tutorial chemical energetics part i solutions

7/28/2019 Tutorial Chemical Energetics Part I Solutions

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Dunman High School

2010 Year 5 H2 CHEMISTRY

Tutorial: Chemical Energetics (Part I)

A Calorimetric Calculations

1 A flame calorimeter was used to find the standard enthalpy change of combustion ofpropan1ol. The following information was obtained during the experiment:

Heat capacity of apparatus = 3.01 kJ K1

Mass of propan1ol burned = 2.40 g

Initial temperature = 27.3C

Maximum temperature = 52.3C

(a) Explain, with the aid of a chemical equation with state symbols, what is meant by thestandard enthalpy change of combustion of propan1ol.

The standard enthalpy change of combustion of CH3CH2CH2OH isthe energy change when one mole of CH3CH2CH2OH iscompletely burnt in oxygen understandard conditions.

CH3CH2CH2OH(l) +2

9O2(g) 3CO2(g) + 4H2O(l)

Steps to calculate H

Step 1: Write the balanced equation (where possible)

Step 2: Calculate apparent amount of heat absorbed/ released by

reaction, Q using Q = mcTStep 3: Calculate actual amount of heat absorbed/ released by reaction,

Q, using Q (no heat loss to surrounding)

or Q = 100

x

Q (x% efficiency)

Step 4: Calculate n where n = no. of moles of water (neutralisation)n = no. of moles of substance (combustion)

Step 5: Calculate enthalpy change using Hrxn = + nQ

where () for exothermic reaction and (+) for endothermic reaction

C

(not c)

Temp H < 0


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(b) Calculate the amount of heat evolved during the combustion. [75250 J]

Amount of heat evolved,Q = CT = Q (assume 100% efficiency)

= 3.01103 (52.3 27.3)= 75250 J

(c) Calculate the standard enthalpy change of combustion of propan1ol.[Specific heat capacity of water is 4.18 J g1 K] [1881250 J mol1]

No. of mol. of CH3CH2CH2OH burnt, n =M

m=

0.60

40.2= 0.04000

Hc

(CH3CH2CH2OH) = OHCHCHCHn

Q

223= 0400.0

75250

= 1881250 J mol1

(d)Why would inadequate supply of air lead to error in the above results?

Inadequate supply of air leads to incomplete combustion ofpropan1ol. This leads to a decrease in the amount of heat

evolved and hence inaccuracy in the Hc

(CH3CH2CH2OH).

Since specific heatcapacity, c notmentioned, use heatcapacity, C

Refers to actual amount of heat Q

Hc(CH3CH2CH2OH)


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2 The standard enthalpy change of combustion of propane is 2200 kJmol1. A 2.40 dm3 sample of propane was burned under a container ofwater originally under r.t.p. conditions.Given that the process was 75% efficient, calculate the mass of water in

the container that could be heated to 80C. [718 g]

No. of mol. of CH3CH2CH3 burnt, n =mV

V=

0.24

40.2= 0.100

Actual amount of heat evolved, Q = Hc(CH3CH2CH3) n

= (2200103) 0.100= 220000 J

Since process is 75 % efficient,

Apparent amount of heat absorbed by water, Q

=100

75Q (Actual amount of heat evolved by burning propane)

Apparent amount of heat absorbed by the water, Q

=10075 Q =

10075 220000 = 165000 J

Since Q = mcT

165000 = m 4.18 (80.0 25.0)

Mass of water heated, m =0.5518.4

165000

= 718 g

Hc(C3H8) = Q / no of moles C3H8

Obtain from Data Booklet


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3 The chemical reaction between 30 cm3

of 1.0 mol dm3

sodium hydroxidesolution and 20 cm

3of 1.0 mol dm

3sulphuric acid solution in a plastic

cup gave rise to a temperature rise of 7.3C due to neutralisation. Theprocess efficiency was expected to be 90%.

Explain, with the aid of a chemical equation with state symbols, what ismeant by the standard enthalpy change of neutralisation.

