5.4 Enthalpy Change, ΔrH of a Reaction – Reaction Enthalpy

The amount of heat evolved or absorbed in a chemical reaction when the number of moles of reactants as represented by the chemical equation have completely reacted, is called heat of reaction or enthalpy of reaction or enthalpy change of reaction \(\left(\Delta_{\mathrm{I}} \mathrm{H}\right)\).

Change in total heat of reaction at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure is called standard heat of reaction.

In a chemical reaction, reactants are converted into products and is represented by,
Reactants \(\rightarrow\) Products
The enthalpy change accompanying a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction is given by the symbol \(\Delta_r H\)
\(\Delta_r H=\) (sum of enthalpies of products) – (sum of enthalpies of reactants)
\(
=\sum_i \mathrm{a}_i H_{\text {products }}-\sum_i b_i H_{\text {reactants }}(6.14)
\)
Here symbol \(\sum\) (sigma) is used for summation and \(\mathrm{a}_i\) and \(\mathrm{b}_i\) are the stoichiometric coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
\(
\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)
\)
\(
\Delta_r H=\sum_i a_i H_{\text {Products }}-\sum_i b_i H_{\text {reactants }}
\)
\(
\begin{aligned}
=\left[H_{\mathrm{m}}\left(\mathrm{CO}_2, \mathrm{~g}\right)+2 \mathrm{H}_{\mathrm{m}}\left(\mathrm{H}_2 \mathrm{O}, l\right)\right]-\left[\mathrm{H}_{\mathrm{m}}\left(\mathrm{CH}_4, \mathrm{~g}\right)\right. \\
\left.+2 \mathrm{H}_{\mathrm{m}}\left(\mathrm{O}_2, \mathrm{~g}\right)\right]
\end{aligned}
\)
where \(H_{\mathrm{m}}\) is the molar enthalpy.
Enthalpy change is a very useful quantity. Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate the temperature dependence of the equilibrium constant.

Exothermic and Endothermic Reactions

Exothermic reactions are those which are accompanied by the evolution of heat. \(\Delta \mathrm{H}\) is negative for exothermic reactions Endothermic reactions are those in which heat is absorbed. \(\Delta \mathrm{H}\) is positive for endothermic reactions.

Different types of heats/ enthalpies of reaction

(a) Standard Enthalpy of Reactions

Enthalpy of a reaction depends on the conditions under which a reaction is carried out. The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.

The standard state of a substance at a specified temperature is its pure form at 1 bar. For example, the standard state of liquid ethanol at \(298 \mathrm{~K}\) is pure liquid ethanol at \(1 \mathrm{bar}\); the standard state of solid iron at \(500 \mathrm{~K}\) is pure iron at 1 bar. Usually, data are taken at \(298 \mathrm{~K}\).

Standard conditions are denoted by adding the superscript \(\ominus\) to the symbol \(\Delta H\), e.g., \(\Delta H^{\ominus}\)

(b) Enthalpy Changes during Phase Transformations
Phase transformations also involve energy changes. Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, the temperature remains constant (at \(273 \mathrm{~K}\) ).

• Enthalpy of fusion or molar enthalpy of fusion: The enthalpy change that accompanies the melting of one mole of a solid substance in a standard state is called standard enthalpy of fusion or molar enthalpy of fusion, \(\Delta_{\text {fus }} H^{\ominus}\). \(\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta_{\text {fus }} H^{\ominus}=6.00 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
  Here \(\Delta_{\text {fus }} H^{\ominus}\) is enthalpy of fusion in standard state. If water freezes, then the process is reversed and an equal amount of heat is given off to the surroundings.
• Enthalpy of vaporization or molar enthalpy of vaporization: The amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, \(\Delta_{v a p} H^{\ominus}\). Melting of a solid is endothermic, so all enthalpies of fusion are positive. Water requires heat for evaporation. At a constant temperature of its boiling point \(T_{\mathrm{b}}\) and at constant pressure: \(\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}); \Delta_{\text {vap }} H^{\ominus}=+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1} \Delta_{\text {vap }} H^{\ominus}\) is the standard enthalpy of vaporisation.
• Standard enthalpy of sublimation: Standard enthalpy of sublimation, \(\Delta_{\text {sub }} H^{\ominus}\) is the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1 bar). Sublimation is the direct conversion of a solid into its vapour. Solid \(\mathrm{CO}_2\) or ‘dry ice’ sublimes at \(195 \mathrm{~K}\) with \(\Delta_{\text {sub }} H^{\ominus}=25.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\); naphthalene sublimes slowly and for this \(\Delta_{\text {sub }} H^{\ominus}=73.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transformations. For example, the strong hydrogen bonds between water molecules hold them tightly in the liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise \(1 \mathrm{~mol}\) of acetone than it does to vaporize \(1 \mathrm{~mol}\) of water. Table 6.1 gives values of standard enthalpy changes of fusion and vaporisation for some substances.

