## Thermodynamics

INTRODUCTION

The word ‘thermodynamics’ implies flow of heat. It deals with energy changes accompanying all types of physical and chemical processes. It helps to lay down the criteria for predicting feasibility or spontaneity of a process, including a chemical reaction, under a given set of conditions. It also helps to determine the extent to which a process, including a chemical reaction, can proceed before attainment of equilibrium. Thermodynamics is based on two generalizations called the first and second law of thermodynamics. These are based on human experience.

SOME BASIC TERMS

System A system is defined as any specified portion of matter under study which is separated from the rest of the universe with a bounding surface. A system may consist of one or more substances.

Surroundings The rest of the universe which might be in a position to exchange energy and matter with the system is called the surroundings.

Types of system

(i)    Isolated system:- A system which can exchange neither energy nor matter with its surrounding is called an isolated system.

(ii)   Open system:-A system which can exchange matter as well as energy with its surroundings is said to be an open system.

(iii)  Closed system:-A system which can exchange energy but not matter with its surroundings is called a closed system.

Macroscopic properties :-The properties associated with a macroscopic system (i.e. consisting of large number of particles) are called macroscopic properties. These properties are pressure, volume, temperature, composition, density etc.

Extensive and Intensive properties :- An extensive property of a system is that which depends upon the amount of the substance present in the system like mass, volume and energy.

An intensive property of a system is that which is independent of the amount of the substance present in the system like temperature, pressure, density, concentration, viscosity, surface tension, refractive index etc.

State of a system:-When macroscopic properties of a system have definite values, the system is said to be in definite state. Whenever there is a change in any one of the macroscopic properties, the system is said to change into a different state. Thus, the state of a system is fixed by its macroscopic properties.

State variables :-Since the state of a system changed with change in any of the macroscopic properties, these properties are called state variables or the thermodynamics parameters which depends only upon the initial and final states of the system and independent of the manner as to how the change is brought are called state functions. Some common state functions are internal energy, enthalpy, entropy, free energy, pressure, temperature, volume etc.

Thermodynamic equilibrium :-A system in which the macroscopic properties do not undergo any change with time is said to be in thermodynamic equilibrium.

Thermodynamic process and their types:- The operation by which a system changes form one state to another is called a process. Whenever a system changes from one state to another it is accompanied by change in energy. In case of open systems, there may be change of matter as well.

The following types of process are known

• Isothermal process:-A process is said to be isothermal if the temperature of the system remains constant during each stage of the process.
• Adiabatic process:-A process is said to be adiabatic if the heat enters or leaves the system during any step of the process.
• Isobaric process:-A process is said to be isobaric if the pressure of the system remains constant during each step of the process.

Question.       Thermodynamics is concerned with

(A)   total energy in a system                            (B)   energy changes in a system

(C)  rate of a chemical change                         (D)  mass changes in nuclear reactions

Solution:              (B)

Isochoric Process

A process is said to be isochoric if the volume of the system remains constant during each step of the process.

Reversible and Irreversible process

A process which is carried out infinitesimally slowly in such a manner that the system remains almost in a state of equilibrium at every stage or a process carried out infinitesimally slowly so that the driving force is only infinitesimally greater than the opposing force is called a reversible process.

Any process which does not take place in the above manner i.e. a process which does not take place infinitesimally slowly, is said to be an irreversible process.

In fact, all the natural processes are irreversible processes.

INTERNAL ENERGY:- Every substance is associated with a definite amount of energy which depends upon its chemical nature as well as upon its temperature, pressure and volume. This energy is known as internal energy. Internal energy of the system is the energy possessed by all its constituent molecules.

Internal energy is a state property i.e. its value depends only upon the state of the substance but does not depend upon how that state is achieved. The absolute value of internal energy of a substance can not be determined. However determining the absolute values of internal energies is neither necessary nor required. It is the change in internal energy accompanying a chemical or a physical process that is of interest and this is a measurable quantity.

The first law of thermodynamics :-The first law of thermodynamics states that energy can neither be created nor destroyed, although it can be transformed from one form to another. This is also known as the law of conservation of energy.

