Upload
lawrence-kok
View
238
Download
5
Tags:
Embed Size (px)
Citation preview
http://lawrencekok.blogspot.com
Prepared by Lawrence Kok
Video Tutorial on Energetics experiments and Hess’s Law.
Average KE same
Heat (q) transfer thermal energy
from hot to cool due to temp diff
2..2
1vmKE
Average translational energy/KE per particle
Heat Temperature
Heat vs Temperature
Symbol Q Unit - Joule
Form of Energy Symbol TUnit – C/K
Not Energy
At 80C
Distribution of molecular speed, Xe, Ar, Ne, He at same temp
2.2
1vmKE
80oC
Diff gas 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 Average KEHigher temp - Higher average KE
2..2
1vmKE
Movement of particle, KE.
Heat energy(energy in transfer)
80oC 50oC
degree of hotness/coldness
Total KE/PE energy of particles in motion
1 liter water 80C2 liter water 80C
Same TempSame Average kinetic energy per particleSame Average speed
Same tempDiff amt heat
More heat energy
Less heat energy
Heat energy(energy in transfer)
Specific heat capacity Amt heat needed to increasetemp of 1g of substance by 1C
Q = Heat transfer
Click here specific heat capacityClick here specific heat capacity
80oC 50oC
Warmer bodyhigher amt average KE
Energy transfer as heat
Gold0.126J/g/K
Silver0.90J/g/K
Water4.18J/g/K
Cold body lower amt average KE
Q = mcθ
Heat
Total KE/PE energy of particles in motion
Symbol Q Unit - Joule
Form of Energy
Amt heat energy Q, need to change temp 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 calorimeterconstant pressure – no gas
Calorimetry - techniques used to measure
enthalpy changes during chemical processes.
Bomb calorimeterConstant vol – gas released
80C
50C
Heat capacity bomb
Heat released∆Hc is calculated.
Combustion - exo - temp water increase.
Specific heat capacity Amount of heat needed to increase temp of 1 g of substance by 1C
Q = Heat transfer
Q = mcθm = mass c = specific
heat capacity
θ = Temp diff
Coffee-cup calorimeterconstant pressure – no gas
Calorimetry - techniques used to measure
enthalpy changes during chemical processes.
Bomb calorimeterConstant vol – gas released
Cup calorimeterDetermine specific heat capacity of X
m = 1000 gHeated 200 C
5000 ml waterm = 5000gc = 4.18Ti = 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 – combustion 1g release 26.38 kJCombustion 0.579 g benzoic acid cause a 2.08°C increase in temp.1.732g glucose combusted, temp increase of 3.64°C. Cal ΔHcomb glucose.
Bomb calorimeter (combustion)Find heat capacity of bomb and ∆Hc X
Bomb sealed, fill with O2.
1g – 26.38kJ0.579g – 26.38 x 0.579 Q = - 15.3kJ
∆Hc glucose = Qbomb Find heat capacity bomb
Q bomb = c∆T
KkJcT
Qc
TcQ
/34.708.2
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 kJ180g – 2.78 x 103 kJmol-1
Click here bomb calorimeter
X
X
1. 2. 3.
System – rxn vessel (rxn take place)
open system closed system isolated system
Enthalpy – Heat content/Amt heat energy in substance /KE + PE - Energy stored as chemical bond + intermolecular force as potential energy
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 rxn = -∆H
2Mg(s) + O2(g) →2MgO(s) ∆H = -1200kJ mol-1 Enthalpy/H(heat content)
2Mg + O2
2MgO
∆H= -1200kJ mol-1
- Energy neither created nor destroyed - Converted from one form into another.- Amt heat lost by system equal to amt heat gain by its surrounding.- Total energy system plus its surrounding is constant, if energy is conserved.
change
Energy Flow to/from System
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORKLose 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 WORKGain energy from surrounding as heat or work
Heat add , q = + 100 JWork 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 JWork 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 JNo work done = 0∆E = q + w∆E = + 100 = + 100 J
Q = Heat lost - 100J
Heat lost, q = - 100 JNo 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
change
Energy Flow to/from System
System – KE and PE energy – Internal Energy (E)
Change Internal energy, ∆E = E final – E initial
Energy transfer as HEAT or WORKLose 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 WORKGain 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 JNo 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
Heat(q) transfer from system to surrounding
↓ Exothermic ∆H < 0
↓HOT
Heat energy ∆H = - ve
Enthalpy Change = Heat of rxn = -∆H
Mg(s) + ½ O2(g) → MgO(s) ∆H = -1200kJ mol-1
Mg + ½ O2
MgO
∆H= -1200
- Energy neither created nor destroyed - Converted from one form into another.- Amt heat lost by system equal to amt heat gain by its surrounding.- Total energy system plus its surrounding is constant, if energy is conserved.
