Chapter 6.10 - Enthalpy of Solution

In the previous section, we saw Lattice enthalpy. In this section, we will see enthalpy of solution

• A solution is formed when a substance (solute) dissolves in another substance (solvent)
• When the solute dissolves in a solvent at constant pressure, two things can happen:
(i) Energy may be absorbed
(ii) Energy may be released
• This absorbed/released energy is called enthalpy of solution
    ♦ It is denoted by the symbol: ΔH^⊖_sol
    ♦ If energy is absorbed, ΔH^⊖_sol will be positive
    ♦ If energy is released, ΔH^⊖_sol will be negative
[Note: For the absorbed/released energy to be designated as ΔH^⊖_sol, one condition must be satisfied:
◼ The solution must be an infinitely dilute solution
• In an infinitely dilute solution, the quantity of solvent will be so large that, addition of more solvent will not result in any further energy absorption /release. We will see more details about this condition, later in this section]
• Our aim is to find the ΔH^⊖_sol for various solutions like:
    ♦ Aqueous solution of NaCl
    ♦ Aqueous solution of MgCl₂ etc.,
• We can devise a general method which can be applied to most of the solutions. It can be written in 26 steps by using the aqueous solution of NaCl as an example

1. The cyan horizontal lines in fig.6.14 below indicates various states in an experiment
• The arrows indicate various processes which transform the system from one state to another

Fig.6.14

2. Consider the thick cyan horizontal line. It is the datum line. It is the state from which we begin our calculations
• At this state, the system consists of:
   ♦ One mol NaCl molecules
         ✰ These molecules are in the solid state
3. From the datum line, we begin our first process
• Our first process is to convert the NaCl molecules into Na⁺(g) ions and Cl^-(g) ions
The equation is: NaCl(s) → Na⁺(g) + Cl^-(g)
4. We have seen this process in the previous section
• The energy required for this process is the ΔH^⊖_lattice(d) of NaCl
   ♦ It's value is 788 kJ mol^-1
• So the thermochemical equation of this process will be:
NaCl(s) → Na⁺(g) + Cl^-(g); ΔH^⊖_lattice(d) = 788 kJ mol^-1
• This process is numbered as I in the fig.6.14 above
5. So when the process I is complete, we will have:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ One mol Cl^- ions in the gaseous state

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6. Our second process is to convert each of the gaseous Na⁺ ions into Na⁺(aq) ions
• Let us see how this is achieved:
7. Consider the H₂O molecules in the liquid state
• We know that, H₂O molecules are polar molecules (Fig.4.225 of section 4.40)
• In each H₂O molecule:
    ♦ The H atoms have a partial positive charge $\mathbf\small{\rm{\delta^+}}$
    ♦ The O atom has a partial negative charge $\mathbf\small{\rm{\delta^-}}$
8. When the NaCl dissolves in water, each of the Na⁺ ions will get surrounded by H₂O molecules as shown in fig.6.15(a) below
    ♦ The red spheres are the O atoms
    ♦ The white spheres are the H atoms
• The partially negative O atoms are attracted towards the Na⁺ ion

Fig.6.15

9. The situation shown in fig.6.15(a) is stable
• Bonds are formed between the H₂O molecules and Na⁺ ions
• These bonds are due to the attraction between $\mathbf\small{\rm{O^{\delta-}}}$ and Na⁺
10. We know that, when bonds are formed, energy is released
◼ The energy released when bonds are formed between ions and H₂O molecules is known as hydration enthalpy
• It's symbol is: ΔH^⊖_hyd
11. The Na⁺ ions, when bonded with water molecules is represented as Na⁺(aq)
• We can represent the process of hydration as:
Na⁺(g) → Na⁺(aq)
12. From the data book, we have:
One mol Na⁺(g) ions release 424 kJ energy when all those ions are converted into Na⁺(aq)
• In other words, ΔH^⊖_hyd for Na⁺(g) = -424 kJ mol^-1
• So the thermochemical equation for this process will be:
Na⁺(g) → Na⁺(aq); ΔH^⊖_hyd = -424 kJ mol^-1
• This process is numbered as II in the fig.6.14 above
13. So when the process II is complete, we will have:
   ♦ One mol Na⁺ ions in the aqueous state
   ♦ One mol Cl^- ions in the gaseous state

