Standard Enthalpy Of Formation, Combustion And Bond Dissociation

Enthalpy of combustion: It is the heat developed when one mole of a substance is completely burnt in oxygen under standard conditions. This quantity becomes very important while assessing the energy content of fuels. Enthalpy of dissociation, ΔH_d, is the energy required to break one mole of a specific type of bond in a molecule into individual atoms in the gas phase. It provides a measure of the strength and stability of chemical bonds. Enthalpy of atomization, ΔH_a, is the energy required to transform one mole of a given compound into its constituent atoms in the gas phase. It, therefore, represents the energy required to break all bonds in a molecule to yield individual atoms.

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This Story also Contains

1. Heat of Combustion
2. Enthalpy Of Dissociation Or Ionization
3. Some Solved Examples
4. Summary

Heat of Combustion

1. It is a change in enthalpy when one mole of a substance is completely oxidized combusted, or burnt.

2. $\Delta \mathrm{H}$ is - ve here as heat is always evolved here that is, exothermic process.

3. Heat of combustion is useful in calculating the calorific value of food and fuels.

4. It is also useful in confirming the structure of organic molecules having C, H, O, N, etc.

5. Enthalpy change by combustion of 1 gm solid,1 gm liquid, or 1 cc gas is called calorific value.

$\begin{aligned} & \text { calorific value }=\frac{\text { Heat of combustion }}{\text { Molecular wt. }} \\ & \Delta \mathrm{H} \text { (heat of reaction) }=-\Sigma \Delta \mathrm{H}_{\mathrm{P}}^{\circ}-\Sigma \mathrm{H}_{\mathrm{R}}^{\circ}\end{aligned}$

transition from liquid to gas.

Enthalpy Of Dissociation Or Ionization

It is defined as, "The quantity of heat absorbed when one mole of a substance is completely dissociated into its ions". Example,

$\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} \quad \Delta \mathrm{H}=13.7 \mathrm{Kcal}$

Heat Of Atomization

It is the enthalpy change (heat required) when bonds of one mole of a substance are broken down completely to obtain atoms in the gaseous phase (isolated) or it is the enthalpy change when one mole of atoms in the gas phase is formed from the corresponding element in its standard state. In the case of diatomic molecules, it is also called bond dissociation enthalpy.

It is denoted by $\Delta \mathrm{H}_{\mathrm{a}}$ or $\Delta \mathrm{H}^*$.
Example,
$
\begin{aligned}
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})-435 \mathrm{~kJ} \\
& \Delta \mathrm{H}=+435 \mathrm{~kJ} / \mathrm{mol} \\
& \mathrm{CH}_4(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})+1665 \mathrm{~kJ} \\
& \Delta \mathrm{H}=+1665 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$

Phase Transition And Transition Energy

• The change of matter from one state (solid, liquid, or gas) to another state is called Phase Transition.

• Such changes occur at definite temperatures such as melting point (solid to liquid), boiling point (liquid to vapors), etc., and are accompanied by absorption or evolution of heat. The enthalpy change during such phase transitions is called heat of transition or transition energy.

Example, C (diamond) $\rightarrow \mathrm{C}$ (amorphous) $\Delta \mathrm{H}=3.3 \mathrm{Kcal}$

Recommended topic video on (Standard Enthalpy Of Formation, Combustion And Bond Dissociation)

Some Solved Examples

Example 1: Calculate the standard enthalpy of combustion (kJ/mol) of glucose

Given, $\Delta H_f^0 \mathrm{CO}_2=-394 \mathrm{~kJ} / \mathrm{mol} \Delta H_f^0 \mathrm{H}_2 \mathrm{O}=-286 \mathrm{~kJ} / \mathrm{mol} \Delta H_f^0 C_6 H_{12} O_6=-1275 \mathrm{~kJ} / \mathrm{mol}$,

1)2805

2)-2805

3)1275

4)-1275

Solution

$\Delta H_f^o O_2$ is equal to Zero because O₂ is in the standard state.

First, write the balanced equation of the reaction,

$\begin{aligned} & \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+6 \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+6 \mathrm{H}_2 \mathrm{O} \\ & \Delta H_c=\left(6 * \Delta H_f^0 \mathrm{H}_2 0+6 * \Delta H_f^0 \mathrm{CO}_2\right)-\left(\Delta H_f^0 C_6 H_{12} O_6+6 * \Delta H_f^0 O_2\right) \\ & \Delta H_c=6 \times(-394)+6 \times(-286)-(-1275)-6 \times(0) \\ & \Delta H_c=-2805 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

Example 2: At 25⁰C 1 atm pressure, the enthalpies of combustion are as given below:

Substance H₂ C (graphite) C₂H₆(g) $\frac{\Delta_c \mathrm{H}^{\ominus}}{\mathrm{kJ} \mathrm{mol}^{-1}}$ -286.0 -394.0 -1560.0

The enthalpy of the formation of ethane is

1) $+54.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(2) $-68.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3) $-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4) $+97.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:

