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 |
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:
◼ 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.
• 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
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
• 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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