# What is Enthalpy? Definition, Equations, Units, Examples

Thermodynamics is a branch of science in which the relation between work, heat, temperature, and energy is studied. Every day we encounter events that include the radiation and absorption of heat for example “When ice melts”. The topic Enthalpy is a state function, so to study what Enthalpy is we need to understand what a state function is.

**What is a State Function?**

Basically, the temperature of matter can be changed by either heating or cooling. So, in the State function, it is important just to change the temperature, and the way or **path **is not considered. Therefore it can be said that the State function is only dependent upon the current state/initial state and the final state._{ }

Some examples of state functions are Temperature, Pressure, Internal energy, **Enthalpy**, etc.

**Internal Energy (U)**

Every substance contains a definite amount of energy stored in them, this energy is nothing but the **Internal energy **of the substance (system) and is denoted by **U**. The energy inside the substance is the sum of all kinetic and potential energy and all the other energy present in the particles of the substance.

So, change in internal energy can be shown as:

**ΔU = Q + W**

where:

**Q**is the heat transfer between the substance and the surrounding and**W**is the work done.

This is also known as the **first law of thermodynamics.**

**Isobaric process**

In thermodynamics, when the pressure of the system stays constant throughout the process is called the Isobaric process. So, the first law of thermodynamics becomes

**ΔU = Q _{p} – P_{ex} × ΔV **

or

**Q _{p = }ΔU + P_{ex} × ΔV**

where:

- Q
_{p}is work done in the isobaric process and - P
_{ex}external pressure.

### Enthalpy (H)

Enthalpy is simply defined as the sum of internal energy and the energy that is resulting due to its pressure and volume.

So, enthalpy can be shown as:

**H = U + PV**

where:

**U**is Internal Energy,**P**is pressure and**V**is volume.

Here, U, P, and V are state functions so H is also a state function. As enthalpy is a state function, change in enthalpy (ΔH) will depend on the initial and the final states of the system

ΔH = H_{2} – H_{1}

Here, H_{1} is the enthalpy of the system in the initial state and H_{2} is the enthalpy of the system in the final state. So if we write the enthalpy’s formula ( H = U + PV ) in a similar form

H_{1} = U_{1 }+ P_{1}V_{1} and H_{2 }= U_{2} + P_{2}V_{2}

So if we put the above equation in the ΔH we get,

ΔH = (U_{2} + P_{2}V_{2}) – (U_{1}+ P_{1}V_{1})

ΔH = U_{2 }+ P_{2}V_{2} – U_{1} -P_{1}V_{1}

ΔH = (U_{2} – U_{1}) + (P_{2}V_{2} -P_{1}V_{1})

**ΔH = ΔU + Δ(PV)**

where:

- ΔU is the change in internal energy and
- Δ(PV) is the change in pressure and volume.

Now, at a constant pressure P1 = P2 = P

ΔH = ΔU + PΔV

Consider pressure inside and outside are the same for this isobaric process (i.e. P_{ex} = P) then the formula for the isobaric process will become,

Q_{p} = ΔU +PΔV

Thus from the above two equations, we get,

**ΔH = Q _{p} **

Thus from this derived formula, we understand that the increase in enthalpy of a system is equal to the heat absorbed by it at a constant pressure.

### Relationship between ΔH and ΔU for chemical reaction

ΔH and ΔU are related by the equation ΔH = ΔU + PΔV, at constant pressure. For reactions between solids and liquids, ΔV is very small because as pressure varies, solids or liquids won`t get affected significantly. So, for these reactions remove PΔV from the equation and write ΔH = ΔU

However, for the reactions involving gases, which are easily affected by the change in pressure, ΔV should strictly be considered.

ΔH = ΔU + PΔV

ΔH = ΔU + P(V_{2} – V_{1})

ΔH = ΔU + PV2 – PV1

where:

- V
_{1}is the initial state ( volume of gas reactants) and - V
_{2}is the final state ( volume of gas products).

Here we consider the reactants and the product to be ideal, so we can use the ideal gas equation (PV = nRT).

Let’s consider there are n_{1} moles of gaseous reactants that produce n_{2} moles of gaseous products. Then the equation becomes

PV_{1} = n_{1}RT and PV_{2} = n_{2}RT

Thus, the equation become

ΔH = ΔU + n_{2}RT – n_{1}RT

ΔH = ΔU + RT (n_{2 }– n_{1})

**ΔH = ΔU + RT Δn**

## Requirements for ΔH to be equal to ΔU

- When the reaction is conducted inside a closed container which prevents the alter of the volume of the system (ΔV = 0). Then change in enthalpy will change as ΔH = ΔU.
- When there is only solids or liquids involved in the reactions then we can neglect ΔV as the change in them due to the pressure is significant. So, ΔH = ΔU.
There reaction in which the moles of gaseous products and reactants are same (i.e. n