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14. Our third process is to convert each of the gaseous Cl^- ions into Cl^-(aq) ions
• Let us see how this is achieved:
15. As before, each of Cl^- will get surrounded by H₂O molecules
• This time, the H atoms with the partial positive charges, get attracted towards the Cl^- ions
• This is shown in fig.6.15(b) above
16. The situation shown in fig.6.15(b) is stable
• Bonds are formed between the H₂O molecules and Cl^- ions
• These bonds are due to the attraction between $\mathbf\small{\rm{H^{\delta+}}}$ and Cl^-
17. As before, hydration enthalpy is released in this case also
18. The Cl^- ions, when bonded with water molecules is represented as Cl^-(aq)
• We can represent this process of this hydration as:
Cl^-(g) → Cl^-(aq)
19. From the data book, we have:
One mol Cl^-(g) ions release 359 kJ energy when all those ions are converted into Cl^-(aq)
• In other words, ΔH^⊖_hyd for Cl^-(g) = -359 kJ mol^-1
• So the thermochemical equation for this process will be:
Cl^-(g) → Cl^-(aq); ΔH^⊖_hyd = -359 kJ mol^-1
• This process is numbered as III in the fig.6.14 above
20. So when the process III is complete, we will have:
   ♦ One mol Na⁺ ions in the aqueous state
   ♦ One mol Cl^- ions in the aqueous state

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Now we can write why it is important to specify 'infinitely dilute solution'. It can be written in 2 steps:
(i) Imagine that, there is not enough water molecules
• Then:
   ♦ All the Na⁺ ions cannot be converted into Na⁺(aq)
         ✰ Some Na⁺ ions will remain as such
         ✰ So all the available ΔH^⊖_hyd for Na⁺ will not be released
   ♦ All the Cl^- ions cannot be converted into Cl^-(aq)
         ✰ Some Cl^- ions will remain as such
         ✰ So all the available ΔH^⊖_hyd for Cl^- will not be released
(ii) If we measure the enthalpies in such a situation:
   ♦ We will be recording a 'lower ΔH^⊖_hyd' than 'actual ΔH^⊖_hyd' for Na⁺
   ♦ We will be recording a 'lower ΔH^⊖_hyd' than 'actual ΔH^⊖_hyd' for Cl^-

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21. The completion of process III was our goal
• Consider the products obtained at the end of this process:
   ♦ One mol Na⁺ ions in the aqueous state
   ♦ One mol Cl^- ions in the aqueous state
22. These products indicate that, one mol NaCl is completely dissolved in water
• We have achieved our goal
23. This goal is indicated by the third cyan line from top
• So from the datum line, we took the path: (I + II + III) to reach the third cyan line
24. From the datum line, we can take another path also
• It is along the red arrow
25. By Hess's law, we can write:
   ♦ Energy along (I + II + III)
   ♦ is equal to
   ♦ Energy along the red arrow
• Thus we get: (788 - 424 - 359) = X
⇒ X = 5
• This is a positive value
• That means, energy should be supplied
• It is an endothermic process
26. Note that, X is related to the process: Na⁺(g) + Cl^-(g) → Na⁺(aq) + Cl^-(aq)
• This indicates the solution of one mol NaCl in water
• So the red arrow represents the same process that we are investigating
• We can write:
To dissolve one mol NaCl in water, we need to supply 5 kJ energy
◼ In other words, ΔH^⊖_sol of NaCl is 5 kJ mol^-1

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Next we will find the ΔH^⊖_sol of CaCl₂. The procedure is same as that for NaCl. So we will write only the minimum required 10 steps:

1. The thick cyan horizontal line in fig.6.16 below, is the datum line. It is the state from which we begin our calculations
• At this state, the system consists of:
   ♦ One mol CaCl₂ molecules
         ✰ These molecules are in the solid state

Fig.6.16

 