MATHEMATICAL EXPRESSION OF FIRST LAW

Let UA be the energy of a system in its state A and UB be the energy in its state B. Suppose the system while undergoing change from state A to state B absorbs heat q from the surroundings and also  performs some work (mechanical or electrical), equal to w. The absorption of heat by the system tends to raise the energy of the system. The performance of work by the system, on the other hand, tends to lower the energy of the system because performance of work requires expenditure of energy. Hence the change of internal energy, Δ U1, accompanying the above process will be given by

$$\displaystyle \Delta U={{U}_{B}}-{{U}_{A}}=q-w$$

In general, if in a given process the quantity of heat transferred from the surrounding to the system is q and work done in the process is w, then the change in internal energy,

Δ U = q + w

This is the mathematical statement of the first law of thermodynamics.

If work is done by the surroundings on the system (as during the compression of a gas), w is taken as positive so that Δ U = q + w. if however work is done by the system on the surroundings (as during the expansion of a gas), w is taken as negative so that Δ U = q – w.

Question.       1 mole of ideal monoatomic gas at 27°C expands adiabatically against a constant external pressure of 1.5 atm from a volume of 4 dm3 to 16 dm3.

Calculate (i) q (ii) w and (iii)

Solution:              (i)    Since process is adiabatic  ∴ q = 0

(ii)    As the gas expands against the constant external pressure.

W = $$\displaystyle -P\Delta V=-1.5\left( {{{V}_{2}}-{{V}_{1}}} \right)$$ = $$\displaystyle -1.5\left( {16-4} \right)=-18\,\,atm\,\,d{{m}^{3}}$$

(iii)          DU = q + w = $$\displaystyle 0+\left( {-18} \right)=-18\,\,atm\,\,d{{m}^{3}}$$

ENTHALPY: It is sum total energy of the system at the constant pressure (Generally).

Heat enthalpy or heat content: Enthalpy ‘H’ is also a state function and independent of path. It is expressed as

H = U + PV

Link U in thermodynamics we deal with change in heat enthalpy ΔH

ΔH = H2-H1

Relation in between and ΔH and  ΔU: The two are related by

ΔH = U + PΔV

at constant volume ΔH= ΔU

First law of thermodynamics: According to this law, mass and energy of an isolated system remains constant. The law is expressed mathematically as

Dq = (dU) + (-dW)                                                                                          ….(1)

(For an infinitesimal change)

where, dU = Change in internal energy

dq = Heat supplied to system

-dW =Work done by the system

Also                            q = ΔU +(-W)                                                 …. (2)

or                                 ΔU= q +W                                                                  (For finite change)

Some useful formulae based upon I law.

1. Isothermal process: A process in which temperature of the system remains constant through the studies

ΔT=0                           ΔU=0

By Eq. (2)                        q= -W

i.e., heat given to a system is used in work done by the system.

1. Adiabatic process: A process during which no exchange of heat takes place in between system and surroundings.

q=0

By Eq. (2)                                   +ΔU = W

or                                                  -ΔU = -W

i.e., work is done by the system on the cost of its internal energy.

1. Cyclic process: A process in which initial state of system is regained after a series of operations.

ΔU= 0

By Eq. (2)                                     q = -W

1. Isochoric process: A process in which volume of the system remains constant throughout the investigations

ΔV=0

By Eq. (2)                                     qv= ΔU

i.e., heat given to a system under constant volume is used up in increasing internal energy.

1. Isobaric process: A process in which pressure of the system remains constant throughout the investigations.

ΔP = 0

Consider a system showing increase in volume from V1 to V2 at a constant pressure P, during absorption of heat qThe expansion work or work done by the system is W = -PΔV

Thus by Eq. (2) $${{q}_{p}}=\Delta U-\left( {-P\Delta V} \right)$$

=$$={{U}_{2}}-{{U}_{1}}-\left[ {P\left( {{{V}_{2}}-{{V}_{1}}} \right)} \right]$$

=$$=\left( {{{U}_{2}}+P{{V}_{2}}} \right)-\left( {{{U}_{1}}+P{{V}_{1}}} \right)$$

$$={{H}_{2}}-{{H}_{1}}$$

$$\left( {\because H=U+PV} \right)$$

$${{q}_{p}}=\Delta H$$

i.e., heat given to a system under constant P is used up in increasing heat enthalpy of system. Work done in isobaric process $$=dU-dq$$ $$=n{{C}_{v}}.dT-n{{C}_{p}}.dT$$

$$=-RdT$$

$$=-nR\left( {{{T}_{{final\,\,}}}-{{T}_{{initial}}}} \right)$$

1. Work done in irreversible isothermal process: Suppose an ideal gas expands against external pressure P and its volume changes by an amount dV then work done W can be given by

