Standard Enthalpy of Formation: Explanation and Bond Dissociation Enthalpy

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The standard enthalpy of formation or standard heat of formation of a compound is referred to as the change of enthalpy during the formation of one mole of the substance from its constituent elements.

  • A pure element in its standard state possesses a standard enthalpy of formation of zero.
  • Standard states are specified for different types of substances.
  • For gas, it is the hypothetical state that it would assume if it followed the ideal gas equation at 1 bar of pressure.
  • For a gaseous or solid solute in a diluted ideal solution, the standard state is the hypothetical solute concentration of exactly one mole per liter (1 M) at 1 bar pressure extrapolated from infinite dilution.
  • The standard state of a pure substance or solvent in a condensed state (a liquid or a solid) is the pure liquid or solid at 1 bar of pressure.

Key Terms: Enthalpy, Bond Enthalpy, Standard formation of enthalpy, Pressure, Pure element, Thermodynamics, Internal energy, Volume, Temperature, Enthalpy change, Chemical reaction


What is Enthalpy?

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The term “enthalpy” is specifically used in science and engineering in order to quantify heat and function. At a constant strain, the enthalpy notifies how much heat and effort has been applied or extracted from the substance.

  • Enthalpy is the sum of internal energy and the product of pressure and volume in a thermodynamic system.
  • Enthalpy is an energy-like characteristic or state function.
  • It has the dimensions of energy and is thus measured in joules or ergs.
  • Its value is determined only by the system's temperature, pressure, and composition, not history.
  • In symbols, the enthalpy, H, is equal to the sum of the internal energy, E, and the product of the system's pressure, P, and volume, V: H = E + PV.

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Standard Enthalpy of Formation

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The typical enthalpy of the composition or the normal heat of the composition is the combination of the enthalpy while forming a single mole of the object from its objects, and all the elements in their common states.

  • In other words, the standard enthalpy is the enthalpy change reaction where all the factors involved are in standard measurements.
  • The enthalpy change of a chemical reaction is denoted by the symbol ΔrH.

A stable integration condition is considered when the temperature is 298.15 K, and the atmospheric pressure is 1 atm. Now, let us understand this reaction by referring to some of the examples -

  1. Formation of methane from hydrogen and oxygen

C (graphite, s) + 2H2 (g) → CH4 (g)

ΔfHo = -74.81kJmol-1

  1. Formation of hydrogen bromide from bromine and hydrogen

H2 (g) + Br2(l) → 2HBr(g)

ΔrHo = -72.81kJmol-1

Here, we can say that by combining two or more reactants we can create a new mole of product. Also, a typical enthalpy of combustion of any object in its state is at zero levels by definition. Moreover, with the help of Hess's Law, we can calculate the enthalpy change of any chemical reaction using a small set of tabulated data.


Standard Enthalpy of Combustion

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Enthalpy combustion of an object is an enthalpy change produced when a single mole molecule is completely burned in the air or oxygen at a given temperature.

  • It can be used to compare which oils or substances produce the most energy when burned.
  • They can be calculated using a calorimeter of a bomb.

Some examples are as follows:

H2(g) + 1/2 O2(g) → H2O(l) ; ΔcHo = - 286 kJmol-1

C4H10 (g) + 13/2 O2(g) → 4CO2(g) + 5H2O(l) ; ΔcHo = -2658 kJmol-1

Significantly, combustion reactions are exothermic so the value for the enthalpy change (is ΔH) is always negative.

  • Although, in the conventional process, the molar enthalpy of combustion is considered a positive value.
  • Therefore, the enthalpy combustion of an object is defined as the heat dissipated when a single mole of an object completely burns in oxygen.
Enthalpy of Combustion
Enthalpy of Combustion

Bond Dissociation Enthalpy

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Enthalpy bond dissociation is defined as an enthalpy transformation in which a single molecule of covalent bonds of the gaseous compound is broken to form products in the gaseous form.

  • In general, enthalpy bond dissociation values differ from bond enthalpy values which is a measure of some of the total bond dissociation strength in an external molecule, in the case of diatomic molecules.
  • It has been said that the concept of bond dissociation enthalpy is found to be highly useful in the field of thermo-chemistry.

Let us take an example

Cl2(g) → 2Cl(g);

ΔCl–ClH0 = 242 kJ mol-1

Bond Dissociation Enthalpy
Bond Dissociation Enthalpy

Things to Remember

  • The standard enthalpy of formation is the change of enthalpy during the formation of one mole of the substance from its constituent elements.
  • Enthalpy is the sum of internal energy and the product of pressure and volume in a thermodynamic system.
  • It has the dimensions of energy and is thus measured in joules or ergs.
  • The standard enthalpy is the enthalpy change reaction where all the factors involved are in standard measurements.
  • Enthalpy combustion is an enthalpy change produced when a single mole molecule is completely burned in the air or oxygen at a given temperature.

