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6. Energetics. 6.1What is Energetics? 6.2Enthalpy Changes Related to Breaking and Forming of Bonds 6.3Standard Enthalpy Changes 6.4Experimental Determination of Enthalpy Changes by Calorimetry 6.5Hess’s Law 6.6Calculations involving Standard Enthalpy Changes of Reactions.

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Energetics

6

Energetics

6.1What is Energetics?

6.2Enthalpy Changes Related to Breaking and Forming of Bonds

6.3Standard Enthalpy Changes

6.4Experimental Determination of Enthalpy Changes by Calorimetry

6.5Hess’s Law

6.6Calculations involving Standard Enthalpy Changes of Reactions


What is energetics

6.1 What is energetics? (SB p.136)

What is energetics?

Energetics is the study of energy changes associated with chemical reactions.

Thermochemistry is the study of heat changes associated with chemical reactions.


Energetics

Internal Energy (U)

U = kinetic energy + potential energy


Energetics

Kinetic Energy

Potential Energy

heat

 T (K)

translational rotational vibrational

Energy

Relative position among particles

Bond breaking  P.E. 

Bond forming  P.E. 


Energetics

H – H(g)  H(g) + H(g)

P.E. 

H(g) + H(g)  H – H(g)

P.E. 

Bond breaking : -

Bond forming : -

Ionization : -

Na(g)  Na+(g) + eP.E. 


Energetics

Internal energy

bond breaking

bond forming

2H2(g) + O2(g)

U1

U = U2 – U1 = -(y-x) kJ

2H2O(l)

U2

Reaction coordinate

Q.1

4H(g) + 2O(g)


Internal energy and enthalpy

enthalpy

Internal energy

6.1 What is energetics? (SB p.137)

Internal energy and enthalpy

H = U + PV


Internal energy and enthalpy1

6.1 What is energetics? (SB p.137)

Internal energy and enthalpy

e.g. Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

qv = U = -473 kJ mol1

qp = H = -470 kJ mol1

Mg

Mg


Energetics

qv = U = -473 kJ mol1

qp = H = -470 kJ mol1

Work done against the surroundings

qv = qp – 3

= qp – w

= qp - PV

PV  (Nm2)(m3)

= Nm

Force  displacement


Internal energy and enthalpy2

6.1 What is energetics? (SB p.138)

Internal energy and enthalpy

H = U + PV

= U + PV at constant P

qp = qv + PV

Heat change at fixed P

Heat change at fixed V

Work done


Internal energy and enthalpy3

6.1 What is energetics? (SB p.138)

Internal energy and enthalpy

qp = qv + PV

On expansion, PV > 0

Work done by the system against the surroundings

System gives out less energy to the surroundings

qp is less negative than qv (less exothermic)


Internal energy and enthalpy4

6.1 What is energetics? (SB p.138)

Internal energy and enthalpy

qp = qv + PV

On contraction, PV < 0

Work done by the surroundings against the system

System gives out more energy to the surroundings

qp is more negative than qv (more exothermic)


Energetics

  • H is more easily measured than H as

  • most reactions happen in open vessels.

  • i.e. at constant pressure.

  • The absolute values of H and H cannot

  • be measured.


Exothermic and endothermic reactions

6.1 What is energetics? (SB p.138)

Exothermic and endothermic reactions

An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)


Exothermic and endothermic reactions1

Check Point 6-1

6.1 What is energetics? (SB p.139)

Exothermic and endothermic reactions

An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)


Law of conservation of energy

6.1 What is energetics? (SB p.136)

Law of conservation of energy

The law of conservation of energy states that energy can neither be created nor destroyed,

but can be exchanged between a system and its surroundings


Energetics

Exothermic : -

P.E. of the system  K.E. of the surroundings

Endothermic : -

K.E. of the surroundings  P.E. of the system


Enthalpy changes related to breaking and forming of bonds

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

Enthalpy changes related to breaking and forming of bonds

CH4 + 2O2 CO2 + 2H2O


Energetics

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

In an exothermic reaction,

E absorbed to break bonds < E released as bonds are formed.


Enthalpy changes related to breaking and forming of bonds1

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

Enthalpy changes related to breaking and forming of bonds

N2(g) + 2O2(g)  2NO2(g)


Energetics

Check Point 6-2

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)

In an endothermic reaction,

E absorbed to break bonds > E released as bonds are formed.


Energetics

For non-gaseous reactions,

PV  0

H = U + PV  U

For gaseous reactions,

H = U + PV = U + (n)RT (PV = nRT)


Energetics

Q.2

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

Given : R = 8.314 J K1 mol1, T = 298 K

U = H – (n)RT

= 885 kJ mol1


Energetics

Enthalpy

bond breaking

bond forming

CH4(g) + 2O2(g)

H1

H = 890 kJ mol1

CO2(g) + 2H2O(l)

H2

Reaction coordinate

C(g) + 4H(g) + 4O(g)


Energetics

373K

373K

H / kJ mol1

At 298K

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) 890

At 373K

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 802

2H2O(l)  2H2O(g) +88 kJ


Energetics

Enthalpy

CH4(g) + 2O2(g)

H = 890 kJ mol1

CO2(g) + 2H2O(l)

Reaction coordinate

C(g) + 4H(g) + 4O(g)

H1

298 K

H2


Energetics

CH4(g) + 2O2(g)

H = 890 kJ mol1

CO2(g) + 2H2O(l)

C(g) + 4H(g) + 4O(g)

Enthalpy

373 K

H1’

H2’

Assume constant H

Reaction coordinate


Energetics

Enthalpy

CH4(g) + 2O2(g)

H = 890 kJ mol1

CO2(g) + 2H2O(l)

Reaction coordinate

C(g) + 4H(g) + 4O(g)

373 K

In fact, H depends on T


Energetics

Enthalpy

CH4(g) + 2O2(g)

H = 802 kJ mol1

CO2(g) + 2H2O(l)

CO2(g) + 2H2O(g)

H = +88 kJ mol1

Reaction coordinate

C(g) + 4H(g) + 4O(g)

373 K


Energetics

6.3

Standard Enthalpy Changes


Standard enthalpy changes

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

H = -802 kJ mol-1at 373 K

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

H = -890 kJ mol-1 at 298 K

6.3 Standard enthalpy changes (SB p.141)

Standard enthalpy changes


Standard enthalpy changes1

Enthalpy change under standard conditions denoted by symbol: H

ø

6.3 Standard enthalpy changes (SB p.141)

Standard enthalpy changes

As enthalpy changes depend on temperature and pressure, it is necessary to definestandard conditions:

1.elements or compounds in their normal physical states;2.a pressure of 1 atm (101325 Nm-2); and3.a temperature of 25oC (298 K)


Energetics

2H2(g) + O2(g) 2H2O(l)

H = 572 kJ mol1

Standard enthalpy change of reaction

The enthalpy change when the molar quantities of reactants as stated in the equation react under standard conditions.

per mole of O2


Energetics

H2(g) + O2(g)  H2O(l)

H = 286 kJ mol1

4H2(g) + 2O2(g)  4H2O(l)

H

H = 1144 kJ

depends on the equation

Standard enthalpy change of reaction

2H2(g) + O2(g) 2H2O(l)

per mole of O2

H = 572 kJ mol1

per mole of H2 or H2O


Energetics

Hf

H2(g) + O2(g)  H2O(l)

Hf [H2O] = 286 kJ mol1

Hf [O2] = 0 kJ mol1

Hf [element] = 0 kJ mol1

Standard enthalpy change of formation

The enthalpy change when one mole of the substance is formed from its elements under standard conditions.