W = -PΔV

(a) Free expansion: W = 0 Since P = 0

(b) Expansion or compression against external pressure

For a finite change V1 to V2

Total work done on the system  is derived by

W = -P(V2-V1)                                                                                                     …. (3)

or                                      Wirr = -P(V2-V1)                                                         …. (4)

Also                                  ΔU =ΔH =0

If V2>V1 then Wirr is –ve, i.e., expansion work done by the system.

7.         Work done in isothermal reversible process: Consider a system under isothermal condition, showing reversible expansion of an ideal gas by a volume dV  then

ΔU = 0

Δq = -dW =+PdV                     (Since dW= -PdV)

The total work done during expansion of gas from V1  to V2

The total work during expansion of gas from  V1  to V2

$$\int{{dW=}}\int\limits_{{{{v}_{1}}}}^{{{{v}_{2}}}}{{-PdV=\int\limits_{{{{v}_{1}}}}^{{{{v}_{2}}}}{{-\frac{{nRT}}{V}.dV}}}}$$

$${{W}_{{rev}}}=-nRT\,{{\log }_{e}}\frac{{{{V}_{2}}}}{{{{V}_{1}}}}=-2.303nRT{{\log }_{{10}}}\frac{{{{V}_{2}}}}{{{{V}_{1}}}}$$

Also, $${{W}_{{rev}}}=-2.303\,\,nRT{{\log }_{{10}}}\frac{{{{P}_{1}}}}{{{{P}_{2}}}}$$

reversible expansion of an ideal gas by a volume dV  , then from I law of thermodynamics,

( Since dq =0)

dU = dW

dU = nCvdT

or                            +Cvn x dT = dW

This equation reveals on further treatment within temperature limits of   and  reveals

dW = +Cv x n x dT

Cp –-Cv = R

reversible expansion of an ideal gas by a volume dV  , then from I law of thermodynamics,

( Since dq =0)

dU = dW

dU = nCvdT

or                            +Cvn x dT = dW

This equation reveals on further treatment within temperature limits of   and  reveals

dW = +Cv x n x dT

Cp –-Cv = R

$$\frac{{{{C}_{p}}}}{{{{C}_{v}}}}-\frac{{{{C}_{v}}}}{{{{C}_{v}}}}=\frac{R}{{{{C}_{v}}}}$$ $$\gamma -1=\frac{R}{{{{C}_{v}}}}$$ $${{C}_{v}}=\frac{R}{{\left( {\gamma -1} \right)}}$$

∴ $$dW=+\frac{R}{{\left( {\gamma -1} \right)}}\times n\times dT$$

$$dW=\frac{{nR}}{{\gamma -1}}dT$$

on integration $$\int{{dW=\frac{{nR}}{{\gamma -1}}\int\limits_{{{{T}_{1}}}}^{{{{T}_{2}}}}{{dT}}}}$$

$${{w}_{{rev}}}=\frac{{nR}}{{\gamma -1}}\left[ {{{T}_{2}}-{{T}_{1}}} \right]$$

If $${{T}_{2}}>{{T}_{1}}$$ then $${{W}_{{rev}}}=+ve,$$ i.e., work done on the system

If $${{T}_{2}}<{{T}_{1}}$$ then $${{w}_{{rev}}}=-ve,$$ i.e., work done by the system

Where γ is Poisson’s ratio =Cp /Cv

Also $$\Delta H=n{{C}_{p}}\left( {{{T}_{2}}-{{T}_{1}}} \right)$$

If final temperature is not known:

$$w=-n{{C}_{v}}{{T}_{1}}\left[ {1-{{{\left( {\frac{{{{P}_{2}}}}{{{{P}_{1}}}}} \right)}}^{{R/{{C}_{p}}}}}} \right]$$

Some important results of adiabatic expansions

1. PVγ =constant
2. TrP1-γ =constant

3.  Vγ-1T =constant

Question       10 gm of Helium at 127°C is expanded isothermally from 100 atm to 1 atm Calculate the work done when the expansion is carried out (i) in single step (ii) in three steps the intermediate pressure being 60 and 30 atm respectively and (iii) reversibly.

