Thermodynamics [02] Section [23 - 53]

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LOCUSLOCUSLOCUSLOCUSLOCUS 24

Chemistry/Thermodynamics

According to Hesss law, the enthalpy change for the overall equation (which is the equation we want) equals the

sum of the enthalpy changes for the two steps. So, if we can find the enthalpy changes for the two steps; their sum

would give us the required enthalpy.

We can determine the enthalpy change for the first step by simply burning graphite in an excess of oxygen. Theresult (found experimentally) is 393.5H kJ = per mole of 2CO formed. For 2 moles of 2 ,CO we multiply

by 2.

( ) ( ) ( ) ( )2 22 2 2 ; 393.5 2.C graphite O g CO g H kJ + = The second step, decomposition ofCO

2, is not an easy experiment. However, the reverse of this decomposition is

simply the combustion ofCO. We could determine the H for that combustion by burning CO in an excess of

oxygen. (In fact, the reaction is similar to the one for the combustion of graphite to2.CO )

2 22 ( ) ( ) 2 ( ); 566.0CO g O g CO g H kJ + = From the properties of thermochemical equations we know that the enthalpy change for the reverse reaction is

simply (1) times the original reaction.

2 22 ( ) 2 ( ) ( ); ( 566.0 ) ( 1)CO g CO g O g H kJ + = If we now add these two steps and all their enthalpy changes, we obtain the chemical equation and the enthalpy

change for the combustion of ,CO which is what we wanted

2 2

2 2

2

2 (graphite) 2 ( ) 2 ( )

2 ( ) 2 ( ) ( )

2 (graphite) ( ) 2 ( )

C O g CO g

CO g CO g O g

C O g CO g

+

+

+

1

2

3

( 393.5 ) (2)

( 566.0 ) ( 1)

221.0

H kJ

H kJ

H kJ

=

=

=

So, we see that the combustion of 2 moles of graphite to give 2 moles ofCO has an enthalpy change of 221.0

kJ. The figure below shows the enthalpy diagram showing the relationship among the enthalpy changes for thiscalculation:

2 ( ) +2 ( )

mol C graphitemol O g 2

H3 = 221.0 kJ

2 mol CO g mol O( ) + 1 (g)2

H kJ2 = + 566.0

2 mol CO (g)2

H

1

=

787.0

kJ

Enthalpy

kJ

(

)

fig.10: Enthalpy diagram illustrating Hess's law. The diagramshows the equality of the enthalpy change for the completecombustion of graphite to the sum of the enthalpy changes for thecombustion of graphite to CO and the combustion of CO to CO .2

The above example illustrated how we can use Hesss law to obtain the enthalpy change for a reaction that is

difficult to determine by direct experiment.


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Hesss law is more generally useful, however, in that it allows us to calculate the enthalpy change for one reaction

from the values for others, whatever their source. (We will illustrate this in the examples that follow.)

Important points

1. Hesss law is a consequence of the fact that enthalpy is a state function.

2. We can calculate H for any process, as long as we find a route for which H is known for each step.

This important fact permits us to use a relatively small number of experimental measurements to calculate

H for a vast number of different reactions( we will illustrate this as we move ahead in this chapter.)

3. Will the final value ofH for a reaction depend on the way in which we break it down to use Hesss law?

BecauseHis a state function, we will always get the same value ofH for an overall reaction, regardless

of how many steps we employ to get to the final products.

4. It should be noted that in applying Hesss law, the individual steps need not be realisable in practice. They

may be hypothetical reactions and the only requirement is that their chemical equations should balance.

Suppose we are given the following data :

2 2( ) ( ) ( ); 297S s O g SO g H kJ + = ...(1)

3 2 22 ( ) 2 ( ) ( ); 198SO g SO g O g H kJ + = + ...(2)

How could we use these data to obtain the enthalpy change for the following equation ?

2 32 ( ) 3 ( ) 2 ( )S s O g SO g + ...(3)

Critical thinking

We need to multiply equations (1) and(2) by factors (perhaps reversing one or both equations) so that

when we add them together we obtain equation (3). We can usually guess what we need to do to the first

two equations to obtain the third one.

Note that Equation (3) has a coefficient of 2for S(s). This suggests that we should multiply equation (1) by

2 (and multiply the H by 2). Note also that 3( )SO g in equation (3) is on the RHS. This suggests that we

should reverse equation (2) (and multiply the H by -1)

Solution: Following the process described above, the whole problem boils down to this :

2 22 ( ) 2 ( ) 2 (S s O g SO g +

2

)

2 ( )SO g 2 3

2 3

( ) 2 ( )

2 ( ) 3 ( ) 2 ( )

O g SO g

S s O g SO g

+

+

( 297 ) (2)

(198 ) ( 1)

792

H kJ

H kJ

H kJ

=

=

=

The next example gives another illustration of how Hesss law can be used to calculate the enthalpy

change for a reaction from the enthalpy values for other reactions. In this case, the problem involves

three equations from which we obtain a fourth. Although this problem is somewhat complicated than

the one we just did, the basic procedure is the same : Compare the coefficients in the equations. See

by what factors we need to multiply the equations whose H s we know to obtain the equation we

want.

Example 6


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What is the enthalpy of reaction, ,H for the formation of tungsten carbide, WC, from the elements ?

( ) (graphite) ( ).W s C WC s+

(The enthalpy change for this reaction is difficult to measure directly, because the reaction occurs at 1400C.

However, the heats of combustion of the elements and of Tungsten Carbide can be measured easily and is

given by :)

(1) 2 32 ( ) 3 ( ) 2 ( ); 1680.6W s O g WO s H kJ + =

(2)2 2

(graphite) ( ) ( ); 393.5C O g CO g H kJ + =

(3) 2 3 22 ( ) 5 ( ) 2 ( ) 2 ( ); 2391.6WC s O g WO s CO g H kJ + + =

Critical thinking

We need to multiply equation (1), (2) and(3) by factors so that when we add the three equations we obtain

the desired equation for the formation of WC(s). To obtain these factors compare equations (1), (2) and(3)

in turn with the desired equation. For instance, note that Equation (1) has 2W(s) on the left side, whereas

the desired equation has W(s). Therefore, we multiply equation (1) (and its H) by1 .

