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Enthalpy of Combustion with Examples for JEE

Last updated date: 21st Feb 2024
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Introduction to Enthalpy

The heat of the reaction is the amount of heat emitted or absorbed in a reaction. Heat changes for various reactions may also be referred to as standard enthalpy changes because enthalpy is the amount of heat evolved or absorbed at constant temperature and pressure. The heat change during a chemical reaction is determined by the amount of substance (number of moles) that has reacted. Other enthalpy changes discussed here include enthalpy of formation and enthalpy of combustion.

Standard Enthalpy of Combustion 

The enthalpy change (or heat change) when 1 mole of a substance is completely burnt in excess of oxygen or air is called enthalpy of Combustion or standard enthalpy of combustion or standard heat of combustion.

 It is generally denoted by $\Delta H_c$.  examples of combustion reactions are:

(i) Combustion of Carbon:

$\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \quad ; \Delta \mathrm{H}_{\mathrm{c}}=-393.5 \mathrm{~kJ}$

(ii) Combustion of Methane:

$\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta \mathrm{H}_{\mathrm{c}}=-890.3 \mathrm{~kJ}$

(iii) Combustion of Benzene:

$C_{6} H_{6}(l)+\frac{15}{2} O_{2}(g) \rightarrow 6 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} \quad \Delta \mathrm{H}_{\mathrm{C}}=-3268 \mathrm{~kJ}$

It may be noted that the enthalpy of combustion $\Delta H_c$ is always negative because combustion reactions are always exothermic and heat evolves during combustion. Enthalpy of combustion of some common fuels are :

The Calorific Value of Some Common Fuels


Calorific Values (kJ/kg)

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Enthalpy of Formation

Enthalpy of formation$\Delta H_f$ refers to the enthalpy change (or heat change) that occurs when one mole of a substance is formed from its elements in their most stable forms.

The subscript 'f' indicates that one mole of the compound has been formed from its elements in their most stable aggregation states. some examples:

$\begin{align} &\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO} ; \Delta \mathrm{H}_{\mathrm{f}}=-393.5 \mathrm{~kJ} \\ &\mathrm{C}_{(\mathrm{s})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_{4 ;} \Delta \mathrm{H}_{\mathrm{f}}=-74.8 \mathrm{~kJ} \end{align}$

$\begin{align} &\mathrm{Hg}_{(\mathrm{g})}+\mathrm{Cl}_{(\mathrm{g})} \rightarrow \mathrm{HgCl}_{2(\mathrm{~g})} ; \Delta \mathrm{H}_{\mathrm{f}}=-224.3 \mathrm{~kJ} \\ &\mathrm{C}_{(\mathrm{s})}+2 \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}) ; \Delta \mathrm{H}_{\mathrm{f}}=-239.9 \mathrm{~kJ} \end{align}$

Because of the definition, the equation for enthalpy of formation should always be written for one mole of the substance to be formed. If we need the coefficients 2, 3,4,5... Etc., in the equation for balancing, then $\Delta \mathrm{H}_{\mathrm{f}}$ values should also be multiplied by the same number. For example, the enthalpy of NH3 formation can be written as:

$\dfrac{l}{2} N_{2(g)}+\dfrac{3}{2} H_{2}(8) \rightleftarrows \mathrm{NH}_{3}(g)\quad \Delta_{f} H=-46.1 \mathrm{~kJ}$

However, if we want to get rid of fractions as $\dfrac{1}{2}$ and $\dfrac{3}{2}$. We may write

$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}$

But, this corresponds to the formation of 2 moles NH3 and $\Delta H$ will be – 92.2 kJ

The enthalpy of formation corresponds to the formation of a compound from its constituent elements.

Standard Enthalpy of Formation

The enthalpy of formation is determined by the temperature, pressure, and physical state of the reactants and products (gas, liquid, or solid). The enthalpy of formation is called standard enthalpy of formation or standard heat of formation. if all the species of the chemical reaction are in their standard state (i.e., stable forms at one bar pressure). Thus, The enthalpy change that occurs when one mole of a compound is formed from its elements, with all of the substances in their standard states, is referred to as standard molar enthalpy of formation. For example

$\begin{align} &\mathrm{H}_{2}(\mathrm{~g})+\dfrac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta \mathrm{H}_{\mathrm{f}}=-285.8 \mathrm{kJmol}^{-1} \\ &2 \mathrm{C}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{~g})+\dfrac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) ; \Delta \mathrm{H}_{\mathrm{f}}=-277.7 \mathrm{kJmol}^{-1} \end{align}$

It should be noted that a standard molar enthalpy of formation is a special caseΔHo when one mole of a compound is formed from its constituent elements. Enthalpy changes for all reactions, however, are not enthalpy of formation. Consider the following exothermic reaction:

$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HBr}(\mathrm{g}) ; \Delta \mathrm{H}^{\circ}=-72.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ 

For HBr, the enthalpy change is not the standard enthalpy of formation $\Delta H^{\circ}$. Two moles of the product are formed from the elements instead of one mole, resulting in

