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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Standard Enthalpy of Reaction

When all the participating substances are in their standard state then the enthalpy of reaction is called the standard enthalpy of reaction. Standard state of a substance is its purest state at 1 bar pressure and 298K temperature. For example standard state of iron is pure solid iron at 1 bar pressure and 298K. Standard state of oxygen is O₂ gas, standard state of water is H₂O liquid and standard state of hydrogen is H₂ gas. Standard conditions are denoted by adding symbol ө to the superscript, e.g. standard enthalpy can be denoted by H^ө.

Phase transfer reactions can be categorized into three categories

1. Fusion or Melting: Involves phase transfer of solid to liquid and enthalpy change is known as enthalpy of fusion Δ_fus H. When 1mole solid substance melts at constant temperature and 1bar pressure then enthalpy change is known as molar or standard enthalpy of fusion Δ_fusH^ө.

H₂O(s) ⟶ H₂O(l)      Δ_fusH^ө = 6kJ mol^-1

The fusion of 1mol of water requires 6kJ mol^-1 heat, this is an endothermic reaction. And same amount of heat will release if we reverse the reaction.

H₂O(l) ⟶ H₂O(s)      Δ_melting H^ө = -6kJ mol^-1

2. Vaporization: involves phase transfer of liquid to gas and enthalpy change is known as enthalpy of vaporization Δ_vap H. Amount of heat required to vaporize 1mole of a liquid substance at constant temperature and 1bar pressure is known as molar or standard enthalpy of vaporization Δ_vap H^ө.

H₂O(l) ⟶ H₂O(g)     Δ_vapH^ө = 40.79kJ mol^-1

3. Sublimation: involves phase transfer of solid to gas and enthalpy change is known as enthalpy of sublimation Δ_sub H. When 1mole solid substance sublimates at constant temperature and 1bar pressure then the enthalpy change is known as molar or standard enthalpy of sublimation Δ_sub H^ө.

CO2(s) ⟶ CO2 (g)       Δ_subH^ө = 25.2kJ mol^-1

You know that enthalpy is an extensive property. Its value depends on external parameters like mass, intermolecular interactions, polarity of bonds etc. For example let's compare the Δ_vap H^ө of water and acetone. Water (H₂O) and acetone (CH₃COCH₃) both are polar molecules, but H bonding in water molecules holds them tightly, so water requires more heat to get vaporized than acetone.

Phase transfer reactions 

Standard enthalpy of combustion Δ_cH^ө

Combustion reactions are very important. These are exothermic reactions. We use them to get energy to run vehicles, machinery, plane and rockets. Our body also uses this reaction to get energy from food. When one mole of a substance undergoes combustion and all the reactant and products are being in their standard state at a specific temperature then the enthalpy change of reaction is called as Standard enthalpy of combustion Δ_cH^ө. Let’s see how much energy we get by combustion of 1mole of glucose:

C₆H₁₂O₆(g) + 6O₂(g) ⟶ 6CO₂(g) + 6H₂O(l)    Δ_cH^ө = -2802 kJ mol^-1

Standard enthalpy of formation Δ_fH^ө

Formation of something means being formed by its constituent elements.  Standard enthalpy of formation Δ_f H^ө is the enthalpy change for the formation of one mole of a compound from its constituent elements in their standard state. For example, reaction for the formation of water will be:

H₂(g) + 1/2O₂(g) ⟶ H₂O(l)     Δ_f H^ө = -285.8 kJ mol^-1

Reaction for the formation of hydrogen bromide will be:

1/2H₂(g) + 1/2Br₂(l) ⟶ HBr(g)    Δ_f H^ө = -36.4 kJ mol^-1

I hope you have understood the concept of formation. Reactant must be in their standard state and only one mole of product should be formed. Only then you get the Δ_f H^ө otherwise it will be just enthalpy of reaction Δ_r H.

