What is Gibbs Free Energy?

In the previous post of thermodynamics we have seen how the change in entropy decides the spontaneity of any reaction. But entropy alone cannot decide spontaneity, enthalpy of the reaction and temperature at which the reaction is being carried out also plays an important role. To sum up all these factors we have another thermodynamic property known as Gibbs free energy “G”.

G = H – TS

‘G’ is an extensive property. It is a state function as it doesn’t depend on path. Units of Gibbs energy is J.

Let’s find change in Gibbs free energy of the system at constant temperature:

ΔG_sys = ΔH_sys – TΔS_sys    ......(1)

This is one of the important equations of thermodynamics. Difference in G decides the spontaneity of any reaction, let’s find out how.

We know that total entropy change for a spontaneous reaction must be more than zero. Total entropy means the entropy of the system and surrounding.

ΔS_total = ΔS_sys  + ΔS_surr       ......(2)

For an exothermic reaction ΔH heat is released from the system to the surrounding. It means when -ΔH_sys heat is lost from the system, it is added to the surrounding as ΔH_sur.

ΔS_sys = -ΔH_sys / T

Then,

ΔS_sur = ΔH_sur / T

Since ΔH_sur = - ΔH_sys

ΔS_sur = ΔH_sur / T = - ΔH_sys / T

On putting the values of ΔS_sur in equation 2 we get,

ΔS_total  = ΔS_sys + (-ΔH_sys / T)

TΔS_total  = TΔS_sys  - ΔH_sys

TΔS_total  =  - (ΔH_sys  - TΔS_sys)

For spontaneity ΔS_total must be greater than zero, so

- (ΔH_sys  - TΔS_sys) > 0

On comparing it with equation 1 we can say that for spontaneity,

-ΔG_sys > 0

- ΔG_sys can be more than zero only when ΔG_sys is negative,

- (-ΔG_sys)  > 0

Or

ΔG_sys  < 0

So, for spontaneity ΔG_sys must be negative.  
 

Gibbs Free Energy and Spontaneity

We know that ΔG_sys = ΔH_sys – TΔS_sys, where ΔH_sys is the enthalpy of the system and TΔS_sys is the energy required to be released or absorbed by the system and by subtracting them you will get the remaining energy ΔG which can be used to do any work. That’s why ΔG is known as free energy which is available to do useful work. For any spontaneous reaction ΔG is negative which means all energy has been used to carry the reaction forward (spontaneous reactions are irreversible). Reactions for which ΔG is negative are known as ‘exergonic reactions’. ‘Endergonic reactions’ have positive ΔG.

Let’s see how we can predict the spontaneity of any reaction by using equation of ΔG:

A + B ⟶ C + D ...... ΔH = (+ve) and ΔS = (+ve)

On applying Gibbs equation

 ΔG_sys = ΔH_sys – TΔS_sys

At lower temperature ΔG will be positive so that above reaction will be non-spontaneous at lower temperature. But if temperature is higher enough to overrule the ΔH, then it can be spontaneous.

Now we understand that if ΔG < 0 then reaction will be spontaneous, and if ΔG > 0 then it will be non-spontaneous. So what happens if ΔG = 0? Reaction will be at equilibrium if ΔG = 0.

ΔG = ΔG^ө + RT lnQ

At equilibrium reaction quotient Q becomes equal to equilibrium constant K and ΔG = 0

ΔG = ΔG^ө + RT lnK = 0

ΔG^ө =- RT lnK

Or

 ΔG^ө =- 2.303RT logK

lnK = -ΔG^ө/ RT

Taking antilog of both side,

K = e ^(-ΔGө/RT)

where ΔG^ө is the standard gibbs energy of the reaction.

When ΔG^ө < 0 then  (-ΔG^ө/ RT) will be positive and e ^{(-ΔGө /RT)} will be more than zero. That means K > 1 which means forward reaction is favoured and equilibrium shifts forward. Here product concentration exceeds reactant concentration.

When ΔG^ө > 0 then  (-ΔG^ө/ RT) will be negative and e ^{(-ΔGө /RT)} will be less than zero. That means K < 1 which means backward reaction is favoured and equilibrium shifts backward and product formation is suppressed.

I hope these posts have been helpful in understanding the thermodynamic concepts. The next post of thermodynamics will focus on the applicability of these concepts where we will try to apply these concepts to solve some numerical problems.