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Sample Questions

Ques. Given,
N2 (g) + 3H2 (g) → 2NH3 (g) ; \(\bigtriangleup\)rH 0 = –92.4 kJ mol-1
What is the standard enthalpy of the formation of NH3 gas? (3 Marks)

Ans. The given equation is:

N2 (g) + 3H2 (g) → 2NH3 (g)

The two moles of ammonia are formed at the time of reaction. Therefore, the reaction for the formation of one mole of ammonia is:

1⁄2N2 (g) + 3⁄2H2 (g) → NH3 (g)

The standard enthalpy of formation of ammonia is calculated as:

Standard Enthalpy of Formation of Ammonia

Thus, the standard enthalpy of the formation of ammonia is -46.2 kJ mol-1

Ques. Enthalpies of formation of CO (g), CO2 (g), N2O(g), and N2O4 (g) are –110,
– 393, 81, and 9.7 kJ mol-1 respectively. Find the value of \(\bigtriangleup\)fH for the
reaction: N2O4 (g) + 3CO(g) → N2O(g) + 3CO2 (g) (3 Marks)

Ans. ΔrH for a reaction is referred to as the difference between the ΔfH value of products and ΔfH value of reactants. 

Enthalpies of Formation of CO

Ques. Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:
CH3OH (l) + 3/2 O2 (g) → CO (g) + 2H2O (l) ; ΔrH° = -726 kJ mol
C(graphite) + O2 (g) → CO2 (g) ; ΔcH° = -393 kJ mol
H2(g) + 1/2 O2 (g) → H2O (l) ; ΔfH° = -286 kJ mol (3 Marks)

Ans. During the formation of CH3OH(l), the reaction that takes place can be written as:

C8 + 2H2O g + ½ O2(g) CH3OHl(1) 

The equation can be obtained from the given reactions by following algebraic calculations:

Equation (i) + 2 x equation (iii) - equation (i)

ΔfHθ [CH3OHl] = ΔcHθ +2ΔfHθ [H2O(l)] - ΔrHθ

= (-393 kJ mol-1) + 2(-286 kJ mol-1) - (-726 kJ mol-1)

Therefore, -239 kJ mol-1

Ques. The enthalpy of combustion of carbon to CO2 is –393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. (3 Marks)

Ans. The formation of CO2 from carbon and dioxygen gas can be written as:

Formation of CO2

(1 mole = 44g)

Heat is released on the formation of 44gCO2 = -393.5 kJ mol-1

Heat is released on the formation of 35.2gCO2

Formation of 35.2gCO2

= -314.8 kJ mol-1

Formation of NH2

Ques. Calculate the enthalpy of the formation of ammonia from the following bond energy data: (N−N) bond =398kJmol-1; (H−H) bond =435kJmol-1 and (N≡N) bond =945.36kJmol-1(2 Marks)

Ans. ΔH = [ΔH(N≡N) + 3 x ΔH(H-H)] - [6ΔH(N-H)]

945 + 3 x 435.0 - 6 x 389.0 = 83.64 kJ

Heat released by NH3 = ΔH/2 = 83.64/2 = 41.82 kJ mol-1

Ques. The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are, –890.3 kJ mol-1 –393.5 kJ mol-1, and –285.8 kJ mol-1 respectively. Enthalpy of formation of CH4 (g) will be (2 Marks)
(i) –74.8 kJ mol-1  (ii) –52.27 kJ mol-1
(iii) +74.8 kJ mol-1  (iv) +52.26 kJ mol-1

Ans. Based on the question:

Enthalpy of Combustion

The desired equation is the one that represents the formation of CH4(g) i.e,

Formation of CH4(g)

Enthalpy of formation of CH4(g) = -74.8 kJ mol-1 

Thus, option (i) is correct. 

Ques. The enthalpies of all elements in their standard states are: (2 Marks)
(i) unity (ii) zero
(iii) < 0 (iv) different for each element

Ans. The enthalpies of all elements in their standard states are zero.

Ques. The heat of combustion of one gram of acetic acid in a bomb colorimeter was found to be -14.57 KJ at 25oC. Calculate the enthalpy of combustion of acetic acid. (2 Marks)

Ans. The combustion reaction of acetic acid is:

Ch3COOH + 2O2 2CO2 +2H2O

The heat of combustion = -14.57 kJ

Number of moles of products = number of moles of reactants 

Thus, the number of moles of gases before and after the reaction is the same

∴ Δ n = 0

∴ The heat of combustion = Enthalpy of combustion

∴ The enthalpy of combustion is: -14.57 kJ

Ques. Define the term enthalpy change of combustion. (2 Marks)

Ans. The enthalpy of combustion of a substance is defined as the change in enthalpy produced when one mole of the substance is completely burnt in air or oxygen at a given temperature.

Ques. Which bond is stronger – covalent or ionic? (2 Marks)

Ans. Ionic bonds are stronger than covalent bonds because there is a greater attraction between ions with opposing charges, which is why separating them, requires a lot of energy. Bonds require the exchange of electron pairs between covalent bonds.

Ques. Which is the strongest bond and why? (1 Mark)

Ans. A covalent bond is the strongest chemical bond.

Ques. What is the enthalpy of the combustion of alcohols? (1 Mark)

Ans. The enthalpy combustion of alcohol or ethanol is 1211 kJ mol-1.

Ques. What is thero-chemistry? (1 Mark)

Ans. Thero-chemistry is defined as the study of heat energy in association with chemical reactions or physical reactions.

Ques. Define in short the Bond Dissociation Enthalpy. (1 Mark)

Ans. It is a concept to calculate the strength of a chemical bond.

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(v) Br2 (aq) and Fe2+(aq).

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