Reactant (Higher energy - Less stable/weaker bond)
Product (Lower energy - More stable/strong bond)
Temp surrounding ↑
Exothermic rxnCombustion C + O2 → CO2
Neutralization H+ + OH- → H2ODisplacement Zn + CuSO4 → ZnSO4 + CuCondensation H2O(g) → H2O(l)
Freezing H2O(l) → H2O(s)
Precipitation Ag+ + CI- → AgCI(s)
Endothermic rxnDissolve 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
EXO
ENDO
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
- Energy neither created nor destroyed - Converted from one form into another.- Amt heat lost by system equal to amt heat gain by its surrounding.- Total energy system plus its surrounding is constant, if energy is conserved.Std Enthalpy Changes ∆Hθ
Std condition
Pressure 100kPa
Conc 1M All substance at std states
Temp 298K
Bond Breaking Heat energy absorbed – break bond
Bond Making Heat energy released – make bond
Std ∆Hcθ
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 ∆Hsolθ
solution∆H when 1 mol gaseous ion is
hydrated
Std ∆Hhydθ
hydration ∆H when 1 mol
metal is displaced from its
sol
Std ∆Hdθ
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 ∆Hpptθ
precipitation Std ∆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)
∆H = - 50 – (+12) = - 62 kJ mol -1
∆H = - 50 kJ mol -1
Cold water = 50g
CuSO4 .100H2O
CuSO4 (s) + 95H2O → CuSO4 .100H2O
CuSO4 (s) + 100H2O → CuSO4 .100H2O
Mass cold water add = 50 gMass warm water add = 50 gInitial Temp flask/cold water = 23.1 CInitial Temp warm water = 41.3 CFinal Temp mixture = 31 C
CuSO4 (s) + 5H2O → CuSO4 .5H2O ∆H = ?
Slow rxn – heat lost huge – flask used.
Heat capacity flask must be determined. CuSO4 (s) + 5H2O → CuSO4 .5H2O
1. Find heat capacity flask
Ti = 23.1C
Hot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask CuSO4
Heat capacity flask, c = 63.5JK-1
Mass CuSO4 = 3.99 g (0.025mol)Mass water = 45 gT initial flask/water = 22.5 CT final = 27.5 C
Mass CuSO4 5H2O = 6.24 g (0.025mol)Mass water = 42.75 gT initial flask/water = 23 CT final = 21.8 C
2. Find ∆H CuSO4 +100H2O → CuSO4 .100H2O3. Find ∆H CuSO4 .5H2O + 95H2O → CuSO4 .100H2O
Hess’s Law
CuSO4 + H2O → CuSO4 .100H2O CuSO4 .5H2O + H2O → CuSO4 .100H2O
Water Flask CuSO4 5H2O
∆Hrxn = Heat absorb water + vacuum (mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + vacuum (mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 5 + 63.5 x 5∆Hrxn= - 1.25 kJ
∆Hrxn= 42.75 x 4.18 x 1.2 + 63.5 x 1.2∆Hrxn= + 0.299 kJ
0.025 mol = - 1.25 kJ 1 mol = - 50 kJ mol-1
0.025 mol = + 0.299 kJ 1 mol = + 12 kJ mol-1
CuSO4 (s) + 5H2O → CuSO4 .5H2O
CuSO4 .100H2O
∆H = + 12 kJ mol -1
AssumptionNo heat lost from system ∆H = 0Water has density = 1.0 gml-1
Sol diluted Vol CuSO4 = Vol H2OAll heat transfer to water + flask
Rxn slow – lose heat to surrounding Plot Temp/time – extrapolation done, Temp correction
limiting
Enthalpy change ∆H - using calorimeter
without temp correction
Lit value = - 78 kJ mol -1
CONTINUE
Time/m
0 0.5
1 1.5
2 2.5
3 3.5
4
Temp/C 22
22 22 22 22
27 28 27 26
∆H = - 60 kJ mol -1
CuSO4 .100H2O
CuSO4 (s) + 95H2O → CuSO4 .100H2O
CuSO4 (s) + 100H2O → CuSO4 .100H2O
CuSO4 (s) + 5H2O → CuSO4 .5H2O ∆H = ?