We have discussed all about feasibility and spontaneity of any reaction but how do we decide the speed of the reaction? On which factors it depends? Can we accelerate its speed? We will find the way to solve this mystery in the coming posts of chemical kinetics.

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Qualitative analysis of Group III(A) cations

In the previous post of salt analysis we have learnt why we have to remove interfering radicals before 3^rd group analysis. After removing the interfering radicals, we will proceed to analyse the 3^rd group cations in the filtrate. To separate the radicals of 3^rdgroup we need to get them precipitated. The group reagent of 3^rd group is ammonium sulphide solution or hydrogen sulphide gas in the presence of ammonia and ammonium chloride. When we add group reagent to the filtrate we will get precipitate of 3^rd gr cations.

Cations of this group are cobalt(II) Co²⁺, Nickel(II) Ni²⁺, iron(II) Fe²⁺, iron(III) Fe³⁺, chromium(III) Cr³⁺, aluminium(III) Al³⁺, zinc(II) Zn²⁺, manganese(II) Mn²⁺, manganese(VII) Mn⁷⁺,  

Iron, aluminium and chromium and sometimes Manganese are precipitated as hydroxide by ammonia solution in presence of ammonium chloride while others are precipitated with ammonium sulphide in the form of sulphides. That’s why 3^rd group is further divided into A and B group.

III(A) is known as Iron group and it consists of iron(III) Fe³⁺, chromium(III) Cr³⁺, aluminium(III) Al³⁺.

III(B) is known as Zinc group and it consists of cobalt(II) Co²⁺, Nickel(II) Ni²⁺, zinc(II) Zn²⁺, manganese(II) Mn²⁺, manganese(VII) Mn⁷⁺.

Now let's continue from the group II filtrate and discuss the steps we have to follow to separate III(A) cations

Qualitative analysis of Group III(A) cations 

Step 1: Precipitation of III(A) cations as Hydroxide - To the filtrate, add few drops of nitric acid HNO₃ to convert ferrous Fe²⁺ into ferric Fe³⁺. Then add solid ammonium chloride NH₄Cl₃ and ammonia NH₃ solution and boil. On boiling you will get precipitate of III(A) cations which may contains Fe(OH)₃, Cr(OH)₃, Al(OH)₃ and a little MnO(OH)₂. Solubility product of iron(III) hydroxide is very small so it precipitates completely.

Fe³⁺ +3NH₃ + 3H₂O ⟶ Fe(OH)₃↓ + 3NH₄⁺

Ammonium chloride should be added in excess otherwise III(B) cations may get precipitated here. But avoid too much excess of ammonium chloride and ammonia otherwise Cr³⁺ may not precipitate and Al³⁺ may form colloidal solution. .

Cr³⁺ + 3NH₃ + 3H₂O ⟶ Cr(OH)₃↓ + 3NH₄⁺

Cr(OH)₃↓ + 6 NH₃ ⟶ [Cr(NH₃)₆]³⁺ + 3OH^-

Al(OH)₃↓ + OH^- ⟶ AlO₂^- + 2H₂O

Step 2: Separation of Fe(III) and Mn(II) from Al(III) and Cr(III) – Wash the precipitate with hot 

water and transfer to a test tube. Add a little water and add 1-2g of sodium peroxide or sodium hydroxide solution and bromine water/ hydrogen peroxide H₂O₂. Boil and filter in hot. On boiling, hydroxides of aluminium and chromium dissolve by forming meta- aluminate and chromate ion, and iron and manganese remain in the solution.

Al(OH)₃↓ + OH^- ⟶ AlO₂^- + 2H₂O

2Cr(OH)₃↓ + 2O₂^2- ⟶ 2CrO₄^2- + 2 OH^- + 2H₂O

Filter the residue to test for iron and manganese, and keep the filtrate for the test of aluminium and chromium.

Step 3: Test for iron and manganese – if residue is black, it means iron and manganese both are present and if it is reddish brown then only iron is present. Divide the residue into two parts.

Part 1 confirmatory test for Fe³⁺- dissolve the precipitate in dil HCl.

Fe(OH)₃ + H⁺ ⟶ Fe³⁺ + 3H₂O

Add few drops of potassium hexacyanoferrate (II) K₄[Fe(CN)₆] solution, intense blue precipitate of iron(III) hexacyanoferrate will appear.