Solution:               (i)    Work done = V.ΔP

V = $$\displaystyle \left( {\frac{{10}}{4}} \right)\times \frac{{8.314\times 400}}{{100\times {{{10}}^{5}}}}=\stackrel{\scriptscriptstyle\to}{\leftarrow}83.14\times {{10}^{{-5}}}$$

So  , W = $$\displaystyle \frac{{83.14}}{{{{{10}}^{5}}}}$$

(100-1)x10=  8230.86 J.

(ii)   In three steps

VI  = 83.14×10-5 m3

WI  = (83.14×10-5)´(100-60)x105       = 3325.6 Jules

V II =  $$\displaystyle \frac{{2.5\times 8.314\times 400}}{{60\times {{{10}}^{5}}}}=138.56\times {{10}^{{-5}}}\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}{{m}^{3}}$$

WII = V. Δ   = 138.56×10-5 (60-30)x105       =    4156.99  ≅4157 J.$$\displaystyle \left( {\frac{{10}}{4}} \right)\times 8.314\times 400\times \log \left( {\frac{{100}}{1}} \right)$$

VIII = $$\displaystyle \frac{{2.5\times 8.314\times 400}}{{30\times {{{10}}^{5}}}}=277.13\times {{10}^{{-5}}}\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}{{m}^{3}}$$

WIII = 277.13 x10-5 (30-1)x105     =    8036.86 J.

W total  =  WI + WII + WIII

= 3325.6 + 4156.909 + 8036.86  =  15519.45 J.

(iii)   For reversible process W = 2.303 nRT log (P1/P2) =  2.303 x  $$\displaystyle \left( {\frac{{10}}{4}} \right)\times 8.314\times 400\times \log \left( {\frac{{100}}{1}} \right)$$

W = 38294.28 Jules

Limitations of the first law. Need for the second law A major limitation of the first law of thermodynamics is that its merely indicates that in any process there is an exact equivalence between the various forms of energies involved, but it provides no information concerning the spontaneity or feasibility of the process. For example, the first law does not indicate whether heat can flow from a cold end to a hot end or not.

The answers to the above questions are provided by the second law of thermodynamics.

Spontaneous and non – spontaneous process

If in the expansion of a gas the opposing pressure is infinitesimally smaller than the pressure of the gas, the expansion takes place infinitesimally slowly i.e. reversible. If however, the opposing pressure is much smaller than the pressure of the gas the expansion takes place rapidly i.e. irreversibly. Natural processes are spontaneous and irreversible.

SECOND LAW OF THERMODYNAMICS :-The second law of Thermodynamics helps us to determine the direction in which energy can be transformed. It also helps us to predict whether a given process or chemical reaction can occur spontaneously or not.

According to Kelvin :- “It is impossible to use a cyclic process to extract heat from a reservoir and to convert it into work without transferring at the same time a certain amount of heat from a hotter to colder part of the body”.

Entropy Change:- Entropy change is the state function and it is the ratio of heat change in a reversible process by the temperature.

Δ S = $$\displaystyle \frac{{{{q}_{{rev}}}}}{T}$$

Thermodynamically irreversible process is always accompanied by an increase in the entropy of the system and its surroundings taken together while in a thermodynamically reversible process, the entropy of the system and its surroundings taken together remains unaltered.

Question.     A system changes its state irreversibly at 300 K in which it absorbs 300 cals of heat. When the same change is carried out reversibly the amount of heat absorbed is 900 cals. The change in entropy of the system is equal to

(A)   1 cal K–1                                                     (B)   3 cals K–1

(C)  2 cal K–1                                                     (D)  1.5 cals K–1

Solution:               (B)

Physical Significance of Entropy: Entropy is the measure of disorderness because spontaneous processes are accompanied by increase in entropy as well as increase in the disorder of the system. Thus, increase in entropy implies increase in disorder.