Example 6.7: A swimmer coming out from a pool is covered with a film of water weighing about \(18 \mathrm{~g}\). How much heat must be supplied to evaporate this water at \(298 \mathrm{~K}\)? Calculate the internal energy of vaporisation at \(298 \mathrm{~K}\).
\(\Delta_{\text {vap }} H^{\ominus}\) for water at \(298 \mathrm{~K}=44.01 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Answer:

We can represent the process of evaporation as
\(
\underset{1 \mathrm{~mol}}{\mathrm{H}_2 \mathrm{O}(\mathrm{l})} \stackrel{\text { vaporisatton }}{\longrightarrow} \underset{1 \mathrm{~mol}}{\mathrm{H}_2 \mathrm{O}(\mathrm{g})}
\)
No. of moles in \(18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}(l)\) is
\(
=\frac{18 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=1 \mathrm{~mol}
\)
Heat supplied to evaporate \(18 \mathrm{~g}\) water at
\(
\begin{aligned}
298 \mathrm{~K} & =\mathrm{n} \times \Delta_{\text {vap }} H^{\ominus} \\
& =(1 \mathrm{~mol}) \times\left(44.01 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\
& =44.01 \mathrm{~kJ}
\end{aligned}
\)
(assuming steam behaving as an ideal gas).
\(
\Delta_{\text {vap }} U=\Delta_{\text {vap }} H^{\ominus}-p \Delta V=\Delta_{\text {vap }} H^{\ominus}-\Delta n_g R T
\)
\(
\Delta_{\text {vap }} H^{\mathrm{V}}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{R} T=44.01 \mathrm{~kJ} -(1)\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(298 \mathrm{~K})\left(10^{-3} \mathrm{~kJ} \mathrm{~J}^{-1}\right)
\)
\(
\begin{aligned}
\Delta_{\mathbf{vap}} U^{\mathrm{V}} & =44.01 \mathrm{~kJ}-2.48 \mathrm{~kJ} \\
& =41.53 \mathrm{~kJ}
\end{aligned}
\)
Therefore,
\(
\begin{aligned}
\Delta \mathrm{H} & =-7.56 \mathrm{~kJ} \mathrm{~mol}^{-1}+\left(-6.00 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\
& =-13.56 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)

There is negligible change in the volume during the change form liquid to solid state.

Therefore, \(\mathrm{p} \Delta \mathrm{v}=\Delta \mathrm{ng} \mathrm{RT}=0\)
\(
\Delta \mathrm{H}=\Delta \mathrm{U}=-13.56 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)

(c) Standard Enthalpy of Formation

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states) is called standard Molar Enthalpy of Formation. Its symbol is \(\Delta_f H^{\ominus}\), where the subscript ‘ \(f\) ‘ indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation. The reference state of an element is its most stable state of aggregation at \(25^{\circ} \mathrm{C}\) and 1 bar pressure.

For example, the reference state of dihydrogen is \(\mathrm{H}_2\) gas and those of dioxygen, carbon and sulphur are \(\mathrm{O}_2\) gas, \(\mathrm{C}_{\text {graphite }}\) and \(\mathrm{S}_{\text {rhombic }}\) respectively. Some reactions with standard molar enthalpies of formation are as follows.

\(
\mathrm{H}_2(\mathrm{~g})+1 / 2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(1) \text {; }
\)
\(
\Delta_f H^{\ominus}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
\(
\mathrm{C} \text { (graphite, s) }+2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{Ch}_4(\mathrm{~g})
\)
\(
\Delta_f H^{\ominus}=-74.81 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
\(
\begin{array}{r}
2 \mathrm{C} \text { (graphite, s) }+3 \mathrm{H}_2(\mathrm{~g})+1 / 2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(1) ; \\
\Delta_f H^{\ominus}=-277.7 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{array}
\)
It is important to understand that a standard molar enthalpy of formation, \(\Delta_f H^{\ominus}\), is just a special case of \(\Delta_r H^{\ominus}\), where one mole of a compound is formed from its constituent elements, as in the above three equations, where \(1 \mathrm{~mol}\) of each, water, methane and ethanol is formed.