CuSO4 (s) + 5H2O → CuSO4 .5H2O
Water Flask CuSO4
Mass CuSO4 = 3.99 g (0.025 mol)Mass water = 45 gT initial mix = 22 CT final = 28 C
Mass CuSO4 5H2O = 6.24 g (0.025 mol)Mass water = 42.75 gT initial mix = 23 CT final = 21 C
Find ∆H CuSO4 + 100H2O → CuSO4 .100H2OFind ∆H CuSO4 .5H2O + 95H2O → CuSO4 .100H2O
Hess’s Law
CuSO4 + H2O → CuSO4 .100H2O CuSO4 .5H2O + H2O → CuSO4 .100H2O
Water Flask CuSO4 5H2O
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 6 + 63.5 x 6∆Hrxn= - 1.5 kJ
∆Hrxn= 42.75 x 4.18 x 2 + 63.5 x 2∆Hrxn= + 0.48kJ
0.025 mol = - 1.5 kJ 1 mol = - 60 kJ mol-1
0.025 mol = + 0.48 kJ 1 mol = + 19 kJ mol-1
CuSO4 (s) + 5H2O → CuSO4 .5H2O
CuSO4 .100H2O
∆H = + 19 kJ mol -1
∆H = - 60 – (+19) = - 69 kJ mol -1
Rxn slow – lose heat to surrounding Plot Temp/time – extrapolation done, Temp correction
limiting
Enthalpy change ∆H using calorimeter
Data collection
Temp correction – using cooling curve for last 5 m
time, x = 2initial Temp = 22 C
final Temp = 28 C
Extrapolation best curve fit y = -2.68x + 33 y = -2.68 x 2 + 33 y = 28 (Max Temp)
Excel plot
CuSO4 + H2O → CuSO4 .100H2O(Exothermic) Heat released
CuSO4 .5H2O + H2O → CuSO4 .100H2O(Endothermic) – Heat absorbed
Temp correction – using warming curve for last 5 m
Time/m
0 0.5
1 1.5
2 2.5
3 3.5 4
Temp/C 23
23 23 23 23
22 22 22.4 22.7
initial Temp = 23 C
time, x = 2
final Temp = 21 C
Excel plot
Extrapolation curve fit y = + 0.8 x + 19.4 y = + 0.8 x 2 + 19.4 y = 21 (Min Temp)
Lit value = - 78 kJ mol -1
with temp correction
∆H displacement using calorimeter
Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C
27
27 27 27 27 27 27 66 71.4 71.8
70
Error AnalysisHeat loss to surrounding
Heat capacity cup / thermometer ignored
Heat capacity sol not 4.18Mass CuSO4 ignored
Excess Zn absorb heatCu form shield Zn from reacting
Zn + CuSO4 → ZnSO4 + Cu
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost rxn = Heat absorb CuSO4 = (mc∆θ) = 25 x 4.18 x (73 – 27) = 25 x 4.18 x 46 = - 4807J
V, CuSO4 = 25mlM, = 1MT initial = 27 CT final = 73 C
Coffee cup calorimeter Data collection
Extrapolation best curve fit y = -2.68x + 81.3 y = -2.68 x 3 + 81.3 y = 73 (Temp)initial Temp = 27 C
final Temp = 73 C
time, x = 3
using Excel plot
Using data logger Min gradient
Extrapolation min fit y = -2.68x + 81.3 y = -2.68 x 3 + 81.3 y = 73 (Min Temp)
Extrapolation max fit y = -3.22x + 85.5 y = -3.22 x 3 + 85.5 y = 76 (Max Temp)
Using data logger Max gradient
Ave Temp = (76 + 73)/2 = 75 C
0.025 mol = - 4807 J 1 mol = - 192 kJ mol-1
Enthalpy displacement = 1 mol of Cu 2+ displaced
Amt Cu 2+ = MV used = 1 x 0.025 = 0.025 mol
Zn + CuSO4 → ZnSO4 + Cu ∆H = -192 kJ mol -1
% Error = (217 – 192) x 100% = 11% 217
Lit value = - 217kJ mol -1
Assump valid if rate of cooling constant If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 m
Measure temp CuSO4, every 0.5m interval then add zinc in excess
Zinc powder (excess)
Min/Max gradient
∆H = - 113 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50 g
MgSO4 .100H2O
MgSO4.7H2O + 93H2O → MgSO4 .100H2O
MgSO4 + 100H2O → MgSO4 .100H2O
Mass cold water add = 50 gMass warm water add = 50 gInitial Temp flask/cold water = 23.1 CInitial Temp warm water = 41.3 CFinal Temp flask/mixture = 31 C
MgSO4 (s) + 7H2O → MgSO4 .7H2O ∆H = ?
Slow rxn – heat lost huge – flask used.