H+(aq) + OH(aq) H2O(l)The standard enthalpy change of neutralisation is the energyevolved when one mole of water is formed from the neutralisationbetween acid and alkali understandard conditions.

Calculate the standard enthalpy change of the neutralisation reaction.

[Assume specific heat capacity of the solution to be 4.18 J g1 K1 and itsdensity to be 1 g cm

3] [56507 J mol

1]

H2SO4(aq) + NaOH(aq) Na2SO4(aq) + H2O(l)

Apparent amount of heat evolved, Q= mcT = (30 + 20) 4.18 7.3 = 1525.7 J

Q = 100

90

Q (given 90% efficiency)

Actual amount of heat evolved, Q =90

100 1525.7 = 1695.2 J

No. of mole of H2SO4 = mol0200.011000

20

No. of mole of NaOH = mol0300.011000

30

NaOH is the limiting reagent.

Hence no. of mole of H2O formed, n = 0.0300

Hn =

OHn

Q

2

= 0300.0

2.1695

= 56507 J mol1


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B Calculations Using Bond Energy

4 Suppose that ethene and steam react to give carbon monoxide and hydrogen with anenthalpy change of 254 kJ mol1 as given.

C2H4(g) + 2H2O(g) 2CO(g) + 4H2(g)

(a) Explain, with the aid of a chemical equation with state symbols, what is meant by the

bond energy of CO.

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

The bond energy of CO is the average energy absorbed when one

mole of CO bonds is broken in the gaseous state to form C andO atoms.

(b) With the aid of appropriate bond energy data from the Data Booklet, calculate the

bond energy of CO bond in carbon monoxide. [+1046 kJ mol1]

reactants products

C2H4(g) + 2H2O(g) 2CO(g) + 4H2(g)

2 2 4

4 (CH) 2(2)(OH)

Hrxn = [BE(C=C)+4BE(CH)+4BE(OH)] [2BE(CO)+4BE(HH)]

BE(CO) =21 {[610+4(410)+4(460)]2544(436)} = + 1046 kJ mol1

Definition:Bond energy of XY bond is the average energy absorbedwhen one mole ofXY bonds is broken in the gaseous state toform X and Y atoms. i.e XY (g) X (g) + Y (g)

Application: A measure of bond strength Used to calculate enthalpies of reactions

Formula

Hrxn = BE of bonds broken in rxts BE of bonds formed in pdts

Energy is absorbed in bond breaking in reactants: endothermic (+)

Energy is released in bond formation in products: exothermic ()

C C

H

H H

HC O

H

O

HH H


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5 Under mild conditions, the catalytic hydrogenation of ethyne, HCCHyields ethane. Under extreme conditions, the catalytic hydrogenation of

nitrogen NN, yields ammonia.

(i) Write equations for the two hydrogenation processes described above.

C2H2 (g) + 2H2(g) C2H6(g)

N2(g) + 3H2(g) 2NH3(g)

(ii) Explain, with the aid of appropriate bond energy values from the DataBooklet, the difference in the behaviour of ethyne and nitrogen.

[Hrxn (HCCH): 278 kJ mol1; Hrxn(NN): 38 kJ mol1]

HCCH + 2HH

Hrxn =[BE(CC)+ 2BE(CH) +2BE(HH)] [BE(CC) + 6BE(CH)]= 278 kJ mol

1

NN + 3HH 2

Hrxn = [BE(NN) + 3BE(HH)] 2[3BE(NH)] = 38 kJ mol1

Hydrogenation of ethyne is more exothermic than that of nitrogen

Hydrogenation of ethyne require mild conditions while that fornitrogen require extreme conditions

Alternative: (kinetics stability of the reactants)

Quote NN bond energy value and CC bond energy value

The NN bond energy value is much higher than that of CC

bond energy. Hence the NN bond is extremely strong hence

large amount of energy required to break the bond in thereactants hence highactivation energy required

C

H

H C H

H

HH

NH

HH


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6 Some bond energy values are given in the table below:

Bond Bond energy / kJmol

1

Bond Bond energy / kJmol

1

HH 436 HH 436

PP 208 ClCl 244PH 322 HCl 431

The PH bond energy is the average of the HH and PP values. Explainwhy the HCl bond energy is not the mean of HH and ClCl values.