Question.       Which of the following statement is true?

(i) A closed system shows exchange of mass and not energy with surroundings.

(ii) Entropy change for fusion reaction is positive.

(iii) Heat is a measure of quantity of energy whereas temperature is a measure of intensity of energy.

Solution:              (i) False                (ii) True                 (iii) True

Some Other State Function: For a spontaneous process entropy change is positive and if it is zero, the system remains in a state of equilibrium. Two other functions are also there to decide the feasibility of the reactions like work function A and free energy change G.

A = E – TS…….(i)

G = H – TS…….(ii)

And ΔA = ΔE – TΔS……(iii)

ΔG = ΔH – TΔS…………(iv)     (for a finite change at constant temperature)

Since, ΔS = qrev./T Hence from eq. (i)

ΔA = ΔE – qrev………………..(v)

and according to first law of Thermodynamics

ΔE – qrev = wrev. …………….(vi)

If during the change, work is done by the system, it would carry a negative sign,

-wrev = ΔE – qrev…………….(vii)

Comparing the equation (v) and (vii)

-ΔA = wrev

Since the process is carried out reversibly where w represents the maximum work. It is thus clear that decrease in function A gives maximum work done that can be done by the system during the given change. The work function A is also called as Helmholtz function.

From equation (iv)

ΔG = ΔH – TΔS

and ΔH = ΔE + PΔV

\ΔG = ΔE + PΔV – TΔS

Comparing it with eq. (iii)

ΔG = ΔA + PΔV

Since, ΔA is equal to – w, hence.

ΔG = – w + PΔV.

– ΔG = w- PΔV

Hence decrease in free energy gives maximum work obtainable from a system other than that due to change of volume at constant temperature and pressure. This is called as Net Work.

Net Work = w-PΔV = -ΔG

The Net Work may be electrical work or chemical work.

Criterion of spontaneity: For a spontaneous process ΔG should be -ve

GIBBS FREE ENERGY This is another thermodynamic quantity that helps in predicting the spontaneity of a process, is called Gibbs energy (G).

It is defined mathematically by the equation.

G = H TS

Where H = heat content, S = entropy of the system, T = absolute temperature

Question.     Which of the following will fit into the blank?

When two phases of the same single substance remain in equilibrium with one another at a constant P and T, their molar _________ must be equal.

(A)   Internal energy                                           (B)   Enthalpy

(C)  Entropy                                                       (D)  Free energy

Solution:              (D)

Free energy change

For isothermal process.

$$\displaystyle \Delta G={{G}_{2}}-{{G}_{1}}=\Delta H-T\Delta S$$

ΔG = change in Gibbs free energy of the system.

It is that thermodynamic quantity of a system the decrease in whose value during a process is equal to the maximum possible useful work that can be obtained from the system.

Question.     Calculate free energy change when one mole of NaCl is dissolved in water at 25°C. Lattice energy = 700 kJ/mol. at 25°C = 26.5, Hydration energy of NaCl = 696 kJ/mol.

(A)   3.9 kJ                                                       (B)   8kJ

(C)  12kJ                                                          (D)  16kJ

Solution:              (A)

RELATIONSHIP BETWEEN FREE ENERGY AND EQUILIBRIUM CONSTANT

The free energy change of the reaction in any state, DG (when equilibrium has not been attained) is related to the standard free energy change of the reaction, DG0 (which is equal to the difference in free energies of formation of the products and reactants both in their standard states) according to the equation.

$$\displaystyle \Delta G=\Delta {{G}^{0}}+RT\ln Q$$

Where Q is the reaction quotient

When equilibrium is attained, there is no further free energy change i.e. ΔG = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes

$$\displaystyle \Delta {{G}^{0}}=-RT\ln {{K}_{{\left( {eq.} \right)}}}$$ or $$\displaystyle \Delta {{G}^{0}}=-2.303\,\,RT\,\,\log K{{ & }_{{\left( {eq.} \right)}}}$$

• In case of galvanic cells. Gibbs energy change DG is related to the electrical work done by the cell.