Heat capacity flask must be determined. MgSO4(s) + 7H2O → MgSO4 .7H2O
1. Find heat capacity flask
Ti = 23.1 C
Hot water = 50 g T i = 41.3 C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Water Flask MgSO4
Heat capacity flask, c = 63.5JK-1
Mass MgSO4 = 3.01 g (0.025mol)Mass water = 45 gT initial mix = 24.1 CT final = 35.4 C
Mass MgSO4 7H2O = 6.16 g (0.025mol)Mass water = 41.8 gT initial mix = 24.8 CT final = 23.4 C
2. Find ∆H MgSO4 +100H2O → MgSO4 .100H2O3. Find ∆H MgSO4 .7H2O + 93H2O → MgSO4 .100H2O
Hess’s Law
MgSO4 + H2O → MgSO4 .100H2O MgSO4 .7H2O + H2O → MgSO4 .100H2O
Water Flask MgSO4 .7H2O
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn= 45 x 4.18 x 11.3 + 63.5 x 11.3∆Hrxn= - 2.83 kJ
∆Hrxn= 41.8 x 4.18 x 1.4 + 63.5 x 1.4∆Hrxn= + 0.3 kJ
0.025 mol = - 2.83 kJ 1 mol = - 113 kJ mol-1
0.025 mol = + 0.3 kJ 1 mol = + 12 kJ mol-1
MgSO4 (s) + 7H2O → MgSO4 .7H2O
MgSO4 .100H2O
∆H = + 12 kJ mol -1
∆H = - 113 - 12 = - 125 kJ mol -1
AssumptionNo heat lost from system ∆H = 0Water has density = 1.00g ml-1
Sol diluted Vol MgSO4 = Vol H2OAll heat transfer to water + flask
Rxn fast – heat lost to surrounding minimizedDont need to Plot Temp/time
No extrapolation
Error AnalysisHeat loss to surroundingHeat capacity sol is not
4.18Mass MgSO4 ignored
Impurity presentMgSo4 already hydrated
limiting
∆H = - 286 kJ mol -1
∆H = - 442 kJ mol -1
Enthalpy change ∆H using calorimeter
Cold water = 50g
Mass cold water add = 50gMass warm water add = 50gInitial Temp flask/cold water = 23.1CInitial Temp warm water = 41.3CFinal Temp flask/mixture = 31C
Mg(s) + ½ O2 → MgO ∆H = ?
Slow rxn – heat lost huge – flask is used.
Heat capacity flask must be determined. Mg + ½ O2 → MgO
1. Find heat capacity flask
Ti = 23.1C
Hot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb Flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask Mg
Heat capacity flask, c = 63.5JK-1
Mass Mg = 0.5 g (0.02 mol)Vol/Conc HCI = 100 g, 0.1MT initial mix = 22 CT final = 41 C
Mass MgO = 1 g (o.o248 mol)Vol/Conc HCI = 100 g, 0.1MT initial mix = 22 CT final = 28.4 C
2. Find ∆H Mg + 2HCI → MgCI2 + H2
3. Find ∆H MgO + 2HCI → MgCI2 + H2O4. Find H2 + ½ O2 → H2O
Hess’s Law
∆H Mg + 2HCI → MgCI2 + H2
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn= 100 x 4.18 x 19 + 63.5 x 19∆Hrxn= - 9.11kJ
∆Hrxn= 100 x 4.18 x 6.4 + 63.5 x 6.4∆Hrxn= -3.1 kJ
0.02 mol = - 9.11 kJ 1 mol = - 442 kJ mol-1
0.0248 mol = - 3.1 kJ 1 mol = - 125 kJ mol-1
∆H = - 125 kJ mol -1
∆H = - 442 – 286 + 125 = - 603 kJ mol -1
AssumptionNo heat lost from system ∆H = 0Water has density = 1.00g ml-1
Sol diluted Vol HCI = Vol H2OAll heat transfer to water + flask
Rxn fast – heat lost to surrounding minimizedDont need to Plot Temp/time
No extrapolation
Error AnalysisHeat loss to surrounding
Heat capacity sol is not 4.18
Mass MgO ignoredImpurity present
Effervescence cause loss Mg
+ 2HCI
MgCI2 + H2
HCI Flask MgO
+ ½ O2
MgCI2 + H2O
+ 2HCI
MgCI2 + H2O
∆H MgO + 2HCI → MgCI2 + H2O
Mg + ½ O2 → MgO
MgCI2 + H2
+ 2HCI
MgCI2 + H2O
+ ½ O2
limiting
MgCI2 + H2O
Data given
2KHCO3(s) → K2CO3 + CO2 + H2O
∆H = + 51.4 kJ mol -1
Enthalpy change ∆H for rxn using calorimeter
Cold water = 50g
Mass cold water add = 50 gMass warm water add = 50 gInitial Temp flask/cold water = 23.1 CInitial Temp warm water = 41.3 CFinal Temp flask/mixture = 31 C
2KHCO3(s) → K2CO3 + CO2 + H2O ∆H = ?
Slow rxn – heat lost huge – vacuum flask is used.
Heat capacity of flask must be determined.