Q.3

O2(g)  O2(g)


Energetics

Hf [diamond] = +1.9 kJ mol1

Most stable allotrope

C(graphite)  C(diamond)


Energetics

(iii)Mg(s) + O2(g)  MgO(s)

(iv)Na(s) + H2(g) + C(graphite) + O2(g)

 NaHCO3(s)

Q.4

(i)C(graphite) + O2(g)  CO2(g)

(ii)C(graphite) + 2H2(g)  CH4(g)

(v)2C(graphite) + 2H2(g) + O2(g)  CH3COOH(l)


Energetics

H2(g) + O2(g)  H2O(l)

Q.5

qv = U = 140.3 kJ per g of H2

n = 0 – 0.496 – 0.248 = -0.744 mol


Energetics

Q.5

H = U + nRT

= -142.1 kJ

Heat released for the formation of 0.496 mol of water

Molar Hf[H2O] =


Energetics

Hc

Hc [C2H5OH(l)] = -1368 kJ mol1

Standard enthalpy change of combustion

The enthalpy change when one mole of the substance undergoes complete combustion under standard conditions.

C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)


Energetics

C(diamond) + O2(g)

Enthalpy

C(graphite) + O2(g)

1.9

395.4

Hf [diamond]

393.5

= +1.9 kJ mol1

CO2(g)

Reaction coordinate

6.3 Standard enthalpy changes (SB p.147)

ø

Substance

Hc(kJ mol-1)

C (diamond)

C (graphite)

-395.4

-393.5


Energetics

(a)C(graphite) + O2(g)  CO(g)

= Hf [CO(g)]

 Hc [graphite]

= 2  Hc [H2(g)]

H

H

= 2  Hf [H2O(l)]

Q.6

Incomplete combustion

(b)2H2(g) + O2(g)  2H2O(l)


Energetics

= Hc [graphite]

= Hf [CO2(g)]

= Hc [CH4(g)]

 Hf [CO2(g)]

H

H

Not formed from elements

 2  Hf [H2O(l)]

Check Point 6-3

Q.6

(c)C(graphite) + O2(g)  CO2(g)

(d)CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)


Standard enthalpy changes of neutralization

Standard enthalpy change of neutralization(Hneut) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under standard conditions.

ø

e.g. H+(aq) + OH-(aq)  H2O(l)

Hneut = -57.3 kJ mol-1

ø

6.3 Standard enthalpy changes (SB p.142)

Standard enthalpy changes of neutralization


Standard enthalpy changes of neutralization1

6.3 Standard enthalpy changes (SB p.142)

Standard enthalpy changes of neutralization

Enthalpy level diagram for the neutralization of a strong acid and a strong alkali


Energetics

Acid

Alkali

Hneu

ø

HCl

HCl

HCl

HF

NaOH

KOH

NH3

NaOH

-57.1

-57.2

-52.2

-68.6

NH3(aq) + H2O(l) NH4+(aq) + OH(aq) H1 > 0

ø

Hneu = H1 + H2 = 52.2 kJ mol1

6.3 Standard enthalpy changes (SB p.142)

H+(aq) + OH-(aq) + Cl(aq) H2O(l)+ Cl(aq)H2 = -57.3

NH3(aq) + HCl(aq)  NH4Cl (aq)


Energetics

Acid

Alkali

Hneu

ø

HCl

HCl

HCl

HF

NaOH

KOH

NH3

NaOH

-57.1

-57.2

-52.2

-68.6

HF(aq) H+(aq) + F(aq) H1 < 0

ø

Hneu = H1 + H2 = 68.6 kJ mol1

6.3 Standard enthalpy changes (SB p.142)

H+(aq) + OH-(aq) + Na+(aq)  H2O(l)+ Na+(aq)H2 = -57.3

HF(aq) + NaOH(aq)  NaF(aq) + H2O(l)


Standard enthalpy change of solution

Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved in a specified number of moles of solvent (e.g. water) under standard conditions.

ø

NaCl(s) + 10H2O(l)  Na+(aq) + Cl-(aq)

Hsoln[NaCl(s)]= +2.008 kJ mol-1

ø

dilution

NaCl(aq) NaCl(aq)

H > 0

6.3 Standard enthalpy changes (SB p.142)

Standard enthalpy change of solution


Standard enthalpy change of solution1

Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved to form an infinitely dilute solution under standard conditions.

ø

concentration  0

NaCl(s) + water  Na+(aq) + Cl-(aq)

Hsoln= +4.98 kJ mol-1

ø

6.3 Standard enthalpy changes (SB p.142)

Standard enthalpy change of solution


Standard enthalpy change of solution2

e.g. NaCl(s) + water  Na+(aq) + Cl-(aq)

Hsoln= +4.98 kJ mol-1

ø

+ 4.98 kJ mol1

Enthalpy level diagram for the dissolution of NaCl

6.3 Standard enthalpy changes (SB p.143)

Standard enthalpy change of solution


Standard enthalpy change of solution3

e.g. LiCl(s) + water  Li+(aq) + Cl-(aq)

Hsoln= -37.2 kJ mol-1

ø

Enthalpy level diagram for the dissolution of LiCl in water

6.3 Standard enthalpy changes (SB p.143)

Standard enthalpy change of solution


Standard enthalpy change of solution4

Salt

Hsoln(kJ mol-1)

ø

NH3

NaOH

HCl

H2SO4

LiCl

NaCl

NaNO3

NH4Cl

35.5

43.1

73.0

74.0

37.2

+4.98

+21.0

+22.6

6.3 Standard enthalpy changes (SB p.143)

Standard enthalpy change of solution


Energetics

6.4

Experimental Determination of Enthalpy Changes by Calorimetry


Experimental determination of enthalpy changes by calorimetry

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)

Experimental determination of enthalpy changes by calorimetry

Calorimeter is any set-up used for the determination of H.

By temperature measurement.

H = qp = (m1c1 + m2c2)T


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)

H = qp = (m1c1 + m2c2)T

where

m1 is the mass of the reaction mixture,

m2 is the mass of the calorimeter,

c1 is the specific heat capacity of the reaction mixture,

c2 is the specific heat capacity of the calorimeter,

T is the temperature change of the reaction mixture.


Determination of enthalpy change of neutralization

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)

Determination of enthalpy change of neutralization


Energetics

T1

T2

t1

t2

Check Point 6-4(a)

If the reaction is fast enough, T1 T2

H (m1c1 + m2c2)(T2 – T0)

H = (m1c1 + m2c2)(T1 – T0)

T0


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.150)

Determination of enthalpy change of combustion

The Philip Harris calorimeter used for determining the enthalpy change of combustion of a liquid fuel


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)

Determination of enthalpy change of combustion

A simple apparatus used to determine the enthalpy change of combustion of ethanol


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)

Heat evolved = (m1c1 + m2c2) ΔT

Where

m1 is the mass of water in the calorimeter,

m2 is the mass of the calorimeter,

c1 is the specific heat capacity of the water,

c2 is the specific heat capacity of the calorimeter,

ΔT is the temperature change of the reaction


Energetics

Check Point 6-4(c)

Q.7(Example)

heat given out


Energetics

6.5

Hess’s Law


Hess s law

6.5 Hess’s law (SB p.153)

Hess’s Law

Hess’s law of constant heat summationstates that the total enthalpy change accompanying a chemical reaction

is independent ofthe route by which the chemical reaction takes place and

depends only on the difference between the total enthalpy of the reactants and that of the products.


