Revision Notes on Chemical Thermodynamics:

Basic Terminology:



Terms

Explanation


System

Part of the universe under investigation.


Open System

A system which can exchange both energy and matter with its surroundings.


Closed System

A system which permits passage of energy but not mass, across its boundary.


Isolated system

A system which can neither exchange energy nor matter with its surrounding.


Surroundings

Part of the universe other than system, which can interact with it.


Boundary

Anything which separates system from surrounding.


State variables

The variables which are required to be defined in order to define state of any system i.e. pressure, volume, mass, temperature, surface area, etc.


State Functions

Property of system which depend only on the state of the system and not on the path.

Example: Pressure, volume, temperature, internal energy, enthalpy, entropy etc.


Intensive properties

Properties of a system which do not depend on mass of the systemi.e. temperature, pressure, density, concentration,


Extensive properties

Properties of a system which depend on mass of the system i.e.volume, energy, enthalpy, entropy etc.


Process

Path along which state of a system changes.


Isothermal process

Process which takes place at constant temperature


Isobaric process

Process which takes place at constant pressure


Isochoric process

Process which takes place at constant volume.


Adiabatic process

Process during which transfer of heat cannot take place between system and surrounding.


Cyclic process

Process in which system comes back to its initial state after undergoing series of changes.


Reversible process

Process during which the system always departs infinitesimally from the state of equilibrium i.e. its direction can be reversed at any moment.






Heat, energy and work:


1) Heat (Q): 

a) Energy is exchanged between system and surround in the form of heat when they are at different temperatures.

b) Heat added to a system is given by a positive sign, whereas heat extracted from a  system is given negative sign.

c) It is an extensive property.

d) It is not a state function.

2) Energy:

a) It is the capacity for doing work.

b) Energy is an extensive property.

c) Unit : Joule.

         3) Work (W):

               a) Work = Force × Displacement i.e. dW = Fdx

b) Work done on the system is given by positive sigh while work done by the system is given negative sign.

c) Mechanical Work or Pressure-Volume Work: work associated with change in volume of a system against an external pressure.

d) Work done in reversible process: W=

                                                          

       e) W = – 2.303 nRT log v2/v1  =  –2.303 nRT log p1/p2 

           Wok done in isothermal reversible contraction of an ideal gas:

           W = – 2.303 nRT log v2/v1  =  –2.303 nRT log p1/p2              

       f) Unit : Joule.

Internal Energy (E or U):



          1) Sum of all the possible types of energy present in the system.

          2) ΔE  = heat change for a reaction taking place at constant temperature and volume.

          3) ΔE is a state function.

          4) For exothermic reactions:


First Law of Thermodynamics:




        1) Energy can neither be created nor destroyed although it can be converted from one form to another.

                                              or

         Energy of an isolated system is constant.

        2) Mathematical Expression

            Heat observed by the system = its internal energy + work done by the system.
            i.e. q = dE + w     

            For an infinitesimal process

            dq = dE + dw 

            Where, q is the heat supplied to the system and w is the work done on the system.

        3) For an ideal gas undergoing isothermal change ΔE =0.

                so q= -w.

        4) For an isolated system, dq=0

             so, dE = - dw

        5) For system involving mechanical work only

            ΔE = q - pdV

            At constant volume

            ΔE = qv  


Enthalpy (H):



       1) H = E+PV

       2) At constant pressure:

           dH = dE + pdV


       3) For system involving mechanical work only

            dH = QP (At constant pressure)

       4) For exothermic reactions:

           dH = -ve

           For endothermic reactions:

           dH = +ve


Relation between dH and dE:



          dH = dE + dng RT

          Where,

          dng = (Number of moles of gaseous products - Number of moles of gaseous reactants)

Heat capacity:



         1) Amount of heat required to rise temperature of the system by one degree.      

             C = q / dT

         2) Specific heat capacity: Heat required to raise the temperature of 1 g of a substance by one dgree.

             Cs = Heat capacity / Mass in grams

         3) Molar heat capacity: Heat required to raise the temperature of 1 g of a substance by one dgree.

             Cm = Heat capacity / Molar mass.

