In this article, we shall study to calculate the change in internal energy and change in enthalpy in a chemical reaction.

Example – 01:

For a particular reaction, the system absorbs 6 kJ of heat and does 1.5 kJ of work on its surroundings. What are the change in internal energy and enthalpy change of the system?

Given: q = + 6 kJ (Heat absorbed), W = -1.5 kJ (Work done on the surroundings).

To Find: Change in internal energy = ΔU =? Enthalpy change = ΔH =?

Solution:

By the first law of thermodynamics

Δ U = q + W

Δ U = 6 k J – 1.5 kJ = 4.5 kJ

ΔH = q_p = Heat supplied at constant pressure = + 6 kJ

Ans: The change in internal energy is 4.5 kJ and enthalpy change is 6 kJ

Example – 02:

An ideal gas expands from a volume of 6 dm³ to 16 dm³ against constant external pressure of 2.026 x 10⁵ Nm^-2. Find Enthalpy change if ΔU is 418 J.

Given:  Initial volume = V₁ = 6 dm³ = 6 × 10^-3 m³, Final volume = V₂ = 16 dm³ = 16 × 10^-3 m³, P_ext = 2.026 x 10⁵ Nm^-2, ΔU = 418 J.

To Find: Enthalpy change = ΔH =?

Solution:

Work done in the process is given by W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 2.026 x 10⁵ Nm^-2 × (16 × 10^-3 m³ – 6 × 10^-3 m³)

∴ W = – 2.026 x 10⁵ × (10 × 10^-3) = – 2026 J

By the first law of thermodynamics

Change in internal energy

Δ U = q_p   + W

∴   418 J = q_p – 2026 J

∴   q_p =   418 J +  2026 J = 2444 J

ΔH = q_p = Heat supplied at constant pressure = 2444 kJ

Ans: enthalpy change is 2444 J

Example – 03:

A sample of gas absorbs 4000 kJ of heat. a) If volume remains constant, what is ΔU? b) Suppose that in addition to absorption of heat by the sample, the surrounding does 2000 kJ of work on the sample. What is ΔU? c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU?

Solution:

Part – a

Given: q = + 4000 kJ (Heat absorbed by sample), ΔV = 0 (Volume remains the same)

To Find: ΔU =?

Work done in the process is given by   W = – P_ext × ΔV = – P_ext × 0 = 0

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = + 4000 kJ   + 0 kJ = + 4000 kJ

Part – b

Given: q = + 4000 kJ (Heat absorbed by sample), W = + 2000 kJ (Work done by surroundings)

To Find: ΔU =?

By the first law of thermodynamics

Δ U = q + W

∴ Δ U = + 4000 kJ   + 2000 kJ = + 6000 kJ

Part – c

Given: q = + 4000 kJ (Heat absorbed by sample), W = – 600 kJ (Work done on the surroundings)

To Find: ΔU =?

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = + 4000 kJ   –   600 kJ = + 3400 kJ

Example – 04:

Calculate the work done in the following reaction when 2 moles of HCl are used at constant pressure and 423 K. State whether work is on the system or by the system.

4 HCl_(g)  + O_2(g)  →  2 Cl_2(g)   +  2 H₂O_(g)

Given: R = 8.314 J K^-1 mol^-1, T = 423 K

To Find: Work done = W =?

Solution:

The reaction is 4 HCl_(g)  + O_2(g)  →  2 Cl_2(g)   +  2 H₂O_(g)

Given 2 moles of HCl are used, hence dividing equation by 2 to get 2 HCl, we get

2 HCl_(g)  + ½O_2(g)  →   Cl_2(g)   +  H₂O_(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (1 + 1) – (2 + ½) = 2 – 5/2 = – ½

Work done in chemical reaction is given by

∴ W = – Δn RT = – (-½) mol × 8.314 J K^-1 mol^-1 × 423 K = 1758 J

∴ W = + 1758 J

Positive sign indicates that work is done by the surroundings on the system

Ans: Work done by the surroundings on the system in the reaction is 1758 J

Example – 05:

Calculate the work done in the following reaction when 1 mol of SO₂ is oxidised at constant pressure at 5o °C. State whether work is on the system or by the system.