2. From the datum line, we begin our first process
• Our process I is to convert the CaCl₂ molecules into Ca²⁺(g) ions and Cl^-(g) ions
The equation is: CaCl₂(s) → Ca²⁺(g) + 2Cl^-(g)
3. From the data book, we have: ΔH^⊖_lattice(d) of CaCl₂ = 2258 kJ mol^-1
• So the thermochemical equation of this process will be:
CaCl₂(s) → Ca²⁺(g) + 2Cl^-(g); ΔH^⊖_lattice(d) = 1158 kJ mol^-1
4. So when the process I is complete, we will have:
   ♦ One mol Ca²⁺ ions in the gaseous state
   ♦ Two mol Cl^- ions in the gaseous state
5. Process II is the conversion of Ca²⁺(g) ions into Ca²⁺(aq)
• Using the data book, we write:
Ca⁺(g) → Ca²⁺(aq); ΔH^⊖_hyd = -1650 kJ mol^-1
• This energy is released when bonds are formed between the water molecules and the Ca²⁺ ions
   ♦ This is similar to the case shown in fig.6.15(a) above
6. Process III is the conversion of Cl^-(g) ions into Cl^-(aq)
• Using the data book, we write:
2Cl^-(g) → 2Cl^-(aq); ΔH^⊖_hyd = 2(-359) kJ mol^-1
• This energy is released when bonds are formed between the water molecules and the Cl^- ions
   ♦ This is the same case shown in fig.6.15(b) above
• Note that in the cas of NaCl, there is only one mol of Cl^- ions
• But in the case of CaCl₂, there are two mol Cl^- ions
   ♦ So we multiply -359 by 2
7. The completion of process III was our goal
• This goal is indicated by the cyan line below the datum line
• So from the datum line, we took the path: (I + II + III) to reach the goal
8. From the datum line, we can take another path also to reach the goal
• It is along the red arrow
9. By Hess's law, we can write:
   ♦ Energy along (I + II + III)
   ♦ is equal to
   ♦ Energy along the red arrow
• Thus we get: (1158 - 1650 - 2(359)) = X
⇒ X = -110
• This is a negative value
• That means, energy will be released
• It is an exothermic process
10. Note that, X is related to the process: Ca²⁺(g) + 2Cl^-(g) → Ca²⁺(aq) + 2Cl^-(aq)
• This indicates the solution of one mol CaCl₂ in water
• So the red arrow represents the same process that we are investigating
• We can write:
When one mol CaCl₂ dissolve in water, 110 kJ energy will be released
◼ In other words, ΔH^⊖_sol of CaCl₂ is -110 kJ mol^-1

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• We have seen the ΔH^⊖_sol of two salts: NaCl and CaCl₂
• We see that:
    ♦ the first is an endothermic process
    ♦ the second is an exothermic process
• We also see that:
    ♦ in the first case, the goal is above the datum line
    ♦ in the second case, the goal is below the datum line
• In the next section, we will see the reason for such differences


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Chapter 6.8 - Application of Bond Enthalpy

In the previous section, we saw Hess's law and bond enthalpy. In this section, we will see some applications of bond enthalpy.