4Fe³⁺ + 3 [Fe(CN)₆]^4- ⟶ Fe₄[Fe(CN)₆]₃ ↓

Excess of reagent dissolves it completely and intense blue solution is obtained and on adding sodium hydroxide solution to this blue solution, a red precipitate of iron(III) hydroxide is obtained.

Fe₄[Fe(CN)₆]₃ + 12OH^- ⟶ 4Fe(OH)₃ ↓ + 3[Fe(CN)₆]^4-

Part 2 confirmatory test for Mn²⁺- dissolve the precipitate in 1ml concentrated nitric acid HNO₃, if necessary add 1-2 drops of H₂SO₃. Add 0.05-0.1g sodium bismuthate NaBiO₃ and shake. A violet solution of permanganate will appear.

2Mn²⁺ + 5NaBiO₃ + 14H⁺ ⟶ 2MnO^-₄ + 5Bi³⁺ + 5Na⁺ + 7H₂O

Step 4: Test for aluminium and chromium – The filtrate from step 2 may contain CrO^-₄ and [Al(OH)₄]^-. If it is yellow, it means chromium might be present but if it is colourless then chromium is absent and you don’t need to test for it.

Divide the filtrate into 3 parts, we will test for chromium in 2 parts and in the third part we will test for aluminium.

Part 1 Test for Cr (III): acidify with dilute acetic acid CH₃COOH and add 0.25M lead acetate Pb(CH₃COO)₂ solution, a yellow precipitate of lead chromate is formed.

2CrO₄^2- + Pb²⁺ ⟶ PbCrO₄↓

This yellow precipitate is soluble in sodium hydroxide solution. it conforms the presence of Cr(III).

PbCrO₄↓ + 4OH^- ⟶ PbO₂^2- + CrO₄^2- + 2H₂O

Part 2 Test for Cr (III): acidify it with dilute nitric acid HNO₃, cool thoroughly and add 1ml amyl alcohol (2-methyl-butane-4-ol) and 3-4 drops of 3% hydrogen peroxide solution H₂O₂. Shake well and allow the two layers to separate. Upper blue layer contains perchromic acid (chromium pentoxide).

2 CrO₄^2- + 2H⁺ + 3 H₂O₂ ⟶ 2CrO₅ + 4 H₂O

Part 3 Test for Al (III): To test for aluminium acidify the filtrate with dilute HCl (test with litmus paper) then add 2M ammonia solution NH₃ until just alkaline. Heat till boiling. White gelatinous precipitate of aluminium hydroxide is obtained.

Al³⁺ + 3NH₃ + 3H₂O ⟶ Al(OH)₃↓ +3NH₄⁺

Filter this precipitate and dissolve a small portion of it into 1ml hot dil HCl. Cool and add 1ml 6M ammonium acetate solution and 0.5ml of 1M aluminon reagent (0.1g tri-ammonium arurine- tricarboxylate O(COONH₄)C₆H₃=C[C₆H₃(OH) COONH₄]₂dissolved in 100ml water), stir it and add ammonium carbonate solution. a red coloured precipitate confirms the presence of Al(III). In this post we discussed analysis of the III(A) group cations. In the next post of salt analysis we will continue with the test test for III(B) cations in the filtrate of III(A).

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Entropy and Spontaneity of The Reaction

Spontaneous reactions are self driven reactions; they take place immediately without any assistance from external source. They have the potential to proceed without any help from outside. What is their driving force? What makes them to proceed by themselves? If we take examples from our day to day experiences, we will be able to understand the concept of spontaneity. When we toss a coin in the air it drops on the ground, here the dropping of coin is spontaneous reaction. Similarly, falling of water in a waterfall is an example of spontaneous reaction. Did you notice that these reactions are spontaneous in one direction only? Coin drops spontaneously but doesn’t jump up spontaneously; water falls from hills but doesn’t climb up spontaneously. In this post we will try to decipher the mystery of spontaneous reactions.

Spontaneity in chemical reactions is also unidirectional. Second Law of Thermodynamics explains it quite well. It also gives us an idea about the direction of energy or heat flow. As per this law heat flows from hotter to colder bodies or you can say that energy flows from higher level to lower level.