2SO_2(g) + O_2(g)  →  2 SO_3(g)

Given:  Temperature = T = 5o °C = 50 + 273 = 323 K, R = 8.314 J K^-1 mol^-1,

To Find: Work done = W =?

Solution:

The reaction is 2SO_2(g) + O_2(g)  →  2 SO_3(g)

Given 1 mole of SO₂ is used, hence dividing equation by 2 to get 1 mol of SO₂

SO_2(g) + ½O_2(g)  → SO_3(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (1 ) -(1 + ½) = 1 – 3/2 = – ½

Work done in chemical reaction is given by

∴ W = – Δn RT = – (-½) mol × 8.314 J K^-1 mol^-1 × 323 K = 1343 J

∴ W = + 1343 J

Positive sign indicates that work is done by the surroundings on the system

Ans: Work done by the surroundings on the system in the reaction is 1343 J

Example – 06:

Calculate the work done in the following reaction when 2 mol of NH₄NO₃ decomposes at constant pressure at 10o °C. State whether work is on the system or by the system.

NH₄NO_3(s) → N₂O_(g)  +  2 H₂O_(g)

Given:  Temperature = T = 10o °C = 100 + 273 = 373 K, R = 8.314 J K^-1 mol^-1 ,

To Find: Work done = W =?

Solution:

The reaction is NH₄NO_3(s) → N₂O_(g)  +  2 H₂O_(g)

Given 2 mol of NH₄NO₃ decomposes, hence multiplying equation by 2

2 NH₄NO_3(s) → 2 N₂O_(g)  +  4 H₂O_(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (2 + 4) -(0) =6

Work done in chemical reaction is given by

∴ W = – Δn RT = – (6) mol × 8.314 J K^-1 mol^-1 × 373 K = – 18607 J

∴ W = – 18.61 kJ

Negative sign indicates that work is done by the system on the surroundings

Ans: Work done by the surroundings on the system in the reaction is – 18.61 kJ.

Example – 07:

CO reacts with O2 according to the following reaction. How much pressure volume work is done and what is the value of ΔU for the reaction of 7.0 g of CO at 1 atm pressure, if the volume change is – 2.8 L.

2CO_(g)  +  O_2(g) → 2CO_2(g)      Enthalpy change = Δ H = – 566 kJ

Given: ΔV = – 2.8 L

To Find: Work done = W =? ΔU = ?

Solution:

Work done in the process is given by

W =  – P_ext × ΔV  = – 1 atm  ×( -2.8) L = 2.8 L atm =  2.8 L atm  × 101.3 J L^-1atm^-1 = 283.6 J

∴ W = 0.2836 kJ

Positive sign indicates the work is done on the system.

From given reaction 2 x (12 + 16) = 56 g of CO on oxidation liberates  566 kJ energy

Hence heat liberated on oxidation of 7.0 g of CO = (7.0/56) × 566 = 70.75 KJ

Hence ΔH = – 70.75 kJ (negative sign as heat is liberated)

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = 0.2836 kJ – 70.75 kJ =  -70.47 kJ

Ans: The work done on the system is 0.2836 kJ and Δ U = -70.47 kJ

Example – 08:

The enthalpy change for the following reaction is – 620 J, when 100 mL of ethylene and 100 mL of H₂ react at 1 atm pressure. Calculate pressure-volume type work and ΔU.

C₂H_4(g)  +  H_2(g) → C₂H_6(g)

Given: ΔH = -620 J

To Find: Work done = W =? ΔU = ?

Solution:

The reaction is C₂H_4(g)  +  H_2(g) → C₂H_6(g)

Thus 1 vol of C₂H_4(g) reacts with 1 vol of H_2(g) to give 1 vol of C₂H_6(g)

Thus 100 mL of C₂H_4(g) reacts with 100 mL of H_2(g) to give 100 mL of C₂H_6(g)

Change in volume during the reaction = ΔV = V_product – V_reactant = (100) – (100 + 100) = – 100 mL = – 0.1 L

Work done in the process is given by

W = – P_ext × ΔV = – 1 atm  ×( -0.1) L = 0.1 L atm =  0.1 L atm  × 101.3 J L^-1atm^-1 = 10.13 J

∴ W = + 10.13 J

Positive sign indicates the work is done on the system.