• The bond enthalpy values given in the data book can be used to find ΔH^⊖_r
• This can be explained in 11 steps using an example
1. Consider the following reaction:
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
• Steam is made to react with carbon monoxide.
2. Let us see the making and breaking of bonds in this reaction:
• When the reaction proceeds:
    ♦ Bonds in the reactant side are broken.
    ♦ As a result, the reactant molecules separate into individual atoms.
    ♦ Those individual atoms rearrange and make new bonds.
    ♦ The new bonds result in product molecules.
3. We want answers to these questions:
• Which all bonds are broken in the reactant side?
    ♦ How many bonds of each type are broken?
• Which all bonds are made in the product side?
    ♦ How many bonds of each type are made?
4. It will be easier to find the answers to the above questions, if we  rewrite the equation in (1) using structural formulae. This is shown below:
C≡O + H-O-H → O=C=O + H-H
5. It is clear that:
• On the reactant side:
    ♦ One C≡O bond is broken.
    ♦ Two O-H bonds are broken.
• On the product side:
    ♦ Two C=O bonds are made.
    ♦ One H-H bond is made.
6. We know the following two facts:
(i) When bonds are broken, energy is absorbed.
    ♦ So, we need to supply energy to break the bonds in the reactant side.
(ii) When bonds are made, energy is released.
    ♦ So we will receive energy when bonds are made in the product side.
7. Let us calculate each of the above two items in our present case.
• From the data book, we have:
    ♦ C≡O has a bond enthalpy of 1072 kJ mol^-1
    ♦ O-H has a bond enthalpy of 463 kJ mol^-1
    ♦ C=O has a bond enthalpy of 799 kJ mol^-1
    ♦ H-H has a bond enthalpy of 436 kJ mol^-1
• So we get:
(i) Total energy required for breaking all the bonds in the reactant side
= (1 × 1072) + (2 × 463) = 1998
(ii) Total energy released when making all the bonds in the product side
= (2 × 799) + (1 × 436) = 2034
8. We see that:
   ♦ Energy required to break the bonds in the reactant side
   ♦ is lesser than
   ♦ Energy released when bonds are made in the product side.
• So we receive a net energy of (2034-1998) = 36 kJ mol^-1
• Since energy is released, it is an exothermic reaction. The enthalpy value should be given a negative sign.
9. Now we can write the thermochemical equation:
CO(g) + H₂O(g) → CO₂(g) + H₂(g); ΔH^⊖_r = -36.0 kJ mol^-1
10. Based on the above discussion, we can write a general equation to find ΔH^⊖_r
Eq.6.10: $\mathbf\small{\rm{\Delta {H^{\ominus }}_r=\sum{Bond\;Enthalpies_{reactants}}-\sum{Bond\;Enthalpies_{products}}  }}$
11. Consider the two terms on the right side of Eq.6.10
• If the first term is larger, it means that, the bonds in the reactant side have greater energy.
   ♦ We will have to supply a net energy to break the bonds.
   ♦ ΔH^⊖_r will have a positive sign.
   ♦ The reaction will be endothermic.
• If the second term is larger, it means that, the bonds in the product side have greater energy.
   ♦ We will receive a net energy.
   ♦ ΔH^⊖_r will have a negative sign.
   ♦ The reaction will be exothermic.
• In our present case, the second term is larger.
• We see that, the Eq.6.10 automatically gives us the appropriate sign.

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Next we will see a simple case. It is called 'simple case' because, only a very few bonds are broken and made. It can be written in steps:
1. Consider the reaction shown in fig.6.11 below:

• It is the reaction between propene and hydrogen to give propane.
2. Let us see the making and breaking of bonds in this reaction:
• When the reaction proceeds:
    ♦ Two bonds in the reactant side are broken.
          ✰ They are marked with red color in fig.6.12 below.
    ♦ Three bonds are newly formed in the product side.
          ✰ They are marked with green color in the fig.6.12

 

