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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Removal of interfering radicals before IIIrd group analysis

What are the interfering radicals? How do they interfere in systematic separation of cationic radicals? Why is it necessary to remove them before III^rd gr analysis? Why don't they interfere in I^st or II^nd group analysis? Interfering radicals are oxalate, tartrate, fluoride, borate and phosphate and they are anionic radicals. They form complex with III^rd gr group reagent ammonium chloride and ammonium hydroxide. This leads to incomplete precipitation of III^rd group cations and causes immature precipitation of IV^th and V^th group cations in alkaline medium. Let’s try to understand it.

Oxalate, tartrate, fluoride, borate, silicate and phosphate of the metals are soluble in acidic medium. 

If you remember, for 1^stand 2^nd analysis medium remain acidic (dilute HCl) that’s why they do not interfere then. But for 3^rd group analysis the medium becomes alkaline by group reagents ammonium chloride and ammonium sulphide. Here interfering radicals come into action and disturb the solubility product of cations which causes their premature or incomplete precipitation.

In acidic medium these salts produce their corresponding acids like oxalic acid, phosphoric acid, hydrofluoric acid, boric acid and tartaric acid. For example, barium oxalate reacts with HCl and produces oxalic acid.

            BaC₂O₄ + 2HCl ⟶ BaCl₂ + H₂C₂O₄

These interfering acids are weak acids so they do not dissociate completely and remain in solution in their unionised form. Equilibrium is developed between dissociated and un-dissociated acid.

            H₂C₂O₄ ⇌ 2H⁺ + C₂O₄^2-

Hydrochloric acid is a strong acid and is ionised completely.

            HCl ⟶ H⁺ + Cl^-

Hydrogen ions acts as common ion among them and higher concentration of H⁺ suppresses the ionization of interfering acid. Therefore, ionic product of C₂O₄^2- and Ba²⁺ doesn’t exceed the solubility product of barium oxalate which is why Ba²⁺ remains in the solution as barium oxalate. That’s how interfering radicals do not interfere as long as the medium remains acidic enough. But when we make the medium alkaline by adding 3^rd group reagent ammonium hydroxide NH₄OH, OH^- ions combine with H⁺and neutralise them. This decreases the concentration of H⁺ ions which shifts the equilibrium of dissociation of interfering acid forward and increases the concentration of C₂O₄^2- . Thus the ionic product of C₂O₄^2- and Ba²⁺ exceeds the solubility product of barium oxalate and Ba²⁺ gets precipitated in the 3^rd group, which actually belongs to the 4^th group.

One or more interfering radicals can be present in the solution. They have to be removed in the following order: first we remove oxalate and tartrate, then borate and fluoride, then silicate and in the last phosphate.

Scheme for the Removal of Interfering Radicals

Procedure for the removal of oxalate and tartrate:  Oxalate and tartrate of metals are soluble in acid and they decompose on heating. Take the filtrate of 2^nd group and boil off H₂S gas from it. Add 4-5ml concentrated nitric acid HNO₃ and heat it till it is almost dry. Repeat this treatment for 2-3 times.

            (COO)₂^2- + H⁺ ⟶  (COOH)₂

                (COOH)₂  ⟶ HCOOH + CO₂↑

            HCOOH ⟶ CO↑ + H₂O↑

Tartrate and tartaric acid decomposes in a complex manner; charring takes place on heating and a smell of burnt sugar develops. Extract the with dilute HCl and filter. Use this filtrate for analysis of 3^rd group or use for removal of other interfering radicals.

Procedure for the removal of borate and fluoride: Take the filtrate and evaporate it to dryness. Add concentrated HCl and again evaporate to dryness.

            F^- + H⁺ ⟶ HF

            CaF₂ + 2HCl ⟶ CaCl2 + 2HF

  

On heating with HCl fluoride forms hydrofluoric acid and Borate forms orthoboric acid which evaporate on heating.

               BO₃^3- + 3H⁺ ⟶ H₃BO₃

            Na₃BO₃ + 3HCl ⟶ 3NaCl + H₃BO₃

Extract the residue with dilute HCl and filter. Use this filtrate for analysis of 3^rd group or use for removal of other interfering radicals.

If fluoride is absent and borate is present then residue use a mixture of 5ml ethyl alcohol and 10ml conc. HCl and evaporate to dryness.