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U =   + 10.13 J – 620 J =  – 609.9 J

Ans: the work done on system is 10.13 J and Δ U = – 609.9 J

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Science > Chemistry > Chemical Thermodynamics and Energetics > Change in Internal Energy and Enthalpy

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The internal energy can be defined as the sum of all possible kinds of energies that a system possesses. Every chemical reaction is associated with heat change. Heat is either evolved or absorbed.  This is possible only when every substance involved in chemical reaction possess certain fixed amount of energy which is called internal energy. Every substance is composed of molecules, atoms and subatomic particles.  The position and motions of the molecules, atoms and subatomic particles is the origin of internal energy.

Constituents of Internal Energy:

Kinetic or Thermal Energy:

The energy that a body possesses due to its movements is called as kinetic energy. Since kinetic energy depends upon the temperature it is called thermal energy.  Thermal energy is directly proportional to the temperature. K.E. is of three types

Translational Energy (E_trans):

The energy associated with the molecules due to the continuous, rapid, random movements along straight line path is called translational energy. Molecules of gases or liquids are in a state of constant random movement. Hence the molecules of gases and liquids have translational energy.

Vibrational Energy (E_vibr):

The energy associated with the molecules due to atomic vibrations is called vibrational energy.

There is a repulsive force between nuclei of two atoms, and electrons of two atoms.  There is an attractive force between the nucleus of one atom and electrons of other atoms and vice versa. As a result of these attractions and repulsions, atoms are in a state of to and fro movement.  Vibrations are of two types viz. stretching and bending vibration.

Rotational Energy  (E_rot):

The energy associated with the molecules by virtue of their rotations about their axes is called as rotational energy. All diatomic and polyatomic molecules rotate about an axis perpendicular to the axis of molecule.

Thus   the total kinetic energy of the system is given by

K.E.  = E_trans + E_vibr +  E_rot

Potential or Bonding Energy:

The energy associated with the body by virtue of its position is called is potential energy. Potential energy is independent of temperature.  It arises due to the bonding between atoms in a molecule. It is classified into two types.

Intermolecular Energy (E_intermole):

The amount of energy required to separate molecules from each other is called as intermolecular energy. In solids and liquids, molecules are held together by means of weak physical forces of attraction called Vander Waals forces. These forces are strong in solids hence intermolecular energy is more in solids than in liquids.

Intramolecular Energy (E_intramole):

The energy required to break the molecule into its constituent atoms is called as intramolecular energy. Atoms are held together in the molecule by certain electrostatics force of attraction called the chemical bond.  So intramolecular energy is nothing but the energy required to break chemical bonds so as to isolate constituent atoms from each other.

Thus the total potential energy of the system is given by

P.E. =   E_intermole + E_intramole.

Total Internal Energy of a System:

Internal energy is the sum of K.E. and P.E. The absolute value of it cannot be determined but the change in it can be determined.

Significance of Internal energy:

• It depends on the quantity of a substance, hence it is extensive property.
• Change in it represents the heat evolved or absorbed in a reaction at constant temperature and constant volume.
• For isothermal process change in it is zero.
• For the process involving the evolution of energy change in it is negative and for the process involving absorption of energy change in it is positive.

Characteristics of Internal energy:

• The internal energy of a system is extensive property.
• It is a state property. The change in internal energy is independent of the path followed.
• Change in it of a cyclic process is zero.

Notes:

• In practice, the absolute value of internal energy (U) is not known and cannot be measured because it is very difficult to determine accurately the most of the quantities that contribute to the internal energy of a system. But in thermodynamics, the quantity internal energy (U) is not important while the change in it (ΔU) is important.
• The quantity change in internal energy (ΔU) is associated with many other thermodynamic quantities by simple mathematical relations. Using those relations the change in internal energy (ΔU) can be determined. Such relations are ΔH = ΔU   +  PΔV  and ΔH = ΔU   + ΔnRT Hence  the change in internal energy (ΔU) can be determined.

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