Fig.6.12

3. Since only a very few bonds are broken and made, we can easily answer our earlier questions:
• Which all bonds are broken in the reactant side?
    ♦ How many bonds of each type are broken?
Answer:
One C=C bond and one H-H bond are broken.
• Which all bonds are made in the product side?
    ♦ How many bonds of each type are made?
Answer:
One C-C bond and two C-H bonds are made.
4. We know the following two facts:
(i) When bonds are broken, energy is absorbed.
    ♦ So, we need to supply energy to break the bonds in the reactant side.
(ii) When bonds are made, energy is released.
    ♦ So we will receive energy when bonds are made in the product side.
7. Let us calculate each of the above two items in our present case.
• From the data book, we have:
    ♦ C=C has a bond enthalpy of 614 kJ mol^-1
    ♦ H-H has a bond enthalpy of 436 kJ mol^-1
    ♦ C-C has a bond enthalpy of 348 kJ mol^-1
    ♦ C-H has a bond enthalpy of 413 kJ mol^-1
• So we get:
(i) Total energy required for breaking bonds in the reactant side
= (1 × 614) + (1 × 436) = 1050
(ii) Total energy released when making all the bonds in the product side
= (1 × 348) + (2 × 413) = 1174
8. We see that:
   ♦ Energy required to break the bonds in the reactant side
   ♦ is lesser than
   ♦ Energy released when bonds are made in the product side.
• So we receive a net energy of (1174-1050) = 124 kJ mol^-1
• Since energy is released, it is an exothermic reaction. The enthalpy value should be given a negative sign.
9. Now we can write the thermochemical equation:
C3H6(g) + H₂(g) → C3H₈(g); ΔH^⊖_r = -124.0 kJ mol^-1
10. Based on the above discussion, we can write a general equation to find ΔH^⊖_r for simple cases
Eq.6.11:
$\mathbf\small{\rm{\Delta {H^{\ominus }}_r=\sum{Broken \; Bond\;Enthalpies}-\sum{Made \; Bond\;Enthalpies}}}$
11. Consider the two terms on the right side of Eq.6.11.
• If the first term is larger, it means that, the 'total energy required to break bonds' is greater.
   ♦ We will have to supply a net energy to break the bonds.
   ♦ ΔH^⊖_r will have a positive sign.
   ♦ The reaction will be endothermic.
• If the second term is larger, it means that, the 'total energy released when bonds are made' is greater.
   ♦ We will receive a net energy.
   ♦ ΔH^⊖_r will have a negative sign.
   ♦ The reaction will be exothermic.
• In our present case, the second term is larger.
• We see that, the Eq.6.11 automatically gives us the appropriate sign.

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• Now we have a basic idea about the application of bond enthalpy
• In the next section, we will see lattice enthalpy

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Chapter 6.7 - Hess's Law of Constant Heat Summation

In the previous section, we saw enthalpies of fusion, vaporization, sublimation and combustion. Next we will see enthalpies of atomization, bond, lattice and solution. But before that, we have to see Hess’s law of constant heat summation. We will see it in this section.

Basics of Hess's law can be written in 10 steps:
1. Consider a chemical reaction between two reactants R₁ and R₂
• Let the products be P₁ and P₂
• We can write:
R₁ + R₂ → P₁ + P₂; ΔH^⊖_r = ±X
    ♦ ‘X’ is the magnitude of the enthalpy change.
    ♦ sign may be ‘+’ or ‘-’
        ✰ sign depends on whether the reaction is exothermic or endothermic.
2. Let state A be the initial state of the reaction.
    ♦ At A, the only molecules present, will be those of R₁ and R₂
3. Let state B be the final state of the reaction.
    ♦ At B, the only molecules present, will be those of the products P₁ and P₂
4. Then we can represent the above reaction as shown in fig.6.10(a) below:

Fig.6.10

Fig.a indicates a transformation from A to B, with an accompanying enthalpy change of ±X

5. Let us consider an alternate path:
• Imagine that, from the initial state A, the system first transforms to another state C.
• And from that state C, it transforms to the final state B.
• We can represent this as shown in fig.6.10(b) above.
6. We see that:
• The first transformation from A to C is accompanied by an enthalpy change of ±Y
• The second transformation from C to D is accompanied by an enthalpy change of ±Z
◼ Then, according to Hess’s law, ±X will be the algebraic sum of ±Y and ±Z
    ♦ That is., (±X) = [(±Y) + (±Z)]
7. In fig.b:
    ♦ the single transformation A → B
    ♦ is replaced by
    ♦ two transformations: A → C and C → B
• The original path is indicated by the white arrow.
• The alternate path is indicated by the two magenta arrows.
8. In some cases
    ♦ the single transformation A → B
    ♦ is replaced by
    ♦ three transformations: A → C, C → D and D → B
• This is shown in fig.6.11(c)
• Applying Hess’s law, we get: (±X) = [(±Y) + (±Z) + (±K)]
• The original path is indicated by the white arrow.
• The alternate path is indicated by the three magenta arrows.
9. In this way, the single transformation A → B can be replaced by any number of intermediate transformations. Whatever be the 'number of intermediate transformations', Hess's law will be valid
• All we need to do is: Find the algebraic sum of the intermediate enthalpies.
10. But before finding the algebraic sum, we have to establish the alternate path.
• With some practice, we will be able to do it easily.
• An example is shown below:

Links to some more examples are given below:

Example 2 

Example 3

Example 4 

Example 5

Example 6  

Example 7   

Example 8  

Example 9

◼  Based on the above discussion, we can write the Hess's Law of Constant Heat Summation:
The law states that, regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law shows that enthalpy is a state function.