BO₃^3- + 3H⁺ ⟶ H₃BO₃

H₃BO₃ + 3C₂H₅OH ⟶ (C₂H₅O)₃B↑ + H₂O

Procedure for the removal of silicate: Evaporate the filtrate of 2^nd group or residue obtained from removal of interfering radicals with concentrated HCl to dryness. Repeat this treatment for 3-4 times.

            SiO₃^2- + 2H⁺ ⟶ H₂SiO₃ ↓

               H₂SiO₃ ↓ ⟶ SiO₂ ↓  + H₂O

On heating with HCl silicate converts to metasilicic acid (H₂SiO₃) which is converted into white insoluble powder silica (SiO₂) on repetitive heating with concentrated HCl.

Test for phosphate HPO₄^2-:  test 0.5ml of the filtrate with 1ml ammonium molybdate reagent and a few drops of concentrated HNO₃, and warm gently, yellow precipitate indicates the presence of phosphate. Its composition is not known exactly.

Procedure for the removal of phosphate: Ferric chloride is generally used for the removal of phosphate. Fe(III) combines with phosphate and removes all phosphate as insoluble FePO₄. Fe(III) is also a member of 3^rd group so first we have to test its presence in the filtrate of 2^nd group then we can proceed for the removal of phosphate.

            HPO₄^2-  + Fe³⁺ ⟶ FePO₄ ↓ + H⁺

Test for Fe: To the filtrate of 2^nd group add ammonium chloride NH₄Cl and a slight excess of ammonia NH₃ solution. If precipitate appears, it indicates the presence of 3^rd group. It may contain hydroxides Fe(OH)₃, Cr(OH)₃, Al(OH)₃, MnO₂.xH₂O, traces of CaF₂ and phosphates of Mg and IIIA, IIIB and IV group metals. Dissolve the precipitate in minimum volume of 2M HCl. Take 0.5ml solution and add potassium hexacyanoferrate (II) K₄[Fe(CN)₆] solution. If iron is present, you will get prussian blue coloured precipitate of iron(III) hexacyanoferrate.

4Fe³⁺ + 3[Fe(CN)₆]^4- ⟶ Fe₄[Fe(CN)₆]₃

Removal of phosphate: To the main solution add 2M ammonia NH₃ solution drop wise, with stirring, until a faint permanent precipitate is just obtained. Then add 2-3ml 9M acetic acid and 5ml 6M ammonium acetate solution. Discard any precipitate if obtained at this stage. If the solution is red or brownish red, sufficient iron Fe(III) is present here to combine with phosphate. If the solution is not red or brownish red in colour then add ferric chloride FeCl₃ solution drop wise with stirring, until the solution gets a deep brownish red coloured. Dilute the solution to about 150ml with hot water, boil gently for 1-2min, filter hot and wash the residue with a little boiling water. Residue may contain phosphate of Fe, Al and Cr. Keep the filtrate for test of IIIB group. Rinse the residue in porcelain dish with 10ml cold water, add 1-1.5g sodium peroxoborate and boil gently until the evolution of O₂ ceases (2-3min). Filter and wash with hot water. Reject the residue to remove phosphate in the form of FePO₄. Keep the filtrate and test for IIIA group.

To test the presence of interfering radicals you need to prepare sodium carbonate extract and then test them separately. Scheme for the test of anionic radicals is not as systematic as cationic radicals. We will study them in coming posts. In the next post we will discuss the analysis of IIIA group cations. 

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Thermodynamics

As the name suggests thermodynamics is the study of transfer of energy/heat. Why do we need to study it? What’s the role it plays in chemistry? We all know energy is involved whenever a reaction takes place. So it is as important as money for any business. When someone starts a business, he calculates the possible profit and loss, and he starts it only after ensuring that it is profitable. Similarly a reaction happens only when it is profitable. In terms of chemistry what does it mean and how can we calculate it?

Internal energy

The profit in a reaction can be calculated in terms of its internal energy. Internal energy is the sum of all types of energies of the molecules which is very difficult to calculate. But there is an easy way to calculate the difference in internal energy of a system using thermodynamics.

If we want to calculate internal energy of a reaction, we have to focus on the reaction and where it is taking place. If the reaction is taking place in a reaction vessel such as a beaker, flask or container then we consider the vessel as a system. And we consider everything other than the vessel as surrounding.