It means that if energy decreases on occurring of a reaction, then it may be a spontaneous reaction. You might think that if a reaction doesn’t require heat to start and is exothermic, it must be spontaneous. Let’s check this assumption:

You must have observed rust on iron rods, pipes or nails. Rust is an oxide of iron which is made by the reaction of Oxygen with Iron.

4Fe (s) + 3O₂ (g) ⟶ 2Fe₂O₃(s)    ..... Δ_rH^ө = -1648 kJ mol^-1

It is an exothermic reaction and spontaneous too. It seems that our assumption is going in right direction. Let’s take another example of exothermic reaction, formation of water.

H₂ (g) + ½ O₂(g) ⟶ H₂O (l)      ..... Δ_rH^ө= -285.8 kJ mol^-1

Even though formation of water is an exothermic reaction, if you mix Hydrogen and Oxygen gas at room temperature you won’t see any identifiable change even after years. Here we are proved wrong. It is not necessary that if a reaction is exothermic then it must be spontaneous too.

Another example of spontaneous reaction from our daily life is dissolution of table salt in water. Let’s see its chemical reaction:

NaCl (s) ⟶ Na⁺ (aq) +  Cl^- (aq) ..... Δ_rH^ө= +4 kJ mol^-1

It’s an endothermic reaction. But it occurs spontaneously. So how can we predict the spontaneity of any reaction if it doesn’t depend on heat of the reaction?

For this, we need another thermodynamic property ‘ENTROPY’ which is denoted by “S”. Entropy is the degree of randomness or you can say it’s a measure of freedom of the molecules. Let’s take the example of different phases of water to understand the concept of entropy.

In ice - molecules are held together by H-bonding and are bound to stay in one place. They have no freedom at all that means they have least entropy.

In water - molecules stay together because of H-bonding but they can move too. They have more freedom than they have in ice. That means entropy of water is more than ice.

In steam - molecules are too far apart to form H-bonds and they can move freely in any direction. So you can see that in vapour phase molecules are most free and therefore out of all phases of water, steam has the highest entropy.

What is Entropy?

Let’s take previous examples of spontaneous reactions and study them from the viewpoint of entropy.

For the formation of water, two gases H₂ and O₂ need to combine to form liquid H₂O. That means entropy of the system decreases in this reaction. That is why it is not spontaneous even if it is an exothermic reaction.

In the case of dissolution of salt, solid NaCl breaks into aqueous ions Na⁺(aq) and  Cl^-(aq) which are now free to move around. That means entropy of the system increases which favours spontaneity.

In both of the examples above, we have seen that there is a relation between the motion of molecules and its entropy. More freedom of motion means more entropy. Entropy doesn’t depend on path which means it’s a state function just like enthalpy and internal energy. Motion of molecules is related to the energy of molecules, and energy of molecules has direct relation to the heat. We can manipulate the energy by controlling the heat supply. So there must be some relation between entropy and heat.

Entropy (S) ∝ Heat (q)

How much motion can be generated by a given amount of heat depends on the temperature of the system. Let’s take an everyday example. A hot cup of coffee on a cold day can energize you but a similar cup of coffee in a hot day can irritate you. Same amount of heat supplied at lower temperate can create more difference in entropy. Because at lower temperate molecules are at rest and a little heat can create chaos and generate more randomness. While at higher temperature molecules are already in chaotic motion and heat can give more energy but doesn’t create more randomness because they are already in that state.  So we can conclude that change in entropy is inversely proportional to the temperature.

ΔS = q_rev/ T

When a system is in equilibrium, it’s entropy is maximum and change in entropy ΔS = 0.

For any spontaneous reaction total entropy change of the system and surrounding must be more than zero.

ΔS_total = ΔS_system + ΔS_surrounding  > 0

Now check the spontaneity of the formation of rust.

4Fe (s) + 3O₂ (g) ⟶ 2Fe₂O₃(s)    ..... Δ_rH^ө = -1648 kJ mol^-1

We have observed it time and again that rust formation is a spontaneous reaction even though entropy change given for this reaction is  -549.4 JK^-1mol^-1at 298K.

