IB Chemistry on Energetics, Enthalpy Change and Thermodynamics

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 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 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

Specific heat capacity Amount of 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 = 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 – 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 X.

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 X = 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 - 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 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 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 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)