http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

Video Tutorial on Energetics, Enthalpy changes, Exothermic and Endothermic reactions.

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 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 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

Specific heat capacity Amount of heat needed to increase temp 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 = 1000gHeated 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  X.

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 X = 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 - 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 reaction = -∆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 bond0

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)

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