Hess s law1

Route 2

C

H2

H3

Route 1

A(HA)

B(HB)

H1

H4

H5

D

Route 3

6.5 Hess’s law (SB p.153)

Hess’s Law

H1 = HB – HA

= H2 + H3

= H4 + H5


Importance of hess s law

6.5 Hess’s law (SB p.155)

Importance of Hess’s law

The enthalpy change of some chemical reactions cannot be determined directly because:

  • the reactions cannot be performed/controlled in the laboratory

  • the reaction rates are too slow

  • the reactions may involve the formation of side products

But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law.


Enthalpy change of formation of co g

Hf [CO(g)]

Hf [CO(g)]

ø

ø

C(graphite) + ½O2(g)

CO(g)

cannot be determined directly

6.5 Hess’s law (SB p.153)

Enthalpy change of formation of CO(g)

due to further oxidation of CO to CO2

The reaction cannot be controlled.


Enthalpy change of formation of co g1

Hc [graphite] = -393 kJ mol-1

ø

Hc [CO(g)] = -283.0 kJ mol-1

ø

Hf [CO(g)]

ø

C(graphite) + ½O2(g)

CO(g)

+ ½O2(g)

+ ½O2(g)

H1

H2

CO2(g)

Hf [CO(g)] +H2=H1

ø

Hf [CO(g)] =H1 -H2

ø

6.5 Hess’s law (SB p.153)

Enthalpy change of formation of CO(g)

= -393 -(-283 )

= -110 kJ mol-1


Enthalpy cycle born haber cycle

Hf [CO(g)]

ø

C(graphite) + ½O2(g)

CO(g)

+ ½O2(g)

+ ½O2(g)

H1

H2

CO2(g)

6.5 Hess’s law (SB p.155)

Enthalpy cycle (Born-Haber cycle)

  • Relate the various equations involved in a reaction


Steps for drawing born haber cycle

Hf [CO(g)]

ø

C(graphite) + ½O2(g)

CO(g)

6.5 Hess’s law (SB p.153)

Steps for drawing Born-Haber cycle

1. Give the equation for the change being considered.


Steps for drawing born haber cycle1

Hf [CO(g)]

ø

C(graphite) + ½O2(g)

CO(g)

+ ½O2(g)

+ ½O2(g)

H1

H2

CO2(g)

Hf [CO(g)]= Hc[graphite] - Hc[CO(g)]

ø

ø

ø

6.5 Hess’s law (SB p.153)

Steps for drawing Born-Haber cycle

2. Complete the cycle by giving the equations for the

combustion reactions of reactants and products.


Steps for drawing born haber cycle2

Hf [CO(g)]

ø

C(graphite) + ½O2(g)

CO(g)

+ ½O2(g)

+ ½O2(g)

H1

H2

CO2(g)

Hf [CO(g)]= Hc[reactant] - Hc[product]

ø

ø

ø

6.5 Hess’s law (SB p.153)

Steps for drawing Born-Haber cycle

2. Complete the cycle by giving the equations for the

combustion reactions of reactants and products.


Calculation of standard enthalpy change of formation from standard enthalpy changes of combustion

Hf =  Hc [reactants] -  Hc [product]

ø

ø

ø

Calculation of standard enthalpy change of formation from standard enthalpy changes of combustion

6B


Energetics

Hf [C4H10(g)]

= 4Hc[C(graphite)] + 5Hc[H2(g)] - Hc[C4H10(g)]

Hf [C4H10(g)]

ø

4C(graphite) + 5H2(g)

C4H10(g)

5Hc [H2(g)]

ø

Hc [C4H10(g)]

ø

+ 2.5 O2(g)

4Hc [graphite]

ø

+ 6.5 O2(g)

4CO2(g)

+ 5H2O(l)

ø

ø

ø

ø

Q.8

+ 4O2(g)

= [4(-393)+ 5(-286) – (2877)] kJ mol1

= 125 kJ mol1


Energetics

Hf [C4H10(g)]

ø

4C(graphite) + 5H2(g)

C4H10(g)

Hc /kJ mol1

Hc

(2) H2(g) + O2(g)  H2O(l) 286

(3) C4H10(g) + O2(g)  4CO2(g) + 5H2O(l) 2877

Hf [C4H10(g)] = [4(-393) + 5(-286)  (2877)] kJ mol1

Q.8Method B :

By addition and/or subtraction of equations with known

(1) C(graphite) + O2(g)  CO2(g) 393

Overall reaction : 4(1) + 5(2) – (3)

= 125 kJ mol1


Enthalpy level diagram

Enthalpy level diagram for the oxidation of C(graphite) to CO2(g)

6.5 Hess’s law (SB p.154)

Enthalpy level diagram

  • Relate substances together in terms of enthalpy changes of reactions


Energetics

C(graphite) + O2(g)

Enthalpy / kJ mol1

Steps for drawing enthalpy level diagram

1. Draw the enthalpy level of elements.


Energetics

Enthalpy / kJ mol1

Steps for drawing enthalpy level diagram

2. Enthalpies of elements are arbitrarily taken as zero.

C(graphite) + O2(g)


Energetics

Enthalpy / kJ mol1

Steps for drawing enthalpy level diagram

3. Higher enthalpy levels are drawn above that of elements

C(graphite) + O2(g)


Energetics

Enthalpy / kJ mol1

Hc[graphite] = 393 kJ

CO2(g)

Steps for drawing enthalpy level diagram

4. Lower enthalpy levels are drawn below that of elements

C(graphite) + O2(g)

Route 1


Energetics

Hf[CO(g)] = 110 kJ

Enthalpy / kJ mol1

CO(g) + O2(g)

Hc[graphite] = 393 kJ

Hc[CO(g)] = 283 kJ

CO2(g)

Steps for drawing enthalpy level diagram

4. Lower enthalpy levels are drawn below that of elements

C(graphite) + O2(g)

Route 1

Route 2


Energetics

  • The formation of ethyne (C2H2(g) can be represented by the following equation:

  • 2C(graphite) + H2(g)  C2H2(g)

  • (i) Draw an enthalpy cycle relating the above equation to carbon dioxide and water.

  • (ii) Calculate the standard enthalpy change of formation of ethyne.

  • (Given: Hc [C(graphite)] = -393.5 kJ mol-1;

  • Hc [H2(g)] = -285.8 kJ mol-1;

  • Hc [C2H2(g)] = -1299 kJ mol-1)

ø

ø

ø

6.5 Hess’s law (SB p.158)

Check Point 6-5


Energetics

Hf

Hf

Hf

2Hc[graphite]

2Hc[graphite]

Hc[H2(g)]

Hc[C2H2(g)]

2CO2(g)

+ Hc[C2H2(g)] =

+ Hc[H2(g)]

= 2Hc[graphite]

- Hc[C2H2(g)]

+ Hc[H2(g)]

6.5 Hess’s law (SB p.158)

2C(graphite) + H2(g)  C2H2(g)

+ 2O2(g)

+ 0.5O2(g)

+ 2.5O2(g)

+ H2O(l)

By Hess’s law,

= [2(393.5) + (285.8) –(1299)] kJ mol1

= +226.2 kJ mol1


Energetics

C2H2(g) + 2.5O2(g)

Hc[C2H2(g)]

Hf[C2H2(g)]

2C(graphite) + H2(g) + 2.5O2(g)

Enthalpy / kJ mol1

2Hc[graphite]

Hc[H2(g)]

2CO2(g) + H2(g) + 0.5O2(g)

2CO2(g) + H2O(l)

(iii). Draw an enthalpy level diagram for the reaction using the enthalpy changes in (ii)

Route 1

Route 2


Energetics

6.6

Calculations involving Standard Enthalpy Changes of Reactions


Calculation of standard enthalpy change of reaction from standard enthalpy changes of formation

H from Hf

6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)

Calculation of standard enthalpy change of reaction from standard enthalpy changes of formation


