The study of thermodynamics is the study of systems that are too large to be extrapolated by mechanics alone. For many generations, thermodynamics was vaguely understood, and many of the results were determined only experimentally. Some of the results posed great theoretical challenges for physicists, who made many unprofitable attempts to explain the origins of the formulas. With the advent of quantum mechanics approached for clarification of the results.

The mechanics of the related particles are immaterially exceedingly complex, nonetheless. For this reason, statistical physics plays an important role in the basis of thermodynamics. Instead of worrying about the exact values of parcels for each particle in a system, we look at statistically averaged values over quantum probabilities. In fact, introductory concepts like the energy of a system are extrapolated as parameters. The conception of enthalpy is consequential for the temperature and pressure necessitated for any chemical reaction.

### What is Enthalpy?

Heat absorbed or evolved at constant pressure is called

heat of the reactionorenthalpy of the reaction. The change in enthalpy accompanying a reaction is called thereaction enthalpy,

Product → Reactant

Enthalpycan also be defined as the total heat content of the system. It’s equal to the sum of internal energy and pressure-volume work. The internal energy change follows the first law of thermodynamics, according to which:

q =ΔU + PΔVwhere,

- U is internal energy and
- H is a static function

Still, ΔV = 0, If the procedure is conveyed out at constant volume. The equation mentioned before additionally reduces to the form:

q_{v }= ΔU

- where the subscript v indicates a constant volume.

Therefore, the internal energy change is a constant amount of soaked or expanded heat.

It can be further stated that since ΔU is a state function, thus, q_{v} is also a state function.

If a process is carried out at constant pressure (as is usually the case, because most of the reactions are studied in vessels open to the atmosphere or if a system consists of a gas confined in a cylinder fitted with a piston, the external pressure acting on the piston is the atmospheric pressure), the work of expansion is given by,

w = – PΔV

- where ΔV is the increase in volume and P is the constant pressure.

According to the first law of thermodynamics,

q = ΔU – w

- where q is the heat soaked by the system, ΔU is the increment in internal energy of the system and w is the work befitted by the system.

Under keeping of constant pressure, emplacing w = -PΔV and depicting the heat absorbed by ap (subscript p denoting constant pressure),

**q _{p }=ΔU + PΔV ** ….(1)

Suppose when the system absorbs q, joules of heat, its internal energy increases from U_{1} to U_{2} and the volume increases from V₁ to V₂. Then, we have

ΔU = U_{2} – U_{1}

and ΔV = V₂ – V₁

Putting these values in equation (1) above, we get

q_{p }= (U₂-U₁) + P(V₂-V₁)

q_{p} = (U₂ + PV₂) – (U₁ + PV₁)

Now, as U, P, and V are the functions of the state, therefore, the quantity U + PV must also be a state function. The thermodynamic quantity U+ PV is called the heat content or enthalpy of the system and is represented by the symbol H, Le, the enthalpy may be defined mathematically by the equation,

**H = U + PV** ….(2)

Thus, if H_{2} is the enthalpy of the system in and putting these values in equation (2), we get

H₂ = U₂ + PV₂

H₁ = U₁ + PV₁

q_{p} = H₂ – H_{1}

or** **

**q = ΔH **….(3)

- where ΔH = H₂-H
_{1}is the enthalpy change of the system.

Hence, the Enthalpy change of a system is equal to the heat absorbed or evolved by the system at constant pressure. It may be remembered that as most of the reactions are carried out at constant pressure (i.e., in the open vessels), the measured value of the heat evolved or absorbed is the enthalpy change.

Further, putting the value of q, from equation (3) in equation (1), we get

**ΔH = ΔU+ PΔV**

### Enthalpy Change of a Reaction

The enthalpy change can be represented as a function of the increase in internal energy of the system and the pressure-volume work done, i.e., expansion, correspondingly as a process.

**The physical concept of enthalpy or heat content –** Enthalpy is defined by the subtle expression, H = U + PV, let us try to understand what this volume is. It has been shown earlier that every substance or system has a certain amount of energy stored in it. called **internal energy**. This energy can be of many types.

The energy stored within the substance or the system that is available for conversion into heat is called the heat content or enthalpy of the substance or the system. Like the internal energy, the absolute value of the heat content or enthalpy of a substance or system cannot be measured and, fortunately, it is not equally necessary. In thermodynamic processes, we are only concerned with enthalpy changes (ΔH), which can be fluently increased experimentally. Subsequently, it can be advertised that U and V are wide parcels, thus the enthalpy is correspondingly a broad property.

**Extensive property-**Value depends on the volume or size of matter in the system. Exemplifications- mass, volume, internal energy, heat capacity, etc.**Intensive property-**Value doesn’t depend on the volume or size of matter in the system. Exemplifications-temperature, viscosity, pressure, etc.