$\Delta \mathrm{H}^{\circ}=2 \Delta \mathrm{H}^{\circ}$ kJ mol-1

Therefore, by dividing all coefficients in the balanced chemical equation by 2, the enthalpy of formation of HBr(g) may be written as:

$\dfrac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\dfrac{1}{2} \mathrm{Br}_{2}(\mathrm{~g}) \rightarrow \mathrm{HBr}(\mathrm{g}) ; \Delta \mathrm{H}_{\mathrm{f}}=-36.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Calculation of Standard Enthalpy of Reaction

The enthalpies of reactions under standard conditions can be calculated using knowledge of standard enthalpies of formation from various substances. In general, the standard enthalpy of any reaction $\Delta H^{\circ}$ is equal to the difference between the $\Delta H^{\circ}$ of all the products and reactants, i.e., heat of formation formula or enthalpy of formation formula:

$\Delta H^{\circ}$ for a reaction =(Sum of the standard enthalpies of formation of products)-( Sum of the standard enthalpies of formation reactants )

$\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=\sum \mathrm{a}_{\mathrm{i}} \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\text { product })-\sum \mathrm{b}_{\mathrm{i}} \Delta \mathrm{H}_{\mathrm{f}^{\circ}}(\text { reactant })$

Where $\sum$ is known as summation and means the sum of, and a and b, are the stoichiometric coefficients of the products and reactants in the balanced equation.

Enthalpy changes of a Reaction:

Consider the reaction 

$\begin{align} &\mathrm{aA}+\mathrm{bB} \rightarrow \mathrm{cC}+\mathrm{dD} \\ &\Delta \mathrm{H}_{f}^{\circ}=\sum \Delta \mathrm{H}^{\circ} \text { ( product) }-\sum \Delta \mathrm{H}^{\circ} \text { ( reactant) } \\ &=\left[\mathrm{c} \Delta \mathrm{H}^{\circ}(\mathrm{C})+\mathrm{d} \Delta \mathrm{H}^{\circ}(\mathrm{D})\right]-\left[\mathrm{a} \Delta \mathrm{H}^{\circ}(\mathrm{A})+\mathrm{b} \Delta \mathrm{H}^{\circ}(\mathrm{B})\right] \end{align}$

For Example:

$\begin{aligned} &\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \\ &\Delta \mathrm{H}^{\circ}=\Delta \mathrm{H}_{\mathrm{f}}[\mathrm{CaC}(\mathrm{s})]+\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left[\mathrm{CO}_{2}(\mathrm{~g})\right]-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left[\mathrm{CaCO}_{3}(\mathrm{~s})\right] \\ &=(-635.1 \mathrm{kJmol}-1)+\left(-393.5 \mathrm{kJmol}^{-1}\right)-\left(-1206.9 \mathrm{kJmol}^{-1}\right) \\ &=178.3 \mathrm{kJmol}^{-1} \end{aligned}$

$\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$

$\begin{align} &\Delta \mathrm{H}^{\circ}=\Delta \mathrm{H}_{\mathrm{f}^{\circ}}[\mathrm{CaC}(\mathrm{s})] \Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}\left[\mathrm{CO}_{2}(\mathrm{~g})\right]-\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left[\mathrm{CaCO}_{3}(\mathrm{~s})\right] \\ &=\left(-635.1 \mathrm{kJmol}^{-1}\right)+\left(-393.5 \mathrm{kJmol}^{-1}\right)-\left(-1206.9 \mathrm{kJmol}^{-1}\right) \\ &=178.3 \mathrm{kJmol}^{-1} \end{align}$ 

The heat of formation table is as follows:

Standard Molar Enthalpy of Various Compound

$\Delta H_f^{\circ}~\dfrac{kJ}{mol}$
































In this, we have discussed what is enthalpy of combustion and enthalpy of formation. The heat of combustion of a substance, also known as its calorific value or energy value, is the amount of heat liberated when a given amount of the substance is burned. The heat of formation is the amount of heat absorbed or evolved  It is also known as enthalpy of formation or standard enthalpy of formation.

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FAQs on Enthalpy of Combustion with Examples for JEE

1. What is the calorie count?

The calorific value of food or fuel is the amount of heat energy present that is determined by the complete combustion of a specified quantity at constant pressure and under normal conditions. It is also known as calorific power. Kilojoule per kilogram (kJ/Kg) is the unit of calorific value.

Water vapor is produced during the combustion process, and heat should be recovered using specific techniques. It has a high calorific value if the heat contained in the water vapor can be recovered. When the heat contained in water vapor cannot be recovered due to its low calorific value.

2. What is Standard reaction enthalpy?

We already know that the enthalpy of any reaction is affected by physical conditions such as temperature, pressure, and so on. The standard enthalpy of any reaction is calculated when all of the components involved in the reaction, namely the reactants and products, are in their standard form. The enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions is thus the standard enthalpy of reaction. The standard state for any substance at a given temperature is its pure form at 1 bar of pressure, according to a convention.