In contrast, the enthalpy change for an exothermic reaction:
\(
\mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \rightarrow \mathrm{CaCo}_3(\mathrm{~s}) ;\Delta_r H^{\ominus}=-178.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
is not an enthalpy of the formation of calcium carbonate, since calcium carbonate has been formed from other compounds, and not from its constituent elements.

Also, for the reaction given below, the enthalpy change is not standard enthalpy of formation, \(\Delta_f H^{\ominus}\) for \(\operatorname{HBr}(\mathrm{g})\).
\(
\mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(l) \rightarrow 2 \mathrm{HBr}(\mathrm{g}); \Delta_r H^{\ominus}=-178.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
Here two moles, instead of one mole of the product is formed from the elements, i.e., .
\(
\Delta_r H^{\ominus}=2 \Delta_f H^{\ominus}
\)

Therefore, by dividing all coefficients in the balanced equation by 2, the expression for enthalpy of formation of \(\mathrm{HBr}(\mathrm{g})\) is written as
\(
1 / 2 \mathrm{H}_2(\mathrm{~g})+1 / 2 \mathrm{Br}_2(1) \rightarrow \mathrm{HBr}(\mathrm{g}); \Delta_f H^{\ominus}=-36.4 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
Standard enthalpies of the formation of some common substances are given in Table 6.2. By convention, standard enthalpy for formation, \(\Delta_f H^{\ominus}\), of an element in the reference state, i.e., its most stable state of aggregation is taken as zero.

Let’s find out how much heat is required to decompose calcium carbonate to lime and carbon dioxide, with all the substances in their standard state.
\(
\mathrm{CaCO}_3(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) ; \Delta_{\mathrm{r}} H^{\ominus}=?
\)
Here, we can make use of the standard enthalpy of formation and calculate the enthalpy change for the reaction. The following general equation can be used for the enthalpy change calculation.
\(
\Delta_r H^{\ominus}=\sum_t \mathrm{a}_t \Delta_f H^{\ominus} \text { (products) }-\sum_t \mathrm{~b}_t \Delta_f H^{\ominus} \text { (reactants) } \dots(6.15)
\)
where \(\mathrm{a}\) and \(\mathrm{b}\) represent the coefficients of the products and reactants in the balanced equation. Let us apply the above equation for the decomposition of calcium carbonate. Here, coefficients ‘ \(a\) ‘ and ‘ \(b\) ‘ are 1 each. Therefore,
\(
\Delta_r H^{\ominus}=\Delta_f H^{\ominus}=[\mathrm{CaO}(\mathrm{s})]+\Delta_f H^{\ominus}\left[\mathrm{CO}_2(\mathrm{~g})\right]-\Delta_f H^{\ominus}=\left[\mathrm{CaCO}_s(\mathrm{~s})\right]
\)
\(
=1\left(-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)+1\left(-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-1\left(-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)=178.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
Thus, the decomposition of \(\mathrm{CaCO}_3\) (s) is an endothermic process and you have to heat it to get the desired products.

(d) Thermochemical Equations

When a balanced chemical equation not only indicates the quantities of different reactants and products but also indicates the amount of heat evolved or absorbed, it is called a thermochemical equation.

A balanced chemical equation together with the value of its \(\Delta_r H\) is called a thermochemical equation. We specify the physical state (along with the allotropic state) of the substance in an equation. For example:
\(
\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)+3 \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l); \Delta_r H^{\ominus}=-1367 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
The above equation describes the combustion of liquid ethanol at constant temperature and pressure. The negative sign of enthalpy change indicates that this is an exothermic reaction.
It would be necessary to remember the following conventions regarding thermochemical equations.

• The coefficients in a balanced thermochemical equation refer to the number of moles (never molecules) of reactants and products involved in the reaction.
• The numerical value of \(\Delta_r H^{\ominus}\) refers to the number of moles of substances specified by an equation. Standard enthalpy change \(\Delta_r H^{\ominus}\) will have units as \(\mathrm{kJ} \mathrm{mol}^{-1}\).

