Chapter 6: Thermodynamics

Ans: We know that the change in internal energy equals the heat absorbed at constant volume, i.e., U= qV.m. However, the majority of chemical reactions are carried out in flasks or test tubes under constant atmospheric pressure U= qp -PV under constant pressure, where qp is the heat absorbed by the system and -PV is the expansion work done by the system.

Let's call the beginning state with subscript 1 and the end state with subscript 2.

The above equation can be rewritten as

U2-U1= P(V2-V1)

On rearranging, we get

qp= (U2+pV2) - (U1 + pV1)...........(1)

Now, we can define another thermodynamic function, the enthalpy H [Greek word enthalpien, to warm or heat content] as:

H = U + pV………(2)

So, equation (1) becomes

qp = H₂- H₁ = ΔH

Despite the fact that q is a path dependent function, H is a state function because it is reliant on state functions U, p, and V. As a result, H is path-independent. As a result, qp is path-independent.

Equation (2) can be written as for finite changes at constant pressure.

ΔH= ΔU + ΔpV

Since, P is constant, we can write:

ΔH = ΔU + PΔV……………(3)

It's worth noting that we're actually measuring changes in the enthalpy when heat is absorbed by the system at constant pressure. Remember  ΔH= qp , heat absorbed by the system at constant pressure. H is negative for exothermic reactions, which produce heat, and positive for endothermic reactions, which absorb heat from the environment.

As a result, equation (3) becomes U = qv at constant volume (V=0).

ΔH= ΔU=qv

Let's have a look at a gaseous reaction. If VA is the total volume of gaseous reactants, VB is the total volume of gaseous products, nA is the number of moles of gaseous reactants, and nB is the number of moles of gaseous products, then we write using the ideal gas law.,

PVA = nART

PVB = nBRT

Thus, PVB- PVA = nBRT - nART = (nB-nA)RT

Or, P(VB- VA) = (nB - nA)RT

Or, pΔV = ΔngRT……….(4)

The number of moles of gaseous products minus the number of moles of gaseous reactants is denoted by ng.

Equation (3) is obtained by substituting the value of PV from equation (4) in equation (3).

ΔH = ΔU + ΔngRT…………(5)

Ans: A contrast is drawn in thermodynamics between extensive and intense qualities. An extensive property is one whose value is proportional to the amount or size of matter in the system. Extensive properties include mass, volume, internal energy, enthalpy, and heat capacity, to name a few.

Properties that are independent of the amount or size of matter present are known.

As though they were intensive properties Temperature, density, and pressure, for example, are intense properties. A molar property, 𝝌m, is the value of an extensive property of the system for 1 mol of the substance. If n is the amount of matter, 𝝌m = 𝝌/n, is independent of the amount of matter. An intensive property is always the ratio of two extensive properties.

Extensive/Extensive = Intensive

E.g., Mole Fraction = Moles / Total number of moles = Extensive/Extensive

Ans:  The enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in a gaseous state is called the lattice enthalpy of an ionic compound.

Na+ + Cl- (s) → Na+ (g) + Cl- (g)

Δlattice H0 → +788 KJmol-1

Let us now calculate the lattice enthalpy of Na+Cl- (s) by following steps given below:

1. Na (s) → Na (g), sublimation of sodium metal, ΔsubH0 = 108.4 KJmol-1

2. Na (g) → Na+ (g) + e- (g) the ionization sodium atoms, ionization enthalpy.

ΔiH0 = 496 KJ/mol 

3. ½ Cl2 (g)  → 2Cl (g), the dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy.

1/2 Δbond H0 = 121 KJ/mol

4. Cl (g) + e- → Cl- (g) electron gained by chlorine atoms. The electron gain enthalpy,  

Δe.g H0 = -348.6 KJ/mol.

5. Na+(g) + Cl- (g) → Na+Cl- (s) 

Ans: We know,

ΔStotal = ΔSsys + ΔSsurr

When a system is in thermal equilibrium with its surroundings, the surroundings' temperature is the same as the system's. Furthermore, a rise in the enthalpy of the surroundings equals a decrease in the system's enthalpy. As a result of the entropy shift in the environment,

ΔSsurr = ΔHsurr/T = -ΔHsys/T

ΔStotal = ΔSsys = (-ΔHsys/T)

Rearranging the above equation:

ΔStotal = TΔSsys - ΔHsys

For spontaneous process,

ΔStotal > 0, so 

TΔSsys - ΔHsys> 0

⇒ ( - ΔHsys - TΔSsys) > 0

The above equation can be written as

- ΔG > 0

ΔG = ΔH - TΔS < 0

ΔHsys is the enthalpy change of a reaction, TΔSsys is the energy which is not available to do

useful work. As a result, G is a measure of 'free energy,' as it is the net energy available to conduct beneficial work. As a result, it's also known as the reaction's free energy. At constant pressure and temperature, G gives a set of conditions for spontaneity,

(i) If ΔG is negative (<0), the process is spontaneous.

(ii) If ΔG is positive (>0), the process is nonspontaneous.

• Unit of ΔG is Joule.

• The reaction will be spontaneous at high temperature.

Ans: (i) Reversible work is represented by the combined areas

(ii) Work against constant pressure, Pf  is represented by the area

• Since the concept of Thermodynamics and the terminologies of Chemistry are a bit new to the students, they should first learn the names of different chemical components and how to write the chemical equations properly by joining the free Classes on Vedantu and by consulting the NCERT Exemplars and Solutions books.

• While writing the chemical equations, they might make some mistakes. To avoid making errors, they should practice writing the equations at home.

• Solving previous years' question papers will help them to manage time and quickly analyze the answers in their minds.

• Comparing their own answers with the solutions provided in the NCERT Exemplar Solutions will help them to be sure of the fact that how efficient they are at answering the questions of Chemistry. It will help them to boost their confidence as well.

• Revising is extremely important when it comes to answering Chemistry questions since a lack of practice and habits of writing can make the students forget some important information that they should include in their answers. The mock papers are beneficial for them to serve this purpose

• Rational thinking plays an important part in answering the questions in Chemistry. The students need to have a clear idea of the basic physical characteristics of different chemical compounds and how they will react depending on the presence of other external factors. Practicing is the only way to sharpen their thinking ability and analysis power.