Electronegativity of P H

HH, PP and PH bonds are nonpolar BE (PH) is the average of BE (HH) and BE (PP)

Electronegativity of Cl > H HH, ClCl are nonpolar while HCl bond is polar due to

difference in electronegativity BE (HCl) is not the average of BE (HH) and BE (ClCl)


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C Other Calculations Methods ofHrxn

7 The standard enthalpy change of the following reaction is 156.1 kJ mol1.MoO2(s) + O2(g) MoO3(s)

(a) The standard enthalpy change of formation of MoO3 is 745.1 kJ mol1

.Explain, with the aid of a chemical equation with state symbols, what ismeant by this statement.

Mo(s) + 3/2 O2(g) MoO3(s)The standard enthalpy change of formation of MoO3 is the energy

change when one mole of MoO3 is formed from its elements i.e.Mo and O2 understandard conditions.

2 Methods

(1) Using enthalpy change of formation, Hf,methodFormula: Hrxn = Hf(pdts) Hf(rxts) Note that Hfofelement (e.g H2) is always zero Use only ifHf(pdts) and Hf(rxts) are provided

(2) Using Hess Law Energy Cycle methodDefinitionHess Law: The enthalpy change accompanying a chemical reactionis the same regardless of the route by which the chemical change

occurs, provided the initial and final conditions are the same.

By Hess Law, H1 =H2 + H3 + H4

Note:

All Hmust have the correct signs Multiply a reaction by a factor multiply Hby the same factor Reverse a reaction Reverse the sign ofH

H1

H3

H2 H4

A

C D

B


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(b) Calculate the standard enthalpy change of formation of MoO2.[589 kJ mol

1]

(Use HfMethod to solve)

MoO2(s) + O2(g) MoO3(s) Hrxn = 156.1 kJ mol1

Hrxn

= Hf

(pdts) Hf

(rxts)

Hrxn = Hf

(MoO3) [Hf(MoO2) +

2

1Hf

(O2)]

Hf(MoO2)= 745.1

2

1(0) (156.1) = 589 kJ mol1

8 The standard enthalpy change of combustion of cyclohexane is3924 kJ mol1. Given that the standard enthalpy changes offormation of water and carbon dioxide are 286 kJ mol

1and

394 kJ mol1

respectively, calculate the standard enthalpychange of formation of cyclohexane. [156 kJ mol1]

(Can use HfMethod to solve Q8)

Formula Method

C6H

12(l) + 9O

2(g) 6CO

2(g) + 6H

2O(l) H

c

(C6H

12)=3924 kJ mol1

Hc(C6H12)=[6Hf

(CO2)+6Hf (H2O)] [Hf

(C6H12)+9Hf (O2)]

Hf

(C6H12)= [6(394) + 6(286)] 9(0)(3924) = 156 kJ mol1

9 The standard enthalpy change of dimerisation of ethene to form but1ene, as follows, is 10 kJ mol

1:

2CH2=CH

2(g) CH

2=CHCH

2CH

3(g)

Calculate the standard enthalpy change of combustion of ethene, given thatthe standard enthalpy change of combustion of gaseous but1ene is 20 kJmol

1. [15 kJ mol

1]

(Can use Hess Law Energy Cycle Method to solve)

Given:

2C2H4(g) C4H8(g) Hrxn

C4H8(g) + 6O2(g)

4CO2(g) + 4H2O(l)

Hc

(C4H8)


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Hess Law Energy Cycle

Hc

(C2H4) = ?Aim: C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)

Hrxn

Hc(C4H8)

2

1C4H8(g) + 3O2(g)

By Hess law:

Hc

(C2H4) =2

1Hrxn

+2

1Hc

(C4H8) =

2

1[10 + (20)]= 15 kJ mol

1

10 The standard enthalpy changes of combustion of hydrogen, cyclohexaneand benzene are 286 kJ mol1, 3920 kJ mol1 and 3268 kJ mol1respectively.

(i) Write a chemical equation to represent the hydrogenation of benzene toform cyclohexane;

C6H6(l) + 3H2(g) C6H12(l)

(ii) Calculate the standard enthalpy change for this reaction using Hess law.[206 kJ mol1]

(Must use Hess Law Energy Cycle Method!)

Given:

H2(g) +2

1O2(g) H2O(l) Hc(H2)

C6H12(l) + 9O2(g) 6CO2(g) + 6H2O(l) Hc(C6H12) C6H6(l) +

2

15O2(g) 6CO2(g) + 3H2O(l) Hc(C6H6)


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Hrxn = ?

Aim: C6H6(l) + 3H2(g) C6H12(l)

Hc(C6H6) 3Hc

(H2) Hc(C6H12)

+2

15O2(g) +2

3 O2(g) + 9O2(g)

6CO2(g) + 6H2O(l)

By Hess law:

Hrxn = Hc

(C6H6)+ 3Hc

(H2) Hc(C6H12)

= 3268 + 3(286) (3920) = 206 kJ mol1

11 The standard enthalpy change of atomisation of carbon and hydrogenare +715 kJ mol1 and +218 kJ mol1 respectively. Given that thestandard enthalpy change of formation of methane is 75 kJ mol

1,

calculate:

(a) the standard enthalpy of atomisation of methane;[+1662 kJ mol

1]

(Can use Hess Law Energy Cycle or Algebraic Method)

Given:

C(s) C(g) Hat (C)

2

1H2(g) H(g) Hat

(H2)

C(s) + 2H2(g) CH4(g) Hf (CH4)

Hat(CH4) = ?

Aim: CH4(g) C(g) + 4H(g)

Hf (CH4) Hat

(C) 4Hat (H2)

C(s) + 2H2(g)

By Hess law:

Hat(CH4) = Hf

(CH4)+ Hat

(C) + 4Hat(H2)

= (75) + 715 + 4(218)= 1662 = +1660 kJ mol1 (3 s.f)


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(b) the CH bond energy. [+415.5 kJ mol1]

Note:

Hat

(CH4) involves breaking 4 mole of CH bond in 1 mol. of CH4to form C and H atoms in the gaseous phase.

Hat(CH4) = 4BE(CH)

BE(CH) =4

1(1662) = + 415.5 = + 416 kJ mol1


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E BornHaber Cycle

Important points to remember when drawing Born Haber Cycle:

Use forendo rxn and forexo rxn Follow the sequence: (label the arrows)

I.EE.A(can be exo rxn or endo rxn)

Hat Hlatt

Hf

IfHat(X2) value is not given, refer to Data Booklet to obtainvalue of BE(XX)

)A(H2)A(BE 2at2

12 The lattice energy of lithium chloride can be calculated from a BornHaber cycle using the following standard enthalpy data:

Enthalpy change of formation of lithium chloride 409 kJ mol1

Enthalpy change of atomisation of lithium 159 kJ mol1

Enthalpy change of atomisation of chlorine 121 kJ mol

1

Electron affinity of chlorine 364 kJ mol1

(a) Construct a BornHaber cycle for lithium chloride and use the cycle, with

the aid of appropriate ionisation data from the Data Booklet, to determinethe lattice energy of lithium chloride. [844 kJ mol

1]

Start:Elements instandard state


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Given:

Hf(LiCl) Li(s) +

2

1Cl2(g) LiCl(s) 409 kJ mol1

Hat(Li) Li(s) Li(g) 159 kJ mol

Hat(Cl2)2

1Cl2(g) Cl(g) 121 kJ mol

1st

IE(Li)

Li(g) Li+(g) + e 519 kJ mol

EA(Cl) Cl(g) + e Cl(g) 364 kJ mol

Hlatt(LiCl) Li

+(g) + Cl

(g) LiCl(s) ?