         4) Heat capacity of system at constant volume:

             Cv  =  (dE/dT)v

         5) Heat capacity of system at constant pressure:

             Cp  =  (dE/dT)p

         6) Cp – Cv = R


Variation Of Heat Of Reaction With Temperature:



           dCP = (dH2 - dH1)/(T2-T1)   &    dCV = (dE2 - dE1)/(T2-T1



Bomb Calorimeter:



                                    

       1) Heat exchange = Z × ΔT

           Z–Heat capacity of calorimeter system

           ΔT– Rise in temp.

       2) Heat changes at constant volumes are expressed in ΔE and Heat changes at constant 

          pressure are expressed in dH.


Enthalpies of Reactions:




Enthalpies

Definitions

Example


Enthalpy of Formation

Enthalpy change when one mole of a given compound is formed from its elements

H2(g) + 1/2O2(g) ->  2H2O(l),              
ΔfH = –890.36 kJ / mol


Enthalpy of Combustion

Enthalpy change when one mole of a substance is burnt in oxygen.

CH4 + 2O2(g) -> CO2  + 2H2O(l),     
ΔcombH  = –890.36 kJ / mol


Enthalpy of Neutralization

Enthalpy change when one equivalent of an acid is neutralized by a base in dilute solution.

H+ (aq) + OH– (aq) -> H2O(l)              
ΔneutH = –13.7 kcal


Enthalpy of Hydration

Enthalpy change when a salt combines with the required number of moles of water to form specific  hydrate.

CuSO4(s) + 5H2O (l) -> CuSO45H2O,       ΔhydH° = –18.69 kcal


Enthalpy of Transition

Enthalpy change when one mole of a substance is transformed from one allotropic form to another allotropic form.

C (graphite) -> C(diamond),                      ΔtransH° = 1.9 kJ/mol


Enthalpy of Sublimation

Enthalpy change when one mole of a solid substance sublime at constant temp. and 1 bar pressure  

CO2(S) ->  CO2(g)
ΔtfusH° = 6.00 kJ/mol


Enthalpy of fusion

Enthalpy change when one mole of a solid melts

H2O(S) ->  H2O (l)
ΔtsubH° = 73.00 kJ/mol



      


Hess’s Law of constant heat summation: The total enthalpy change of a reaction is the same, regardless of whether the reaction is completed in one step or in several steps.



                                                           
           According to Hess’s law: ΔH = ΔH1 + ΔH2

Born–Haber Cycle
:Question:1) A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.00 C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be:

(R = 8.314 J/mol K) (ln 7.5 = 2.01) (IIT-JEE 2013)

(1) q = – 208 J, w = – 208 J

(2) q = – 208 J, w = + 208 J

(3) q = + 208 J, w = + 208 J

(4) q = + 208 J, w =– 208 J 

Answer: 4

Solution:

For isothermal process, ∆ U = 0

Process is isothermal reversible expansion, hence dE =0

U = 0.

So, Using 1st law of thermodynamics

q = -W

or

w = -q = -208 J

Question : 2) For an ideal gas, consider only P-Vwork in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is(are) correct?

[Take ∆S as change in entropy and w as work done]  (IIT-JEE 2012)

 

1) ΔS x→z = ΔS x→y + ΔS y→z

2) w x→z = w x→y + w y→z

3) w x→y→z = w x→y

4) ΔS x→y→z = ΔS x→y


Answer:

A & C

Solution:

Entropy  being a state function shows additive property.

Hence,

ΔS x→z = ΔS x→y + ΔS y→z

For the process Y →Z pressure is constant.

W = -pΔV

At constant pressure

W =0

Hence, w x→y→z = w x→y

Question: 3 ) The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 25 0C are–400 kJ/mol, –300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 250C is  (IIT JEE 2013)

a. +2900kJ

b. -2900 kJ

c. +16.11 kJ

d. -16.11 kJ

Answer: C.

Solution:

Formation of CO2:

C( graphite) + O2 (g)→ CO2 (g)    ΔH1= -400 kJ ………….(1)

Formation of H2O:

H2( g) + O2 (g) → H2O (l)    ΔH2= -300 kJ…………..(2)

Formation of C6H12O6:

6C( graphite) + 3O2 (g) + 6H2( g)  → C6H12O6 (s)    ΔH3= -1300 kJ ………(3)

6×Equation (1) + 6×Equation (2) - Equation...