To illustrate the concept, let us consider the calculation of heat of reaction for the following reaction:
\(
\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}),
\)
From the Table (6.2) of standard enthalpy of formation \(\left(\Delta_f H^{\ominus}\right)\), we find :
\(
\Delta_f H^{\ominus}\left(\mathrm{H}_2 \mathrm{O}, l\right)=-285.83 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {; }
\)
\(
\Delta_f H^{\ominus}\left(\mathrm{Fe}_2 \mathrm{O}_3, \mathrm{~s}\right)=-824.2 \mathrm{~kJ} \mathrm{~mol}^{-1}
\)
Also \(\Delta_f H^{\ominus}(\mathrm{Fe}, \mathrm{s})=0\) and \(\Delta_f H^{\ominus}\left(\mathrm{H}_2, \mathrm{~g}\right)=0\) as per convention
Then,
\(
\Delta_f H_1^{\ominus}=3\left(-285.83 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-1\left(-824.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)
\)
\(
\begin{aligned}
& =(-857.5+824.2) \mathrm{kJ} \mathrm{mol}^{-1} \\
& =-33.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
\)
Note that the coefficients used in these calculations are pure numbers, which are equal to the respective stoichiometric coefficients. The unit for \(\Delta_r H^{\ominus}\) is \(\mathrm{kJ} \mathrm{mol}{ }^{-1}\), which means per mole of reaction. Once we balance the chemical equation in a particular way, as above, this defines the mole of the reaction. If we had balanced the equation differently, for example,
\(
\frac{1}{2} \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{Fe}(\mathrm{s})+\frac{3}{2} \mathrm{H}_2 \mathrm{O}(\mathrm{l})
\)
then this amount of reaction would be one mole of reaction and \(\Delta_r H^{\ominus}\) would be
\(
\begin{aligned}
\Delta_f H_2^{\ominus} & =\frac{3}{2}\left(-285.83 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\
& -\frac{1}{2}\left(-824.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\right) \\
= & (-428.7+412.1) \mathrm{kJ} \mathrm{mol}^{-1} \\
= & -16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}=1 / 2 \Delta_r H_1^{\ominus}
\end{aligned}
\)
\(
\text { It shows that enthalpy is an extensive quantity. }
\)

• When a chemical equation is reversed, the value of \(\Delta_r H^{\ominus}\) is reversed in sign. For example
  \(
  \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g}) ;\Delta_r H^{\ominus}=-91.8 \mathrm{~kJ} . \mathrm{mol}^{-1}
  \)
  \(
  2 \mathrm{NH}_3(\mathrm{~g}) \rightarrow \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \text {; }\Delta_r H^{\ominus}=+91.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
  \)

(e) Hess’s Law of Constant Heat Summation

We know that enthalpy is a state function, therefore the change in enthalpy is independent of the path between the initial state (reactants) and final state (products). In other words, enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps. If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature. 

Hess’s Law states that the total amount of heat evolved or absorbed in a reaction is the same whether the reaction takes place in one step or in a number of steps. In other words, the total amount of heat change in a reaction depends only upon the nature of initial reactants and the nature of final products and is independent of the path or manner by which this change is brought about.

Let us understand the importance of this law with the help of an example.
Consider the enthalpy change for the reaction
\(
\mathrm{C} \text { (graphite,s) }+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta_r H^{\ominus}=?
\)
Although \(\mathrm{CO}(\mathrm{g})\) is the major product, some \(\mathrm{CO}_2\) gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. However, if we can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction.
Let us consider the following reactions:
\(
\mathrm{C} \text { (graphite, } \mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2 \text { (g); }\Delta_r H^{\ominus}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { (i) }
\)
\(
\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\Delta_r H^{\ominus}=-283.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { (ii) }
\)
We can combine the above two reactions in such a way so as to obtain the desired reaction. To get one mole of \(\mathrm{CO}(\mathrm{g})\) on the right, we reverse equation (ii). In this, heat is absorbed instead of being released, so we change the sign of \(\Delta_r H^{\ominus}\) value
\(
\mathrm{CO}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) ;\Delta_r H^{\ominus}=+283.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { (iii) }
\)
Adding equation (i) and (iii), we get the desired equation,
\(
\mathrm{C}(\text { graphite }, \mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ;
\)
for which \(\Delta_r H^{\ominus}=(-393.5+283.0)\) \(=-110.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
In general, if enthalpy of an overall reaction \(\mathrm{A} \rightarrow \mathrm{B}\) along one route is \(\Delta_r H\) and \(\Delta_r H_1, \Delta_r H_2\), \(\Delta_r H_3 \ldots \ldots\) representing enthalpies of reactions leading to same product, B along another route, then we have
\(
\Delta_r H=\Delta_r H_1+\Delta_r H_2+\Delta_r H_3 \cdots(6.16)
\)
It can be represented as: