Introduction to Thermochemistry & Second Law of Thermodynamics | Physical Chemistry PDF Download

THERMOCHEMISTRY

Change in Internal energy in a chemical reaction:

From first law dq = dU – w

If a chemical reaction takes place at constant temperature and constant volume. Then W = 0

Hence ΔU = q_v = heat exchange at constant volume

Suppose V_R & U_P be the internal energy of reactant and products.
then, ΔU = U_P - U_R
= q_v = Heat or enthalpy of reaction at constant volume

Change in Enthalpy a chemical reaction: Let qp be the heat exchange in the reaction at constant pressure. Then
ΔH = qP
If H_R & H_P be the enthalpy of reactant and product then

ΔH = H_P - H_R
= qp = heat or enthalpy o f reaction at constant pressure

Relation between Enthalpy of reaction at constant volume and a constant pressure: 

We know that
ΔH = qP & ΔU = qV
then qP = qV + PΔV
Forn moles of an ideal gas
PV = nRT

Let n₁ & n₂ be the number of moles of gaseous reactants & products.
then                  Δn = n₂ - n₁ = increase in the number of gaseous moles
Then corresponding increase in volume ΔV is given by

 
Hence,         

 
Then           q_P = q_V + Δn_g RT

Variation of enthalpy of reaction with temperature i.e., Kirchhoff equation.
The enthalpy change for the reaction
aA + bB → cC + dD

 is given by  

The temperature dependence of enthalpy at constant pressure is:
then 

or d(_DH) = _DCPdT ...(1) The temperature dependence of enthalpy of constant volume is:

or d(ΔU) = ΔC_VdT ...(2)
Integrating equation (1) & (2) we get

or

Similarly  

 or ΔU₂ - ΔU₁ = ΔC_V (T₂ - T₁ )
Above equations are known as Kirchhoff equations.
Problem.  Calculate ΔH^o373 for the reaction

             ΔH298 K = -33.18 KJ mol^-1
 Given, C_P (NO₂, g) = 37.20 J K^-1 mol^-1
             C_P (N₂, g) = 29.13 JK^-1 mol^-1
             C_P (O₂, g) = 29.36 JK^-1 mol^-1s
 Sol. 

 
 = - 6.73 x 10^-3JK mol^-1 
Now,  

ΔH⁰_{373 K} - ΔH⁰ _{298 K} = ΔC_P (373 - 298)

•  ΔH⁰ _{373 K} = [-33.18 + (-6.73 × 10^-3) (373 - 298)] KJ mol^-1
• ΔH⁰ _{373 K} = -33.68 KJ mol^-1

The second law of Thermodynamics: The second law of thermodynamics identifies a new state function called the entropy which provides a criterion for identifying the equilibrium state of a system.
Entropy of the universe (system + surrounding) increases for irreversible process whereas it remains constant for reversible process.

Carnot cycle: Consider the system is contained in a frictionless piston and cylinder arrangement. We also use two thermal reservoirs one at higher temperature T₂ and other one at a lower temperature T₁. 

Four successive operations as given below:
(1)  The isothermal reversible expansion from volume V₁ to volume L at the higher temperature T₂ .
Since dT = 0, therefore, ΔU₁ = 0 According to first law of thermodynamics, we have q₂ = 2 

(2)  The adiabatic reversible expansion from volume V₂ to volume V₃.
Temperature of the system after expansion is T₁. Since for adiabatic process q = 0, therefore, it follows that   

(3)  The isothermal reversible compression from V₃ to V₄ at temperature T₁.
Here             ΔU₃ = 0
and

(4)  Adiabatic reversible compression from volume V₄ to V₁.
Temperature of the system returns T₂. Therefore,

Newt work involved in the cyclic process is: 

W_total = W₂ + W + W₁ + W’

Relationship between V₁, V₂, V₃ & V₄: Second and fourth process are adiabatic expansion and compression process then from poisson equation we get

                                 (for second process)

From equation (A) & (B) we get

then


W_total =

Efficiency of Carnot cycle: It is denoted by η.

Thus, the efficiency of a heat engine operating in a Carnot cycle depends only on the two temperatures and primarily, on the difference of the two temperatures between which the engine operators.
The difference is greater, the greater the efficiency.
 

Problem. An engine operator in a Carnot cycle absorb 3.347 kJ at 400°C, how much work is done on the engine per cycle and how much heat is involved at 100°C in each cycle?
 Sol. The efficiency of Carnot cycle is 

 
Thus,

the heat involved is:
q₁ = 1.855 kJ                                             (heat is negative sign)
q₁ = –1.855 kJ
and work done is :
                                                       W = q₂ – q₁ = –3.347 + 1.855 = –1.492 kJ

Coefficient of performance: The coefficient of performance of a refrigerator is defined as the ratio of heat transferred from a lower temperature to a higher temperature to the work done on the machine to cause this removal i.e.

Relation between η & β :

⇒
⇒
⇒
⇒
⇒
i.e.