BornHaber cycle to determine Hlatt(LiCl):

Li+(g) + Cl(g) + e

1st IE(Li)

Li(g) + Cl(g)

EA(Cl) Li+(g) + Cl(g)

Hat(Cl2 )

Li(g) +21Cl2(g)

Hat

(Li)Li(s) +

2

1Cl2(g)

Hlatt(LiCl)

Hf

(LiCl)

LiCl(s)

By Hess law:Hf

(LiCl) = Hat

(Li) + Hat

(Cl2) + 1

stIE(Li) + EA(Cl) + Hlatt

(LiCl)

Hlatt(LiCl)= 409 [159 + 121 + 519 + (364)

= 844 kJ mol1

From DataBooklet


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(b) Explain how you would expect the numerical magnitude of the latticeenergy of sodium chloride to compare with that of lithium chloride.

Hlatt

rr

qq

Ionic radius, r+: Na+

> Li+

Magnitude ofHlatt

: NaCl < LiCl

13 Suppose calcium can form two ionic compounds with chlorine: calcium(I)chloride (CaCl) and calcium(II) chloride (CaCl2).

The following information about some standard enthalpy changes is provided:

Lattice energy of calcium(I) chloride 700 kJ mol1

Lattice energy of calcium(II) chloride 2197 kJ mol1Enthalpy change of atomisation of calcium 178 kJ mol1Electron affinity of chlorine 362 kJ mol1

(a) Draw a BornHaber cycle in each case to determine the standardenthalpy change of formation of (i) calcium(I) chloride; (ii) calcium(II)chloride. [759 kJ mol

1]

Given:(i) For CaCl;

Hf(CaCl) Ca(s) + Cl2(g) CaCl(s) ?

Hat(Ca) Ca(s) Ca(g) 178 kJ mol

Hat(Cl2) 1/2Cl2(g) Cl(g) Hat

(Cl) = BE(ClCl)

= (244)

1st IE(Ca) Ca(g) Ca+(g) + e 590 kJ mol1

EA(Cl) Cl(g) + e Cl(g) 362 kJ mol Hlatt

(CaCl) Ca+(g) + Cl(g) CaCl(s) 700 kJ mol

Note:

Smallerionic radius

Oppositely charged ions are closer

Electrostatic force ofattraction becomes strongerHence the largerthe magnitude of lattice energy

(Stability of ionic compound is greater)


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BornHaber cycles to determine (i) Hf(CaCl)

Ca+(g) + Cl(g) + e

1st IE(Ca)Ca(g) + Cl(g)

EA(Cl)Ca+(g) + Cl(g)

Hat(Cl2 )

Ca(g) +2

1Cl2(g)Hlatt

(CaCl)

Hat(Ca)

Ca(s) +2

1Cl2(g)

Hf(CaCl)

CaCl(s)

By Hess law:Hf

(CaCl)=Hat(Ca)+Hat

(Cl2)+1st IE(Ca) + EA(Cl) + Hlatt

(CaCl)

= 178 +2

1 (244) + 590 + (362) + (700)

= 172 kJ mol1

(ii) Hf

(CaCl2)

Hf(CaCl) Ca(s) + Cl2(g) CaCl2(s) ?

Hat(Ca) Ca(s) Ca(g) 178 kJ mol

Hat(Cl2) 1/2Cl2(g) Cl(g) Hat

(Cl) = BE(ClCl)= 244

1st IE(Ca) Ca(g) Ca+(g) + e 590 kJ mol

2n IE(Ca) Ca+(g) Ca+(g) + e 1150 kJ mol

EA(Cl) Cl(g) + e Cl

(g) 362 kJ mol1

Hlatt

(CaCl2) Ca+(g) + 2Cl

(g) CaCl2(s) 2197 kJ mol


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BornHaber cycles to determine (ii) Hf(CaCl2)

