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 T Unit – 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 KE Higher 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 80C 2 liter water 80C

Same Temp Same Average kinetic energy per particle Same Average speed

Same temp

Diff amt heat

More heat energy

Less heat energy

Heat energy (energy in transfer)

Specific heat capacity Amt heat needed to increase temp of 1g of substance by 1C

Q = Heat transfer

Click here specific heat capacity Click 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 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 calorimeter

constant pressure – no gas

Calorimetry - techniques used to measure enthalpy changes during chemical processes.

Bomb calorimeter

Constant 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 = 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 chemical processes.

Bomb calorimeter

Constant vol – gas released

Cup calorimeter Determine specific heat capacity of X

m = 1000 g

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 – combustion 1g release 26.38 kJ Combustion 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 – 26.38kJ 0.579g – 26.38 x 0.579 Q = - 15.3kJ

∆Hc glucose = Qbomb Find heat capacity bomb

Q bomb = c∆T

KkJc

T

Qc

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.

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

reaction system

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 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

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 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

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 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

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 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1 C Initial Temp warm water = 41.3 C Final 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 g T initial flask/water = 22.5 C T final = 27.5 C

Mass CuSO4 5H2O = 6.24 g (0.025mol) Mass water = 42.75 g T initial flask/water = 23 C T final = 21.8 C

2. Find ∆H CuSO4 +100H2O → CuSO4 .100H2O 3. 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

Assumption No heat lost from system ∆H = 0 Water has density = 1.0 gml-1

Sol diluted Vol CuSO4 = Vol H2O All 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 g T initial mix = 22 C T final = 28 C

Mass CuSO4 5H2O = 6.24 g (0.025 mol) Mass water = 42.75 g T initial mix = 23 C T final = 21 C

Find ∆H CuSO4 + 100H2O → CuSO4 .100H2O 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 + 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 = 2 initial 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 Analysis Heat loss to surrounding

Heat capacity cup / thermometer ignored Heat capacity sol not 4.18

Mass CuSO4 ignored Excess Zn absorb heat

Cu 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 = 25ml M, = 1M T initial = 27 C T 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 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1 C Initial Temp warm water = 41.3 C Final 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 g T initial mix = 24.1 C T final = 35.4 C

Mass MgSO4 7H2O = 6.16 g (0.025mol) Mass water = 41.8 g T initial mix = 24.8 C T final = 23.4 C

2. Find ∆H MgSO4 +100H2O → MgSO4 .100H2O 3. 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

Assumption No heat lost from system ∆H = 0 Water has density = 1.00g ml-1

Sol diluted Vol MgSO4 = Vol H2O All heat transfer to water + flask

Rxn fast – heat lost to surrounding minimized Dont need to Plot Temp/time

No extrapolation

Error Analysis Heat loss to surrounding

Heat capacity sol is not 4.18 Mass MgSO4 ignored

Impurity present MgSo4 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 = 50g Mass warm water add = 50g Initial Temp flask/cold water = 23.1C Initial Temp warm water = 41.3C Final 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, 0.1M T initial mix = 22 C T final = 41 C

Mass MgO = 1 g (o.o248 mol) Vol/Conc HCI = 100 g, 0.1M T initial mix = 22 C T final = 28.4 C

2. Find ∆H Mg + 2HCI → MgCI2 + H2 3. Find ∆H MgO + 2HCI → MgCI2 + H2O 4. 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

Assumption No heat lost from system ∆H = 0 Water has density = 1.00g ml-1

Sol diluted Vol HCI = Vol H2O All heat transfer to water + flask

Rxn fast – heat lost to surrounding minimized Dont need to Plot Temp/time

No extrapolation

Error Analysis Heat loss to surrounding

Heat capacity sol is not 4.18 Mass MgO ignored

Impurity 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 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1 C Initial Temp warm water = 41.3 C Final 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, 2M T initial mix = 25 C T final = 20 C

Mass K2CO3 = 2.75 g (0.02 mol) Vol/Conc HCI = 30 g, 2M T initial mix = 25 C T final = 28 C