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• Next we will see enthalpy change for atomization. It can be explained in 5 steps:
1. A sample of hydrogen gas will consist of H₂ molecules. By supplying enough energy, we can separate each of those H₂ molecules into to individual H atoms.
2. Using calorimetry, scientists have calculated that:
    ♦ 1 mole of H₂ gas requires
    ♦ 435.0 kJ energy
3. Based on this, we can write the thermochemical equation:
H₂(g) → 2H (g); ΔH^⊖_a = 435.0 kJ mol^-1
• Note that, instead of 'r', the subscript is 'a'. This is to indicate 'atomization'.
• Another example is the atomization of chlorine. The thermochemical equation is:
Cl₂(g) → 2Cl (g); ΔH^⊖_a = 243.0 kJ mol^-1
4. The above two examples are diatomic molecules. We can consider polyatomic molecules also:
CH₄(g) → C(g) + 4H(g); ΔH^⊖_a = 1665.0 kJ mol^-1
• In the product side, there are only individual atoms:
    ♦ One C atom and four H atoms.
5. In the case of sublimation, we know that, the products are in gaseous form.
• If that gaseous form consists only of individual atoms, we can write:
Enthalpy of atomization will be same as the enthalpy of sublimation.
• An example is the sublimation of Na
    ♦ Na(s) → Na (g); ΔH^⊖_sub = 108.4 kJ mol^-1
    ♦ Na(s) → Na (g); ΔH^⊖_a = 108.4 kJ mol^-1 