The discussion of spontaneity we had so far contradicts this observation. We discussed that for spontaneity total entropy must be greater than zero. How can this reaction be spontaneous when the entropy of the system is negative? Well, we did forget something, didn't we? Let’s work out the entropy change of the surrounding. This reaction is exothermic which means 1648 kJ mol^-1 amount of heat is absorbed by the surrounding.

ΔS_surr = Δ_rH^ө/ T

ΔS_surr = 1648 kJ mol^-1/ 298K

ΔS_surr = 5.53 kJ mol^-1K^-1= 5530 Jmol^-1K^-1

Now calculate the change in total entropy

ΔS_total = ΔS_system + ΔS_surrounding

ΔS_total = -549.4 JK^-1mol^-1+ 5530 Jmol^-1K^-1

ΔS_total = 4980.6 JK^-1mol^-1

The total entropy increases by the rusting which favours spontaneity. This is in accordance with the Second Law of Thermodynamics which tells us that on the transformation of energy from one form to another form entropy always increases and energy always decreases.

You have seen that enthalpy alone isn’t the deciding factor for a process to be spontaneous; you have to consider the entropy as well. As we saw in the example of rusting, it is quite confusing to decide the spontaneity of the reaction by just seeing the enthalpy and entropy of the system. To ease this problem, one more thermodynamic term has been introduced which is called the “Gibbs free energy”. In the next post we will discuss Gibbs free energy and will try to find out how we can predict spontaneity of any reaction by using it.

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Born Haber Cycle

Lattice enthalpy Δlattice Hө is the enthalpy change when one mole of ionic compound dissociates into its constituent atoms in the form of gaseous ions. According to this definition lattice enthalpy is always positive which means it will be endothermic process.

            MX(s) ⟶ M⁺(g) + X^-(g)            Δ_lattice H^ө = “+”

But you can also say that lattice energy is the energy which is released when gaseous ions bind together to form ionic compound. By this definition lattice enthalpy is always negative which means it will be exothermic process.

            M⁺(g) + X^-(g) ⟶ MX(s)            Δ_lattice H^ө = “-”

In general terms we can say that lattice energy is the energy difference between separate gaseous ions and ionic solid. In the previous post of thermodynamics you have seen that there is no direct way to calculate Δ_lattice H^ө experimentally. Two German scientists Max Born and Fritz Haber developed an indirect way to calculate the lattice energy.

They applied Hess’s law of constant heat summation in the enthalpy diagram. They designed a cycle of reactions which has 5 steps.

Step 1: Formation of NaCl from its constituent elements in their standard stage.

            Na(g) + Cl(g)  ⟶ NaCl(s)         ...........Δ_f H^ө = -411.2 kJ mol^-1

Step 2: Sublimation of Na(s) to Na(g) or you can also call it atomization

            Na(s) ⟶ Na(g)             ...........Δ_a H^ө = 108.4 kJ mol^-1

Step 3: Atomization of Cl₂ to Cl(g). This process involves the dissociation of Cl-Cl bond, so it will be equal to the bond dissociation enthalpy of Cl-Cl.

            Cl₂(g) ⟶ Cl(g)              ...........Δ_Cl-Cl H^ө = 242 kJ mol^-1

For the formation of NaCl we need only one Cl and the value given abov is for Cl₂, so we will take half of its value (242/2) = 121 kJ mol^-1.

Step 4: (a) Ionization of Na(g) to Na⁺

                    Na(g) ⟶ Na⁺(g) + e^-                 ...........Δ_ionization H^ө = 496 kJ mol^-1

(b) Ionization of Cl(g) to Cl^-(g). In this process Cl gains an electron to get converted to Cl^-(g). So we need to calculate the electron gain enthalpy.

               Cl(g) + e^- ⟶ Cl^-(g)        ...........Δ_eg H^ө = -348.6 kJ mol^-1

Step 5: Na⁺ and Cl^- ions come closer and arrange themselves in lattice to form NaCl.

            Na⁺(g) + Cl^-(g) ⟶ NaCl(s)         ...........Δ_lattice H^ө = -? kJ mol^-1

Cycle starts from the formation of NaCl from its constituent elements and ends by the union of Na⁺ and Cl^- to form NaCl. We know that the enthalpy change in a cyclic process is zero.