2

Solution: Multiplying Equation (1) by 1 ,2 we obtain

2 3

3 1( ) ( ) ( ); ( 1680.6 )22

W s O g WO s H kJ + = = 840.3 kJ

Compare Equation (2) with the desired equation. Both have C(graphite) on the left side; therefore, we

leave equation (2) as it is. Now, compare equation (3) with the desired equation. Equation (3) hasWC(s) on the left side, whereas the desired equation has WC(s) on the right side. Hence we reverse

equation (3) and multiply it (and its H ) by 1 .2

3 2 2

5 1( ) ( ) ( ) ( ); ( 2391.6 )22

WO s CO g WC s O g H kJ + + = 1195.8kJ=

Example 7


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Note that the H is obtained by multiplying the value for equation (3) by 1 .2 Now these threeequations and the corresponding H s are added together.

2

3( ) ( )

2W s O g + 3 ( )WO s

2

; 840.3

(graphite) ( )

H kJ

C O g

=

+ 2 ( )CO g

3

; 393.5

( )

H kJ

WO s

=

2 ( )CO g+ 25

( ) ( )2

WC s O s + ; 1195.8

( ) (graphite) ( ) 38.0

H kJ

W s C WC s H kJ

+ =

+ =

To give you more practice for mastering how to manipulate thermochemical equations in applying Hesss

law we include one more example with only its critical thinking. You are requested to provide the solution

yourself. (The answer is given.)

Use thermochemical equations given below to determine H (at 25C) for the following reaction :

2 4(graphite) 2 ( ) ( )C H g CH g +

Given:

2 2(graphite) ( ) ( ) ; 395.3C O g CO g H kJ + = ...(1)

2 2 2

1( ) ( ) ( ) ; 285.8

2H g O g H O l H kJ+ = ...(2)

4 2 2 2( ) 2 ( ) ( ) 2 ( ) ; 890.3CH g O g CO g H O l H kJ + + = ...(3)

Example 8


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Critical thinking

1. We want one mole of C (graphite) as reactant, so we write down equation (1) as it is.

2. We want two moles of 2 ( )H g as reactants, so we multiply equation (2) by 2.

3. We want one mole of4 ( )CH g as product, so we reverse equation (3).

4. We do the same operations on each H value.

5. Then we add these equations term by term. The result is the desired thermochemical equation; all

unwanted substances cancel. The sum of the H values is theH for the desired reaction.

Answer:2 4

(graphite)+2 ( ) ( ) 74.8C H g CH g H kJ =

(You are requested to provide the solution yourself)

STANDARD STATE

Before we delve into standard enthalpies of formation, we will have to study and understand clearly, so as to what

does the standard state of a substance means and why is it needed. We have seen earlier in this chapter that the

magnitude of enthalpy change of any reaction depends upon the state (gas, liquid or solid) of reactants and products.

However, this is not the only factor on which enthalpy change depends. It depends also upon the conditions of

temperature and pressure of the reactants and products. So, we can make out from this that depending upon

variousvarying conditions we will get different enthalpy changes for the same reaction.

To understand so as to what problem this poses in our study of reactions, consider two arbitrary reactions.

Suppose H for the first reaction at 2 bar, 30C is 150kJ and that for the second reaction at 5 bar , 200C

is 250 kJ. Then can we say that the second reaction is more exothermic than the first one ? The answer is NO.

We can say that a certain reaction is more ( or less)exothermic (or endothermic) than another reaction only when

we are comparing theirH's under same conditions of temperature and pressure.

So, this means that, in order to compare the enthalpy change of different reactions we need to have a certain

standard condition of temperature and pressure.

The term standard state refers to the standard thermodynamic conditions chosen for substances whenlisting or comparing thermodynamic data : 51 (1 10 )bar Pa and a specified temperature ( if the

temperature is not specified we assume it to be 25C.)

So, thisshouldmean that the physical state of a substance at a pressure of 1 bar and a specified temperature is the

standard state of that substance at that particular temperature. But we still have an anomaly to deal with. Many

substances exist in the same physical state (gas, liquid or solid) in two or more distinct forms. For example, oxygen

in any of the physical states occurs both as dioxygen (commonly called simply oxygen.) with2O molecules, and as

ozone, with 3O molecules. So, at a specified temperature and pressure of 1 bar, which one of the two should be

considered as the standard state of oxygen ? We resolve this anomaly as follows :


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Oxygen with its two different forms under same conditions is said to exist in different allotropic forms. Anallotrope

is one of two or more distinct forms of an element in the same physical state.

Further, we define reference form of an element as the stablest form (physical state and allotrope) of the

element under standard thermodynamic conditions.

With these two terms defined we have all that we need to define standard state without any ambiguity. If at a

particular temperature and pressure of 1 bar, a substance exists in two or more allotropic forms, we choose the

stablest (i.e. reference form) of them as the standard state of that substance at that particular temperature.

So, now, we can say that the standard state of a substance is the physical state of the substance (at 1 bar

and a specified temperature) in its reference form. Reference form of some substances (at 25C) are mentioned

here :

Substance Reference form

Hydrogen H2, gas

Oxygen O2, gas

Carbon C, graphite

Sulphur S, rhombic.

Important Points

1. It is a general misconception to confuse standard thermodynamic conditions with STP for gases. They are

not identical.

2. Another widespread misconception is regarding the temperature in standard thermodynamic conditions.

Please read the definition and note that although the standard pressure has been fixed at 1 bar, no fixed

temperature is mentioned. So, substances can have various standard states at different temperatures. For

example, standard state of2H O at 5C is solid; at 5C it is liquid and at 105C it is gas.

3. However, if no temperature is mentioned, we assume it to be 25C.

4. The standard enthalpy change, reactionH (abbreviated generally as rH ), for a reaction,

reactantsproducts

refers to the H when the specified no. of moles of reactants, all at standard states, are convertedcompletely to the specified number of moles of products, all at standard states. We allow the reaction to

take place, with changes in temperature or pressure if necessary; when the reaction is complete, we return

the products to the same conditions of temperature and pressure that we started with, keeping track of

energy or enthalpy changes as we do so. When we describe a process as taking place at constant Tand

P, we mean that the initial and final conditions are the same. Because we are dealing with state functions,

the net change is the same as the change we would have obtained hypothetically with TandPheld

constant.


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STANDARD ENTHALPIES OF FORMATION

If a compound is formed from its constituent elements, then the enthalpy change for the reaction is calledenthalpy

of formation, . fH For example,

2 2 1(diamond) ( ) ( ); 395.4C O g CO g H kJ + =

1H is the enthalpy of formation of 2CO here as it is being formed from the elements that constitute it. But the

following reaction gives enthalpy of formation of2CO as well :

2 2 2(graphite) ( ) ( ); 393.5C O g CO g H kJ + =

Under same thermodynamic conditions we can see that1 2.H H Besides for a formation reaction ( i.e. a

reaction where a substance is formed from its constituent elements), the enthalpy of formation will vary with

changing number moles of the reactants.