Types of System

In order to calculate internal energy we have to observe properties of a system. First it is necessary to understand the system. System is the place where the reaction takes place and any place in this universe other than system is the surrounding. System and the surrounding are separated by a boundary. You can understand it by taking an example of your room. Suppose you are playing with a ball in your room. Playing with ball is a reaction and it is taking place in your room so your room is the system and the walls of the room are the boundary. All the other places in the house are surrounding. You and ball may or may not go out of the room; it depends on the design of your room. If you keep all of the windows and doors open, you or the ball can go outside but if you close all the windows and doors, no one can go out of the room, only the sound of the ball hitting the walls can be heard from outside. If your room is closed and sound proof too, then neither sound nor the ball or you can go outside. Similarly a system can be of three types.

• Open system where energy and matter can be exchanged with the surrounding.
• Closed system where energy can be exchanged with the surrounding but there is no exchange of matter.
• Isolated system is the one which doesn’t allow any exchange of matter or energy.

Types of Thermodynamic System

State Variables

How can we define the state of the system? We can define it by measurable quantities like pressure, temperature, volume and amount. If you have to define the state of the room in above example, you will define it by its present pressure, temperature, volume, you and ball. How this room gets that temperature or pressure doesn’t affect the state of room. Such variables are called as state variables; those values depend on the state of system not on the path how it is reached. Internal energy “U” is also a state variable.

Possibility of any reaction happening depends on the benefit of internal energy it gets. If a reaction gets benefit of internal energy then it happens. That means to check the feasibility of any reaction we have to calculate the benefit of U or ΔU. Let’s see how internal energy of the system changes. If we add or remove any kind of energy to/from the system we can increase/ decrease the internal energy of the system. Energy deference can be created as a result of some work just like when you workout you lose energy or when someone gives you massage your body gets relaxed and you feel energetic again. If you add some heat to the system it also increases the internal energy of the system and vice versa. Removal or addition of matter also affects the internal energy of the system. Now we will discuss different ways to create difference in internal energy of the system.

Internal Energy of a Closed System

Let’s take hot water in a beaker and cover it with a lid. It’s an example of closed system. Now define the state of the system at time t1 by pressure, volume and temperature. Note the state of the system after time t2. You will find only difference in temperature of initial state and final state of the system. So the change in internal energy of the system can be given by the temperature difference. The energy which is a result of temperature difference is called heat q.

   

         ΔU = T_final - T_initial

            ΔU = q

In above example T_final is lesser than T_initial because heat is transferred from the system to the surrounding and you will get q negative. Similarly when you give heat to the system, q will get a positive sign.

Internal Energy of a Adiabatic System

The other way to bring change in the internal energy of the system is work. Let’s see how you can change the U by the work done on the system or by the system. To observe the energy change by doing work we have to keep other factors constant, that means there should be no heat transfer between system and surrounding. To make this sure we can take adiabatic system. Take some water in a flask with thermally insulated walls. Make some arrangement to add paddle and thermometer to it. Note the initial temperature of the water.  Now stir up water for some time with the help of paddles and note the temperature at the final state. This change in temperature will give us change in internal energy.

ΔU = T_final - T_initial

Here you will find T_final is greater than T_initial so the difference in internal energy will be positive. Here mechanical work is done on the system which adds some more energy to the system and you get positive change in U. Difference in internal energy is equal to the work done in an adiabatic system.

ΔU = w

When work is done on the system it will get positive sign and when it is done by the system it will get the negative sign.

First law of thermodynamics

We have studied two types of systems one is closed system which allows heat transfer and the other is adiabatic system which doesn’t allow heat transfer. In the first system, change in internal energy is carried out by heat and in the latter it is done by doing work on the system. What happens when we put both in order to change the internal energy?

First Law of Thermodynamics

In that case internal energy difference is calculated by the equation:

 ΔU = q + w

This equation is the mathematical expression of “First law of thermodynamics” it states that “The energy of an isolated system is constant.”

Internal energy is a state function; its value depends only on initial and final state. It will be independent of the way the change is carried out. In the above example you did mechanical work on the system but if you replace paddle with an emersion rod and do electrical work on the system you will get the same results. In the next post we will study its applications.

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