0 = Δ_f H^ө + Δ_a H^ө+ ½ Δ_Cl-Cl H^ө + Δ_ionization H^ө + Δ_egH^ө + Δ_lattice H^ө

Now let's put the sign of all enthalpies. Δ_eg H^өis always negative since energy is required to add an electron to the neutral atom. Δ_lattice H^ө also is negative because energy is released when two ions come together to form ionic solid.

0 = Δ_f H^ө + Δ_a H^ө+ ½ Δ_Cl-Cl H^ө + Δ_ionization H^ө - Δ_egH^ө - Δ_lattice H^ө

Enthalpy of formation, bond dissociation, ionization and electron gain enthalpy can be measured experimentally. As we want to calculate lattice enthalpy, we will transfer it to the left side of the equation.  

Δ_lattice H^ө = Δ_f H^ө+ Δ_a H^ө + ½ Δ_Cl-Cl H^ө + Δ_ionizationH^ө - Δ_eg H^ө

Now let's put in the values:

Δ_lattice H^ө = 411.2 kJ mol^-1+ 108.4 kJ mol^-1 + 121 kJ mol^-1 + 496 kJ mol^-1- 348.6 kJ mol^-1

Δ_lattice H^ө = + 788 kJ mol^-1

Born Haber Cycle  

That means 788 kJ mol^-1 energy is required to break one mole of NaCl into Na⁺ and Cl^- ions. Lattice energy defines the stability of ionic solids. The high value of lattice energy is the reason why ionic solids have such high melting and boiling point.

As you have seen that ionization enthalpy (IE) and electron gain enthalpy (EG) play an important role in determining the value of lattice energy and we can compare the lattice energy of given ionic compounds by observing the trends of IE and EG. Let’s try:

Compare the lattice energy of NaCl and MgCl₂

IE of Mg  > Na so the lattice energy of MgCl₂(2326 kJ mol^-1) will be higher than NaCl (788 kJ mol^-1).

Compare the lattice energy of NaCl and LiCl.

Li has higher IE than Na, so LiCl has higher lattice energy (860 kJ mol^-1) than NaCl (788 kJ mol^-1).

Compare the lattice energy of MgCl₂ and MgO

Electron gain enthalpy: first Eg of O is lesser than Cl, but the second Eg will be much higher as it becomes difficult to add one more electron to the negatively charged ion.

So the lattice energy of MgO (3795 kJ mol^-1) will be higher than that of MgCl₂ (2326 kJ mol^-1).

Now you have understood the first law of thermodynamics. You have learnt about internal energy and enthalpy. But we need one more thermodynamic property to understand the spontaneity of a reaction. You may think that if a reaction doesn’t require heat to start and is exothermic, it must be spontaneous. But it doesn’t happen in every reaction. In the next post of thermodynamics we will discuss this in detail.

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Hess’s Law of Constant Heat Summation

Russian chemist Germain Henri Hess gave an important law about the enthalpy change. He said that enthalpy change of one compound to the other compound is always same, even if it occurs in one step or more steps.

Enthalpy is a state function that means it doesn’t depend on the path taken by the system. Let’s take an example to understand it better:

A + B ⟶ D      Δ_rH

This reaction may take place into more steps like:

A + B ⟶ C         Δ_rH₁       .....equation 1                 

C ⟶ D             Δ_rH₂       .....equation 2

Then you can get Δ_rH by simply applying mathematical operations by adding equation 1 and 2.

[A + B ⟶ C Δ_rH₁] + [ C ⟶ D Δ_rH₂] = A + B ⟶ D  

  

 So, Δ_rH = [Δ_r H₁ + Δ_rH₂]

Let’s take another example: if you have given two different reactions:

C(graphite) + O₂(g) ⟶ CO₂(g) Δ_r H= -393.5 kJ mol^-1    ........equation 1                 

CO(g) + ½ O₂(g) ⟶ CO₂(g)      Δ_r H= -283 kJ mol^-1           .........equation 2                 

And asked you to calculate the Δ_r H of the following reaction

C(graphite) + ½ O₂(g) ⟶ CO(g)           Δ_r H =?

It’s puzzle time; try to rearrange the above two equation so that you can get the desired equation. We want the carbon and oxygen on the reactant side and CO on the product side. We can take equation 1 as it is given, but we have to reverse the equation 2 to get the CO on the product side. When you reverse the equation 2, sign of Δ_rH also gets reversed. And you get a new equation 3.