To resolve these anomalies, we define standard enthalpy of formation : The standard enthalpy of formation

(also called the standard heat of formation) of a substance is the enthalpy change for the formation ofone mole

of the substance in a specified state from its elements in their reference form and in their standard states. It is

denoted by .fH

To understand this definition, consider the standard enthalpy of formation of liquid water. Note that the stablest

forms of hydrogen and oxygen at 1 atm and 25C are2 ( )H g and 2 ( ),O g respectively. These are therefore the

reference forms of the elements. We can write the formation reaction for liquid water as follows :

2 2 22 ( ) ( ) 2 ( ); 571.6H g O g H O l H kJ+ =

The standard enthalpy change for this reaction is 571.6 kJper two moles of2H O formed. But according to

the definition of fH we need the enthalpy change for only 1 mole of the substance being formed. So, we

divide the thermochemical equation above by 2 to get :

2 2 21( ) ( ) ( ); 285.82 fH g O g H O l H kJ+ =

So, the standard enthalpy of formation of liquid water at 25C is 285.8 kJ.

The standard enthalpies of formation of the elements in their reference state are zero, by definition. Therefore

fH of pure graphite is zero. Note that the standard enthalpy of formation of an element will depend upon the

form of the element. For example, the fH for diamond equals the enthalpy change from the stablest form of

carbon (graphite) to diamond. The thermochemical equation is :

(graphite) (diamond); 1.9fC C H kJ

=


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On the other hand, the fH for graphite equals zero. Here, we list fH

of some compounds and elements. You

are not expected to memorise this table. But it is expected that you study this table and at least memorise the

reference states of elements (when there 0).fH

= It will be useful later. (These are marked in bold italics.)

Please also note that the fH of an electron and ( )H aq+ are zero, by definition.

Standard Enthalpies of formation (at 25 C)

Formula ( )/fH kJ mole

Formula ( )/fH kJ mole

e(g) 0 S8 (rhombic) 0

H+(aq) 0 S8(monoclinic) 2H(g) 218.0 F2(g) 0H2(g) 0 Cl2(g) 0

Na(g) 107.8 Br2(g) 30.91

Na(s) 0 Br2(l) 0NaCl (s) -411.1 I2(g) 62.44

Ca(s) 0 I2(s) 0C(g) 715.0 B(-rhombohedral) 0C(graphite) 0 Sn (gray) 3C(diamond) 1.9 Sn (white) 0

Si (s) 0 P(g) 333.9Pb (s) 0 P(red) 0

N(g) 473 P(white) 69.8

N2(g) 0 P2(g) 146.2O(g) 249.2 P4(g) 128.9

O2(g) 0 Cu(s) 0O3(g) 143 Cu(g) 341.1

H2O(g) -241.8 Hg(g) 61.30

H2O (l) -285.8 Hg(l) 0S(g) 279 Fe(s) 0S2(g) 129 Fe(l) 13.13

Listing out standard enthalpies of formation, actually helps us calculate standard enthalpy of any reaction. Let us

see how to use standard enthalpies of formation to find the standard enthalpy change for a reaction. We will first

look at this problem from the point of view of Hess law. But when we are done we will note a pattern in the result,

which will allow us to state a simple formula for solving this type of problem.


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Consider the equation

4 2 4( ) 4 ( ) ( ) 4 ( ); ?CH g Cl g CCl l HCl g H + + =

The enthalpies of formation of4 4( ), ( )CH g CCl l and ( )HCl g are :

2 4(graphite) 2 ( ) ( ); 74.9fC H g CH g H kJ + = ...(1)

2 4(graphite) 2 ( ) ( ); 139fC Cl g CCl g H kJ + = ...(2)

2 21 1( ) ( ) ( ); 92.3

2 2 fH g Cl g HCl g H kJ+ = ...(3)

We now apply Hesss law. Since we want4CH to appear on the left and 4CCl and 4HCl on the right, we

reverse equation (1) and add equation (2) and 4 Equation (3).

4 ( ) (graphite)CH g C 22 ( )H g+ ( 74.9 ) ( 1)

(graphite)

kJ

C

2 4

2 2

2 ( ) ( ) ( 139 ) (1)

2 ( ) 2 ( )

Cl g CCl l kJ

Cl g H g

+

+

4 2 4

4 ( ) ( 92.3 ) (4)

( ) 4 ( ) ( ) 4 ( ) 433

HCl g kJ

CH g Cl g CCl l HCl g H kJ

+ + =

The set up of this calculation can be greatly simplified if we closely examine what we are doing. Note that thef

H

of each compound has been multiplied by its coefficient in the chemical equation whose H we are calculating.

Moreover, the fH for each reactant is multiplied by a negative sign. For example, the coefficient ofHCl in the

original equation is 4 and we see that we have multiplied the fH ofHClby 4 itself.

We can symbolise the enthalpy of formation of a substance by writing the formula in parantheses following .fH

Then our calculation of H can be written as follows :

4 4 2

( ) 4. ( ) ( ) 4. ( )f f f f

H H CCl H HCl H CH H Cl = + +

[ ] [ ]( 139) 4.( 92.3) ( 74.9) 4.(0)kJ kJ = + +

433kJ=

Extrapolating on this observation, in general, we can calculate the H for a reaction by the equation.

(products) (reactants)f fH n H m H =

Here represents sum of and m and n are the coefficients of the substances in the chemical equation.


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Important Points

1. The superscript zero in fH signifies standard pressure, 1 bar. Negative values of fH

describe

exothermic formation reactions, whereas positive values for fH describe endothermic formation reactions.

2. Remember that all our calculations are based on enthalpy changes, not on actual enthalpy values. So,

defining fH of elements (in their reference forms) as zero does not mean that the enthalpy of the elements

is zero. It only means that we are establishing a thermochemical reference point, from which all changes

are measured.

VARIOUS TYPES OF HEAT OF REACTIONS

Depending upon the type of reaction, heat of reactions are named after these types. We study the important oneshere :

(a) Heat of combustion

Enthalpy change accompanying the complete combustion of1 mole of a substance in an excess of

oxygen is known as the heat of combustion of that substance.

Example :

4 2 2 2( ) 2 ( ) ( ) 2 ( ); 887CH g O g CO g H O l H kJ + + =

(b) Heat of formation

We have already studied this under standard enthalpies of reaction.