Energetics

NH3(g) + HCl(g) NH4Cl(s)

H

Hf[HCl(g)]

Hf[NH4Cl(s)]

Hf[NH3(g)]

Hf[HCl(g)]

Hf[NH3(g)]

H

H

Hf[NH4Cl(s)]

Hf[NH4Cl(s)]

Hf[NH3(g)]

Hf[HCl(g)]

0.5N2(g) + 1.5H2(g)

+

+

=

-

=

-

Q.9

+ 0.5H2(g) + 0.5Cl2(g)

By Hess’s law,

= [314 –(46) –(92)] kJ mol1 = 176 kJ mol1


Energetics

NH3(g) + HCl(g) NH4Cl(s)

H

Hf[HCl(g)]

Hf[NH3(g)]

Hf[NH3(g)]

Hf[HCl(g)]

Hf[HCl(g)]

Hf[NH4Cl(s)]

Hf[NH4Cl(s)]

Hf[NH4Cl(s)]

Hf[NH3(g)]

H

H

0.5N2(g) + 1.5H2(g)

+

+

=

-

=

-

ø

Hreaction= Hf [products]-  Hf [reactants]

ø

Q.9

+ 0.5H2(g) + 0.5Cl2(g)

By Hess’s law,


Energetics

5Hf[N2O4(l)]

4Hf[CO2(g)]

4Hf[CH3NHNH2(l)]

12Hf[H2O(l)]

4C(graphite) + 12H2(g) + 4N2(g)

+ 5N2(g) + 10O2(g)

H

Q.10

H

4CH3NHNH2(l) + 5N2O4(l) 4CO2(g) + 9N2(g) + 12H2O(l)

By Hess’s law,

= [4(393) + 12(286) - 4(+53) – 5(20)] kJ

= 5116 kJ

Highly exothermic/ignites spontaneously

Used as rockel fuel in Apollo 11


Energetics

C3H6(g) + H2(g) C3H8(g)

H /kJ mol1

H

H

Q.11

(1)C3H6(g) + 4.5O2(g)  3CO2(g) + 3H2O(l) 2058

(2)H2(g) + 0.5O2(g)  H2O(l) 286

(3)C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) 2220

Overall reaction : (1) + (2) – (3)

= [(2058) + (286) – (2220)] kJ mol1

= 124 kJ mol1


Energetics

Hf[H2O(l)]

Hc[C3H6(g)]

H

H

Hc[C3H8(g)]

3H2O(l) + 3CO2 (g)

+ H2O(l)

+ (2220) =(2058) +(286)

Q.11

H

C3H6(g) + H2(g) C3H8(g)

By Hess’s law,

= [(2058) + (286) – (2220)] kJ mol1

= 124 kJ mol1


Energetics

Hc[H2(g)]

Hc[C3H6(g)]

Hc[C3H8(g)]

3H2O(l) + 3CO2 (g)

Hrx =  Hc [reactants] -  Hc [products]

Hf =  Hc [reactants] -  Hc [product]

ø

ø

ø

ø

ø

ø

Q.11

H

C3H6(g) + H2(g) C3H8(g)

+ H2O(l)


Energetics

Relative stability of compounds and Hf

Hf indicates the energetic stability of the compound with respect to its elements

Hf

Elements  Compound

Hf

Hf

< 0  energetically more stable than its elements

> 0  energetically less stable than its elements


Energetics

H2O2(l) H2O(l) + 0.5O2(g)

Hrx = 98 kJ mol1

H2(g) + O2(g)

Enthalpy / kJ mol1

Hf[H2O2(l)] = 188 kJ

H2O2(l)

Hrx= 98 kJ

H2O(l) + 0.5O2(g)

Hf [H2O2(l)] = 188 kJ mol1

H2(g) + O2(g)  H2O2(l)


Energetics

Enthalpy / kJ mol1

Hf[H2O2(l)] = 188 kJ

Hrx= 98 kJ

H2O2 is energetically stable w.r.t. H2 and O2

H2O2 is energetically unstable w.r.t. H2O and 0.5O2

H2(g) + O2(g)

H2O2(l)

H2O(l) + 0.5O2(g)


Energetics

Enthalpy / kJ mol1

Hf[H2O2(l)] = 188 kJ

Hc[H2(g) ] = 286 kJ

Hrx= 98 kJ

H2 and O2 are energetically unstable w.r.t. H2O2 and H2O

H2(g) + O2(g)

H2O2(l)

H2O(l) + 0.5O2(g)


Energetics

H = 2 kJ mol1

C(diamond)  C(graphite)

H = 2 kJ

Enthalpy / kJ mol1

Hc[diamond] = 395 kJ

Hc[graphite] = 393 kJ

Energetically unstable

C(diamond) + O2(g)

C(graphite) + O2(g)

CO2(g)


Energetics

H = 2 kJ mol1

C(diamond)  C(graphite)

H = 2 kJ

Enthalpy / kJ mol1

Hc[diamond] = 395 kJ

Hc[graphite] = 393 kJ

Kinetically stable

The rate of conversion is extremely low

C(diamond) + O2(g)

C(graphite) + O2(g)

CO2(g)


Energetics

C(g) + O(g) + O(g)

bond breaking

bond forming

Enthalpy / kJ mol1

Hc[diamond] = 395 kJ

C(diamond) + O2(g)

C(graphite) + O2(g)

Hc[graphite] = 393 kJ

CO2(g)


Energetics

Enthalpy / kJ mol1

Hc[diamond] = 395 kJ

Rate of reaction depends on the ease of bond breaking in reactants (e.g. diamond)

The minimum energy required for a reaction to start is known as the activation energy, Ea

C(diamond) + O2(g)

H = 2 kJ

C(graphite) + O2(g)

Hc[graphite] = 393 kJ

CO2(g)


Energetics

H = 2 kJ mol1

C(diamond)  C(graphite)

H = 2 kJ

Enthalpy / kJ mol1

Hc[diamond] = 395 kJ

Hc[graphite] = 393 kJ

The extremely low rate is due to high Ea

C(diamond) + O2(g)

C(graphite) + O2(g)

CO2(g)


Energetics

H = 2 kJ mol1

C(diamond)  C(graphite)

H = 2 kJ

Enthalpy / kJ mol1

Hc[diamond] = 395 kJ

Hc[graphite] = 393 kJ

Ea is always > 0 as bond breaking is endothermic

C(diamond) + O2(g)

C(graphite) + O2(g)

CO2(g)


Energetics

tells how far a reaction can proceed

Hf

Ea tells how fast a reaction can proceed.

Rate of reaction / kinetics / kinetic stability

Equilibrium / energetics / energetic stability


Energetics

 Energetic stability

< 0  higher energetic stability w.r.t. its elements

> 0  lower energetic stability w.r.t. its elements

Hf

Hf

Hf

Ea kinetic stability (rate of reaction)

Higher Ea higher kinetic stability of reactants w.r.t. products

lower rate to give products

Lower Ea lower kinetic stability of reactants w.r.t. products

higher rate to give products


Energetics

diamond

graphite

Diamond  energetically unstable w.r.t. graphite

 kinetically stable w.r.t. graphite

Graphite  stable w.r.t. diamond

energetically and kinetically


Energetics

diamond

graphite

extremely slow

diamond graphite

extremely slow


Energetics

6.7

6.7 Entropy change (SB p.164)

Spontaneity of Changes


Energetics

external

boundary

external

external

external

  • Spontaneity : The state of being spontaneous

  • Spontaneous :

  • self-generated

  • natural

  • happening without external influence

internal


Energetics

6.7 Entropy change (SB p.164)

  • A process is said to be spontaneous

  • If no external “forces” are required to keep the process going,

  • although external “forces” may be required to get the process started (Ea).