_{2}= n_{1}). So, ΔH =ΔU

### Work done in the Chemical Reactions

The work done at constant pressure and temperature is given by

**W = – P _{ex} × ΔV**

Assume P_{ex} = P, then the equation becomes

W = -P( V_{2} – V_{1})

W = PV_{1 }– PV_{2}

Using the Ideal gas equation,

W = n_{1}RT – n_{2}RT

W = -RT (n_{2} – n_{1})

**W= – RT Δn**

**Sample Problems**

**Problem 1: For a reaction, the system absorbs 10 kJ of heat and does 3 kJ of work on its surroundings. What are the changes in the Internal energy and Enthalpy of the system? **

**Solution: **

According to the First law of thermodynamics,

ΔU = Q + W

Q = +10 kJ and W = -3 kJ

(W = -3 kJ because the work is done on the surrounding by the system so the system has lose that energy)

ΔU = 10 kJ – 3 kJ

∴ ΔU = +7 kJand, Q

_{p}= ΔH

∴ Q_{p }= +10kJ

Thus, the Internal energy increases by 7 kJ and Enthalpy by 10 kJ.

**Problem 2: For a reaction, 5 kJ of heat is released from the system and 10 kJ of work was done on the system. What are the changes in the Internal energy and Enthalpy of the system?**

**Solution: **

According to first law of thermodynamics,

ΔU = Q + W

Q = -5 kJ, W = +10 kJ

ΔU = -5 kJ +10 kJ =

+5 kJand Q

_{p}= ΔH

∴Q_{p }= -5 kJ

Thus, the Internal energy increases by 5 kJ and Enthalpy decreases by -5 kJ.

**Problem 3: An Ideal gas expands from a volume of 5 dm ^{3 }to 15 dm^{3} against a constant external pressure of 3.036 x 10^{5}Nm^{-2}. Find ΔH if ΔU is 400 J.**

**Solution: **

ΔH = ΔU + PΔV

ΔH = ΔU + P(V

_{2}– V_{1})Assume that P

_{ex}= P, P =3.036 *10^{5}N m^{-2}ΔU = 400 J

V

_{1 }= 5 dm^{3}= 5 × 10^{-3 }m^{3}V

_{2}= 10 dm^{3 }= 10 × 10^{-3 }m^{3}Substituting the values in the equation

ΔH = 400 J + 3.036 × 10

^{5}Nm^{-2}* (10 × 10^{-3}m^{3}– 5 × 10^{-3}m^{3})ΔH = 400 J + 3.036 ×10

^{5}Nm^{-2}* (15 – 5) × 10^{-3}m^{3}ΔH = 400 J + 3.036 × 10

^{3}J

ΔH = 3436 J.

**Problem 4: Calculate the work done in the following reaction when 2 moles of HCl are used at Constant pressure at 420 K.**

**4HCl (g) + O _{2} (g) **

**→**

**2Cl**

_{2}(g) + 2H_{2}O (g)**State, whether the work done, is by the system or on the system.**

**Solution : **

According to the Formula to calculate the work done in chemical reactions,.

W = – Δn RT

W = – RT ( n

_{2}– n_{1})2 moles of HCl react with 0.5 mole of O

_{2}to give 1 mole of Cl_{2}and 1 mole of H_{2}OHence, n

_{1}= 2.5,n

_{2}= 2,R = 8.314 JK

^{-1}mol^{-1}T = 420 K

Substituting the values in the equation,

W = – 8.314 J K

^{-1}mol^{-1}× 420 K × (2 – 2.5) molW = -8.314 × 420 × (-0.5) J

W = 1745.94 J

**Conceptual Questions**

**Question 1: Explain the state function with an example.**

**Answer: **

Any property of a system whose value depends on the current state of the system and is independent of the path followed to reach that state is called the state function.

Example, Temperature.

**Question 2: Explain in brief Internal Energy.**

**Answer: **

Every substance is associated with a definite amount of energy. This energy is stored in a substance (System) is called its internal energy and is denoted by U. The internal energy is the sum of kinetic energies of all the molecules, ions and atoms of the system, the potential energies associated with the forces between the particles, the kinetic and potential energies of nuclei and electrons in the particles and the energy associated with existence of mass of the system.

**Question 3: How to calculate the change in enthalpy?**

**Answer: **

The change in enthalpy ( ΔH) can be obtained by

ΔH = ΔU + RT Δn

**Question 4: What is an Isobaric process and give any 2 examples.**

**Answer: **

Most chemical reactions are run in open containers under constant pressure. In such reactions the volume of the system is allowed to change, such kind of processes are called as Isobaric processes.

Examples:

- Boiling of water and its conversion into steam,
- Freezing of water into ice.

**Question 5: Explain the First Law of Thermodynamics and give its mathematical equation.**

**Answer: **

First law of

thermodynamics is simply the law of conversation of energy. According to this law the total energy of a system and its surroundings remain constant when the system changes from initial state to final state. The law is stated in different ways but the meaning is the same that the energy is conserved in all the changes.Mathematical expression for the First Law of Thermodynamics,

ΔU = Q + W

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