(c) Heat of solution

The heat of solution of a substance is the heat evolved or absorbed when 1 mole of that substance is

dissolved in a stated quantity of solvent (usually the solvent is water)

Example:

( ) ( ) ( ); 5NaCl s Na aq Cl aq H kJ+ + =

(d) Heat of ionisation

The heat of ionisation of a substance is the amount of heat absorbed when 1 mole of a compound

completely dissociates into ions.

Example :

( ) ( ) ( ); 46HCN aq H aq CN aq H kJ+ + = +


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(e) Heat of neutralisation

Heat of neutralisation is the heat change accompanying an acid neutralising a base forming 1 mole of

2H O and salt. (another definition of heat of neutralisation is : it is the amount of heat liberated when 1 gm-equivalent of an acid is completely neutralised by 1 gm-equivalent of base)

Some important points about neutralisationH are :

(i) When a strong acid reacts with a strong base (i.e. they are completely ionised in water by themselves)

then the neutralisationH is same for any pair of strong acid and strong base. We can understand this by

the following example:

Neutralisation of a strong acid by a strong base should be written in ionic form as the actual reaction

takes place in ionic form. Consider neutralisation of4HClO ( a strong acid) by NaOH ( a strong

base) written in ionic form :

4( )H aq ClO+ + ( ) ( )aq Na aq++ OH Na ++ 4( ) ( )aq ClO aq

+ 2 ( )H O l+

After cancelling we get the net ionic equation :

2( ) ( ) ( ); 57.27H aq OH aq H O l H kJ+ + =

The 57.27H kJ = per mole of liquid water formed. We can see that the reaction of any strong

acid with any strong base boils down to the same net ionic equation as above. So, H accompanying

any strong acid and base will be same i.e.neutralisation

57.27H kJ = for reaction between a strong

acid and strong base.

(ii) However, when any of the acid or base is weak (i.e. they dont completely ionise in the water), the

accompanying heat released is less than 57.27 .kJ This is because some of the heat is utilised in

ionising the weak acid or base. Some examples of weak acids are 3, ,HCN CH COOH etc. and

examples of weak bases are4 ,NH OH etc. (A detailed discussion on acids and bases will be taken up

in the unit on ionic equilibrium.)

(f) Heat of vapourisation

Vapouration is the change of a liquid to the vapour state. For example :

2 2 vap( ) ( ); 44H O l H O g H kJ =

the H accompanying such a process is called heat of vapourisation. Here,

vapH (at 25C) = 44kJ.

(g) Heat of fusion

Fusion (or melting) is the change of a solid to the liquid state. For example,

2 2( ) ( ); fusH O s H O l H

the H accompanying such a process is called heat of fusion.


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(h) Heat of sublimation

Sublimation is the change of solid directly to the vapour state. For example ,

2 2 sublimation

dry ice

( ) ( );CO s CO g H

the H accompanying such a process is called heat of sublimation.

Please attempt the following basic problems before seeing their solution.

The heat of fusion (also called heat of melting), ,fusH of ice is the enthalpy change for

2 2( ) ( ); .fusH O s H O l H

Similarly, the heat of vaporisation, ,vapH of liquid water is the enthalpy change for

2 2( ) ( ); .vapH O l H O g H

How is the heat of sublimation, ,subH the enthalpy change for the reaction

2 2( ) ( ); subH O s H O g H

related to fusH and ?vapH

Solution: You can think of the sublimation of ice as taking place in two stages. First, the solid metals to liquid,

then the liquid vaporises. The first process has an enthalpy .fusH The second process has an enthalpy

.vapH Therefore, the total enthalpy, which is the enthalpy of sublimation, is the sum of these two

enthalpies:

sub fus vapH H H = +

What is the enthalpy change for the preparation of one mole of liquid water from the elements, given the

following equations?

2 2 2( ) (1/ 2) ( ) ( ); fH g O g H O g H+

2 2( ) ( ); vapH O l H O g H


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Solution: You can imagine this process taking place in two steps : first, the preparation of water vapour from the

elements, and second, the change of the vapour to liquid. Here are the equations :

2 2 21( ) ( ) ( );2 fH g O g H O g H+

2 2( ) ( ); vapH O g H O l H

The last equation is the reverse of the vapourisation of water, so the enthalpy of the step is the negative

of the enthalpy of vapourisation. The enthalpy change for the preparation of one mole of liquid water,

H is the sum of the enthalpy changes for these two steps :

( )f vap f vapH H H H H = + =

BOND ENTHALPY AND BOND DISSOCIATION ENERGY

Consider the breaking of an H2molecule into two hydrogen atoms:

2( ) ( ) ( )H g H g H g + ....(1)

What has been essentially done in this process is that anHHbond has been broken to produce twoHatoms.

This must have required energy. This energy has helped form atoms by dissociating a bond. We define this as:

Energy required to break a particular type ofcovalent bondis called Bond dissociation energy.

Let us consider the corresponding thermochemical equation of (1):

0

2 ( ) ( ) ( ); 435.8H g H g H g H kJ + =

i.e. the correspending standard molar enthalpy change of (1) is + 435.8 kJ. This is Bond dissociation enthalpy of

(1). The Bond dissociation enthalpy is defined as enthalpy change accompanying breaking of 1 mole of a bond of

particular type ingaseous state.

Now, consider the experimentally determined enthalpy changes for the dissociation of a CHbond in CH4and

C2H

6(both in gas phase):

H C H

H

H

( )g H C

H

H

(g) + H g( );H kJ= +435

H C C H

H

H

( )g H C C

H

H

H kJ= +410

H

H

H

H

(g) +H g( );


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We see that bond dissociation enthalpy forCHbond in both the cases is different. This is owing to the different

molecules from which the bonds have been broken.

So, we cannot specify the exact bond dissociation enthalpy of a C H bond. However, we see that the two

values are pretty close to each other. This suggests that the enthalpy change for the dissociation of a CHbond

may be about the same in other molecules. Comparisons of this sort lead to the conclusion that we can obtain

approximate values of energies of various bonds.

We define theA Bbond energy as the averageenthalpy change for the breaking of 1 mole of A B bondin a molecule in the gas phase to form A and B in gas phase. It is also sometimes referred to as bond enthalpy.

It is denoted by ( )A B . We make the distinction between bond dissociation energy and bond energy moreclear in the following example:

Given the following bond dissociation energies, calculate the average bond enthalpy for the Ti Cl

bond.