  • The process may be a physical change or a chemical change


Energetics

6.7 Entropy change (SB p.164)

  • Spontaneous physical change:

  • E.g. condensation of steam at 25°C

  • Spontaneous chemcial change:

  • E.g. burning of wood once the fire has started

Exothermic  spontaneous ?

Endothermic  not spontaneous ?

Q.12


Energetics

Yes

Yes

Yes

Yes

Change

Exothermic ? (Yes / No)

Spontaneous ?

(Yes / No)

Burning of CO at 25°C

Yes

Condensation of steam at 25°C

Yes

Melting of ice at 25°C

No

Dissolution of NH4Cl in water at 25°C

No


Energetics

6.7 Entropy change (SB p.164)

  • Exothermicity is NOT the only reason for the spontaneity of a process

  • Some spontaneous changes are endothermic

  • E.g.Melting of ice

  • Dissolution of NH4Cl in water


Energetics

Melting of ice

Dissolution of ammonium nitrate in water

6.7 Entropy change (SB p.164)


Energetics

Solid

Gas

Liquid

Entropy increases

6.7 Entropy change (SB p.165)

Entropy

  • Entropy is a measure of the randomness or the degree of disorder (freedom) of a system


Energetics

6.7 Entropy change (SB p.166)

Entropy change (S)

  • Entropy change means the change in the degree of disorder of a system

  • S = Sfinal - Sinitial


Energetics

6.7 Entropy change (SB p.166)

Positive entropy change (S > 0)

  • Increase in entropy

  • Sfinal > Sinitial

  • Example:

  • Ice (low entropy)  Water (high entropy)

  • S = Swater – Sice = +ve


Energetics

6.7 Entropy change (SB p.166)

Negative entropy change (S < 0)

  • Decrease in entropy

  • Sinitial > Sfianl

  • Example:

  • Water (high entropy)  Ice (low entropy)

  • S = Sice – Swater = -ve

Q.13


Energetics

2

2


Energetics

Consider an isolated system which has no exchange of energy and matter with its surroundings


Energetics

S1

S2

S2

S1

S = S2 – S1 > 0

Which one is spontaneous ?

S = S1 – S2 < 0


Energetics

S1

S2

S2

S1

Spontaneous, S = S2 – S1 > 0

Not spontaneous, S = S1 – S2 < 0


Energetics

vacuum

Slit is open

Probability

A molecular statistical interpretation

Free adiabatic expansion of an ideal gas into a vacuum.


Energetics

vacuum

1 mole

Probability

A molecular statistical interpretation

Free adiabatic expansion of an ideal gas into a vacuum.

Slit is open


Energetics

A spontaneous process taking place in an isolated system is always associated with an increase in entropy (I.e. S > 0)

6B

In closed system,

The spontaneity of a process depends on both H and S.

The driving force of a process is a balance of H and S.


Energetics

6.7 Entropy change (SB p.166)

Ice (less entropy)  Water (more entropy)

H is +ve  not favourable

S is +ve  favourable

Considering both S & H ,

the process is spontaneous

Q.14


Energetics

Favourable


Energetics

Not favourable


Energetics

Spontaneity depends on temperature


Energetics

Favourable


Energetics

Not favourable


Energetics

Spontaneity depends on temperature


Energetics

J. Willard Gibbs (1839 – 1903)

Spontaneity of a process depends on

H, S & T

G = H –TS

G is the (Gibbs’) free energy


Energetics

G = H –TS

Spontaneity depends on G

‘Free’ means the energy free for work


Energetics

A spontaneous process is always associated with a decrease in the free energy of the system.

G < 0  spontaneous process

G > 0  not spontaneous process

Q.15


Energetics

G = H –TS


Energetics

high

+ve

Not spontaneus

low

-ve

Spontaneous

G = H –TS

H

S

T

G

Results

+ve

+ve

high

-ve

Spontaneous

+ve

+ve

low

+ve

Not spontaneous

-ve

-ve

-ve

-ve


Energetics

G = H –TS


Energetics

G = H –TS

H

S

T

G

Results

-ve

+ve

high

-ve

Spontaneous

-ve

+ve

low

+ve

-ve

high

+ve

Not spontaneous

+ve

-ve

low


Energetics

G = H –TS

Q.16


Energetics

Diamond Graphite

H < 0

S > 0

G = H –TS < 0

The process is spontaneous, although activation energy is required to start the conversion.


Energetics

S1

S2

Spontaneous S = S2 – S1 > 0

The entropy of a system can be considered as a measure of the availability of the system to do work.

Before expansion, the system is available to do work.

After expansion, the system is not available to do work.


Energetics

S1

S2

Spontaneous S = S2 – S1 > 0

The lower the entropy of a system(before expansion), the more available is the system to do work.

Thus, entropy is considered as a measure of the useless energy of a system.


Energetics

Useless energy

Total energy

Useful energy

G = H –TS

G = H – TS

H = G + TS


Energetics

Useless energy

Total energy

Useful energy

If the universe is an isolated system

H is a constant and H is always zero

H = G + TS


Energetics

cosmic background radiation = 4K

Useful energy

Useless energy

H = G + TS

H = G + TS = 0

= G + TS = 0

S always > 0,

S always ,

useless energy always 

G always < 0,

G always ,

useful energy always 


Energetics

In an isolated system, entropy will only increase with time, it will not decrease with time.

The second law of thermodynamics


Energetics

If the universe is an isolated system,

Suniverse = Ssystem + Ssurroundings > 0

the total entropy (randomness) of the universe will tend to increase to a maximum;

the total free energy of the universe will tend to decrease to a minimum.


Energetics

As time increases, the universe will always become more disordered.

Entropy is considered as a measure of time.

Entropy can be used to distinguish between future and past.


Energetics

Time can only proceed in one direction that results in an increase in the total entropy of the universe.

This is known as the arrow of time.


Energetics

minimum entropy, maximum free energy(singularity)

Big bang

expanding

maximum entropy, minimum free energy

Big chill

The history of the universe

T  1.41032 K

H = G +TS

T  0 K


Energetics

h = Planck’s constant G = gravitational constant c = speed of light in vacuum k = Boltzmann constant

Planck’s units

Planck’s temperature

1.416785(71) × 1032 K

Planck’s length

Planck’s time


Energetics

Absolute hot beyond which all physical laws break down

Planck’s units

Planck’s temperature

1.416785(71) × 1032 K

Planck’s length

Planck’s time


Energetics

1.616252(81) × 10−35 m

Planck’s units

Physical significance not yet known

Planck’s temperature

1.416785(71) × 1032 K

Planck’s length

Planck’s time


Energetics

the diameter of proton

1.616252(81) × 10−35 m

Planck’s units

Planck’s temperature

1.416785(71) × 1032 K

Planck’s length

Planck’s time


Energetics

1.616252(81) × 10−35 m

  • = 7.310-37 m

Planck’s units

22 times of Mr Chio’s wavelength

Planck’s temperature

1.416785(71) × 1032 K

Planck’s length

Planck’s time


Energetics

5.39124(27) × 10−44 s

1.616252(81) × 10−35 m

It is the time required for light to travel, in a vacuum, a distance of 1 Planck length.

Planck’s units

Planck’s temperature

1.416785(71) × 1032 K

Planck’s length

Planck’s time


Energetics

5.39124(27) × 10−44 s

1.616252(81) × 10−35 m

10-15 s  femtosecond(飛秒)

10-18 s  attosecond(阿托秒)

Planck’s units

Time taken for light to travel the length of 3 H atoms

Planck’s temperature

1.416785(71) × 1032 K

Planck’s length

Planck’s time


Energetics

Q.17(a)

The drop in temperature of the system is accompanied by the rise in temperature of its surroundings.