## Relationship between Heat of Reaction at Constant Pressure and Constant Volume

It has already been discussed that,

q

_{p}= ΔHand

q….(1)_{v}= ΔUIt has also been derived already that at constant pressure,

ΔH = AU + PΔV….(2)

- where ΔV is the change in volume
Equation. (2) can be rewritten as:

ΔH = ΔU + P(V₂-V₁) = ΔU + (PV₂-PV₁)….(3)

- where V
_{1}is the initial volume and V₂ is the final volume of the system.But for ideal gases, PV=nRT so that,

PV₁ = n

_{1 }RTand

PV₂=n₂ RT

- where n
_{1}is the number of moles of the gaseous reactants and n_{2}is the number of moles of the gaseous products.Substituting these values in the equation. (3),

ΔH = ΔU+ (n

_{2}RT- n_{1}RT) = ΔU+(n₂-n_{1}) RT

ΔH = ΔU + Δn RT….(4)where Δn

_{g }= n_{2 }– n_{1}is the difference between the number of moles of the gaseous products and those of the gaseous reactants. Putting the values of AU from equation (1), equation (4) becomes:

q_{p}= q_{v}+Δn RT

**Conditions under which q _{p}=q_{v} or ΔH = ΔU **

- When the reaction is carried out in a closed vessel so that the volume remains constant, that is, ΔV=0
- When the reaction involves only solids or liquids or solutions but no gaseous reactants or products. This is because the change in the volume of solids and liquids during a chemical reaction is negligible.
- When reaction involves gaseous reactants and products but their number of moles are equal (ie, = n
_{p = }n_{r}), e.g., in the reactions

**H₂ (g) + Cl₂ (g)→ 2HCl(g) **

**C(s) + O₂(g) → CO₂ (g)**

Thus, q_{p} is different from q_{v} only in those reactions which involve gaseous reactants and products and (n_{p})_{gaseous} ≠ (n_{r}) _{gaseous.}

### Sample Problems

**Problem 1: The heat of combustion of benzene in a bomb calorimeter (ie constant volume) at 25°C was found to be 3263.9 kJ mol ^{-1}. Calculate the heat of combustion of benzene at constant pressure.**

**Solution:**

The reaction is:

C_{6}H_{6}(l) + 7 ½O₂ (g) → 6CO₂ (s) + 3H₂O(l)In this reaction, O

_{2}is the only gaseous reactant and CO_{2}is the only gaseous product.Δn

_{g}= n_{p}-n_{r}= 6-7 ½ =-1½ = -2(3/4)Also, given ΔU (or q

_{v})=-3263.9 kJ mol^{–}¹T=25°C = 298 K

R=8314 J K

^{-1}mol^{-1}= (8.314/ 1000) KJ K^{-1 }mol^{-1}ΔH (or q

_{p}) = ΔU +Δn_{g}RT= -3263.9 kJ mol

^{–}¹ + (-3/2 mol) (8.314/1000 KJ K^{-1}mol^{-1}=-3263.9-3.7 kJ mol-

=-3267.6 kJ mol

^{-1}=

8.314 kJ K^{-1}mol^{-1}

**Problem 2: If water vapour is assumed to be a complete gas, the molar enthalpy change at 1 bar and 100°C is 41 kJ mol. Calculate the internal energy change when:**

**1 mole of water vaporizes at 1 bar pressure and 100 °C.****1 mol of water turns into ice.**

**Solution:**

- For vaporization of water, the change is: H₂O (l)→ H₂O(g)
Δn

_{g}= n_{p}-n_{r}= 1-0 = 1

ΔH = ΔU+Δn

_{g}RTΔU=ΔH-Δn

_{g}RT= 41.00 kJ mol

^{–}¹- (1 mol ) × (8.314 × 10^{-3}kJ K^{–}¹ mol^{-1}) (373 K)= 41.00 – 3.10 kJ mol

^{-1}=

37.90 kJ mol^{-1}

- For conversion of water into ice, the change is H₂O (l) → H₂O (s) this case, the volume change is negligible.
Hence, ΔH – ΔU =

41.00 kJ mol^{-1}.

**Problem 3: A swimmer coming out of a pool is covered with a film of internal energy of 18 g. How much heat must be supplied to make this water evaporate at 298 K? Calculate the evaporation at 100°C. Δvap H ^{0} for water 373 K = 40.66 kJ mol^{–}¹. **

**Solution: **

The process of evaporation is: 18 g H₂O (l) 18 g H₂O (g)

No. of moles in 18 g H₂O = 18 g / 18 g mol

^{-1}= 1 molΔn

_{g }=1-0=1 molΔvap U

^{0}= Δvap H^{0}-Δn_{g}RT= 40.66 kJ mol

^{–}¹- (1 mol) (8.314 x 10^{–}³ JK^{-1}mol^{–}¹) (298 k)= 40.66 kJ mol

^{-1 }– 3.10 kJ mol^{-1}=

37.56 kJ mol^{-1}

**Problem 4: Find the internal energy change for the reaction A(l) → A(g) at 373 K. The heat of vaporization is 40.66 kJ/mol and R=8.3 J mol K ^{–}¹.**

**Solution:**

A(l)→ A (g),

ΔH – ΔU + Δn

_{g}RTΔU=ΔH- Δn

_{g}RT= 40660 J-1 mol x 8.314J K

^{–}¹ mol^{–}¹ x 373 K= 40660 J-3101 J

= 37559 3 mol

=

37.56 kJ mol^{–}¹