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Enthalpy | Equation, Relations and Numericals

Enthalpy is the heat content of the system at constant pressure. It is donated by H. Most of the processes are performed at constant pressure, and enthalpy or heat content is one of them. Its unit is KJ/mol. It is defined as follows:

H = U + PV


  • U = Internal energy of the system(E)
  • P = External pressure of the system
  • V = Volume of the system

H is a state function because U, P, and V all are state functions. The change of state function depends only upon the initial and final state of the system and is not affected by the path of the process. Similarly, ΔH is also a state function. The ΔH of a cyclic process is zero (ΔHcyclic = 0). H is an extensive property

Enthalpy is a state function and does't depend on the path

Heat Content (H)

First to understand the concept of heat content (H), let us a system containing some matter (solid or liquid, or gas). The molecules of matter have some internal energy (the energy due to the interaction of molecules). To establish a system at pressure P and volume V from empty space.

Heat content = Internal energy + to establish system at pressure P, volume V from empty space

In thermodynamics, Is there any significance of H?

No. There is no significance of H because the exact value of H cannot be calculated. Only the change in H can be measured.

Change in Enthalpy

The change in enthalpy is equal to the change in internal energy plus the change in pressure P, and volume V.

Let H1 = Enthalpy of the system in an initial state

H2 = Enthalpy of the system in an initial state

∆H = H2 – H1

H1 = U1 + P1V1 and H2 = U2 + P2V2

∆H = (U2 + P2V2) – (U1 + P1V1)

∆H = (U2 – U1) + ( P2V2 – P1V1)

Or ∆H = ∆U + ∆(PV)

Both the pressure and volume are variable and this above definition is 100% correct for any process.

At constant Volume:

∆H = ∆U + V ∆P

At constant Pressure:

∆H = ∆U + P ∆V

The majority of the reactions occur at constant pressure.

A non-ideal gas goes from state 1 to state 2. The pressure, volume, and temperature of state 1 are 2 atm, 3 L, and 95 K, and the P, V, and T of state 2 are 4 atm, 5 L, and 245 K respectively. Find the value of ∆H, if ∆U is 30 atm L.

∆H = ∆U + ∆(PV)

∆H = ∆U + ( P2V2 – P1V1)

∆H = 30 + ( 4×5 – 2×3)

∆H = 30 + 14

∆H = 44 atm L

An ideal gas expands isothermally from 1 L to 10 L at 25°C. Find the value of change in H.

∆H = ∆U + ∆(PV)

The internal energy of an ideal gas depends only upon temperature and isotherm mean at the same temperature so the value of ∆U = 0.

∆H = 0 + ∆(PV)

According to Boyle’s law condition,

P1V1 = P2V2

∆H = 0

Enthalpy in terms of the First law of Thermodynamics


H = U + PV

∆H = ∆U + ∆(PV)

At constant pressure,

∆H = ∆U + P ∆V ———- (1)

First law of Thermodynamics

According to the first law of Thermodynamics, the change in internal energy is equal to the heat given to the system plus the work done on the system.

∆U = ∆Q + ∆W

At constant pressure,

∆U = ∆Qp – Pext ∆V

∆Qp = ∆U + P ∆V ———- (2)

From equations (1) and (2), we get

∆H = ∆Qp (at constant P) ———- (3)

The change in H is equal to the heat content of the system at constant pressure. Now the question is that the change in enthalpy is always equal to heat. No, the change in H is equal to heat content only at constant P.

As we know that

Q = nC∆T


Q = heat

n = no. of moles of gas

C = molar specific heat capacity

∆T = Change in T

At constant pressure,

∆Qp = nCp∆T

Put the value of ∆Qp in eq (3), and we get

∆H = nCp∆T

Cp = molar specific heat capacity at constant P

How to find molar-specific heat capacity Cp?

The formula to calculate the molar specific heat capacity at constant pressure is:

Cp = (f/2 + 1) R

f is the degree of freedom and it depends upon the nature of the gas. The value of the degree of freedom for monoatomic, diatomic, and polyatomic gases is 3, 5, and 6 respectively. R is the general gas constant.

2 moles of helium is heated at constant pressure from 0°C to 50°C. Find ∆H in J.

∆H = nCp∆T

∆H = 2 × (f/2 +1) R × 50

∆H = 2 mol × (3/2 +1) 8.314 J/mol K × 50 k

∆H = 5 × 8.314 × 50

∆H =2078.5 J

Relationship between ∆H and ∆U

Here we discussed the relationship between ∆H and ∆U in a chemical reaction containing gaseous reactants or products or both. In the case of liquids or solids, changes in the state do not have significant changes in the volume (∆V = 0). For such purposes, ∆H and ∆U are approximately the same. However, in the gaseous state the reactants and products ∆H and ∆U are different. For a particular reaction in gaseous state R P

∆H = ∆U + ∆(PV)

For the gaseous state, PV = nRT

∆H = ∆U + ∆(nRT)

At constant pressure and temp,

∆H = ∆U + RT ∆(n)

∆H = ∆U + ∆ng RT ———- (4)


∆ng = n2 – n1 (gaseous)

n1 = moles of gaseous reactant

n2 = moles of gaseous product

The above equation expresses the relationship between ∆H and ∆U.

Prove that Qp = Qv + ∆ng RT or What is the difference between heat content at constant pressure and heat content at constant volume?

∆H = ∆U + ∆ng RT

From the first law of thermodynamics,

∆U = ∆Q + ∆W

∆U = ∆Q – P ∆V

At constant volume, ∆V = 0

∆U = ∆Q – 0

∆U = ∆Qv (at constant V) ———- (5)

From eq—(3) and eq—(5), we can write the eq—(4)

Qp = Qv + ∆ng RT

Find ∆H > ∆U or ∆H < ∆U or ∆H = ∆U for the given reactions.

1) N2 + 3H2 ⇌ 2NH3

∆H = ∆U + ∆ng RT

∆ng = n2 – n1

∆ng = 2 – 4 = -2

so, ∆U > ∆H

2) H2 + Br2 ⇌ 2HBr

∆H = ∆U + ∆ng RT

∆ng = n2 – n1

∆ng = 2 – 2 = 0

so, ∆U = ∆H

3) MgCO3 (s)MgO(s) + CO2 (g)

∆H = ∆U + ∆ng RT

∆ng = n2 – n1

∆ng = 1 – 0 = 1

so, ∆U < ∆H

Find the value of ∆H – ∆U for the reaction at 25°C in KJ.

2C6H6 + 15O2 → 12CO2 + 6H2O

∆H = ∆U + ∆ng RT

∆H – ∆U = ∆ng RT

∆ng = n2 – n1

∆ng = 12 – 15 = -3

∆H – ∆U = (-3) (8.314) (298)

∆H – ∆U = 7.4

Bilal kamboh

A pioneer in the Chemistry space, Bilal is the Content writer at UO Chemists. Driven by a mission to Success, Bilal is best known for inspiring speaking skills to the passion for delivering his best. He loves running and taking fitness classes, and he is doing strength training also loves outings.

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