Ssystem < 0

Ssurroundings > 0

Suniverse = Ssystem + Ssurroundings > 0


Energetics

Q.17(b)

The drop in entropy of the system is at the cost of the rise in entropy of its surroundings.

Ssystem < 0

Ssurroundings > 0

Suniverse = Ssystem + Ssurroundings > 0


Energetics

Let's Think 1

Check Point 6-8

6.8 Free energy change (SB p.170)


Energetics

The END


Energetics

(a) Endothermic

(b) Endothermic

(c) Exothermic

6.1 What is energetics? (SB p.140)

Back

Check Point 6-1

State whether the following processes are exothermic or endothermic.

(a) Melting of ice.

(b) Dissolution of table salt.

(c) Condensation of steam.

Answer


Energetics

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)

Check Point 6-2

  • State the difference between exothermic and endothermic reactions with respect to

  • (i)the sign of H;

  • (ii)the heat change with the surroundings;

  • (iii) the total enthalpy of reactants and products.

Answer

  • (i) Exothermic reactions: H = -ve; endothermic reactions: H = +ve

  • (ii)Heat is given out to the surroundings in exothermic reactions whereas heat is taken in from the surroundings in endothermic reactions.

  • (iii)In exothermic reactions, the total enthalpy of products is less than that of the reactants. In endothermic reactions, the total enthalpy is greater than that of the reactants.


Energetics

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)

Check Point 6-2

  • Draw an enthalpy level diagram for a reaction which is

  • (i) endothermic, having a large activation energy.

  • (ii)exothermic, having a small activation energy.

Answer


Energetics

  • (i)

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)

Check Point 6-2


Energetics

  • (ii)

6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)

Check Point 6-2

Back


Energetics

6.3 Standard enthalpy changes (SB p.147)

Check Point 6-3

  • Why must the condition “burnt completely in oxygen” be emphasized in the definition of standard enthalpy change of combustion?

Answer

  • If the substance is not completely burnt in excess oxygen, other products such as C(s) and CO(g) may be formed. The enthalpy change of combustion measured will not be accurate.


Energetics

  • Standard enthalpy change of combustion of CO

  • =  (-566.0) kJ

  • = -283.0 kJ

6.3 Standard enthalpy changes (SB p.147)

Check Point 6-3

  • (b) The enthalpy change of the following reaction under standard conditions is –566.0 kJ.

  • 2CO(g) + O2(g)  2CO2(g)

  • What is the standard enthalpy change of combustion of carbon monoxide?

Answer


Energetics

  • What terms may be given for the enthalpy change of the following reaction?

  • N2(g) + O2(g)  NO2(g)

6.3 Standard enthalpy changes (SB p.147)

Check Point 6-3

(c) ½ Enthalpy change of combustion of nitrogen or enthalpy change of formation of nitrogen dioxide.

Answer

Back


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)

Example 6-4A

Determine the enthalpy change of neutralization of 25 cm3 of 1.25 M hydrochloric acid and 25 cm3 of 1.25 M sodium hydroxide solution using the following data:

Mass of calorimeter = 100 g

Initial temperature of acid = 15.5 oC (288.5 K)

Initial temperature of alkali = 15.5 oC (288.5 K)

Final temperature of the reaction mixture = 21.6 oC (294.6 K)

The specific heat capacities of water and calorimeter are 4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively.

Answer


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)

Example 6-4A

Assume that the density of the reaction mixture is the same as that of water, i.e. 1 g cm-3.

Mass of the reaction mixture = (25 + 25) cm3 1 g cm-3 = 50 g = 0.05 kg

Heat given out = (m1c1 + m2c2) T

= (0.05 kg  4200 J kg-1 K-1 + 0.1 kg  800 J kg-1 K-1)  (294.6 – 288.5) K

= 1769 J

H+(aq) + OH-(aq)  H2O(l)

Number of moles of HCl = 1.25 mol dm-3  25  10-3 dm3 = 0.03125 mol

Number of moles of NaOH = 1.25 mol dm-3  25  10-3 dm3 = 0.03125 mol

Number of moles of H2O formed = 0.03125 mol


Energetics

Heat given out per mole of H2O formed

=

= 56608 J mol-1

The enthalpy change of neutralization is –56.6 kJ mol-1.

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)

Example 6-4A

Back


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)

Example 6-4B

Determine the enthalpy change of combustion of ethanol using the following data:

Mass of spirit lamp before experiment = 45.24 g

Mass of spirit lamp after experiment = 44.46 g

Mass of water in copper calorimeter = 50 g

Mass of copper calorimeter without water = 380 g

Initial temperature of water = 18.5 oC (291.5 K)

Final temperature of water = 39.4 oC (312.4 K)

The specific heat capacities of water and copper calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1 K-1 respectively.

Answer


Energetics

Heat evolved by the combustion of ethanol

= Heat absorbed by the copper calorimeter

= (m1c1 + m2c2) T

= (0.05 kg  4200 J kg-1 K-1 + 0.38 kg  2100 J kg-1 K-1)  (312.4 – 291.5)K

= 21067 J

C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)

Mass of ethanol burnt = (45.24 – 44.46) g = 0.78 g

Number of moles of ethanol burnt = = 0.017 mol

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)

Example 6-4B


Energetics

Heat given out per mole of ethanol

=

= 1239235 J mol-1

= 1239 kJ mol-1

The enthalpy change of combustion of ethanol is –1239 kJ mol-1.

There was heat loss by the system to the surroundings, and incomplete combustion of ethanol might occur. Also, the experiment was not carried out under standard conditions. Therefore, the experimentally determined value (-1239 kJ mol-1) is less than the theoretical value of the standard enthalpy change of combustion of ethanol (-1371 kJ mol-1).

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)

Example 6-4B

Back


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152)

Example 6-4C

0.02 mol of anhydrous ammonium chloride was added to 45 g of water in a polystyrene cup to determine the enthalpy change of solution of anhydrous ammonium chloride. It is found that there was a temperature drop from 24.5 oC to 23.0 oC in the solution.

Given that the specific heat capacity of water is 4200 J kg-1 K-1 and

NH4Cl(s) + aq  NH4Cl(aq)

Calculate the enthalpy change of solution of anhydrous ammonium chloride.

(Neglect the specific heat capacity of the polystyrene cup.)

Answer


Energetics

Heat absorbed = m1c1T ( c2 0)

= 0.045 kg  4200 J kg-1 K-1  (297.5 – 296) K

= 283.5 J (0.284 kJ)

Heat absorbed per mole of ammonium chloride =

= 14.2 kJ mol-1

The enthalpy change of solution of anhydrous ammonium chloride is +14.2 kJ mol-1.

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152)

Example 6-4C

Back


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

Check Point 6-4

(a) A student tried to determine the enthalpy change of neutralization by putting 25.0 cm3 of 1.0 M HNO3 in a polystyrene cup and adding 25.0 cm3 of 1.0 M NH3 into it. The temperature rise recorded was 3.11 oC. Given that the mass of the polystyrene cup is 250 g, the specific heat capacities of water and the polystyrene cup are 4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively. Determine the enthalpy change of neutralization of nitric acid and aqueous ammonia. (Density of water = 1 g cm-3)

Answer


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

Check Point 6-4

  • Heat evolved = m1c1T + m2c2 T

  • = 0.050 kg  4200 J kg-1 K-1  3.11 K + 0.25 kg  800 J kg-1 K-1  3.11 K

  • = (653.1 + 622) J

  • = 1275.1 J

  • No. of moles of HNO3 used = 1.0 M  25  10-3 dm3

  • = 0.025 mol


Energetics

Heat evolved per mole of H2O formed

=

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

Check Point 6-4

Back

  • No. of moles of NH3 used = 1.0 M  25  10-3 dm3

  • = 0.025 mol

  • No. of moles of H2O formed = 0.025 mol

= 51.004 kJ mol-1

The enthalpy change of neutralization of nitric acid and aqueous ammonia is –51.004 kJ mol-1.


