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 6.9 - Lattice Enthalpy

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

Lattice enthalpy can be explained in 7 steps:
1. Consider the ionic compound NaCl
   ♦ We know that, NaCl has a crystal structure
   ♦ The crystal is made up of Na⁺ and Cl^- ions
   ♦ Those ions are arranged in a symmetric 3D form
   ♦ This 3D form is called crystal lattice
   ♦ (some images can be seen here)
2. The crystal lattice of NaCl is very stable because of the strong electrostatic attractions between the Na⁺ and Cl^- ions
• We want to know how much energy will be required for the following process:
   ♦ Separating all the Na⁺ and Cl^- ions in one mol NaCl
◼  The resulting individual ions should be scattered far away from each other so that, there will be no attractive or repulsive forces between the resulting ions. In other words, the resulting ions must be in the gaseous state
3. It is obvious that:
• When the lattice is turned into gaseous ions, we will need to supply energy. In other words, it will be an endothermic process
• For the reverse process, that is., when the gaseous ions come together and form the lattice, we will receive energy. In other words, it will be an exothermic process  
◼  This energy that we supply/receive is called lattice enthalpy
4. Lattice enthalpy can be defined in two ways:
◼  Lattice dissociation enthalpy
The lattice dissociation enthalpy is the enthalpy change needed to convert 1 mole of solid crystal lattice into its scattered gaseous ions
   ♦ We will denote it by the symbol: ΔH^⊖_lattice(d)
◼  Lattice formation enthalpy
The lattice formation enthalpy is the enthalpy change when 1 mole of solid crystal lattice is formed from its separated gaseous ions
   ♦ We will denote it by the symbol: ΔH^⊖_lattice(f)
5. It is obvious that:
   ♦ Both ΔH^⊖_lattice(d) and ΔH^⊖_lattice(f) will have the same magnitude
   ♦ The sign of ΔH^⊖_lattice(d) will be positive
   ♦ The sign of ΔH^⊖_lattice(f) will be negative
6. To find ΔH^⊖_lattice(d) of NaCl means, to find X in the following thermochemical equation:
   ♦ NaCl(s) → Na⁺(g) + Cl^-(g); ΔH^⊖_lattice(d) = X
7. To find ΔH^⊖_lattice(f) of NaCl means, to find X in the following thermochemical equation:
   ♦ Na⁺(g) + Cl^-(g) → NaCl(s); ΔH^⊖_lattice(f) = X

---

• So we will see how ΔH^⊖_lattice(d) or ΔH^⊖_lattice(f) is determined
• It is not possible to carry out the experiment in (6) above. Neither is it possible to carry out the experiment in (7)
◼  That means, we cannot measure lattice enthalpy from any experiments
• So we use an indirect method. It can be written in 22 steps:
1. The cyan horizontal lines in fig.6.13 below indicates various states in an experiment
• The arrows indicate various processes which transform the system from one state to another

Fig.6.13

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 Na atoms
         ✰ These atoms are in the solid state
   ♦ Half mol Cl₂ molecules
         ✰ These molecules are in the gaseous state
◼  The above solid and gaseous states are obvious because:
   ♦ At standard temperature and pressure
         ✰ a sample of sodium consists of Na atoms in the solid state
         ✰ a sample of chlorine consists of Cl₂ molecules in the gaseous state
3. From the datum line, we begin our first process
• Our first process is to convert the sodium into gaseous state
• So it will be a sublimation process
4. From the data book, we have:
One mol Na(s) requires 108.4 kJ energy, for complete sublimation
• In other words, ΔH^⊖_sub for Na(s) = 108.4 kJ mol^-1
• So the thermochemical equation for this process will be:
Na(s) + 1/2Cl₂(g) → Na(g) + ½ Cl₂(g); ΔH^⊖_sub = 108.4 kJ mol^-1
• This process is numbered as I in the fig.12.13 above
5. So when the process I is complete, we will have:
   ♦ One mol Na atoms in the gaseous state
   ♦ Half mol Cl₂ molecules in the gaseous state

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6. Our second process is to convert each of the gaseous Na atoms into gaseous Na⁺ ions
• So it will be an ionization process
7. From the data book, we have:
One mol Na(g) requires 495.6 kJ energy for complete ionization
• In other words, ΔH^⊖_i for Na(g) = 495.6 kJ mol^-1
• So the thermochemical equation for this process will be:
Na(g) + 1/2Cl₂(g) → Na⁺(g) + ½ Cl₂(g); ΔH^⊖_i = 495.6 kJ mol^-1
• This process is numbered as II in the fig.12.13 above
8. So when the process II is complete, we will have:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ Half mol Cl₂ molecules in the gaseous state