( / )H kJ mol

4 3( ) ( ) ( )TiCl g TiCl g Cl g + 335

3 2( ) ( ) ( )TiCl g TiCl g Cl g + 423

2( ) ( ) ( )TiCl g TiCl g Cl g + 444

( ) ( ) ( )TiCl g Ti g Cl g + 519

Solution: If the above dissociation process was written as one equation, it would be:

4 ( ) ( ) 4 ( ); 1721TiCl g Ti g Cl g H kJ + =

Where the 1721H kJ = is sum of all the bond dissociation energies for breaking each ofTi Cl bonds. The bond disssociation energies are different for subsequent bond breaking owing to different

entities from which the bonds are broken.

So, the enthalpy change accompanying this bond dissociation is actually a sum of various bond

dissociation energies. But the Bond enthalpy is defined for 1 mole of bonds broken. But here 4 molesof Ti Cl bonds have been broken. To obtain average bond enthalpy for 1 mole ofTi Cl bondswe must divide 1721 by 4 to get the required value.

, the average bond enthalpy1721

430 254

kJkJ= =

So, we may conclude that the bond energy of a diatomic molcule likeH2, O

2,HCl, etc. is same as their

bond dissociation energy but for bonds which are found only in polyatomic molecules, the bond

dissociation energies will be different from the bond energy. (as illustrated in the example above).


16/31



We can use Bond energies to estimate heats of reaction or enthalpy changes, ,H for gaseous

reactions. To illustrate this, let us find the H for the following reaction:

4 2 3( ) ( ) ( ) ( )CH g Cl g CH Cl g HCl g + +We can imagine that the reaction takes place in steps involving the breaking and forming of bonds.

Starting with the reactants, we see that one CHbond and the ClClbond break. i.e.

H C H + Cl Cl

H

H

H C + H + Cl + Cl

H

H

The enthalpy change is ( ) ( )C H Cl Cl + . Now we reassemble the fragments to give theproducts:

H C + H + Cl + Cl

H

H

H C Cl + H Cl

H

H

In this case, CClandHClbonds are formed, and the enthalpy change equals the negative of

the bond energies i.e. ( )C Cl and ( )H Cl . So, the net H for the overall reactionmay be written as:

( ) ( ) ( ) ( )H C H Cl Cl C Cl H Cl + !

Because the bond energy concept is only approximate, the H obtained above will also be approximate.(In fact the experimental value is 101 kJand the value obtained by above method is 104 kJfor the

above reaction. Approximate, but close! Isnt it?)

In general the enthalpy of a reaction is obtained (approximately) using:

reactants products(Bond energy) (Bond energy)H =

The above method of calculating reactionH using bond enthalpies is used only when the thermochemical

data are not known by another approach.

Important points

1. The definition of bond energies is limited to the bond breaking process only and doesnt include any

provision for changes of state. Thus it is valid only for substances in the gaseous state. Therefore the

calculations of this section apply only when all substances in the reaction are gases. If liquids or solids were

involved, then additional information such as heats of vapourisation and fusion would be needed to account

for phase changes.

2. Bond energies are perhaps of greatest value when we try to explain heats of reaction to understand the

relative stabilities of compounds. In general a reaction is exothermic (gives off heat) if weak bonds are

replaced by strong bonds. In the reaction we just discussed, two bonds were broken and replaced by two

new, stronger bonds.


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BORN-HABER CYCLE

Lattice energy cannot be determined directly by experiment. It can, however, be calculated by envisioning the

formation of an ionic compound as occurring in a series of well-defined steps. We can then use Hesss law to put

these steps together in a way that gives us the lattice energy for the compound. By so doing, we construct a

Born-Haber cycle.

Let us try to understand the Born-Haber cycle using the example ofNaCl.

In the Born-Haber cycle forNaClwe consider the formation ofNaCl(s) from the elementNa(s) and Cl2(g) by

two different routes, as shown in the Figure:

LattieenergyofNaC

l

LattieenergyofNaCl

Na g + e + Cl g+

( ) ( )

E Cl( )

Na g +Cl g+

( ) ( )I Na1( )

Na g +Cl g( ) ( )

1( ) ( )

22Na g Cl g+

1( ) ( )2

2Na s Cl g+[ ]( )H Na gf

NaCl s( )

[ ]( )H NaCl sf

[ ]H Cl g( )f

Energy

A Born-Haber cycle shows the energetic relationships inthe formation of ionic solids from the elements. Theenthalpy of formation of NaCl s from elemental sodiumand chlorine Equation is equal to the sum of theenergies of several individual steps Equation through

by Hess's law.

( )(1)

( (2)(6))

The enthalpy change for the direct route is the heat of formation ofNaCl(s):

2

1( ) ( ) ( )

2Na s Cl g NaCl s+ [ ]( ) 410.9fH NaCl s kJ

= ...(1)

,The indirect route consists of five steps. First we generate gaseous atoms of sodium by vapourising sodium metal.

Then we form gaseous atoms of chlorine by breaking the bond in the Cl2molecule. The enthalpy changes for these

processes are available to us as enthalpies of formation :

[ ]( ) ( ) ( ) 107.7fNa s Na g H Na g kJ = ...(2)

[ ]21

( ) ( ) ( ) 121.7

2

fCl g Cl g H Cl g kJ = ...(3)

fig. 11


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Notice that both of these processes are endothermic; energy is required to generate gaseous sodium and chlorine

atoms.

In the next two steps we remove the electron fromNa(g) to formNa+(g) and then add the electron to Cl(g) to

form Cl (g). The enthalpy for these processes equal the first ionisation energy ofNa,I1(Na), and the electron

affinity ofCl, denoted byE(Cl), respectively.:

( ) ( )Na g Na g e+ + 1( ) 496H I Na kJ = = ...(4)

( ) ( )Cl g e Cl g + ( ) 349H E Cl kJ = = ...(5)

Finally, we combine the gaseous sodium and chloride ions to form solid sodium chloride. Because this process is

just the reverse of the lattice energy (breaking a solid into ions), the enthalpy change is the negative of the lattice

energy, the quantity that we wish to determine:

( ) ( ) ( )Na g Cl g NaCl s+ + lattice ?H H = = ...(6)

The sum of the five steps in the indirect path gives usNaCl(s) fromNa(s) and 21

( ).2

Cl g Thus, from Hesss law

we know that the sum of the enthalpy changes for these five steps equals that for the direct path, indicated by the

bold arrow, Equation (1) :

[ ] [ ] [ ] 1

lattice

( ) ( ) ( ) ( ) ( ) .

411 108 122 496 349

f f fH NaCl s H Na g H Cl g I Na E Cl H

kJ kJ kJ kJ kJ H

= + +

= + +

Solving for lattice :H

lattice 108 122 496 349 411H kJ kJ kJ kJ kJ = + + +

788 kJ=

Thus the lattice energy ofNaClis 788 kJ/ mol.