ΔG = -nFE(cell)  where n = no. of moles of electrons involved

F = the Faraday constant   ;  E = emf of the cell

If reactants and products are in their standard states $$\displaystyle \Delta {{G}^{0}}=-nFE_{{cell}}^{0}$$

THIRD LAW OF THERMODYNAMICS

The entropy of a pure crystalline substance increases with increase of temperature, because molecular motion increases with increase of temperature and vice – versa.

Or the entropy of a perfectly crystalline solid approaches zero as the absolute zero of temperature is approached. This is third law of thermodynamics.

Change of internal energy in a chemical reaction

Let us consider a chemical reaction taking place at constant temperature and at constant volume. In such a case, w = 0 and hence from the first law

ΔU = qv

Where qv is the heat exchanged at constant volume, or heat or enthalpy of reaction at constant volume.

Change of Enthalpy in a chemical reaction

Let qP be the heat exchanged in the chemical reaction taking place at constant pressure, Then evidently,

ΔH = qP = Heat or Enthalpy of reaction at constant pressure.

Exothermic and Endothermic reaction

Reaction that give out heat, i.e. which are accompanied by evolution of heat, are called exothermic reaction. In such reactions DH is negative. On the other hand, reaction that intake heat, i.e. which are accompanied by absorption of heat are called endothermic reactions. In these reactions DH is positive.

Illustration 10. Fill in the blanks with appropriate word in following:

(i) Combustion of reactions are usually ……………         (ii) Combustion of F2 in oxygen is …………………

Solution:              (ii) Exothermic              (iii) Endothermic

Enthalpy of reaction

It is the enthalpy change taking place during the reaction when the number of moles of reactants and products are same as the stoichiometric coefficient indicates in the balanced chemical equation. The enthalpy change of the reaction depends upon the conditions like temperature, pressure etc under which the chemical reaction is carried out. Therefore, it is necessary to select the standard state conditions. According to thermodynamics conventions, the standard state refers to 1 bar pressure and 298 K temperature. The enthalpy change of a reaction at this standard state conditions is called standard enthalpy of the reaction ΔH0.

Different types of enthalpy

(i)    Enthalpy of formation: Enthalpy change when one mole of a given compound is formed from its elements.

H2(g) + 1/2O2(g) ¾®→ 2H2O(l),                        ΔH = –890.36 kJ / mol

(ii)    Enthalpy of combustion: Enthalpy change when one mole of a substance is burnt in oxygen.

CH4 + 2O2(g) ¾→ CO2  + 2H2O(l),          ΔH  = –890.36 kJ / mol

(iii)  Enthalpy of Neutralization: Enthalpy change when one equivalent of an acid is neutralized by a base or vice – versa in dilute solution. This is constant and its value is –13.7 kcal for neutralization of any strong acid by a base since in dilute solutions they completely dissociate into ions.

H+ (aq) + OH (aq) ¾→H2O(l),                             ΔH = –13.7 kcal

For weak acids and bases, heat of neutralization is different because they are not dissociated completely and during dissociation some heat is absorbed. So total heat evolved during neutralization will be less.

e.g. HCN + NaOH ¾→ NaCN + H2O,                                   ΔH = –2.9 kcal

Heat of ionization in this reaction is equal to (–2.9 + 13.7) kcal = 10.8 kcal

(iv)  Enthalpy of hydration: Enthalpy of hydration of a given anhydrous or partially hydrated salt is the enthalpy change when it combines with the requisite no.of mole of water to form a specific hydrate. For example, the hydration of anhydrous copper sulphate is represented by

CuSO4(s) + 5H2O (l) ¾→CuSO45H2O(s),                      ΔH0= –18.69 kcal

(v)   Enthalpy of Transition:  Enthalpy change when one mole of a substance is transformed from one allotropic form to another allotropic form.

C (graphite) ¾→  C(diamond),                                                      ΔH0= 1.9 kJ/mol

HESS’S LAW :- This law states that the amount of heat evolved or absorbed in a process, including a chemical change is the same whether the process takes place in one or several steps.