CO₂(g) ⟶ CO(g) + ½ O₂(g)     Δr H= +283 kJ mol^-1          .........equation 3                 

Now if we add the equation 1 and 3 we can get the desired equation.

Δ_r H = [Δ_rH₁ -393.5 kJ mol^-1] + [Δ_r H₃ + 283 kJ mol^-1]

Δ_r H = -110.5 kJ mol^-1

So you can state the law of constant heat summation as ‘If a reaction takes place in several steps then its Δ_r H^ө is the sum of standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.’

 Δ_r H^ө= Δ_r H^ө₁ + Δ_r H^ө₂+ Δ_r H^ө₃ + ....Δ_r H^ө_n

Hess’s Law of Constant Heat Summation

Bond Dissociation Enthalpy Δbond H^ө

Now you have learnt to calculate enthalpy of reaction by using standard enthalpy of formation and by using standard enthalpies of intermediate reactions. There is one more way to calculate the enthalpy of reaction. When any reaction occurs, some bonds are broken and some new bonds are formed. Energy is needed to break the bonds and energy is released when new bonds are formed. It means if we calculate the bond enthalpies of reactant and products we can get the enthalpy of reaction.

Δ_r H = Ʃ Δbond H^ө(reactants) - Ʃ Δbond H^ө(products)

Bond enthalpy or bond dissociation enthalpy is the enthalpy change when one mole of covalent bond of a gaseous covalent compound is broken into gaseous products.

H₂(g) ⟶ 2H(g) ....... Δ_H-H H^ө = 435.0 kJ mol^-1

Cl₂(g) ⟶ 2H(g) ....... Δ_H-H H^ө = 435.0 kJ mol^-1

The above examples are of simple diatomic molecules. What happens in polyatomic molecule? Let’s take an example of methane CH₄. It has four C-H bond. C is sp³ hybridised, all C-H bonds are identical and have same energy. But each successive step requires different energy to break C-H bond. Because for each C-H bond there is different environment. In first step there are three more C-H bonds while in second step there are two more C-H bonds and so on.

CH₄(g) ⟶ CH₃(g) + H(g)          ........ Δ_C-HH^ө = 427.0 kJ mol^-1

CH₃(g) ⟶ CH₂(g) + H(g)          ........ Δ_C-HH^ө = 439.0 kJ mol^-1

CH₂(g) ⟶ CH(g) + H(g)           ........ Δ_C-HH^ө = 452.0 kJ mol^-1

CH(g) ⟶ C(g) + H(g)               ........ Δ_C-H H^ө = 347.0 kJ mol^-1

In such cases we take the mean value for the bond dissociation enthalpy. So Δ_C-H H^ө will be:

Δ_C-H H^ө = ¼ (427 + 439 + 452 + 347) kJ mol^-1

Δ_C-H H^ө = 416 kJ mol^-1

We have calculated Δ_C-H H^ө for methane. It can be different for other molecules but it will be nearer to the mean value so we can take it as a reference for any C-H bond.

Lattice enthalpy

Bond dissociation enthalpy is useful for covalent bonds but not for ionic compounds. In ionic compounds, ions are trapped in lattice and we need to overcome the lattice energy to set these ions free. You know that ionic compounds dissociate into its constituent ions only when they get dissolved in a solvent. Let’s try to understand the process of dissolution of any ionic compound in a solvent. This process occurs in two steps, in the first step lattice gets broken and ions are freed, and in the second step solvationoccurs. If the solvent is water then solvation is known as hydration.

MX(s) ⟶ M⁺(g) + X^-(g) ⟶ M⁺(aq) + X^-(aq)

So the enthalpy change for the overall reaction is called the enthalpy of solution Δ_solution H^ө which is the sum of lattice enthalpy Δ_lattice H^ө and solvation enthalpy Δ_hydrationH^ө.

Δ_solution H^ө = Δ_lattice H^ө+ Δ_hydration H^ө

It means if you want to know the lattice enthalpy of NaCl you must know the values of Δ_solution H^ө and Δ_hydrationH^ө.  Two German scientists Max Born and Fritz Haber developed an indirect way to calculate the lattice enthalpy of NaCl. They constructed an enthalpy diagram by applying Hess’s law of constant heat summation. We will learn about it in the next post of Thermodynamics. 

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