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# Introduction to Thermochemistry - Thermochemistry Chemistry Notes | EduRev

## Physical Chemistry

Created by: Asf Institute

## Chemistry : Introduction to Thermochemistry - Thermochemistry Chemistry Notes | EduRev

The document Introduction to Thermochemistry - Thermochemistry Chemistry Notes | EduRev is a part of the Chemistry Course Physical Chemistry.
All you need of Chemistry at this link: Chemistry

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 = qv = heat exchange at constant volume

Suppose V& UP be the internal energy of reactant and products.
then, ΔU = UP - UR
= qv = 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& HP be the enthalpy of reactant and product then

ΔH = HP - HR
= 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           qP = qV + Δng 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) = ΔCVdT ...(2)
Integrating equation (1) & (2) we get

or Similarly or ΔU2 - ΔU1 = ΔCV (T2 - T1 )
Above equations are known as Kirchhoff equations.
Problem.  Calculate ΔHo373 for the reaction ΔH298 K = -33.18 KJ mol-1
Given, CP (NO2, g) = 37.20 J K-1 mol-1
CP (N2, g) = 29.13 JK-1 mol-1
CP (O2, g) = 29.36 JK-1 mol-1s
Sol.   = - 6.73 x 10-3JK mol-1
Now, ΔH0373 K - ΔH0 298 K = ΔCP (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:

Wtotal = 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  Wtotal = 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 = q– 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. Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Physical Chemistry

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