1. Find heat capacity vacuum
Ti = 23.1C
Hot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O=Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
HCI Flask KHCO3
Heat capacity vacuum, c = 63.5JK-1
Mass KHCO3 = 3.5 g (0.035 mol)Vol/Conc HCI = 30 g, 2MT initial mix = 25 CT final = 20 C
Mass K2CO3 = 2.75 g (0.02 mol)Vol/Conc HCI = 30 g, 2MT initial mix = 25 CT final = 28 C
2. Find ∆H 2KHCO3 + 2HCI → 2KCI + 2CO2 + 2H2O3. Find ∆H K2CO3 + 2HCI → 2KCI + CO2 + H2O
Hess’s Law
∆Hrxn = Heat absorb water + vacuum (mc∆θ) + (c∆θ)
∆Hrxn = Heat absorb water + vacuum (mc∆θ) + (c∆θ)
∆Hrxn= 30 x 4.18 x 5 + 63.5 x 5∆Hrxn= + 0.9 kJ
∆Hrxn= 30 x 4.18 x 3 + 63.5 x 3∆Hrxn= -0.56 kJ
0.035 mol = + 0.9 kJ 1 mol = + 25.7 kJ mol-1
0.02 mol = - 0.56 kJ 1 mol = - 28 kJ mol-1
∆H = - 28 kJ mol -
1
∆H = +51.4 – (-28) = + 79 kJ mol -1
AssumptionNo heat lost from system ∆H = 0Water has density = 1.00g ml-1
Sol diluted Vol HCI = Vol H2OAll heat transfer to water + vacuum
Rxn fast – heat lost to surrounding minimizedDont need to Plot Temp/time
No extrapolation
Error AnalysisHeat loss to surrounding
Heat capacity sol is not 4.18
Mass of MgO ignoredImpurity present
Effervescence cause loss Mg
+ 2HCI
HCI Flask K2CO3
2KCI + 2CO2 + 2H2O
+ 2HCI
+ 2HCI
limiting
2KHCO3(s) → K2CO3 + CO2 + H2O
2KCI + CO2 + H2O
2KHCO3 + 2HCI → 2KCI + 2CO2 + 2H2O K2CO3 + 2HCI → 2KCI + CO2 + H2O
x 2
2KCI + 2CO2 + 2H2O 2KCI + CO2 + H2O
+ 2HCI
Vol/ml 0 5 10 15 20 25 30 35 40 45 50
Temp/C
22
24. 26 28 30 31 30 30 29 28 27
Error AnalysisHeat loss to surrounding
Heat capacity cup / thermometer ignored
Heat capacity solution is not 4.18Mass of HCI/NaOH ignored
∆H neutralization rxn
HCI + NaOH → NaCI + H2O
2.M HCI – 5 ml added and temp recorded
∆H rxn = Heat absorb water = (mc∆θ) = 75 x 4.18 x (31 – 22) = 75 x 4.18 x 9 = - 2821J
V, NaOH = 50 mlM, = 1MT initial = 22 CT final = 31 C by extrapolation.
Coffee cup calorimeter Data collection
Vol, acid = 25
Max Temp = 31∆T = ( 31 – 22) = 9 C
0.05 mol = - 2821 J 1 mol = - 56 kJ mol-1
Amt OH- = MV used = 1 x 0.05 = 0.05 mol
% Error = (57.3 – 56) x 100% = 2 % 57.3
Lit value = - 57.3 kJ mol -1
Assump valid if rate of cooling constant If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding Plot Temp/vol – extrapolation done, Temp correction
Temp correction – using heating/cooling curve
Thermometric Titration
HCI + NaOH → NaCI + H2O ∆H = - 56 kJ mol -1
initial Temp = 22 C
final Temp = 31 C
Excel plot
Heating curvey = 0.362x + 22.4
Cooling curvey = - 0.16 x + 35.7
Heating curvey = 0.362x + 22.4
Cooling curvey = - 0.16 x + 35.7
Solving for , x and y
y = + 0.362 x + 22.4 y = - 0.16 x + 35.7
+0.362 x + 22.4 = -0.16 x + 35.7 x = 25 (vol acid) y = 31 (Temp)
Strong acid with Strong alkali
∆H neutralization = 1 mol H+ react 1 mol OH- form 1 mol H2O
Strong acid vs Strong alkali
Weak acid vs strong alkali
0.05 mol = - 2541 J 1 mol = - 51 kJ mol-1
Vol/ml 0 5 10 15 20 25 30 35 40 45 50
Temp/C
21
22. 24 26 27 28 28 28 27 27 26
∆H n always same (-57.3) regardless strong or weak acidH+ + OH- → H2O ∆H = 57.3 kJmol-1
∆H n weak acid vs strong alkali is lower (less –ve)Weak acid - molecule doesn’t dissociate completelyHeat absorb to ionize/dissociate molecule – less heat released
∆H neutralization rxn
CH3COOH + NaOH → CH3COONa + H2O
1.9M CH3COOH – 5 ml added and temp recorded
∆H rxn = Heat absorb water = (mc∆θ) = 76 x 4.18 x (29 – 21) = 76 x 4.18 x 8 = - 2541J