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

Check Point 6-4

  • When 0.05 mol of silver nitrate was added to 50 g of water in a polystyrene cup, a temperature drop of 5.2 oC was recorded. Assuming that there was no heat absorption by the polystyrene cup, calculate the enthalpy change of solution of silver nitrate.

  • (Specific heat capacity of water = 4200 J kg-1 K-1)

Answer


Energetics

  • Energy absorbed = mcT

  • = 0.05 kg  4200 J kg-1 K-1  5.2 K

  • = 1092 J

  • No. of moles of AgNO3 used = 0.05 mol

  • Energy absorbed per mole of AgNO3 used =

  • = 21.84 kJ mol-1

  • The enthalpy change of solution of silver nitrate is +21.84 kJ mol-1.

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

Check Point 6-4

Back


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

Check Point 6-4

(c) A student used a calorimeter as shown in Fig. 6-15 to determine the enthalpy change of combustion of methanol. In the experiment, 1.60 g of methanol was used and 50 g of water was heated up, raising the temperature by 33.2 oC. Given that the specific heat capacities of water and copper calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1 K-1 respectively and the mass of the calorimeter is 400 g, calculate the enthalpy change of combustion of methanol.

Answer


Energetics

6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)

Check Point 6-4

  • Heat evolved = m1c1T + m2c2 T

  • = 50 g  4.18 J g-1 K-1  33.2 K + 400g  2.10 J g-1 K-1  33.2 K

  • = (6939 + 27888) J

  • = 34827 J

  • No. of moles of methanol used =

  • = 0.05 mol

  • Heat evolved per mole of methanol used =

  • = 696.5 kJ mol-1

  • The enthalpy change of combustion of methanol is –696.5 kJ mol-1.


Energetics

  • Given the following thermochemical equation:

  • 2H2(g) + O2(g)  2H2O(l)

  • (i)Is the reaction endothermic or exothermic?

  • (ii)What is the enthalpy change for the following reactions?

  • (1) 2H2O(l)  2H2(g) + O2(g)

  • (2) H2(g) + O2(g)  H2O(l)

  • (iii) If the enthalpy change for the reaction H2O(l)  H2O(g) is +41.1 kJ mol-1, calculate the H for 2H2(g) + O2(g)  2H2O(g).

6.5 Hess’s law (SB p.158)

Check Point 6-5

Answer


Energetics

  • (i) Exothermic

  • (ii) (1) +571.6 kJ mol-1

  • (2) –285.8 kJ mol-1

  • (iii)

  • H = [-571.6 + 2  (+41.1)] kJ mol-1 = -489.4 kJ mol-1

6.5 Hess’s law (SB p.158)

Check Point 6-5


Energetics

  • Given the following information about the enthalpy change of combustion of allotropes of carbon:

  • Hc [C(graphite)] = -393.5 kJ mol-1

  • Hc [C(diamond)] = -395.4 kJ mol-1

  • (i) Which allotrope of carbon is more stable?

  • (ii) What is the enthalpy change for the following process?

  • C(graphite)  C(diamond)

ø

ø

6.5 Hess’s law (SB p.158)

Check Point 6-5

Answer


Energetics

  • (i) Graphite

  • (ii)

  • H = [-393.5 – (-395.4)] kJ mol-1 = +1.9 kJ mol-1

6.5 Hess’s law (SB p.158)

Check Point 6-5


Energetics

  • The formation of ethyne (C2H2(g) can be represented by the following equation:

  • 2C(graphite) + H2(g)  C2H2(g)

  • (i) Draw an enthalpy level diagram relating the above equation to carbon dioxide and water.

  • (ii) Calculate the standard enthalpy change of formation of ethyne.

  • (Given: Hc [C(graphite)] = -393.5 kJ mol-1;

  • Hc [H2(g)] = -285.8 kJ mol-1;

  • Hc [C2H2(g)] = -1299 kJ mol-1)

ø

ø

ø

6.5 Hess’s law (SB p.158)

Check Point 6-5

Answer


Energetics

Given the following information, find the standard enthalpy change of the reaction:

C2H4(g) + H2(g)  C2H6(g)

Hf [C2H4(g)] = +52.3 kJ mol-1

Hf [C2H6(g)] = -84.6 kJ mol-1

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)

Example 6-6A

Answer


Energetics

Note:H1 = [Hf (reactants)] = Hf [C2H4(g)] + Hf [H2(g)]

H2 = [Hf (products)] = Hf [C2H6(g)]

Applying Hess’s law,

H1 + H = H2

H = H2 - H1

= Hf [C2H6(g)] – (Hf [C2H4(g)] + Hf [H2(g)])

= [-84.6 – (+52.3 + 0)] kJ mol-1 =-136.9 kJ mol-1

The standard enthalpy change of the reaction is –136.9 kJ mol-1.

ø

ø

ø

ø

ø

ø

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)

Example 6-6A

Back


Energetics

Given the following information, find the standard enthalpy change of the reaction:

6PbO(s) + O2(g)  2Pb3O4(s)

Hf [PbO(g)] = -220.0 kJ mol-1

Hf [Pb3O4(g)] = -737.5 kJ mol-1

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)

Example 6-6B

Answer


Energetics

Note:H1 = [Hf (reactants)] = 6  Hf [PbO(s)] + Hf [O2(g)]

H2 = [Hf (products)] = 2  Hf [Pb3O4(s)]

Applying Hess’s law,

H1 + H = H2

H = H2 - H1

= 2  Hf [Pb3O4(s)] – (6  Hf [PbO(s)] + Hf [O2(g)])

= [2  (-737.5) – 6  (-222.0) – 0] kJ mol-1 =-155.0 kJ mol-1

The standard enthalpy change of the reaction is –155.0 kJ mol-1.

ø

ø

ø

ø

ø

ø

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)

Example 6-6B

Back


Energetics

Given the following information, find the standard enthalpy change of the reaction:

Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)

Hf [Fe2O3(s)] = -822.0 kJ mol-1

Hf [CO(g)] = -110.5 kJ mol-1

Hf [CO2(g)] = -393.5 kJ mol-1

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)

Example 6-6C

Answer


Energetics

Note:H1 = [Hf (reactants)] = Hf [Fe2O3(s)] + 2  Hf [CO(g)]

H2 = [Hf (products)] = 2  Hf [Fe(s)] + 3  Hf [CO2(g)]

Applying Hess’s law,

H1 + H = H2

H = H2 - H1

= 2  Hf [Fe(s)] + 3  Hf [CO2(g)] - Hf [Fe2O3(s)] - 3  Hf [CO(g)]

= [2  (0) + 3  (-393.5) –(-822.0) – 3  (-110.5)] kJ mol-1

=-27.0 kJ mol-1

The standard enthalpy change of the reaction is –27.0 kJ mol-1.