Suppose in a process the system changes from state A to state B in one step and the heat exchanged in this change is q. Now suppose the system changes from state A to state B in three steps involving a change from A to C, C to D and finally from D to B. If q1, q2 and q­3 are the heats exchanged in the first, second and third step, respectively then according to Hess’s law

q1 + q2 + q3 = q

Hess’s law is simply a corollary of the first law of thermodynamics. It implies that enthalpy change of a reaction depends on the initial and final state and is independent of the manner by which the change is brought about.

Question.             H2O (l) H2(g) + 1/2O2(g) ΔH = + 890.36 kJ / mole

What is ΔH for H2O (l) from its constituent elements

Solution:              H2O(l) → H2(g) + ½ O2(g)          ΔH = + 890.36 kJ / mole

H2(s) + 1/2O2(g) → H2O(l)          ΔH = – 890.36 kJ / mole

(ΔHf)H2O  = –890.36 kJ / mole

APPLICATION OF HESS’S LAW

1. Calculation of enthalpies of formation

There are large number of compounds such as C6H6, CO, C2H6 etc whose direct synthesis from their constituent element is not possible. Their ΔH0f values can be determined indirectly by Hess’s law. e.g. let us consider Hess’s law cycle for CO2 (g) to calculate the DH0f of CO(g) which can not determined otherwise.

2C + O2 2 CO+2 O2   ΔH1  = ?

2CO + O2 2CO2   ΔH2  = -283 KJ

C + O2 → CO2         ΔH3  = -0393 KJ

According to Hess’s law,

$$\displaystyle \Delta {{H}_{{ & 3}}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}$$  or  $$\displaystyle \Delta {{H}_{{ & 1}}}=\Delta {{H}_{3}}-\Delta {{H}_{2}}$$

=   -393 – (-283)  ⇒ -110 KJ/mole

1. Calculation of standard Enthalpies of reactions

From the knowledge of the standard enthalpies of formation of reactants and products the standard enthalpy of reaction can be calculated using Hess’s law.

According to Hess’s law

$$\displaystyle \sum{{\Delta H_{f}^{0}\left( P \right)}}=\sum{{\Delta H_{f}^{0}\left( R \right)+\Delta {{H}^{0}}}}$$

$$\displaystyle \therefore \Delta {{H}^{0}}=\sum{{\Delta H_{f}^{0}\left( P \right)}}-\sum{{\Delta H_{f}^{0}\left( R \right)}}$$

$$\displaystyle \therefore \,\Delta {{H}^{0}}=\left[ \begin{array}{l}sum\,\,of\,\,s\tan dard\,\,enthalpies\,\,of\,\,\\formation\,\,of\,\,products\end{array} \right]-\left[ \begin{array}{l}sum\,\,of\,\,s\tan dard\,\,enthalpies\,\,\\of\,\,formation\,\,of\,\,reac\tan ts\end{array} \right]$$

1. In the calculation of bond energies

BOND ENERGY

Bond energy for any particular type of bond in a compound may be defined as the average amount of energy required to dissociate one mole, viz Avogadro’s number of bonds of that type present in the compound. Bond energy is also called the enthalpy of formation of the bond.

Calculation:

For diatomic molecules like H2, O2, N2, HCl, HF etc, the bond energies are equal to their dissociation energies. For polyatomic molecules, the bond energy of a particular bond is found from the values of the enthalpies of formation. Similarly the bond energies of heteronuclear diatomic molecules like HCl, HF etc can be obtained directly from experiments or may be calculated from the bond energies of homonuclear diatomic molecules.

Question.     Calculate the bond energy of HCl. Given that the bond energies of H2 and Cl2 are 430 KJmol-1 and 242 KJ mol-1 respectively and ΔH0f  for HCl is -91 KJ mol-1

Solution             H2(g) → 2H(g)                                      ΔH= 430 KJ/mol                                                ….(i)

Cl2(g) → 2Cl(g)                                      ΔH= 242 KJ/mol                                                ….(ii)

HCl(g) → H(g) +Cl(g)                           ΔH= ?                                                                    ….(iii)

For the reaction (iii)

ΔH = $$\displaystyle \sum{{\Delta H_{f}^{0}\left( {product} \right)}}-\sum{{\Delta H_{f}^{0}\left( {reac\tan t} \right)}}$$

= $$\displaystyle \left[ {\Delta H_{f}^{0}\left( H \right)+\Delta H_{f}^{0}\left( {Cl} \right)} \right]-\left[ {\Delta H_{f}^{0}\left( {HCl} \right)} \right]$$ = $$\displaystyle \frac{1}{2}\times 430+\frac{1}{2}\times 242-\left( {-91} \right)$$

ΔH = 427 KJ mol-1

LATTICE ENERGY OF AN IONIC CRYSTAL (BORN–HABER CYCLE)

The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from its gaseous ions, is known as Lattice energy.