V, NaOH = 50 mlM, = 1MT initial = 21 CT final = 29 C
Coffee cup calorimeter Data collection
Vol, acid = 26
Max Temp = 29∆T = (29– 21) = 8 C
∆H neutralization = 1 mol H+ react 1 mol OH- form 1 mol H2O
Amt OH- = MV used = 1 x 0.05 = 0.05 mol
% Error = (57.3 – 51) x 100% = 11 % 57.3
Lit value = - 57.3 kJ mol -1
Assump valid if rate of cooling constant If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding Plot Temp/vol – extrapolation done, temp correction
Temp correction – using heating/cooling curve
Thermometric Titration
initial Temp = 21 C
final Temp = 29 C
Excel plot
Heating curvey = 0.31 x + 21.1
Cooling curvey = - 0.104 x + 31.7
Heating curvey = 0.31 x + 21.1
Cooling curvey = - 0.104 x + 31.7
Solving for , x and y
y = + 0.31 x + 21.1 y = - 0.104 x + 31.7
+ 0.31 x + 21.1 = -0.104 x + 31.7 x = 26 (vol acid) y = 29 (Temp)
Weak acid with Strong alkali
CH3COOH + NaOH→CH3COONa + H2O ∆H = - 51 kJ mol-1
Strong acid vs Strong alkali
Weak acid vs strong alkali
Vol acid = 25 mlM x V (acid) = M x V (alkali)
M x 25 = 1 x 50M (acid) = 2 M
Vol/ml 0 5 10 15 20 25 30 35 40 45 50
Temp/C
22
24. 26 28 30 31 30 30 29 28 27
∆H neutralization rxn
HCI + NaOH → NaCI + H2O ∆H = ?
???? M HCI – 5 ml added and temp recorded
∆H rxn = Heat absorb water = (mc∆θ) = 75 x 4.18 x (31 – 22) = 75 x 4.18 x 9 = - 2821J
V, NaOH = 50 mlM, = 1MT initial = 22 CT final = 31 C by extrapolation.
Coffee cup calorimeter Data collection
Vol, acid = 25
Max Temp = 31∆T = ( 31 – 22) = 9 C
0.05 mol = - 2821 J 1 mol = - 56 kJ mol-1
Amt OH- = MV used = 1 x 0.05 = 0.05 mol
% Error = (57.3 – 56) x 100% = 2 % 57.3
Lit value = - 57.3 kJ mol -1
Assump valid if rate of cooling constant If not constant - extrapolation is wrong
Temp correction – using heating/cooling curve
Thermometric Titration
HCI + NaOH → NaCI + H2O
initial Temp = 22 C
final Temp = 31 C
Excel plot
Heating curvey = 0.362x + 22.4
Cooling curvey = - 0.16 x + 35.7
Heating curvey = 0.362x + 22.4
Cooling curvey = - 0.16 x + 35.7
Solving for , x and y
y = + 0.362 x + 22.4 y = - 0.16 x + 35.7
0.362 x + 22.4 = -0.16 x + 35.7 x = 25 (vol acid) y = 31 (Temp)
Strong acid with Strong alkali Thermometric titration – find conc unknown acid
1 mol HCI = 1 mol NaOH
limiting
HCI + NaOH → NaCI + H2O ∆H = - 56 kJ mol -1
Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C
28
28 28 28 28 28 28 30 30.2 29 28
25 ml, 0.5M KCI
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost rxn = Heat absorb solution = (mc∆θ) = 50 x 4.18 x (31 – 28) = 50 x 4.18 x 3 = - 627J
Coffee cup calorimeter Data collection
Extrapolation best curve fit y = -2.68x + 39 y = -2.68 x 3 + 39 y = 31 (Max Temp)
initial Temp = 28 C
final Temp = 31 C
time, x = 3
using Excel plot
Using data logger Min gradient
Extrapolation min gradient y = -2.68x + 39 y = -2.68 x 3 + 39 y = 31 (Min Temp)
Extrapolation max gradient y = -3.22x + 44 y = -3.22 x 3 + 44 y = 34 (Max Temp)
Using data logger Max gradient
Ave Max Temp = (34 + 31)/2 = 32 C
0.0125 mol = - 627 J 1 mol = - 50 kJ mol-1
Enthalpy precipitation = 1 mol of AgCI precipitated
Amt Cu 2+ = MV used = 0.5 x 0.025 = 0.0125 mol
Ag+ + CI- → AgCI ∆H = -50 kJ
mol -1
% Error = (65 – 50) x 100% = 23 % 65
Lit value = - 65 kJ mol -1
Assump valid if rate of cooling constant If not constant - extrapolation is wrong
Rxn slow – lose heat to surrounding Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 mins
∆H precipitation rxn