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9. Our third process is to separate each of the gaseous Cl₂ molecules into two gaseous Cl atoms
• So it will be a bond dissociation process
10. From the data book, we have:
One mol Cl₂(g) requires 242 kJ energy for complete bond dissociation
• In other words, ΔH^⊖_Cl-Cl = 242 kJ mol^-1
   ♦ In our present case, we have half mol Cl₂ molecules
   ♦ It will give one mol Cl atoms
   ♦ It will require (242/2) = 121 kJ
• So the thermochemical equation for this process will be:
Na⁺(g) + ½ Cl₂(g)  → Na⁺(g) + Cl(g); 1/2ΔH^⊖_Cl-Cl = 121 kJ mol^-1
• This process is numbered as III in the fig.12.13 above
11. So when the process III is complete, we will have:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ One mol Cl atoms in the gaseous state

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12. Our fourth process is to convert each of the gaseous Cl atoms into gaseous Cl^- ions
• So it will be an electron accepting process
• We have seen electron gain enthalpy in an earlier chapter (details here)
13. From the data book, we have:
One mol Cl(g) releases 348.6 kJ energy when all the atoms accept one electron each
• In other words, ΔH^⊖_eg for Cl(g) = 348.6 kJ mol^-1
• So the thermochemical equation for this process will be:
Na⁺(g) + Cl(g) → Na⁺(g) + Cl^-(g); ΔH^⊖_eg = -348.6 kJ mol^-1
• This process is numbered as IV in the fig.12.13 above
14. So when the process IV 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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15. The completion of process IV is a milestone
• Consider the products obtained at the end of this process:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ One mol Cl^- ions in the gaseous state
◼  It is from these products that, we calculate the of ΔH^⊖_lattice(f) NaCl. This fact is clear from the 'definition of ΔH^⊖_lattice(f)' that we wrote in (4) at the beginning of this section
16. So, our fifth process is to make the system release an energy equal to ΔH^⊖_lattice(f)
• When this energy is released, we will get NaCl(s)
• The ΔH^⊖_lattice(f) value of NaCl is available in the data book
   ♦ But remember that, our aim itself is to find this ΔH^⊖_lattice(f)
    ♦ For the time being, we will ignore the value given in the data book, and find it ourselves
• The thermochemical equation of this process will be:
Na⁺(g) + Cl^-(g) → NaCl(s); ΔH^⊖_lattice(f) = X kJ mol^-1
• This process is numbered as V in the fig.12.13 above
• Our aim is to find X

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17. To find X, we split it into X₁ and X₂
• This is shown in fig.6.13
◼  It is clear that:
After completing process IV, if the system releases X₁, the datum line can be reached
• The process in which X₁ is released, is marked as V₁
◼  When this X₁ is released, we can say:
The net energy absorbed/released by the system becomes zero
• Thus we get:
108.4 + 495.6 + 121 - 348.6 - X₁ = 0
⇒ X₁ = 376.4
18. So when the process V₁ is complete, we will have:
   ♦ One mol Na(s)
   ♦ Half mol Cl₂(g)

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19. The completion of process V₁ is another milestone
• We have reached back at the datum
• From here, we have to release some more energy
   ♦ This energy is marked as X₂
• The process in which X₂ is released, is marked as V₂
20. We have to find the magnitude of X₂
   ♦ For this, the red arrow gives us a valuable clue. It can be explained in 4 steps:
(i) Consider the products obtained at the end of process V₁:
   ♦ One mol Na(s)
   ♦ Half mol Cl₂(g)
◼  It is from these products that, we calculate the of ΔH^⊖_(f) NaCl(s)
◼  This fact is clear from the definition of ΔH^⊖_(f) that we wrote in an earlier section
   ♦ It is the enthalpy of formation from constituent elements in the standard states
   ♦ It is available in the data book: -411.2 kJ mol^-1
◼  It is clear that:
From the datum line, if the system releases 411.2 kJ, the bottom most cyan line can be reached
(ii) The process in which this ΔH^⊖_(f) is released, is marked with the downward red arrow
   ♦ The down ward red arrow reaches upto the bottom most cyan line
   ♦ After completing process V₁ also, we have to reach the bottom most cyan line
(iii) So we can write:
   ♦ The process represented by the downward red arrow
   ♦ is equivalent to
   ♦ The process V₂
(iv) Thus we get: X₂ = 411.2 kJ
21. So we have calculated both X₁ and X₂
   ♦ Then X = (X₁ + X₂) = (376.4 + 411.2) = 787.6 kJ
   ♦ Thus we can write: ΔH^⊖_lattice(f) of NaCl(s)= -787.6 kJ mol^-1
22. The cyclic process shown in fig.6.13 is known as the Born-Haber cycle
◼  It is based on the Hess's law of constant heat summation
◼  Whatever be the path, the energy of the system at a particular state, will be the same
◼  This method can be used to determine those ΔH values which are impossible to find experimentally