H₂(g) + Br₂(l) ⟶ 2HBr(g)     Δ_r H^ө = -72.8 kJ mol^-1

Here 2mole of HBr is formed. So the change in enthalpy you will get here is the Δ_r H^ө not Δ_f H^ө. If you divide the whole reaction by 2 then you will get the Δ_f H^ө = Δ_r H^ө /2.

By convention, standard enthalpy of formation of an element in its standard state is taken as zero. For example Δ_f H^ө of H₂(g), Δ_f H^ө of Br₂(l) and  Δ_f H^ө of O₂(g) all will be zero.

How to calculate standard enthalpy of reaction Δ_rH^өwith the help of Δ_f H^ө?

Enthalpy change of a reaction gives us valuable information about the reaction. With the help of enthalpy change of formation you can calculate exact value of Δ_rH^ө. Let’s see how to calculate the energy needed to decompose calcium carbonate (CaCO₃) into lime (CaO). First write the balance equation for it:

CaCO₃(s) ⟶ CaO(s) + CO₂(g)

Δ_r H^ө = Ʃ ai Δ_f H^ө (products) - Ʃ bi Δ_f H^ө (reactants)

Here a and b are coefficients of the products and reactants for a balanced chemical equation.

Get the values of Δ_f H^ө of compounds and put them in formula,

Δ_f H^ө (CaO) = -635.1 kJ mol^-1

Δ_f H^ө (CO₂) = -393.5 kJ mol^-1

Δ_f H^ө (CaCO₃) = -1206.9 kJ mol^-1

Δ_rH^ө = [1(-635.1 kJ mol^-1) + 1(-393.5 kJ mol^-1)] -1(-1206.9 kJ mol^-1)

Δ_rH^ө = + 178.3 kJ mol^-1

Positive value of Δ_rH^ө shows that it is an endothermic reaction and you will have to give 178.3 kJ mol^-1 heat to decompose calcium carbonate into lime.

Let’s take another example:

Fe₂O₃(s) + 3H₂(g) ⟶ 2Fe(s) + 3H₂O(l)

Δ_f H^ө (Fe) and Δ_f H^ө (H2) is taken as 0 kJ mol^-1by convention as they are elements in their standard state.

Δ_f H^ө (H₂O) = -285.83 kJ mol^-1

Δ_f H^ө (Fe₂O₃) = -824.2 kJ mol^-1

Δ_r H^ө = [2(Δ_f H^ө Fe) + 3(Δ_f H^ө H₂O)] – [3(Δ_f H^ө H₂) + (Δ_f H^ө Fe₂O₃)]

Δ_r H^ө = [2(0) + 3(-285.83 kJ mol^-1)] – [3(0) + (-824.2 kJ mol^-1)]

Δ_r H^ө = -857.49 kJ mol^-1 - (-824.2 kJ mol^-1)

Δ_r H^ө = -33.29 kJ mol^-1

Negative value of Δ_r H^өshows that it is an exothermic reaction and 33.29 kJ mol^-1 heat will be released from this reaction.

Enthalpy is a state function and it is independent of path taken. On the basis of this fact scientist Germain Henri Hess gave a law which is known as Hess’s law of heat summation. We will discuss it in this next post.

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What is Enthalpy?

Enthalpy is the heat absorbed by the system at constant pressure. Why do we need this thermodynamic function?  What is its significance? Enthalpy is introduced to study profit and loss of heat of reactions those are carried out at constant pressure. In this post we will study how you can make a balance sheet of a reaction with the help of enthalpy of reactants and products and predict the heat status of a reaction, whether it will be endothermic or exothermic.

What is the Enthalpy of a reaction?

