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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 n_{1} & n_{2} be the number of moles of gaseous reactants & products.
then Δn = n_{2}  n_{1} = 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(_{D}H) = _{D}CPdT ...(1) The temperature dependence of enthalpy of constant volume is:
or d(ΔU) = ΔC_{V}dT ...(2)
Integrating equation (1) & (2) we get
or
Similarly
or ΔU_{2}  ΔU_{1} = ΔC_{V} (T_{2}  T_{1} )
Above equations are known as Kirchhoff equations.
Problem. Calculate ΔH^{o}373 for the reaction
ΔH298 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,
ΔH^{0}_{373 K } ΔH^{0}_{ 298 K }= ΔC_{P} (373  298)
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, ΔU_{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 ΔU_{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} + W’
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 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:
q_{1} = 1.855 kJ (heat is negative sign)
q_{1} = –1.855 kJ
and work done is :
W = q_{2 }– q_{1} = –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.
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