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• Note:
The above result of -787.6 kJ mol^-1 can be obtained using another approach also. It can be explained in 4 steps:
1. From the datum line, we travel along the path: I - II - III - IV - V
• We reach the bottom most cyan line
• The net energy along this path = (108.4 + 495.6 + 121 - 348.6 - X) = (376.4 - X)
2. From the datum line, we travel along the red arrow
• We reach the bottom most cyan line
• The net energy along this path = -411.2
3. In both the paths:
   ♦ The starting states are the same: The datum line
   ♦ The ending states are the same: The bottom most cyan line
4. So, according to Hess's law, the energies must be the same
• Thus we can write: (376.4 - X) = -411.2
⇒ X = 787.6 kJ mol^-1

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• In the next section, we will see enthalpy of solution


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Chapter 6.5 - Enthalpy Change of Reaction

In the previous section, we saw how 'enthalpy changes' can be determined using calorimetry. In this section, we will see 'enthalpy change of reaction'.

Enthalpy change of reaction can be explained in 13 steps. While writing those steps, we will see enthalpy change of formation also:
1. Consider the equation of a simple reaction: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
• This is a balanced equation. We know that, a balanced equation will give the number of moles of each of the reactants and products involved in the reaction.
• For example, in this reaction, one mol zinc reacts with two mol HCl to give one mol ZnCl₂ and one mol H₂
2. We can write the general form of a balanced equation:
a₁R₁ + a₂R₂ + a₃R₃ + . . . →  b₁P₁ + b₂P₂ + b₃P₃ + . . .   
• R₁, R₂, R₃, . . . are the reactants
    ♦ a₁, a₂, a₃, . . . are the ‘number of mol’ of each of those reactants.
• P₁, P₂, P₃, . . . are the products
    ♦ b₁, b₂, b₃, . . . are the ‘number of mol’ of each of those products.
3. Now we apply the concept of enthalpy to each reactant and product:
• Let ΔH_f[R₁] be the enthalpy change when one mole of R₁ is formed.
    ♦ The subscript 'f' denotes 'formation'.
• This enthalpy change must be determined only after carefully considering two aspects:
(i) R₁ can be formed in many ways.
For example:
• H₂O can be formed by the combination of H₂ and O₂
    ♦ 2H₂ + O₂ → 2H₂O
• H₂O can be formed as one of the products in a reaction
    ♦ HCl + NaOH → NaCl + H₂O
• We must consider only that reaction in which R₁ is formed from it's elements.
• So in the case of H₂O, we must consider only: 2H₂ + O₂ → 2H₂O
(ii) While carrying out the reaction between H₂ and O₂, different labs may use different temperatures and pressures. This will give different ΔH_f values.
• In order to avoid such a confusion, scientists have given a set of rules:
    ♦ The elements from which R₁ is formed must be
          ✰ Under a pressure of 1 bar
          ✰ At a temperature of 298.15 K
          ✰ At the most stable state
          ✰ For example, oxygen is most stable when it exist as O₂ molecules, not as individual O atoms.
4. When the above two aspects are obeyed, the ΔH_f value obtained is written as: ΔH_^⊖f
• The superscript '⊖' indicates that, it is the standard value.
• ΔH_^⊖f is known as the standard enthalpy change of formation.
5. It is interesting to note that, ΔH_^⊖f values for elements is zero.
• For example: ΔH_^⊖f[O₂] = 0,  ΔH_^⊖f[C] = 0   etc.,
    ♦ This is because:
          ✰ O₂ is formed from O₂. There is no enthalpy change.
          ✰ C is formed from C. There is no enthalpy change.
• We can look up the ΔH_^⊖f values of most compounds from the data book or text book.
6. Now consider the general reaction mentioned in (2)
• Let us write down the ΔH_^⊖f values of the reactants R₁, R₂, R₃, . . .