You have learnt the first law of thermodynamics in previous post. Let’s write it for an expansion reaction at constant pressure:

      

      ΔU = q + w

Now consider the previous example of work done on ideal gas by piston. This time we keep pressure constant. What will happen now? Gas will expand. Work will be done by the gas so it will get negative sign. Let’s write the equation for expansion reaction at constant pressure:

            ΔU = q - w

            ΔU = q_p - pΔV

            U₂- U₁ = q_p – p(V₂- V₁)

            q_p = U₂- U₁ + p(V₂- V₁)

            q_p = (U₂+ pV₂)-(U₁ + pV₁)

Enthalpy is the heat absorbed by the system at constant pressure. It is denoted by symbol “H”.

            q_p = (H₂)-(H₁)

            q_p = ΔH

           ΔH = ΔU + p ΔV

How is enthalpy different from heat?

You have seen that enthalpy is a special kind of heat (heat absorbed at constant pressure q_p). Heat depends on the conditions of the reaction. When reaction occurs at constant volume is becomes q_v and it becomes q_p when reaction occurs at constant pressure. That means it depends on the path taken by the system. So it is a path dependent function.

Enthalpy depends on internal energy, pressure and volume. Internal energy U, pressure p and volume V are all state functions and so is the enthalpy H.

What is enthalpy change of a reaction?

You can get the enthalpy change of a reaction by subtracting the sum of enthalpy of reactants from the sum of enthalpy of products.

aA + bB ⟶ cC + dD

ΔH_reaction = (cH_mC+ dH_mD) - (aH_mA+ bH_mB)

Where H_m is the molar enthalpy. Molar enthalpy is the heat of one mole of substance at constant pressure.

H_m = (q / n) _p

Where n is the number of moles.

ΔH_reaction = ƩH_products– ƩH_reactants

Enthalpy of a reaction gives us important information about the reaction. If it is positive it means that the reaction will be endothermic and negative enthalpy results in exothermic reaction.

ΔH_reaction = ‘-ve’ ⟶ ƩH_reactants > ƩH_products⟶products liberate excess of heat = exothermic reaction.

ΔH_reaction = ‘+ve’ ⟶ Ʃ_Hreactants < ƩH_products ⟶ reactants absorb heat to get converted into products = endothermic reaction.

Now you can predict the effect of temperature change to the reaction. And by applying Le-Chatelier’s principleyou can adjust the yield of products.

Exothermic and Endothermic Reaction

Extensive and Intensive Properties

Enthalpy of a reaction depends on the number moles of reactants and products. It means its value depends on the quantity of matter present in the system; it isn’t the property of the system. Such properties are called extensive properties, for example enthalpy; internal energy, mass, volumes, heat capacity all are extensive properties.

Properties which are independent of quantity or size of system are known as intensive properties e.g. Temperature, density, pressure.

But when you divide an extensive property by number of moles then it becomes the property of a mole of substance which is independent of size or amount of the matter.

Extensive property/ n = Intensive property

For example enthalpy H is an extensive property while molar enthalpy H_m is an intensive property. Heat capacity is an extensive property and molar heat capacity is an intensive property.

Now you can understand why we use molar enthalpy to calculate the enthalpy of a reaction. Although enthalpy of reaction is an extensive property which varies with the amount of matter and reaction conditions but by putting some special conditions we can get the standard value of enthalpy of reaction.

In the next post we will study standard enthalpy of reaction of different kind of reactions, learn about enthalpy of formation and see how we can calculate heat of reaction with the help of enthalpy of formation.

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Applications of First Law of Thermodynamics

In the previous physical chemistry post named thermodynamics we learned the First law of thermodynamics. We know that we can calculate the change in internal energy by adding heat change in the process and work done by/on the system (ΔU = q + w). There are two types of processes, reversible or irreversible. Measurement of heat q can be done by calorimeter but we have to calculate the work. In this post we will learn how to calculate work.

Let’s take an example of a cylinder which contains one moleof ideal gas and is fitted with a frictionless piston. The initial volume of the gas is V_i and pressure of the gas inside is p. If the external pressure p_ex is greater than the p, then the piston will move inward till p_ex becomes equal to the p and final volume of gas will become V_f.

We can calculate the volume change by multiplying the distance travelled by the piston with the cross section of the cylinder.