The standard molar enthalpies of combustion ofC2H

2(g), C(graphite) andH

2(g) are 1299.63, 393.51,

285.85 kJmol1

, respectively. Calculate the standard enthalpy of formation ofC2H2(g).

Critical thinking

In order to proceed in such questions, it is suggested that you write down the target equation that will

lead us to finding the required quantity. In this case, the target equation will be the equation representing

the formation of C2H

2(g).Thereafter, see how the given data can be used in conjunction with the target

equation to obtain the required quantity.

Example 9


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Solution: Our target equation is,

2 2 22 (graphite) ( ) ( );C H g C H g H + ...(1)

and we have to determine its Has its His actually the enthalpy of formation C2H2.

Now, the given data can be written as:

2 2 2 2 2

5( ) ( ) 2 ( ) ( )

2C H g O g CO g H O l + + 1299.63 /H kJ mol = ...(2)

2 2(graphite) ( ) ( )C O g CO g + 393.51 /H kJ mol = ...(3)

2 2 21( ) ( ) ( )

2H g O g H O l+ 285.85 /H kJ mol = ...(4)

We can use Hesss law to get Eq.(1) from Eq (2), (3) and (4) as follows:

2(3) 2 2 (graphite) 2C O + ( ) 2g CO 2

2

( ) 2 393.51 /

1(4) 1 ( )2

g H kJ mol

H g O

=

+ 2 2( )g H O ( ) 285.85 /

(2) 1 2

l H kJ mol

CO

=

2 2( )g H O+5

( )2

l O 2 2 2

2 2 2

( ) ( ) 1299.63 /

2 (graphite) ( ) ( ) 226.76 /

g C H g H kJ mol

C H g C H g H kJ mol

+ =

+ =

Hence, ( )2 2 , 226.76 / .fH C H g kJ mol =

Use the following information to determine fH forPbO(s, yellow)

PbO (s, yellow) + CO (g) Pb(s) + CO2(s); 65.69 .rH kJ

=

Given:

fH forCO2 (g) = 393.5 kJ/mol.

fH forCO(g) = 110.5 kJ/mol.

Critical thinking

We can use Hesss law in the form (products) (reactants).r f fH n H m H = Now we are given

reactionH and the fH

for all substances except PbO (s, yellow). We can solve for this unknown.

Example 10


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Solution: (products) (reactants)r f fH n H m H =

2( , ) ( , ) ( ( , yellow) ( , )

r f f f f H H Pb s H CO g H PbO s H CO g = + +

65.69 0 ( 393.5) ( ( , yellow) ( 110.5)fH PbO s = + +

Rearranging to solve for ( ( , yellow)),fH PbO s we have,

( )( , ) 65.69 393.5 110.5fH PbO s yellow = +

= 217.3 kJ/molofPbO.

When Aluminium metal is exposed to atmospheric oxygen, it is oxidised to form Aluminium oxide. How much heat

is released by the complete oxidation of 24.2 gram of Aluminium at 25C and 1 atm ? The thermochemical

equation is :

4Al(s) + 3O2

(s) 2Al2O

3; H= 3352 kJ/mol

Critical thinking

The thermochemical equation tells us that 3352 kJ of heat is released for every mole of reaction i.e. for

every 4 moles of Al that reacts. We convert 24.2 g of Al to moles, and then calculate the number of

kilojoules corresponding to that number of moles of Al.

Solution: For 24.2 grams ofAl, heat released can be obtained by,

24.2g Al1mol Al

27g Al

1 reactionmol

4mol Al

3352

1 reaction

kJ

mol

= 751 kJ

This tells us that 751 kJof heat is released to the surroundings during the oxidation of 24.2 grams of

aluminium.

Given the following standard enthalpies of reactions.

(i) Enthalpy of formation of water = 285.8 kJmol1.

(ii) Enthalpy of combustion of acetylene = 1299.6 kJmol1.

(iii) Enthalpy of combustion of ethylene = 1410.8kJ mol1.

Calculate the heat of reaction for the hydrogenation of acetylene to ethylene at constant volume (25C).

Example 11

Example 12


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Critical thinking

Hydrogenation of acetylene to ethylene can be written as:

2 2 2 2 4( ) ( ) ( )C H g H g C H g +

We have to find the heat of reaction for this reaction. But under what conditions? And under these condi-

tions which state function change will you calculate to obtain the heat of the reaction?

Solution: Our target equation is:

(1) : C2H

2(g) +H

2(g) C

2H

4(g)

We can obtain the Haccompanying this reaction from the given data by applying Hesss law. But

that would give us heat of the reaction under constant pressure. The conditions under which we haveto obtain the heat of reaction is constant volume. This can be obtained by calculating Ufor thereaction. First, let us calculate the Hby applying Hesss law as follows:

Given data may be written as :

(2) : 2 2 21( ) ( ) ( )

2H g O g H O l+ 285.8 /H kJ mol =

(3) : 2 2 2 2 25( ) ( ) 2 ( ) ( )

2C H g O g CO g H O l + + 1299.6 /H kJ mol =

(4) : 2 4 2 2 2( ) 3 ( ) 2 ( ) 2 ( )C H g O g CO s H O l + + 1410.8 /H kJ mol =

We can modify Equation (2), (3), (4) as follows to obtain Equation (1)

21(1) 1 : ( )

2H g O + 2 2( )g H O

2 2

( ) 285.8 /

5(2) 1 : ( )2

l H kJ mol

C H g O

=

+ 2 ( ) 2g CO 2 2( )g H O+ ( ) 1299.6 /

(3) 1: 2

l H kJ mol

CO

=

2 2( ) 2s H O+ 2 4( ) ( ) 3l C H g O + 2

2 2 2 2 4

( ) 1410.8 /

( ) ( ) ( ) 174.6 /

g H kJ mol

H g C H g C H g H kJ mol

= +

+ =

Now, to obtain Uwe use the following equation:

H= U+ (n)RT.

U= H (n)RT. ...(5)

Here, change in number of moles gaseous substances n is: 12 = 1.

Substituting the appropriate values in the (5) above, we get,


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U= [(174.6 kJ/mol) (1) (8.314 103kJ/mol.K) (298K)]

= (174.6 + 2.48) kJ/mol

= 172.12 kJ/mol

(Note : In the calculation ofUabove, we have taken the value ofR as 8.314 103kJ/mol.Kin order to keepthe unit of energy uniform throughout i.e. kJ).