Step 1:   Conversion of metal to gaseous atoms   ;   M(s) ¾→ M(g) ,              ΔH1 = sublimation

Step 2:   Dissociation of X2 molecules to X atoms  ; X2(g) ¾→2X (g),             ΔH2 = Dissociation energy

Step 3:   Conversion of gaseous metal atom to metal ions by losing electron

M(g) ¾→ M+ (g) +  e,        ΔH3 = (Ionization energy)

Step 4:   X(g) atoms gain an electron to form X ions   ; X(g) + e ¾→ X(g),   ΔH4 = Electron affinity

Step 5:   M+ (g) and X (g) get together and form the crystal lattice

M+ (g) + X (g) ¾→ MX(s)         ΔH5 = lattice energy

Applying Hess’s law we get ;  ΔH1 + 1/2  ΔH2 + ΔH3 + ΔH4 + ΔH5 = ΔHf (MX)

On putting the various known values, we can calculate the lattice energy.

BOMB CALORIMETER

The bomb calorimeter used for determining change in internal energy at constant volume if reaction for the combustion is known than enthalpy of combustion can be estimated by using formula:  ΔH = ΔE + ΔnRT

This apparatus was devised by Berthelot (1881) to measure the heat of combustion of organic compounds. A modified form of the apparatus shown in Figure consists of a sealed combustion chamber, called a bomb, containing a weighed quantity of the substance in a dish alongwith oxygen under about 20 atm pressure. The bomb is lowered in water contained in an insulated copper vessel. This vessel is provided with a stirrer and a thermometer reading up to 1/100th of a degree. It is also surrounded by an outer jacket to ensure complete insulation from the atmosphere. The temperature of water is noted before the substance is ignited by an electric current. After combustion, the rise in temperature of the system is noted on the thermometer and heat of combustion can be calculated from the heat gained by water and the calorimeter. By knowing the heat capacity of calorimeter and also the rise in temperature, the heat of combustion can be calculated by using the expression

Heat exchange = Z x ΔT

Z  =  Heat capacity of calorimeter system

ΔT= rise in temp

Heat changes at constant volumes are expressed in DE and Heat changes at constant pressure are expressed in DH. Also, ΔH = ΔE + ΔnRT  ;  Δn = Moles of gaseous product– Moles of gaseous reactant.

HEAT CAPACITY AND SPECIFIC HEAT

The heat capacity (C) of a sample of substance is the quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or Kelvin). Then, q = cΔt

Heat capacity is directly proportional to the amount of substance.

The specific heat capacity is the quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius  at constant pressure. Then, q = s x m x Δt

where q is the heat required to raise temperature

m = mass in grams ; s = specific heat of the substance ; Δt = temperature difference

VARIATION OF HEAT OF REACTION WITH TEMPERATURE

The heat of reaction depends on the temperature. The relation between the two is known as Kirchoff’s equation.

i) $$\displaystyle \frac{{\Delta {{\text{H}}_{\text{2}}}-\Delta {{\text{H}}_{\text{1}}}}}{{{{\text{T}}_{\text{2}}}-{{\text{T}}_{\text{1}}}}}$$ = ΔCP

ii) $$\displaystyle \frac{{\Delta {{\text{E}}_{\text{2}}}-\Delta {{\text{E}}_{\text{1}}}}}{{{{\text{T}}_{\text{2}}}-{{\text{T}}_{\text{1}}}}}$$ = ΔCV

ΔCP = molar heat capacity of products – molar heat capacity of reactants (at constant pressure)

ΔCv = molar heat capacity of products – molar heat capacity of reactants (at constant volume)