ø

ø

ø

ø

ø

ø

ø

ø

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)

Example 6-6C

Back


Energetics

Given the following information, find the standard enthalpy change of the reaction:

4CH3 · NH · NH2(l) + 5N2O4(l)

 4CO2(g) + 12H2O(l) + 9N2(g)

Hf [CH3 · NH · NH2(l)] = +53 kJ mol-1

Hf [N2O4(l)] = -20 kJ mol-1

Hf [CO2(g)] = -393.5 kJ mol-1

Hf [H2O(l)] = -285.8 kJ mol-1

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.161)

Example 6-6D

Answer


Energetics

Note:H1 = [Hf (reactants)] = 4  Hf [CH3·NH ·NH2(l)] + 5  Hf [N2O4(l)]

H2 = [Hf (products)] = 4  Hf [CO2(g)] + 12  Hf [H2O(l)] + 9  Hf [N2(g)]

Applying Hess’s law,

H1 + H = H2

H = H2 - H1

= (4  Hf [CO2(g)] + 12  Hf [H2O(l)] + 9  Hf [N2(g)] – (3  Hf [CH3·NH ·NH2(l)] + 5 Hf [N2O4(l)]

= [4  (-393.5) + 12  (-285.8) + 9  (0) – 4  (+53) – 5  (-20)] kJ mol-1

=- 5115.6 kJ mol-1

The standard enthalpy change of the reaction is –5115.6 kJ mol-1.

ø

ø

ø

ø

ø

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.161)

Example 6-6D

Back


Energetics

Given the following information, find the standard enthalpy change of formation of methane gas.

C(graphite) + O2(g)  CO2(g)

Hc [C(graphite)] = -393.5 kJ mol-1

H2(g) + O2(g)  H2O(l) Hc [H2(g)] = -285.8 kJ mol-1

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

Hc [CH4(g)] = -20 kJ mol-1

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)

Example 6-6E

Answer


Energetics

Direct measurement of ΔHf [CH4(g)] is impossible because carbon(graphite) and hydrogen do not combine directly, and methane does not decompose directly to form carbon(graphite) and hydrogen. Since methane contain carbon and hydrogen only, they can be related to carbon dioxide and water by the combustion of methane and its constituent elements as shown in the diagram below.

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)

Example 6-6E


Energetics

Note: H1 = Hc [C(graphite)]

H2 = 2  Hc [H2(g)]

H3 = Hc [CH4(g)]

Applying Hess’s law,

Hf [CH4(g)] + H3 = H1 + H2

Hf [CH4(g)] = H1 + H2 - H3

= Hc [C(graphite)] + 2  Hc [H2(g)] - Hc [CH4(g)]

= [-393.5 + 2  (-285.8) –(-890.4)] kJ mol-1

= -74.7 kJ mol-1

The standard enthalpy change of formation of methane gas is –74.7 kJ mol-1.

ø

ø

ø

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)

Back

Example 6-6E


Energetics

Given the following information, find the standard enthalpy change of formation of methanol.

C(graphite) + O2(g)  CO2(g)

Hc [C(graphite)] = -393.5 kJ mol-1

H2(g) + O2(g)  H2O(l) Hc [H2(g)] = -285.8 kJ mol-1

C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)

Hc [C2H5OH(l)] = -1371 kJ mol-1

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)

Example 6-6F

Answer


Energetics

6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)

Example 6-6F


Energetics

ø

Note: H1 = 2  Hc [C(graphite)]

H2 = 3  Hc [H2(g)]

H3 = Hc [C2H5OH(l)]

Applying Hess’s law,

Hf [C2H5OH(l)] + H3 = H1 + H2

Hf [C2H5OH(l)] = H1 + H2 - H3

= 2  Hc [C(graphite)] + 3  Hc [H2(g)] - Hc [C2H5OH(l)]

= [2  (-393.5) + 3  (-285.8) –(-1371)] kJ mol-1

= -273.4 kJ mol-1

The standard enthalpy change of formation of ethanol is –273.4 kJ mol-1.

ø

ø

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)

Back

Example 6-6F


Energetics

(a) Find the standard enthalpy change of formation of butane gas (C4H10(g)).

Given: Hc [C(graphite)] = -393.5 kJ mol-1

Hc [H2(g)] = -285.8 kJ mol-1

Hc [C4H10(g)] = -2877 kJ mol-1

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)

Check Point 6-6

Answer


Energetics

Hf [C4H10(g)]

= Hc [C(graphite)]  4 + Hc [H2(g)]  5 - Hc [C4H10(g)]

= [(-393.5)  4 + (-285.8)  5 – (-2877)] kJ mol-1

= -126 kJ mol-1

ø

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)

Check Point 6-6


Energetics

(b) Find the standard enthalpy change of the reaction:

Br2(l) + C2H4(g)  C2H4Br2(l)

Given: Hf [C2H4(g)] = +52.3 kJ mol-1

Hf [C2H4Br2(l)] = -80.7 kJ mol-1

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)

Check Point 6-6

Answer


Energetics

H

= [Hf (products)] - [Hf (reactants)]

= [-80.7 – (+52.3) – 0)] kJ mol-1

= -133 kJ mol-1

ø

ø

ø

6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)

Back

Check Point 6-6


Energetics

(a) Increase

(b) Decrease

(c) Increase

(d) Decrease

(e) Decrease

6.7 Entropy change (SB p.167)

Back

Check Point 6-7

  • Predict whether the following changes or reactions involve an increase or a decrease in entropy.

  • Dissolving salt in water to form salt solution

  • Condensation of steam on a cold mirror

  • Complete combustion of carbon

  • Complete combustion of carbon monoxide

  • Oxidation of sulphur dioxide to sulphur trioxide

Answer


Energetics

6.8 Free energy change (SB p.170)

Back

Let's Think 1

In the process of changing of ice to water, at what temperature do you think G equals 0?

Answer

G equals 0 means that neither the forward nor the reverse process is spontaneous. The system is therefore in equilibrium. Melting point of ice is 0 oC (273 K) at which the process of changing ice to water and the process of water turning to ice are at equilibrium. At 0 oC, G of the processes equals 0.


Energetics

  • 100 oC

(b) Enthalpy and entropy

6.8 Free energy change (SB p.170)

Check Point 6-8

  • At what temperatures is the following process spontaneous at 1 atmosphere?

  • Water  Steam

  • What are the two driving forces that determine the spontaneity of a process?

Answer


Energetics

6.8 Free energy change (SB p.170)

Check Point 6-8

Back

  • State whether each of the following cases is spontaneous at all temperatures, not spontaneous at any temperature, spontaneous at high temperatures or spontaneous at low temperatures.

  • (i) positive S and positive H

  • (ii)positive S and negative H

  • (iii) negative S and positive H

  • (iv) negative S and negative H

  • Spontaneous at high temperatures

  • Spontaneous at all temperatures

  • Not spontaneous at any temperature

  • Spontaneous at low temperatures

Answer


Enthalpy change of formation of caco 3 s

Ca(s) + C(graphite) + O2

CaCO3(s)

H2

H1

CaO(s) + CO2(g)

ø

Hf [CaCO3(s)] = H1 + H2

= -1028.5 kJ mol-1 + (-178.0) kJ mol-1

= -1206.5 kJ mol-1

ø

Hf [CaCO3(s)]

6.5 Hess’s law (SB p.153)

Enthalpy change of formation of CaCO3(s)


Enthalpy change of hydration of mgso 4 s

ΔH

ø

MgSO4(s) + 7H2O(l)

MgSO4·7H2O(s)

aq

ΔH1

ΔH2

Mg2+(aq) + SO42-(aq) + 7H2O(l)

ø

ΔH = enthalpy change of hydration of MgSO4(s)

ΔH1 = molar enthalpy change of solution of anhydrous magnesium sulphate(VI)

ΔH2 = molar enthalpy change of solution of magnesium

sulphate(VI)-7-water

ΔH = ΔH1 - ΔH2

6.5 Hess’s law (SB p.153)

Enthalpy change of hydration of MgSO4(s)

aq

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