An intimate mixture of ferric oxide,Fe2O

3, and aluminium,Al, is used in solid fuel rockets. Calculate the fuel value

per gram and fuel value percm3 of the mixture. Given:

f

H(Al2

O3

,s) = 1699 kJ mol1; f

H(Fe2

O3

,s) = 833 kJ mol1.

Density of 32 3 5.2 ;Fe O g cm

= Density of 32.7Al g cm=

Critical thinking

We are given that a mixture of Fe2O

3and Al are used as solid rocket fuel. What is its fuel value ? It is the

heat released when Fe2O

3and Al react to form products according to the following thermochemical equa-

tion:

2 3 2 3( ) 2 ( ) ( ) 2 ( ); ?Fe O s Al s Al O s Fe s H+ + =

Solution: The target equation is:

(1) 2 3 2 3( ) 2 ( ) ( ) 2 ( ); reactionFe O s Al s Al O s Fe s H+ +

And we are given,

2 3( , ) 1669 /fH Al O s kJ mol =

2 3( , ) 833 /fH Fe O s kJ mol =

Using this we can calculate the Hreaction

of (1) as follows:

2 3 2 3( , ) 2 ( , ) ( , ) ( , )reaction f f f f H Al O s H Fe s H Fe O s H Al s = +

= (1669 + 0 (833) 0) kJ/mol

= 836 kJ/mol.

Now, in order to calculate fuel value per gram of the fuel (i.e. the reactants) we must divide this Hbythe molar mass of reacting species (each multiplied by appropriate stoichiometric coefficients.) i.e.

Total mass of the reactants = molar mass (Fe2O

3) + 2 molar mass (Al)

Example 13


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= 160g+ 2 27g

= 214g

fuel value per gram 836 /214

kJ mol

g=

= 3.906 kJ/mol.

In order to calculate the fuel value percm3 of the mixture, we must determine the volume of the

mixture,

molar volume of 2 32 3

2 3

molar mass of( )

density of

Fe OFe O s

Fe O=

3160

5.2

g

g cm=

= 30.77 cm3

molar value ofmolar mass of

( )density of

AlAl s

Al=

327

2.7

g

g cm=

= 10 cm3.

the volume of the mixture = molar volume (Fe2O3) + 2 molar volume (Al)

= (30.77 + 2 10) cm3.

= 50.77 cm3.

Therefore fuel value percm3 of fuel 3836

50.77

kJ

cm=

= 16.47 kJ/cm3.

From the following data, calculate the enthalpy change for the combustion of cyclopropane at 298K. The enthalpy

of formation ofCO2(g),H

2O(l) and propene (g) are 393.5, 285.8 and 20.42 kJ mol1, respectively. The

enthalpy of isomerisation of cyclopropane to propene is 33.0 kJ mol1.

Example 14


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Critical thinking

Our target Equation is

CH2

CH2H2C

2 2 29

( ) ( ) 3 ( ) 3 ( ); ?2

g O g CO g H O l H+ + = ...(1)

The H accompanying this reaction is the required enthalpy of combustion. We have to get to this equation

from the given equations using Hesss law.

Solution: Given :

2( , ) 393.5 /fH CO g kJ mol =

2( , ) 285.8 /fH H O l kJ mol =

3 2( ) 20.42 /fH CH CH CH kJ mol = =

isomerisation CH CH CH g H kJ mol 3 2 ( ); = 33.0 /r

CH2

CH2H2C

...(2)

It is not possible to get the Hfor equation (1) from this data directly. However, if we analyse closely,we see that H

combustionfor propene is obtainable and the combustion reaction for propene could be

combined with equation (2) to obtain the equation (1) and hence its Hcombustion

.

So, combustion reaction for propene is :

3 2 2 2 2

9( ) ( ) 3 ( ) 3 ( )

2

CH CH CH g O g CO g H O l = + + ...(3)

Hfor this reaction can be obtained as :

2 2 3 2( ) 3 ( , ) 3 ( , ) ( )c f f f H propene H CO g H H O l H CH CH CH = + =

[3 ( 393.5) 3( 285.8) (20.42)] /kJ mol = +

2058.32 /kJ mol =


25/31



Now, if we add reactions (2) and (3) we get equation (1):

CH2

CH2H2C

isomerisationCH CH CH 3 2= (g) H kJ mol= 33 /

3 2CH CH CH = 2 2 29

( ) ( ) 3 ( ) 3 ( )2

g O g CO g H O l+ + 2058.3 /H kJ mol =

____________________________________________________________________________________

2 2 29

( ) 3 ( ) 3 ( ); 2091.32 /2

O g CO g H O l H kJ mol + + =

Hence, enthalpy of combustion of cyclopropane is 2091.32 kJ/mol

Use the given bond energies to estimate the heat of reaction at 298Kfor the following reaction. (All bonds are

single bonds).

2 2 3( ) 3 ( ) 2 ( )Br g F g BrF g+

Given: 192 /Br Br kJ mol =

159 /F F kJ mol =

197 /Br F kJ mol =

Critical thinking

Each BrF3

molecule contains three Br F bonds. So, two moles of BrF3contain six moles of Br F bonds.

Three moles of F2

contain a total of three moles of F F bonds, and one mole of Br2

contains one mole of

Br Br bonds. We only have to use bond energy form of Hesss law i.e. (reactants) (products). . . .H B E B E =

(note: this can be applied in gas phase reactions only.)

Example 15


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Solution: [ 3. ] [6. ]reaction Br Br F F Br F H = +

= 192 + 3 (159) 6(197)

= 513 kJ/mol

The enthaplies of the following reactions are shown alongwith.

2 2

1 1( ) ( ) ( )

2 2H g O g OH g+ 142.09H kJ mol =

2( ) 2 ( )H g H g1435.89H kJ mol =

2( ) 2 ( )O g O g 1495.05H kJ mol =

Calculate the O Hbond energy for the hydroxyl radical.

Critical thinking

As discussed in the theory, we have to calculate the enthalpy change for the following reaction (which will

be same as bond energy of the hydroxyl radical):

( ) ( ) ( );OH g O g H g H +

As pointed out in example - 9 above, this is our target equation. How will you proceed to obtain the

required H now?

Solution: Our target equation is:

(1) ( ) ( ) ( );OH g O g H g H + = ?

