QUESTION NO. 3

(a) Calculate enthalpy of combustion of methane. If enthalpies of formation of CH4, CO₂ and H₂O are 213 kcals, 94 kcals and 68 kcals respectively.

To calculate the enthalpy of combustion of methane (( \Delta H_{\text{comb}} )), we can use the enthalpies of formation (( \Delta H_f )) of the reactants and products involved in the combustion reaction. The combustion of methane is given by:

[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) ]

The enthalpy change for the reaction (( \Delta H_{\text{reaction}} )) can be calculated using the formula:

[ \Delta H_{\text{reaction}} = \sum \Delta H_f (\text{products}) – \sum \Delta H_f (\text{reactants}) ]

Given:

• (\Delta H_f ) of (\text{CH}_4 (g) = -213 \, \text{kcal/mol})
• (\Delta H_f ) of (\text{CO}_2 (g) = -94 \, \text{kcal/mol})
• (\Delta H_f ) of (\text{H}_2\text{O} (l) = -68 \, \text{kcal/mol})

(Note: The enthalpy of formation of ( \text{O}_2(g) ) is zero because it is in its standard state.)

Now, we apply these values to our formula. First, let’s find the sum of the enthalpies of formation for the products and reactants:

[ \sum \Delta H_f (\text{products}) = \Delta H_f (\text{CO}_2(g)) + 2 \times \Delta H_f (\text{H}_2\text{O}(l)) ]

[ \sum \Delta H_f (\text{products}) = -94 \, \text{kcal/mol} + 2 \times (-68 \, \text{kcal/mol}) ]

[ \sum \Delta H_f (\text{products}) = -94 \, \text{kcal/mol} – 136 \, \text{kcal/mol} ]

[ \sum \Delta H_f (\text{products}) = -230 \, \text{kcal/mol} ]

Next, we find the sum of the enthalpies of formation for the reactants:

[ \sum \Delta H_f (\text{reactants}) = \Delta H_f (\text{CH}_4(g)) + 2 \times \Delta H_f (\text{O}_2(g)) ]

[ \sum \Delta H_f (\text{reactants}) = -213 \, \text{kcal/mol} + 2 \times 0 \, \text{kcal/mol} ]

[ \sum \Delta H_f (\text{reactants}) = -213 \, \text{kcal/mol} ]

Now we can calculate the enthalpy change for the reaction:

[ \Delta H_{\text{reaction}} = \sum \Delta H_f (\text{products}) – \sum \Delta H_f (\text{reactants}) ]

[ \Delta H_{\text{reaction}} = -230 \, \text{kcal/mol} – (-213 \, \text{kcal/mol}) ]

[ \Delta H_{\text{reaction}} = -230 \, \text{kcal/mol} + 213 \, \text{kcal/mol} ]

[ \Delta H_{\text{reaction}} = -17 \, \text{kcal/mol} ]

Thus, the enthalpy of combustion of methane is (-17 \, \text{kcal/mol}).

(b) What is Bond energy? How is it determined?

Bond Energy:

Bond energy (also known as bond enthalpy or bond dissociation energy) is the amount of energy required to break one mole of bonds in gaseous molecules, resulting in the separation of the bonded atoms into individual atoms in the gas phase. It is typically expressed in kilojoules per mole (kJ/mol) or kilocalories per mole (kcal/mol).

Bond energy is a measure of bond strength in a chemical bond. Higher bond energy indicates a stronger bond, meaning more energy is needed to break the bond.

Determination of Bond Energy:

1. Experimental Methods:

• Calorimetry: One of the direct ways to determine bond energy is by measuring the heat changes in chemical reactions using a calorimeter. However, determining the bond energy directly for individual bonds in complex molecules can be challenging.

1. Thermochemical Data:

• Bond energies are often calculated using thermochemical data from a variety of reactions. This involves using Hess’s Law and enthalpies of formation (( \Delta H_f )).

Using Hess’s Law to Determine Bond Energy:

Hess’s Law states that the total enthalpy change for a reaction is the same, regardless of the number of steps in the reaction. This principle can be used to calculate bond energies.

1. Breaking and Forming Bonds:

• When a chemical reaction occurs, bonds in the reactants are broken and new bonds are formed in the products. The energy required to break bonds is absorbed, and the energy released when new bonds form is released.

1. Enthalpy Change of Reaction:

• The overall enthalpy change (( \Delta H )) of a reaction can be determined from the enthalpy of formation of reactants and products.