    ♦ They can be written as: ΔH_^⊖f[R₁], ΔH_^⊖f[R₂], ΔH_^⊖f[R₃], . . .
• Let us write down the ΔH_^⊖f values of the products P₁, P₂, P₃, . . .
    ♦ They can be written as: ΔH_^⊖f[P₁], ΔH_^⊖f[P₂], ΔH_^⊖f[P₃], . . .
7. The above ΔH_^⊖f values that we obtain from the data book, are related to one mole.
• For example, the ΔH_^⊖f value of Al₂O₃ is -1675.7 kJ mol^_-1
    ♦ That means, when one mol Al₂O₃ is formed from Al and O₂, the enthalpy change is -1675.7 kJ
• But in our present case, we have:
    ♦ a₁ mol of reactant R₁
    ♦ a₂ mol of reactant R₂ . . . so on
• So we must multiply the values by the corresponding mol number.
• Thus the step (6) can be modified as:
    ♦ ΔH_^⊖f values of the reactants: a₁ΔH_^⊖f[R₁], a₂ΔH_^⊖f[R₂], a₃ΔH_^⊖f[R₃], . . .
    ♦ ΔH_^⊖f values of the products: b₁ΔH_^⊖f[P₁], b₂ΔH_^⊖f[P₂], b₃ΔH_^⊖f[P₃], . . .
8. Now we find two sums:
(i) Sum for the reactants:
$\mathbf\small{\rm{\sum\limits_{i}{a_i \Delta {H^\circleddash}_f[R_i]}}}$ = a₁ΔH_^⊖f[R₁] + a₂ΔH_^⊖f[R₂] + a₃ΔH_^⊖f[R₃] . . .
(ii) Sum for the products:
$\mathbf\small{\rm{\sum\limits_{i}{b_i \Delta {H^\circleddash}_f[P_i]}}}$ = b₁ΔH_^⊖f[P₁] + b₂ΔH_^⊖f[P₂] + b₃ΔH_^⊖f[P₃] . . .
9. Next we subtract the first sum from the second sum.
• The result is: standard enthalpy change of reaction.
    ♦ It is denoted as: ΔH_^⊖r
10. So for the general reaction mentioned in (2), we can write:
Eq.6.9: $\mathbf\small{\rm{\Delta {H^\circleddash }_r=\sum\limits_{i}{b_i \Delta {H^\circleddash}_f[P_i]}-\sum\limits_{i}{a_i \Delta {H^\circleddash}_f[R_i]}}}$
11. Using this equation, let us find the ΔH_^⊖r of the reaction mentioned in (1), which is:
Zn (s) + 2HCl (aq) → ZnCl₂ (s) + H₂ (g)
It can be done in 2 steps:
(i) From the data book (values can be obtained online also) , we have:
    ♦ ΔH_^⊖f[Zn(s)] = 0 kJ mol^_-1
    ♦ ΔH_^⊖f[HCl(aq)] = -167.16 kJ mol^_-1
    ♦ ΔH_^⊖f[ZnCl₂(aq)] = -488.2 kJ mol^_-1
    ♦ ΔH_^⊖f[H₂(g)] = 0 kJ mol^_-1
(ii) Applying Eq.6.9, we get:
ΔH_^⊖r = [-415.1 -(2 × -167.16)] = -153.88 kJ mol^_-1
12. Now we will see the significance of the sign (+ve or -ve) of the ΔH value.
• It can be written in 4 steps:
(i) We have: U_B = U_A + Q_P  – W
(Here we asume that, the system absorbs Q_P. So it is given a positive sign)
⇒ U_B = U_A + Q_P – P(V_B – V_A)
⇒ (U_B + PV_B) – (U_A + PV_A) = Q_P
⇒ ΔH = H_B – H_A = Q_P
(ii) We see that, if H_B is greater than H_A, ΔH will be positive.
• Q_P will also be positive
    ♦ That means our assumption is correct.
    ♦ That means, the system absorbs heat.
(iii) So we can conclude that:
A positive ΔH indicates that the system absorbs heat. It is an endothermic reaction.
(iv) We can write the converse also:
A negative ΔH indicates that the system releases heat. It is an exothermic reaction.
13. We have seen how ΔH_^⊖r is calculated for the reaction between Zn and HCl.
• Let us see another example. This time we want the ΔH_^⊖r of the following reaction:
CaCO₃(s) → CaO(s) + CO₂(g)
• This is the decomposition reaction of calcium carbonate. The answer can be written in 3 steps:
(i) From the data book, we have:
    ♦ ΔH_^⊖f[CaCO₃(s)] = -1206.92 kJ mol^_-1
    ♦ ΔH_^⊖f[CaO(aq)] = -635.09 kJ mol^_-1
    ♦ ΔH_^⊖f[ZnCl₂(aq)] = -393.51 kJ mol^_-1
(ii) Applying Eq.6.9, we get:
ΔH_^⊖r = [-635.09 - 393.51 - (-1206.92)] = 178.32 kJ mol^_-1
• We get a positive value. That means, we have to supply 178.32 kJ mol^_-1 for the reaction to take place.
(iii) From the balanced equation, it is clear that, one mol CaCO₃ is undergoing decomposition.
• So we can write: 178.32 kJ is required for the decomposition of one mol CaCO₃
• If we know the mass (in grams) of CaCO₃ at the beginning of the reaction, we can calculate the number of moles present in that mass.
• Using that ‘number of moles’, the heat energy required can be calculated.
• This is the advantage of knowing the ΔH_^⊖r value of a reaction.

