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**THERMOCHEMISTRY**

Change in Internal energy in a chemical reaction:

From first law dq = dU â€“ w

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

Hence Î”U = q_{v} = heat exchange at constant volume

Suppose V_{R }& U_{P} be the internal energy of reactant and products.

then, Î”U = 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

Î”H = qP

If H_{R }& H_{P} be the enthalpy of reactant and product then

Î”H = 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

Î”H = qP & Î”U = qV

then qP = qV + PÎ”V

Forn moles of an ideal gas

PV = nRT

Let n_{1} & n_{2} be the number of moles of gaseous reactants & products.

then Î”n = n_{2} - n_{1} = increase in the number of gaseous moles

Then corresponding increase in volume Î”V is given by

Hence,

Then q_{P} = q_{V} + Î”n_{g} RT

**Variation of enthalpy of reaction with temperature i.e., Kirchhoff equation.**

The enthalpy change for the reaction

aA + bB â†’ cC + dD

is given by

The temperature dependence of enthalpy at constant pressure is:

then

or d(_{D}H) = _{D}CPdT ...(1) The temperature dependence of enthalpy of constant volume is:

or d(Î”U) = Î”C_{V}dT ...(2)

Integrating equation (1) & (2) we get

or

Similarly

or Î”U_{2} - Î”U_{1} = Î”C_{V} (T_{2} - T_{1} )

Above equations are known as Kirchhoff equations.**Problem. Calculate Î”H ^{o}373 for the reaction**

Î”H298 K = -33.18 KJ mol^{-1}

Given, C_{P} (NO_{2}, g) = 37.20 J K^{-1} mol^{-1}

C_{P} (N_{2}, g) = 29.13 JK^{-1} mol^{-1}

C_{P} (O_{2}, g) = 29.36 JK^{-1} mol^{-1}s

Sol.

= - 6.73 x 10^{-3}**JK mol ^{-1} **

Now,

Î”H^{0}_{373 K }- Î”H^{0}_{ 298 K }= Î”C_{P} (373 - 298)

- Î”H
^{0}_{373 K}= [-33.18 + (-6.73 Ã— 10^{-3}) (373 - 298)] KJ mol^{-1} - Î”H
^{0}_{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 T_{2} and other one at a lower temperature T_{1}.

Four successive operations as given below:

(1) The isothermal reversible expansion from volume V_{1} to volume L at the higher temperature T_{2} .

Since dT = 0, therefore, Î”U_{1} = 0 According to first law of thermodynamics, we have q_{2} = 2

(2) The adiabatic reversible expansion from volume V_{2} to volume V_{3}.

Temperature of the system after expansion is T_{1}. Since for adiabatic process q = 0, therefore, it follows that

(3) The isothermal reversible compression from V_{3} to V_{4} at temperature T_{1}.

Here Î”U_{3} = 0

and

(4) Adiabatic reversible compression from volume V_{4} to V_{1}.

Temperature of the system returns T_{2}. Therefore,

**Newt work involved in the cyclic process is: **

W_{total }= W_{2} + W + W_{1} + Wâ€™

**Relationship between** V_{1}, V_{2}, V_{3} & V_{4}: 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 Î·.

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 400Â°C, how much work is done on the engine per cycle and how much heat is involved at 100Â°C in each cycle? Sol.** The efficiency of Carnot cycle is

Thus,

the heat involved is:

q_{1} = 1.855 kJ (heat is negative sign)

q_{1} = â€“1.855 kJ

and work done is :

W = q_{2 }â€“ q_{1} = â€“3.347 + 1.855 = â€“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 **Î·** & Î² :**

â‡’

â‡’

â‡’

â‡’

â‡’

i.e.

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