[ \Delta H_{\text{reaction}} = \sum \Delta H_f (\text{products}) – \sum \Delta H_f (\text{reactants}) ]

3. Bond Energy Calculations:

• To find the bond energy, consider a specific reaction where bonds are broken and formed. The general approach is:

[ \Delta H_{\text{reaction}} = \sum (\text{Bond Energies of bonds broken}) – \sum (\text{Bond Energies of bonds formed}) ]

Example Calculation:

Let’s consider a simple example to calculate the bond energy of the H-H bond in the dissociation of hydrogen gas:

[ H_2(g) \rightarrow 2H(g) ]

• The enthalpy change for this reaction is the bond dissociation energy of ( H_2 ), denoted as ( D(H-H) ).
• If we know the enthalpy change (( \Delta H )) for this reaction from experimental data, it directly gives the bond energy.

For example, if the enthalpy change for the dissociation of ( H_2 ) is known to be ( 436 \, \text{kJ/mol} ), then:

[ D(H-H) = 436 \, \text{kJ/mol} ]

Average Bond Energies:

In many cases, average bond energies are used because the exact bond energy can vary depending on the molecular environment. For example, the C-H bond energy in methane (CH(_4)) may be slightly different from the C-H bond energy in ethane (C(_2)H(_6)). Therefore, average values are compiled from a variety of compounds to provide a more general estimate.

Conclusion:

Bond energy is a critical concept in understanding chemical reactions and bond strengths. It is determined through experimental data and thermochemical calculations, often utilizing Hess’s Law and enthalpies of formation. This knowledge is essential in fields such as thermodynamics, kinetics, and material science.

(c) Explain with suitable examples:
(i) Enthalpy of neutralisation
(ii) Enthalpy of solution

(i) Enthalpy of Neutralization

Enthalpy of neutralization is the enthalpy change that occurs when one mole of an acid reacts with one mole of a base to form water under standard conditions. It is usually measured in kilojoules per mole (kJ/mol). This reaction is typically exothermic, meaning it releases heat.

The general reaction for neutralization is:

[ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} ]

For strong acids and strong bases, the enthalpy of neutralization is nearly constant, about (-57.1 \, \text{kJ/mol}), because the reaction essentially involves the combination of hydrogen ions (( \text{H}^+ )) from the acid and hydroxide ions (( \text{OH}^- )) from the base to form water.

[ \text{H}^+ (aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O} (l) ]

Example:

When hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), the enthalpy of neutralization can be measured:

[ \text{HCl} (aq) + \text{NaOH} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2\text{O} (l) ]

For strong acid and strong base, the enthalpy of neutralization is approximately (-57.1 \, \text{kJ/mol}).

However, for weak acids and bases, the enthalpy of neutralization can be different due to the partial ionization of weak acids or bases. For example, the neutralization of acetic acid (a weak acid) with NaOH:

[ \text{CH}_3\text{COOH} (aq) + \text{NaOH} (aq) \rightarrow \text{CH}_3\text{COONa} (aq) + \text{H}_2\text{O} (l) ]

The enthalpy of neutralization for this reaction might be less than (-57.1 \, \text{kJ/mol}) due to the energy required to ionize the acetic acid.

(ii) Enthalpy of Solution

Enthalpy of solution (or enthalpy of dissolution) is the enthalpy change that occurs when one mole of a solute is dissolved in a specified amount of solvent to form a solution. It can be either endothermic (absorbing heat) or exothermic (releasing heat), depending on the nature of the solute-solvent interactions.

The enthalpy of solution (( \Delta H_{\text{sol}} )) is composed of three main steps:

1. Breaking the solute-solute interactions (endothermic).
2. Breaking the solvent-solvent interactions (endothermic).
3. Forming solute-solvent interactions (exothermic).

The overall enthalpy change is the sum of these steps.

Example:

When potassium chloride (KCl) is dissolved in water, the process involves breaking the ionic bonds in the solid KCl and the hydrogen bonds in water, followed by the formation of new interactions between the K(^+) and Cl(^-) ions and water molecules.

[ \text{KCl} (s) \rightarrow \text{K}^+ (aq) + \text{Cl}^- (aq) ]

The enthalpy of solution for KCl can be endothermic, meaning the solution absorbs heat and the temperature of the solution decreases.

Another example is the dissolution of ammonium nitrate (NH(_4)NO(_3)) in water, which is also endothermic:

[ \text{NH}_4\text{NO}_3 (s) \rightarrow \text{NH}_4^+ (aq) + \text{NO}_3^- (aq) ]

The enthalpy of solution is positive, indicating that the solution process absorbs heat from the surroundings, often resulting in a cooling effect.

On the other hand, dissolving sodium hydroxide (NaOH) in water is exothermic:

[ \text{NaOH} (s) \rightarrow \text{Na}^+ (aq) + \text{OH}^- (aq) ]

The enthalpy of solution is negative, indicating that the process releases heat and the temperature of the solution increases.

Summary

• Enthalpy of Neutralization: The heat change when an acid and a base react to form water and a salt. For strong acids and bases, it is typically (-57.1 \, \text{kJ/mol}).
• Enthalpy of Solution: The heat change when one mole of solute dissolves in a solvent. It can be endothermic or exothermic depending on the interactions between solute and solvent.