AgNO3 + KCI → AgCI + KNO3
V, AgNO3 = 25 mlM, = 0.5 MT initial = 28 CT final = 31 C
Measure temp AgNO3, every 0.5m interval then add KCI (stirring)
Using excel plot ∆T = (31 - 28) = 3 C
Using max/min plot ∆T = (32 - 28) = 4 C
% Error = (65 – 66) x 100% = 2 % 65
CONTINUE
Min/Max gradient
Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C
28
28 28 28 28 28 28 30 30.2 29 28
25 ml, 0.5M KCI
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost rxn = Heat absorb solution = (mc∆θ) = 50 x 4.18 x (31 – 28) = 50 x 4.18 x 3 = - 627J
Coffee cup calorimeter Data collection
Extrapolation best curve fit y = -2.68x + 39 y = -2.68 x 3 + 39 y = 31 (Max Temp)
initial Temp = 28 C
final Temp = 31 C
time, x = 3
using Excel plot
0.0125 mol = - 627 J 1 mol = - 50 kJ mol-1
Enthalpy precipitation = 1 mol of AgCI precipitated
Amt Cu 2+ = MV used = 0.5 x 0.025 = 0.0125 mol
Rxn slow – lose heat to surrounding Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 mins
∆H precipitation rxn
AgNO3 + KCI → AgCI + KNO3
V, AgNO3 = 25 mlM, = 0.5 MT initial = 28 CT final = 31 C
Measure temp AgNO3, every 0.5m interval then add KCI (stirring)
Silver Halides
Enthalpy Precipitati
on
AgCI -65 kJ mol-1
AgBr - 85 kJ mol-
1
AgI - 111 kJ mol-1
Enthalpy precipitation AgCI lowest↓
Size CI- – small– strong lattice enthalpy– hydrated by H2O↓
Lot energy need to break to release CI-
↓Free CI- will precipitate with Ag+ form AgCI
Anion
Enthalpy Hydration
CI- -359 kJ mol-1
Br- - 328 kJ mol-1
I- - 287 kJ mol-1
Smaller size--↑ -ve ∆H hydration More heat released (-ve)
CI- Br- I-
CI-
Br-
I-
Ag+ + CI- → AgCI ∆H = -50 kJ
mol -1
Bigger size--↑ - Weaker attraction bet I- with H2O↓
Easier to release I- to form ppt
∆H hydration
C2H5OH + 3O2 → 2CO2 + 3H2O
Cold water = 50g
Mass cold water add = 50 gMass warm water add = 50 gInitial Temp flask/cold water = 23.1CInitial Temp warm water = 41.3CFinal Temp flask/mixture = 31C
1. Find heat capacity calorimeter
Ti = 23.1CHot water = 50g T i = 41.3C
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost hot H2O = Heat absorb cold H2O + Heat absorb flask (mc∆θ) (mc∆θ) (c ∆θ)
50 x 4.18 x (41.3 – 31) = 50 x 4.18 x (31-23.1) + c x (31-23.1)
Heat capacity flask, c = 63.5JK-1
∆Hrxn = Heat absorb water + flask (mc∆θ) + (c∆θ)
∆Hrxn= 250 x 4.18 x (58 – 30) + 63.5 x (58 – 30)∆Hrxn= - 31 kJ
Lit value = - 1368 kJ mol -1
∆Hc combustion using calorimeter
Mass water = 250 gInitial Temp water = 30 C
Final Temp water/flask = 58 C
2. Find ∆Hc combustion data
Mass initial spirit lamp/alcohol = 218.0 gMass final spirit lamp/alcohol = 216.6 gMass alcohol combusted = 1.4 g
0.03 mol = - 31 kJ 1 mol = - 1033 kJ mol-1
Error AnalysisHeat loss to surrounding
Heat capacity sol not 4.18 JK -1
Incomplete combustion (black soot)
Evaporation alcoholDistance too farWater not stir
(Heat distribution uneven)Use bomb calorimeter
% Error = (1368 – 1033) x 100% = 24 % 1368
Mol alcohol = 1.4/46 = 0.03 mol
Hydrocarbon
∆Hc
kJ mol-1
CH3OH - 726
C2H5OH - 1368
C3H7OH - 2021
C4H9OH - 2676
C5H11OH - 3331
Longer hydrocarbon chain↓
∆H combustion more -veC- OH
C-C- OH
C-C-C- OH
C-C-C-C- OH
C-C-C-C-C- OH
Hydrocarbon
∆Hc
kJ mol-1
CH3OH - 726
C2H5OH - 1368
C3H7OH - 2021
C4H9OH - 2676
C5H11OH -3331
H H H H ᴵ ᴵ ᴵ ᴵH – C – C – C – C - OH ᴵ ᴵ ᴵ ᴵ H H H H
H H OH H ᴵ ᴵ ᴵ ᴵH – C– C – C – C -H ᴵ ᴵ ᴵ ᴵ H H H H
OH ᴵ CH3 – C – CH3
ᴵ CH3
Will structural isomer have same ∆Hc
∆H = -1676kJ mol-1
Research Question ?