Given,
$
\begin{aligned}
& \mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \quad \Delta_{\mathrm{c}} \mathrm{H}=-286 \mathrm{~kJ} / \mathrm{mol}----(1) \\
& \mathrm{C}_{(\mathrm{s})}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \quad \Delta_{\mathrm{c}} \mathrm{H}=-394 \mathrm{~kJ} / \mathrm{mol}----(2)
\end{aligned}
$
$
\mathrm{C}_2 \mathrm{H}_6+\frac{7}{2} \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O} \quad \Delta_{\mathrm{C}} \mathrm{H}=-1560 \mathrm{~kJ} / \mathrm{mol}-----
$

The enthalpy of the formation of ethane :
$
2 \mathrm{C}_{(\mathrm{s})}+3 \mathrm{H}_2 \rightarrow \mathrm{C}_2 \mathrm{H}_6 \quad \Delta_{\mathrm{C}} \mathrm{H}=\text { ? }
$

The formation reaction of ethane can be obtained by $2 \times(2)+3 \times(1)-(3)$
Then,
$
\begin{aligned}
\Delta_f \mathrm{H} \text { of } \mathrm{C}_2 \mathrm{H}_6 & =2 \times(-394)+3 \times(-286)-(-1560) \\
& =-1646+1560 \\
& =-86 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$

Example 3: At 25^oC and 1 atm pressure, the enthalpy of combustion of benzene (l) and acetylene (g) are $-3268 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-1300 \mathrm{~kJ} \mathrm{~mol}^{-1}$.respectively. The change in enthalpy for the reaction $3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$, is

1) $+324 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2) $+632 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3) $-632 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4) $-732 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution
Given, $\mathrm{C}_6 \mathrm{H}_6(l)+\left(\frac{15}{2}\right) \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2+3 \mathrm{H}_2 \mathrm{O}, \Delta_{\mathrm{c}} \mathrm{H}=-3268 \mathrm{~kJ} / \mathrm{mol}$
$
\underset{\text { (Acetylene) }}{\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})}+\left(\frac{5}{2}\right) \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}, \Delta_{\mathrm{C}} \mathrm{H}=-1300 \mathrm{~kJ} / \mathrm{mol}
$

We need to find the enthalpy change of the following reaction
$
3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6 \text { (l), } \Delta_{\mathrm{r}} \mathrm{H}=\text { ? }
$

So, by doing $3 \times(2)-(1)$, we get the required reaction.
Thus, the enthalpy change of the reaction is given as
$
\Delta_{\mathrm{r}} \mathrm{H}=3 \times(-1300)-(-3268)
$

$\Delta_{\mathrm{r}} \mathrm{H}=-632 \mathrm{~kJ} / \mathrm{mol}$

Example 4: For the combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $\mathrm{e}_{\mathrm{C}} \mathrm{H}^{\ominus}=-601.70 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the magnitude of change in internal energy for the reaction is _____________kJ. (Nearest integer)

(Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

1) 600

2)400

3)300

4)0

Solution

Given,
$
\mathrm{T}=300 \mathrm{~K}
$
$\mathrm{P}=1$ bar
$
\begin{aligned}
& \Delta_{\mathrm{C}} \mathrm{H}^0=-601.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$

Reaction of Combustion-
$
\begin{aligned}
& \mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{MgO}(\mathrm{s}) \\
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT}
\end{aligned}
$

$
-601.7=\Delta \mathrm{U}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)-\frac{1 \times 8.3 \times 300}{2 \times 1000}
$$

On solving-
$
\Delta \mathrm{U}=-600.455 \mathrm{~kJ} \approx-600 \mathrm{~kJ}
$

Hence, the answer is (600).

Example 5: An average person needs about 10000 kJ of energy per day. The amount of glucose (molar mass = 180.0 g mol^–1) needed to meet this energy requirement is _________ g.

(Use : $\Delta_{\mathrm{C}} \mathrm{H}$ (glucose) $=-2700 \mathrm{~kJ} \mathrm{~mol}^{-1}$ )

1) 667

2)687

3)565

4)45

Solution
Total energy required $=10^4 \mathrm{~kJ}$
Energy obtained from 1 mole of Glucose $=2700 \mathrm{~kJ}$
$\therefore 2700 \mathrm{~kJ}$ is obtained from 180 g Glucose.
$\therefore 10^4 \mathrm{~kJ}$ will be obtained from $\frac{180}{2700} \times 10^4$
$
=667 \mathrm{~g}
$

Hence, the answer is the option (1).

Summary

The heat of combustion is the heat evolved during the complete oxidation of a substance in oxygen, as in a bomb calorimeter, and gives its energy content as fuel. Enthalpy of dissociation provides the energy required to break specific chemical bonds in a molecule and hence allows estimation of the strength of bonds. The enthalpy of atomization expresses the energy required to break all bonds in a compound to form individual atoms. Finally, enthalpy of change refers to the associated energy of the phase change of the substance; either a melt or boil. Knowing these enthalpies gives an estimate of the changes in energy during the progress of a chemical reaction.