Chapter 6.12 - Spontaneity

In the previous section, we saw details about endothermic or exothermic nature of enthalpy of solution. In this section, we will see spontaneity

• Basics of spontaneity can be written in 15 steps:
1. Consider the open vessel in fig.6.20(a) below. It contains some water at room temperature 25 ^०C

Fig.6.20

2. We know that, a mass of water is nothing but a collection of H₂O molecules
• All those molecules are in random motion
• So all those molecules have kinetic energies
• The temperature of the water mass is the average kinetic energy of the molecules
3. Due to the random motion, the molecules are constantly colliding with each other
• Imagine that, due to the collisions, some molecules near the surface of the water, looses considerable kinetic energies
    ♦ Then the temperature of those molecules will fall
• Let the temperature fall to such a low level that, those molecules freeze and become ice
4. Remember that, loss of kinetic energy is due to collisions. The kinetic energies lost by the molecules will be gained by the surrounding molecules
• So, when the molecules freeze and become ice, the surrounding molecules become warm
• We will get some ice at the center of the vessel and that ice will be surrounded by warm water. This is shown in fig.6.20(b)
5. Formation of ice in this manner, does not defy the law of conservation of energy because,
    ♦ there is no destruction of energy
    ♦ also, no energy is created
• The energy lost by some molecules is gained by some other molecules
6. Even though the law of conservation is obeyed, we never see such a spontaneous formation of ice at room temperature

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• A Spontaneous process is the one which do not require external source of energy to proceed
• For a process to be called spontaneous, it is not necessary that it occurs at a high speed
• A slow process can also be called spontaneous if it does not require external supply of energy
• The reaction between hydrogen and oxygen is an example of a slow spontaneous process
    ♦ A mixture of H₂ and O₂ can be left undisturbed in a container for many years
    ♦ There will not be any noticeable effects
    ♦ But the reaction will be taking place all the time, with out any aid of external energy

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7. The spontaneous formation of ice in this manner requires heat to flow (without external help) from a cold object to a hot object
◼ We never observe such a flow. What we observe is the flow of heat from hot object to cold object
• We can say:
    ♦ The spontaneous process proceeds only in one direction: Ice → Water
    ♦ We never see the spontaneous process: Water → Ice
        ✰ For this process, we have to supply energy through a refrigerator
◼ In fact, all naturally occurring processes (physical or chemical) will tend to proceed in one direction only
8. Let us see another example:
• If a canister containing some gas is opened, the gas molecules will spontaneously spread out into the whole volume of the room
• We never see the gas molecules in the room to enter back spontaneously into the canister
9. One more example:
• Consider the burning of carbon
• During the process, carbon combines with oxygen to give carbon dioxide
    ♦ That is: C + O₂ → CO₂
    ♦ This is a spontaneous process
    ♦ Once the carbon is ignited, no external energy is required to keep the process going
• In the reverse process, carbon and oxygen is obtained from carbon dioxide
    ♦ That is: CO₂ → C + O₂
    ♦ This reverse process is not spontaneous
    ♦ Energy is required to accomplish this reverse process
10. We see that, all spontaneous processes proceed in one direction only
• We want to answer this question:
Why all spontaneous processes proceed only in one direction?
11. To find the answer, we consider some common phenomena that we see in our day to day life
(i) Flowing of water
• The flow of water starts from top of hill and ends at the ground level
• During the flow, the potential energy stored in the water is continuously released in the form of kinetic energy
• When the water reaches the ground level, it’s potential energy will be zero. This is because, all the potential energy is released (in the form of kinetic energy) into the surroundings
(ii) Stone falling from a height
• The fall of stone starts from a higher level and ends at the ground level
• During the fall, the potential energy stored in the stone is continuously released in the form of kinetic energy
• When the stone reaches the ground level, it’s potential energy will be zero. This is because, all the potential energy is released (in the form of kinetic energy) into the surroundings
12. The flow of water and fall of stone are spontaneous processes. They do not require any aid of external energies
• So we are inclined to think that:
All processes in which there is a ‘release of stored energy’ will be spontaneous
13. We know that, in exothermic reactions, the stored chemical energy is released as heat energy
• So we are inclined to think that:
    ♦ All exothermic reactions are spontaneous
    ♦ The reverse of an exothermic reaction is endothermic
        ✰ It involves absorbtion of energy
    ♦ So a spontaneous reaction proceeds in the exothermic direction only
• Let us examine whether this is true
14. Some thermochemical equations are given below:
(i) 1/2N₂(g) + 3/2H₂(g) → NH₃(g); ΔH^⊖_r = – 46.1 kJ mol^-1
(ii) 1/2H₂(g) + 1/2Cl₂(g) → HCl(g); ΔH^⊖_r = – 92.32 kJ mol^-1
(iii) H₂ + 1/2O₂(g) → H₂O(l); ΔH^⊖_r = –285.8 kJ mol^-1
• The above three reactions are spontaneous
• The negative sign of ΔH^⊖_r shows that, they are exothermic reactions
◼ So we become even more inclined to think that:
    ♦ All exothermic reactions are spontaneous
    ♦ Being exothermic is the only criterion for spontaneity
15. But before making a decision, let us see two more thermochemical equations:
(i) 1/2N₂(g) + O₂(g) → NO₂ (g); ΔH^⊖_r = +33.2 kJ mol^-1
(ii) C(graphite, s) + 2S(l) → CS₂(l); ΔH^⊖_r = +128.5 kJ mol^-1
• The above two reactions are spontaneous
• The positive sign of ΔH^⊖_r shows that, they are endothermic reactions

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◼ So it is impossible to conclude that:
Being exothermic is the only criterion for a reaction to be spontaneous
• Scientists became convinced that, there are some other factors also playing major roles
• Researches in this direction lead to the discovery of entropy and the second law of thermodynamics
• We will see them in the next section


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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 6.5 - 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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