О
ОО
Time/m 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Temp/C
25
25 25 25 25 25 25 29 29.4 29 28
0.848 g LiCI
No heat loss from system (isolated system)
↓ ∆H system = O
Heat lost rxn = Heat absorb solution = (mc∆θ) = 36 x 4.18 x (29.8 – 25) = 36 x 4.18 x 4.8 = - 0.72 kJ
Coffee cup calorimeter Data collection
Extrapolation best curve fit y = -2.68x + 38 y = -2.68 x 3 + 38 y = 29.8 (Temp)
initial Temp = 25 C
final Temp = 29.8 C
time, x = 3
using Excel plot
0.02 mol = - 0.72 kJ 1 mol = - 36 kJ mol-1
Amt LiCI = 0.848/42.4 used = 0.02 mol
LiCI(s) + H2O → LiCI(aq) ∆H = -36 kJ mol -1
Rxn slow – lose heat to surrounding Plot Temp/time – extrapolation done, Temp correction
Temp correction – using cooling curve for last 5 mins
∆H solution rxn
LiCI (s) + H2O → LiCI
(aq)
Mass water = 36 mlT initial = 25 CT final = 29.8 C
Lithium salts
Enthalpy solution
LiF +4.7 kJ mol-1
LiCI -37 kJ mol-1
LiBr - 48 kJ mol-
1
LiI - 63 kJ mol-
1
∆H solution
CI-
Br-
I-
limiting
F-
Lithium salt
Lattice enthalpykJ mol-1
LiF + 1050
LiCI + 864
LiBr + 820
LiI + 764
∆H(sol) = ∆HLE + ∆Hhyd(cation) + ∆Hhyd(anion)
∆Hlattice ∆Hhydration
Size anion ↑
depends
Cation
Enthalpy HydkJ mol-1
Li+ - 538
Anion
Enthalpy HydkJ mol-1
F- - 504
CI- -359
Br- - 328
I- - 287 CONTINUE
0.848 g LiCINo heat loss from system (isolated system)
↓ ∆H system = O
Heat lost rxn = Heat absorb solution = (mc∆θ) = 36 x 4.18 x (29.8 – 25) = 36 x 4.18 x 4.8 = - 0.72 kJ
0.02 mol = - 0.72 kJ 1 mol = - 36 kJ mol-1
Amt LiCI = 0.848/42.4 used = 0.02 mol
LiCI(s) + H2O → LiCI(aq) ∆H = -36 kJ mol -1
∆H solution rxn
LiCI (s) + H2O → LiCI
(aq)
Mass water = 36 mlT initial = 25 CT final = 29.8 C
Lithium salt
Enthalpy solutionkJ mol-1
LiF + 4.7
LiCI - 37
LiBr - 48
LiI - 63
Smaller size -↑ -ve ∆H hyd More heat released (-ve)
CI- Br- I-
CI-
Br-
I-
limiting
F-
Std ∆H sol = 1 mol solute dissolved form infinitely dilute sol
∆Hsolution
∆Hlattice ∆Hhydration
∆H(sol) = ∆HLE + ∆Hhyd(cation) + ∆Hhyd(anion)
depends
Lithium salt
Lattice enthalpykJ mol-1
LiF + 1050
LiCI + 864
LiBr + 820
LiI + 764
Anion
Enthalpy HydkJ mol-1
F- - 504
CI- -359
Br- - 328
I- - 287
Size anion ↑↓
∆H lattice decrease ↓ (less +ve)↓
∆H sol = ∆H latt + ∆H hyd
∆H sol (–ve) = ∆H hyd > ∆H latt↓
∆H = -ve (more soluble)
∆H latt > ∆H hyd = +ve ∆H (Insoluble)∆H hyd > ∆H latt = -ve ∆H (Soluble)
Soluble Insoluble
- ve energy
Coffee cup calorimeter
Cation
Enthalpy HydkJ mol-1
Li+ - 538
∆H hydration ∆H lattice
+ ve absorbto break bonds
- ve releasedto make bonds
Lattice (+)
Hydration (-)
Hydration > lattice (-ve ∆H)
Lattice > Hydration (+ve ∆H)
Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com