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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