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Next we will see enthalpy change in the following two cases:
    ♦ Existing bonds between atoms are broken.
    ♦ New bonds between atoms are formed.
• It can be explained in 8 steps:
1. A sample of hydrogen gas will consist of H₂ molecules. By supplying enough energy, we can separate each of those H₂ molecules into to individual H atoms. For such a separation, we will have to break the H-H bond in each of those H₂ molecules.
2. Using calorimetry, scientists have calculated that:
    ♦ 1 mole of H₂ gas requires
    ♦ 435.0 kJ energy
• If one mol H₂ molecules are converted completely into H atoms, we can be sure that, one mol 'H-H bonds' are broken.
3. Based on this, we can write the thermochemical equation:
H₂(g) → 2H(g); ΔH^⊖_H-H = 435.0 kJ mol^-1
• Note that, instead of 'r', the subscript is 'H-H'. This is to indicate 'the making or breaking of the single bond between two H atoms'.
('Making of bonds' is opposite of 'breaking of bonds'
    ♦ When bonds are broken, a certain energy will be absorbed.
    ♦ If the same bonds are made, the same energy will be released)
• Another example is the enthalpy of Cl-Cl bond. The thermochemical equation is:
Cl₂(g) → 2Cl (g); ΔH^⊖_Cl-Cl = 243.0 kJ mol^-1
4. It is clear that, for diatomic molecules, bond enthalpy will be equal to ΔH^⊖_a
• This is because,
    ♦ breaking all the bonds of diatomic molecules
    ♦ is same as
    ♦ changing the molecules into individual atoms
5. But for polyatomic molecules, bond enthalpy will not be same as ΔH^⊖_a
• The reason can be explained in 8 steps, using CH₄ as an example:
(i) We have the thermochemical equation for the atomization of CH₄:
CH₄(g) → C(g) + 4H(g); ΔH^⊖_a = 1665.0 kJ mol^-1
• That means, to convert one mol CH₄ into individual C and H atoms, we need to supply 1665 kJ energy.
(ii) We know that, in one CH₄ molecule, there are four C-H bonds.
• So in one mol CH₄, there will be four mol C-H bonds
• That means, to break four mol C-H bonds, we need to supply 1665.0 kJ
(iii) So it seems that, to break one mol C-H bonds, we need to supply (¹⁶⁶⁵⁄₄) = 416 kJ
• It seems that, we can write: ΔH^⊖_C-H = 416.0 kJ mol^-1
• But this result is not valid. Let us see the reason:
(iv) Consider the following thermochemical equation:
CH₄(g) → CH₃(g) + H(g); ΔH^⊖_r = 427.0 kJ mol^-1
• It indicates that
    ♦ One mol CH₄ molecules is taken
    ♦ In each of those molecules, one of the four C-H bonds is broken.
    ♦ Thus one mol H atoms are set free.
    ♦ In effect, one mol C-H bonds are broken.
• But we see that, the energy absorbed is 427.0 kJ. This is different from the result in (iii)
(v) There are even more differences:
• Consider the following thermochemical equation:
CH₃(g) → CH₂(g) + H(g); ΔH^⊖_r = 439.0 kJ mol^-1
• It indicates that
    ♦ One mol CH₃ molecules is taken.
    ♦ In each of those molecules, one of the three C-H bonds is broken.
    ♦ Thus one mol H atoms are set free.
    ♦ In effect, one mol C-H bonds are broken.
• But we see that, the energy absorbed is 439.0 kJ. This is different from the result in (iii) and (iv)
(vi) Scientists have discovered the reason for the difference between the results in (iv) and (v):
    ♦ In (iv), the first H atom is set free.
    ♦ Once that H atom leaves, the remaining three are held more tightly by the C atom.
    ♦ So to release a second H atom, more energy will be required.
    ♦ Thus the energy in (iv) is greater than that in (iii)
(vii) We can continue like this until the remaining two H atoms are also set free from the C atom:
• CH₂(g) → CH(g) + H(g); ΔH^⊖_r = 452.0 kJ mol^-1
• CH₃(g) → CH₂(g) + H(g); ΔH^⊖_r = 347.0 kJ mol^-1
(viii) So there are five possible values for ΔH^⊖_C-H. They are:
    ♦ The value obtained by dividing ΔH^⊖_a by 4: 416.0
    ♦ The value when the first H atom is released: 427.0
    ♦ The value when the second H atom is released: 439.0
    ♦ The value when the third H atom is released: 452.0
    ♦ The value when the fourth H atom is released: 347.0
6. To make matters worse, the above five values do not complete the 'list of possible values'. There are even more. The reason can be explained in 3 steps:
(i) Consider the compound CH₃CH₂Cl
• We know that, there are some C-H bonds in this compound.
   ♦ The energy required to break them are different from the five values that we saw.
(ii) Consider the compound CH₃NO₂
• We know that, there are some C-H bonds in this compound.
   ♦ The energy required to break them are different from the five values that we saw.
   ♦ The energy required to break them are different from the values in CH₃CH₂Cl also.
(iii) So it is clear that, the value for C-H bond differs from compound to compound also.
7. But there is nothing to worry about. After considering the different possible values, scientists have agreed upon an average value. It is: 413 kJ mol^-1
• That means, we can write: ΔH^⊖_C-H = 413 kJ mol^-1
• This is the value that we will find in the data book.
• Whenever we encounter a C-H bond, we can use '413 kJ mol^-1' in the calculations.
8. Just like in C-H, we will get different values in other bonds like C-Cl, N-O etc., also.
• Scientists have given the appropriate values that can we can use in such cases also. They are available in the data book. Some examples are given below:
   ♦ For C-Cl, the value is 328 kJ mol^-1
   ♦ For N-O, the value is 201 kJ mol^-1

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• Now we have a basic idea about bond enthalpy.
• Different text books may use different terms for bond enthalpy.
• Bond dissociation enthalpy, bond strength and bond energy are all same as bond enthalpy.
• In the next section, we will see how this bond enthalpy can be put to practical use.


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