrounding as heat or work
Heat add , q = + 100 J
Work done by gas, w = - 20 J
∆E = + 100 – 20 = + 80 J
Q = Heat gain
+ 100J
w = work done by system = -20 J
w = work done on system = +20 J
Q = Heat lost
- 100J
Heat lost , q = - 100 J
Work done on gas, w = + 20 J
∆E = - 100 + 20 = - 80 J
∆E universe = ∆E sys + ∆E surrounding = 0
System – KE and PE energy – Internal Energy (E)
Heat and work Heat only
Q = Heat gain
+ 100 J
No work – no gas produced
Heat add , q = + 100 J
No work done = 0
∆E = q + w
∆E = + 100 = + 100 J
Q = Heat lost
- 100J
Heat lost, q = - 100 J
No work done = 0
∆E = q + w
∆E = - 100 = - 100 J
No work – no gas produced
H = E + PV ∆H = ∆E + P∆V
Enthalpy change w = work done
by/on gas
1st Law Thermodynamics
∆E = q + w ∆E = q
∆E = q + w
+ q = sys gain heat
- q = sys lose heat
+ w = work done on sys
- w = work done by sys
∆E = q + w
Work done by gas
No gas – No work

6. change
Energy Flow to/fromSystem
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Lose energy to surrounding as heat or work
E = sum kinetic energy/motion of molecule, and potential
energy represented by chemical bond bet atom
∆E = q + w
∆E = Change internal
energy
q = heat
transfer
w = work done
by/on system
Thermodynamics
Study of work, heat and energy on a system
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORK
Gain energy from surrounding as heat or work
No work done by/on system
∆E = q + w w = 0
∆E = + q (heat flow into/out system)
∆H = ∆E = q (heat gain/lost)
∆E universe = ∆E sys + ∆E surrounding = 0
System – KE and PE energy – Internal Energy (E)
Heat only – Exothermic and Endothermic reaction
Q = Heat gain
+ 100 J
No work – no gas produced
Heat add , q = + 100 J
No work done = 0
∆E = q + w
∆E = + 100 J
∆E = ∆H = + 100 J
Q = Heat lost
- 100J
Heat lost, q = - 100 J
No work done = 0
∆E = q + w
∆E = - 100 J
∆E = ∆H = - 100J
No work – no gas produced
H = E + PV ∆H = ∆E + P∆V
Constant pressure
Enthalpy change w = work done
by/on gas
1st Law Thermodynamics
P∆V = 0
∆E = q + w∆H = ∆E + P∆V
∆E = q + 0
↓
∆E = q
No gas produced
V = 0
∆H = ∆E + 0
↓
∆H = ∆E
At constant pressure/no gas produced
∆H = q
∆Enthalpy change = Heat gain or lost
No work done
w = 0
H
E
E
∆H = + 100J
H ∆H = - 100J
Enthalpy Change

7. Heat(q) transfer from system to surrounding
↓
Exothermic ∆H < 0
↓
HOT
Heat
energy ∆H = - ve
EnthalpyChange = Heat of rxn = -∆H
Mg(s) + ½ O2(g) → MgO(s) ∆H = -1200kJ mol-1
Mg + ½ O2
MgO
∆H= -1200
- Energyneithercreatednor destroyed - Converted from oneform into another.
- Amt heat lost by system equalto amt heat gain by its surrounding.
- Total energysystem plus its surrounding isconstant,if energyis conserved.
Reactant (Higher energy - Less stable/weaker bond)
Product (Lower energy - More stable/strong bond)
Temp surrounding ↑
Exothermic rxn
Combustion C + O2 → CO2
Neutralization H+ + OH- → H2O
Displacement Zn + CuSO4 → ZnSO4 + Cu
Condensation H2O(g) → H2O(l)
Freezing H2O(l) → H2O(s)
Precipitation Ag+ + CI- → AgCI(s)
Endothermic rxn
Dissolve NH4 salt NH4CI (s) → NH4
+ + CI -
Dissolve salt MgSO4. 7H2O(s) → MgSO4(aq
CuSO4. 5H2O(s) → CuSO4(aq)
Na2CO3.10H2O(s) → Na2CO3(aq
Evaporation/Boiling H2O(l) → H2O(g)
Melting H2O(s) → H2O(l)
Heat(q) transfer to system from surrounding
↓
Endothermic. ∆H > 0
↓
COLD
Heat
energy
Reactant (Lower energy - More stable/strong bond)
Product (Higher energy - Less stable/weak bond)
∆H = + ve Temp surrounding ↓
Click here thermodynamics
∆H= + 16
NH4CI (s)
NH4CI (aq)
Enthalpy Change = Heat of rxn = -∆H
NH4CI (s) → NH4CI (aq) ∆H = + 16 kJ mol-1
Click here enthalpy
3000
1800
116
100
E
X
O
E
N
D
O

8. NaCI (s)
Na(s) + ½CI2 (g))
LiCI (s)
Li+
(g) + CI–
(g)
AgCI
Ag+ + CI-
NaCI + H2O
HCI + NaOH
ZnSO4 + Cu
Zn + CuSO4
Li+
(aq)
Li+
(g) + H2O
LiCI(aq)
LiCI+ H2O
2CO2 + 3H2O
C2H5OH + 3O2
- Energyneithercreatednor destroyed - Converted from oneform into another.
- Amt heat lost by system equalto amt heat gain by its surrounding.
- Total energysystem plus its surrounding isconstant,if energyis conserved.
Std Enthalpy Changes ∆Hθ
Std condition
Pressure
100kPa
Conc 1M All substance
at std states
Temp
298K
Bond Breaking
Heat energy absorbed – break bond0
Bond Making
Heat energy released – make bond
Std ∆Hc
θ combustion
Std Enthalpy Changes ∆Hθ
∆H for complete combustion
1 mol sub in std state in excess O2
∆H when 1 mol solute
dissolved form infinitely dilute sol
Std ∆Hsol
θ solution
∆H when 1 mol
gaseous ion is hydrated
Std ∆Hhyd
θ hydration
∆H when 1 mol metal
is displaced from its sol
Std ∆Hd
θ displacement
∆H when 1 mol
H+ react OH- to form 1 mol H2O
C2H5OH+ 3O2 → 2CO2 + 3H2O LiCI(s) + H2O → LiCI(aq)
Ag+ + CI - → AgCI (s)
Zn + CuSO4 → ZnSO4 + Cu
Std ∆Hlat
θ lattice
∆H when 1 mol
precipitate form from its ion
Std ∆Hppt
θ precipitationStd ∆Hn
θ neutralization
∆H when 1 mol crystalline
sub form from its gaseous ion
HCI + NaOH→ NaCI + H2O Li+
(g) + CI–
(g) → LiCI (s)
Li+
(g) + H2O→Li+
(aq)
∆H = - ve ∆H = - ve ∆H = - ve ∆H = - ve
∆H = - ve ∆H = - ve ∆H = - ve
∆H when 1 mol form from
its element under std condition
∆H = - ve
Std ∆Hf
θ formation
Na(s) + ½CI2 (g)→ NaCI (s)