We are given

(2) 2 21 1

( ) ( ) ( )2 2H g O g OH g+ 42.09 /H kJ mol =

(3)2( ) 2 ( )H g H g 435.89 /H kJ mol =

(4)2( ) 2 ( )O g O g 495.05 /H kJ mol =

Example 16


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We can obtain equation (1) from (2), (3), (4) by applying Hesss law as follows:

2 2

2

2

1 1(1) 1: ( ) ( ) ( ) 42.09 /

2 2

1 1 1(2) : ( ) ( ) 435.89 /

2 2 2

1 1 1(3) : ( ) ( ) 495.05 /

2 2 2

( ) ( ) ( ) 423.38 /

OH g H g O g H kJ mol

H g H g H kJ mol

O g O g H kJ mol

OH g H g O g H kJ mol

+ =

=

=

+ =

Use the given bond energies to estimate the heat of reaction at 298 K for the following reaction:

H C C C H +Cl Cl

H H H

H H H

H C C C + ClCl H

H H H

H H H

C H s +Cl g C H Cl g +HCl g3 8 2 3 7( ) ( ) ( ) ( )

Given: 414 /C H kJ mol =

243 /Cl Cl kJ mol =

330 /C Cl kJ mol =

431 /H Cl kJ mol =

Critical thinking

Two moles of C C bonds and seven moles of C H bonds are the same before and after the reaction, so

we do not need to include them in the bond energy calculation. The only reactant bonds that are broken are

one mole of C H bonds and one mole of Cl Cl bonds. On the product side, the only new bonds formed

are one mole of C Cl bonds and one mole of H Cl bonds. We need to take into account only the bonds

that are different on the two sides of the equation. As in the last example, we add and subtract the appro-

priate bond energies.

Solution: 0 [ ] [ ]r C H Cl Cl C Cl H Cl H = + +

[414 243] [330 431]= + +

104 /kJ mol =

When 2.1gof iron combines with sulphur, 3.77 kJare evolved. Calculate the heat of formation of iron sulphide.

Example 17

Example 18


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Critical thinking

The reaction involved can be written as:

reaction( ) ( ) ( ); ?Fe s S s FeS s H+ =

Recall that this is a thermochemical equation and has a molar interpretation. Now,. H for the reactionwill be:

reaction ( , ) ( , ) ( , )f f fH H FeS s H Fe s H S s =

, the last two terms are fH of elements, they are zero.

, ( , )fH Fe S s is same as reaction.H

The thermochemical equation above can be interpreted as : 1 mol of Fe gives 1 mol of FeS and is

accompanied by reactionH (which is same as ( , )).fH FeS s

How will you use this information with the

given data in the problem to obtain reaction ?H

Solution: 1 mol ofFe reacts to release reaction.H in this reaction

i.e. 56 grams ofFe reacts to release reaction.H

But we are given that,

2.1 grams ofFe reacts to release 3.77 kJ

1 gram ofFe reacts to release3.77

2.1kJ

56 grams ofFe reacts to release3.77

56 100.5

2.1

kJ kJ =

Hence, reaction ( , ) 100.5 / .fH H FeS s kJ mol = =


29/31



Here is a reference list of fH values that you will need to solve the various problems of Try Yourself - II and

Exercises at end.

Substance fH (at 298 K) in kJ /mol

2 2 ( )C H g 226.8

2 4 ( )C H g 52.3

2 6 ( )C H g 89.7

2( )CO g 393.5

( )CO g 110.5

2 ( )H O g 241.8

2 ( )H O l 285.8

CuO (s) 162.0


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Q. 1 Does the value of H for a reaction depend on the presence of catalysts in a system ? Substantiate youranswer.

Q. 2 Upon the complete combustion of ethylene, C2H

4(with the formation of liquid water), 6226 kJwere

evolved. Find the volume of the oxygen that entered into the reaction in standard conditions.

Q. 3 In the reduction of 12.7gof copper (II) oxide with coal (with the formation ofCO), 8.24 kJare absorbed.

Determine298H of formation ofCuO.

Q. 4 Water gas is a mixture of equal volumes of hydrogen and carbon monoxide. Find the amount of heat

evolved in the combustion of 112 litres of water gas taken in standard conditions.

Q. 5 Find 298H of formation of ethylene using the following data:

2 4 2 2 2( ) 3 ( ) 2 ( ) 2 ( ); 1323C H g O g CO g H O g H kJ + + =

2 2(graphite) ( ) ( ); 393.5C O g CO g H kJ + =

2 2 2

1( ) ( ) ( ); 241.8

2H g O g H O g H kJ+ =

Q. 6 Ethylamine undergoes an endothermic gas phase dissociation to produce ethylene (or ethene) and ammonia.

C C

H H

H C C N

H H

| |

| |

H

H

H

HH

H

+ NH|

HH

H kJ mol= + 54.68 /0

reaction

The following average bond energies per mole of bonds are given : CH= 414 kJ; CC= 347 kJ

C C = 611 kJ;NH= 389 kJ. Calculate the CNbond energy in ethylamine.

Q. 7 The standard enthalpies of formation at 298 K for 4 2 2( ), ( ), ( )CCl g H O g CO g and ( )HCl g

are 106.7, 241.8, 393.7 and 92.5 kJmol1, respectively. Calculate 298H for the reaction,

4 2 2( ) 2 ( ) ( ) 4 ( )CCl g H O g CO g HCl g + +

TRY YOURSELF - II


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Q. 8 Diborane is a potential rocket fuel which undergoes combustion according to the reaction

2 6 2 2 3 2( ) 3 ( ) ( ) 3 ( )B H g O g B O s H O g+ +

From the following data, calculate the enthalpy change for the combustion of diborane.

12 2 32 ( ) (3 / 2) ( ) ( ) 1273 .B s O g B O s H kJ mol

+ =

12 2 2( ) (1/ 2) ( ) ( ) 286H g O g H O l H kJ mol

+ =

12 2( ) ( ) 44H O l H O g H kJ mol

=

12 2 62 ( ) 3 ( ) ( ) 36B s H g B H g H kJ mol

+ =

Q. 9 Bond energies of 2F and 2Cl are respectively 36.6 and 58.0 kcalper mole. If the heat liberated in the

reaction2 2 2F Cl FCl+ is 26.6 kcal, calculate the bond energy ofFClbond.

Q. 10 From the following data, calculate the heat evolved in the formation of 2.5 litres of carbon monoxide from

its elements:

2 2 94,500C O CO H cal + =

2 22 2 136,000 .CO O CO H cal + =

Q. 11 Calculate fH

of 6 12 6 ( )C H O s from the following data:

combustionH of 6 12 6( ) 2816 /C H O s kJ mol =

Thermodynamics [02] Section [23 - 53]

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Transcript of Thermodynamics [02] Section [23 - 53]