Ca2+(g) + 2Cl(g) + 2e

2nd IE(Ca) Ca

+(g) + 2Cl(g) + e

2EA(Cl)Ca

2+(g)+2Cl

(g)

1st IE(Ca) Ca(g) + 2Cl(g)

Hlatt(CaCl2)

2Hat(Cl2 )

Ca(g) + Cl2(g)

Hat(Ca)

Ca(s) + Cl2(g)

Hf(CaCl2)

CaCl2(s)

By Hess law:Hf

(CaCl2)= Hat

(Ca)+ 2Hat

(Cl2)+ 1

stIE(Ca)+2

ndIE(Ca)+

2EA(Cl)+Hlatt(CaCl)

= 178 + 244 + 590 + 1150 + 2(362) + (2197)=759 kJ mol1

(b) Comment on the relative numerical magnitudes of the standard enthalpychange of formation of calcium(I) chloride and calcium(II) chloride.

Formation of CaCl is less exothermic than that of CaCl2.Since magnitude ofHf

(CaCl) is less than that ofHf(CaCl2),

Hence CaCl is energetically less stable than CaCl2.


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E Energetics of Aqueous Solution

Definitions:

Hsol = the energy change when one mole of the substance is

completely dissolved in a solvent to form an infinitely dilute

solution under standard conditions.

Hhyd = the energy evolved when one mole of the gaseous ion

is hydrated under standard conditions.

Hlatt = the energy evolved when one mole of the solid ionic

compound is formed from its constituent gaseous ions understandard conditions.

Memorise how to draw the energy cycle for calculation of

Hsol of an ionic compound

Hsol = Hhyd HlattMemorise how to draw the energy level diagram for calculation

ofHsol of an ionic compoundExothermic dissolution Endothermic dissolution

From lecture notes

From lecture notes


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14 The lattice energy of lithium chloride is 848 kJ mol1

and that ofsodium chloride is 776 kJ mol

1. Some enthalpy changes of hydration

are listed below:

Ions Hhyd /kJ mol1

Li+ 499

Na

+

390Cl

381

(a) Draw two energy cycles using data in this question to determine theenthalpy change of solution for these two salts. [36 kJ mol1]

Energy cycle to determine Hsoln(LiCl):

LiCl(s) Li+(aq) + Cl

(aq)

Hlatt(LiCl) Hhyd

(Li+) Hhyd(Cl)

Li+(g) + Cl

(g)

By Hess law:Hsoln

(LiCl) = Hlatt(LiCl)+ Hhyd

(Li+) + Hhyd(Cl)

= (845) + (499) + (381)

= 32 kJ mol1

Energy cycle to determine Hsoln(NaCl):

NaCl(s) Na+(aq) + Cl(aq)

Hlatt(NaCl)

Hhyd

(Na

+) Hhyd

(Cl

)

Na+(g) + Cl(g)

By Hess Law:Hsoln

(NaCl) = Hlatt

(NaCl)

+ Hhyd

(Na

+) + Hhyd

(Cl

)

= + 5 kJ mol1


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(b) Comment on the relative solubilities of sodium chloride and lithiumchloride.

The dissolution process for LiCl is exothermic compared to thedissolution process for NaCl which is endothermic.

Hence, LiCl is more soluble in water than NaCl

(c) State the significance of the relative numerical magnitude of the latticeenergy of sodium chloride compared to lithium chloride

Hlatt

rr

qq

Ionic radius, r+: Na+ > Li+

Magnitude ofHlatt: NaCl < LiCl

LiCl is more stable than NaCl since Hlatt LiCl is more

exothermic.

(d) Explain how you would expect the numerical magnitude of standardenthalpy change of hydration of bromide ion to compare with that ofchloride ion.

Hhyd (anion)

rq

Ionic radius of Cl < Br

Charge density of Cl > Br

Strength of interaction with water molecules : Cl > Br

I H hyd I : Cl

> Br

tutorial chemical energetics part i solutions

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