2. Find ∆H 2KHCO3 + 2HCI → 2KCI + 2CO2 + 2H2O 3. 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

Assumption No heat lost from system ∆H = 0 Water has density = 1.00g ml-1

Sol diluted Vol HCI = Vol H2O All heat transfer to water + vacuum

Rxn fast – heat lost to surrounding minimized Dont need to Plot Temp/time

No extrapolation

Error Analysis Heat loss to surrounding

Heat capacity sol is not 4.18 Mass of MgO ignored

Impurity 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 Analysis Heat loss to surrounding

Heat capacity cup / thermometer ignored Heat capacity solution is not 4.18

Mass 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 ml M, = 1M T initial = 22 C T 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 curve y = 0.362x + 22.4

Cooling curve y = - 0.16 x + 35.7

Heating curve y = 0.362x + 22.4

Cooling curve y = - 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 acid H+ + OH- → H2O ∆H = 57.3 kJmol-1

∆H n weak acid vs strong alkali is lower (less –ve) Weak acid - molecule doesn’t dissociate completely Heat 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 ml M, = 1M T initial = 21 C T 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 curve y = 0.31 x + 21.1

Cooling curve y = - 0.104 x + 31.7

Heating curve y = 0.31 x + 21.1

Cooling curve y = - 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 ml M x V (acid) = M x V (alkali)

M x 25 = 1 x 50 M (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 ml M, = 1M T initial = 22 C T 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 curve y = 0.362x + 22.4

Cooling curve y = - 0.16 x + 35.7

Heating curve y = 0.362x + 22.4

Cooling curve y = - 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 ml M, = 0.5 M T initial = 28 C T 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 ml M, = 0.5 M T initial = 28 C T final = 31 C

Measure temp AgNO3, every 0.5m interval then add KCI (stirring)

Silver Halides

Enthalpy Precipitation

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 g Mass warm water add = 50 g Initial Temp flask/cold water = 23.1C Initial Temp warm water = 41.3C Final Temp flask/mixture = 31C

1. Find heat capacity calorimeter

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)

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 g Initial Temp water = 30 C

Final Temp water/flask = 58 C

2. Find ∆Hc combustion data

Mass initial spirit lamp/alcohol = 218.0 g Mass final spirit lamp/alcohol = 216.6 g Mass alcohol combusted = 1.4 g

0.03 mol = - 31 kJ 1 mol = - 1033 kJ mol-1

Error Analysis Heat loss to surrounding

Heat capacity sol not 4.18 JK -1

Incomplete combustion (black soot) Evaporation alcohol

Distance too far Water 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 -ve

C- 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 ᴵ

H – C – C – C – C - OH

ᴵ ᴵ ᴵ ᴵ

H H H H

H H OH H

ᴵ ᴵ ᴵ ᴵ

H – 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

О

О

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 ml T initial = 25 C T 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 enthalpy kJ mol-1

LiF + 1050

LiCI + 864

LiBr + 820

LiI + 764

∆H(sol) = ∆HLE + ∆Hhyd(cation) + ∆Hhyd(anion)

∆Hlattice ∆Hhydration

Size anion ↑

depends

Cation Enthalpy Hyd

kJ mol-1

Li+ - 538

Anion Enthalpy Hyd

kJ mol-1

F- - 504

CI- -359

Br- - 328

I- - 287 CONTINUE

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

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 ml T initial = 25 C T final = 29.8 C

Lithium salt Enthalpy solution kJ 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 enthalpy

kJ mol-1

LiF + 1050

LiCI + 864

LiBr + 820

LiI + 764

Anion Enthalpy Hyd

kJ 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 Hyd

kJ mol-1

Li+ - 538

∆H hydration ∆H lattice

+ ve absorb to break bonds

- ve released to make bonds

Lattice (+)

Hydration (-)

Hydration > lattice

(-ve ∆H)

Lattice > Hydration

(+ve ∆H)