            

             ΔV = V_f-V_i = l × A

How do we calculate the work done by the piston? Work can be calculated by force multiplied by distance and force can be calculated by pressure multiplied by the area.

Work done on Ideal gas by Piston

Force on the piston = pressure ×Area

             F = p_ex × A

Work = force × distance

             w = (p_ex × A) × l

             w = p_ex × ΔV = p_ex(V_f -V_i) 

In our example, V_i is greater than V_fbecause compression of gas is done by the piston. That means work is done on the system so its sign has to be positive. To get the correct sign we do a little adjustment in the work equation by adding a negative sign with ΔV.

         
            w = p_ex × (-ΔV)                                     ...............1

For the compression, w gets a positive sign but w will be negative for expansion as the work is done by the system.

What happens in reversible process? How does it differ from irreversible process? In reversible process the changes occur in infinite number of steps and there is infinite number of equilibrium stages so that system and the surrounding always remain in equilibrium. Even a minor difference can manipulate the direction of these reactions or you can say that these reactions can be reversed at any point of the reaction.

Irreversible and Reversible Compression

For a reversible reaction, the piston will take infinite number of steps to cover the distance l. In that case we will get a new set of pressure and ΔV for every single step. Then we have to get the sum of all the steps to calculate the total work done.

        
            w_rev = - Ʃ p ΔV                                     ...............2

In the case of compression, external pressure is always infinitesimally greater than the pressure of the gas and at each step; volume is decreased by an infinitesimal amount ∂V. For compression, p_ex for each step is equal to (p_in +∂p) and for expansion, p_ex = (pin -∂p) and volume is increased by ∂V.

For compression p_ex = (p_in+∂p)

For expansion p_ex = (p_in -∂p)

In generalised way we can write it as p_ex= (p_in ±∂p)

Internal pressure p_in is the pressure of gas filled in the cylinder. So we can apply the ideal gas equation.

          pV = nRT

          p = nRT/V

If the process is carried out at a constant temperature then the equation of w for isothermal process is,

         w_rev = -2.303nRT log (V_f  / V_i)

Calculations of Reversible Work Done

Applications of First Law of Thermodynamics

Now we have learnt to calculate the work done in isothermal expansion or compression for irreversible and reversible processes. Let’s calculate the internal energy change for these processes.

           ΔU = q + w

• Isothermal process: for isothermal process internal energy remain constant that means ΔU = 0. So,

            q = - w

• For isothermal irreversible process:

            q = -w = - [p_ex × (-ΔV)]

            q = p_ex × ΔV

• For isothermal reversible process:

            q = - w_rev = - [-2.303nRT log (V_f/V_i)]

            q = 2.303nRT log (V_f /V_i)

• For Isochoric process: If the process is carried out at constant volume (ΔV=0), then work done will be zero (w =0) and

            ΔU = q_v

         Where q_v is the heat supplied at constant volume.

• Free expansion: When expansion of gas occurs in vacuum (p_ex= 0) then it is called free expansion. That means no work is done during the free expansion of ideal gas neither in irreversible nor reversible process.

         ΔU = q

• Isothermal free expansion: for isothermal free expansion of ideal gas, no work is done during the process because p_ex= 0. And there is no change in internal energy since temperature is kept constant. So,

          ΔU = 0, q = 0, w = 0

• Adiabatic process: for adiabatic changes no heat transfer is allowed between the system and surrounding so q = 0.

         ΔU = q + w

         ΔU = w_ad

We have discussed changes in different conditions like in vacuum or at constant temperature but when we work in a laboratory; we carry out the reactions mostly at atmospheric pressure, which means pressure remains constant.

        ΔU = q + w

        ΔU = q_p + pΔV

Here q_p denotes that heat is supplied at a constant pressure and here new thermodynamic function comes in action which is called the enthalpy H.

            q_p = ΔH

In the next post we will discuss this new and very important function of thermodynamics and its applications. ​

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