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• It is clear that, the ΔH_^⊖f values of individual reactants and products play an important role in the ΔH_^⊖r value of the overall reaction.
• In step (3), we have seen two important aspects about ΔH_^⊖f
    ♦ Now we will see them in some more detail
• It can be written in 5 steps:
1.We have seen that, the ΔH_^⊖f values can be looked up from the data book.
• Those values in the data book are determined by scientists using precision instruments in the lab.
• Those experiments are carried out under standard conditions (1 bar pressure and 298.15 K)
    ♦ So whenever the experiments are repeated in different labs, the same ΔH_^⊖f values will be obtained.
2. The experiments are chosen in such a way that the required compound is formed from the constituent elements only.
• This point can be explained using an example:
(i) The following reaction → CaCO₃ as the product:
CaO(s) + CO₂(g) → CaCO₃(s)
    ♦ ΔH for the reaction is: -178.3 kJ mol_^-1
(iii) CaCO₃ is the only product. Also, only one mol CaCO₃ is formed.
• So it appears that ΔH_^⊖f of CaCO₃ is -178.3 kJ mol_^-1
• But it is not the acceptable value because, in this reaction, CaCO₃ is formed from other compounds.
• For the ΔH to be acceptable as ΔH_^⊖f, there must be only Ca, C and O (in their pure forms) in the left side of the equation in (i)
3. Consider the reaction given below:
H₂(g) + Br₂(l) → 2HBr(g)
• The ΔH for this reaction is -72.8 kJ mol^-1
• Can we take -72.8 kJ mol^-1 as the ΔH_^⊖f value of HBr?
   ♦ On the left side only H₂ and Br₂ is present.
   ♦ On the right side, only HBr is present.
   ♦ So at a first glance, it appears that, -72.8 kJ mol-1 is indeed the ΔH_^⊖f value of HBr.
• But remember that, value is related to the formation of one mole of a compound.
   ♦ In our present case, two moles of HBr is formed
   ♦ So 72.8 kJ is the energy released when two moles of HBr is formed.
   ♦ Thus, 72.8 kJ mol^-1 is not acceptable.
• However, if we divide the equation through out out by 2, we will get:
¹⁄₂H₂(g) + ¹⁄₂Br₂(l) → HBr(g)
   ♦ This time, only one mol HBr is formed. So the energy released will be half of 72.8 kJ
   ♦ That means, when one mol HBr is formed, the energy released is (¹⁄₂ × 72.8) = 36.4
   ♦ So we can write: ΔH_^⊖f of HBr is -36.4 kJ mol^-1
4. Let us compare ΔH_^⊖f and ΔH_^⊖r
• We  know that, both are 'enthalpy changes' taking place during reactions.
• But ΔH_^⊖f is a special case of ΔH_^⊖r
   ♦ Because, for an enthalpy change to be acceptable as ΔH_^⊖f, the three rules mentioned above must be satisfied.
• The three rules can be summarized as follows:
(i) standard conditions should be adopted.
(ii) On the left side of the reaction equation, there must be the constituent elements (in standard form) only.
• On the right side, there must be only one compound
(iii) On the right side, there must be only one mol of the compound.
5. The reader must try and become convinced that:
    ♦ All ΔH_^⊖f values are ΔH_^⊖r values
    ♦ But all ΔH_^⊖r values are not ΔH_^⊖f values

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In the next section we will see thermochemical equations

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