Problem 29 - Chemistry in 30 days

Question

Given that bond energies of \(\mathrm{H}-\mathrm{H}\) and \(\mathrm{Cl}-\mathrm{Cl}\) are \(430 \mathrm{~kJ} / \mathrm{mol}\) and \(240 \mathrm{~kJ} / \mathrm{mol}\), respectively. \(\Delta H_{f}\) for \(\mathrm{HCl}\) is \(-90 \mathrm{~kJ} / \mathrm{mol}\). Bond enthalpy of \(\mathrm{HCl}\) is (a) \(380 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(425 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(245 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(290 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The bond enthalpy of \( \mathrm{HCl} \) is \( 425 \mathrm{~kJ/mol} \), option (b).

Step 1: Understanding the Problem

We need to find the bond enthalpy of \( \mathrm{HCl} \). We know the bond energies of \( \mathrm{H}-\mathrm{H} \) and \( \mathrm{Cl}-\mathrm{Cl} \), and \( \Delta H_{f} \) for \( \mathrm{HCl} \) is given. We will use these values to calculate the bond enthalpy of \( \mathrm{HCl} \).

Step 2: Writing the Reaction Equation

The reaction to form \( \mathrm{HCl} \) is as follows: \[ \frac{1}{2} \mathrm{H}_2 + \frac{1}{2} \mathrm{Cl}_2 \rightarrow \mathrm{HCl}. \] In this equation, \( \frac{1}{2} \mathrm{H}-\mathrm{H} \) bonds and \( \frac{1}{2} \mathrm{Cl}-\mathrm{Cl} \) bonds are broken, and one \( \mathrm{H}-\mathrm{Cl} \) bond is formed.

Step 3: Applying Bond Enthalpy Formula

The formula for bond enthalpy involves the bonds broken and formed: \[ \Delta H_{reaction} = \text{Bonds broken} - \text{Bonds formed}. \] Using the given enthalpy change of formation, \( \Delta H_f(\mathrm{HCl}) = -90 \mathrm{~kJ/mol} \), we have: \[ -90 = \frac{1}{2}(430) + \frac{1}{2}(240) - E_{\mathrm{HCl}}, \] where \( E_{\mathrm{HCl}} \) is the bond enthalpy of \( \mathrm{HCl} \).

Step 4: Solving for \( E_{\mathrm{HCl}} \)

Substitute the bond energies into the equation: \[ -90 = \frac{1}{2}(430) + \frac{1}{2}(240) - E_{\mathrm{HCl}} \] which simplifies to: \[ -90 = 215 + 120 - E_{\mathrm{HCl}}. \] Then, simplify further: \[ -90 = 335 - E_{\mathrm{HCl}}. \] Solving for \( E_{\mathrm{HCl}} \), we get: \[ E_{\mathrm{HCl}} = 335 + 90 = 425 \mathrm{~kJ/mol}. \]

Step 5: Selecting the Correct Answer

From our calculation, the bond enthalpy of \( \mathrm{HCl} \) is \( 425 \mathrm{~kJ/mol} \). Therefore, the correct answer is (b) \( 425 \mathrm{~kJ/mol} \).

Key Concepts

Enthalpy Change

Enthalpy change is an essential concept in chemistry, representing the heat absorbed or released during a chemical reaction at constant pressure. It's denoted as \( \Delta H \) and tells us if a reaction is endothermic (absorbs heat, \( \Delta H > 0 \)) or exothermic (releases heat, \( \Delta H < 0 \)). In this exercise, the enthalpy change \( \Delta H_f \) for the formation of hydrogen chloride \( \text{HCl} \) is given as \(-90 \text{ kJ/mol} \). This negative value indicates that the formation of \( \text{HCl} \) from hydrogen gas \( \text{H}_2 \) and chlorine gas \( \text{Cl}_2 \) releases energy, making it an exothermic reaction. Understanding the enthalpy change helps us predict energy requirements or production in chemical processes.

Bond Energy

Bond energy, also known as bond enthalpy, refers to the amount of energy required to break one mole of a specific type of bond in gaseous molecules. It is a measure of bond strength. For example, in this exercise, the bond energies for \( \text{H}-\text{H} \) and \( \text{Cl}-\text{Cl} \) are \( 430 \text{ kJ/mol} \) and \( 240 \text{ kJ/mol} \), respectively. The bond energy reflects the stability of a molecule: the higher the bond energy, the stronger the bond, and the more energy required to break it. Thus, knowing bond energies can help us determine the stability of molecules and the energy changes that occur when bonds break and form during a reaction.

Chemical Reaction Energy Calculation

Chemical reaction energy calculation involves determining the overall energy change when a reaction occurs. This is calculated by subtracting the energy of formed bonds from the energy of broken bonds. The formula is given by:

• \( \Delta H_{\text{reaction}} = \text{Bonds broken} - \text{Bonds formed} \)

Understanding this concept is crucial for predicting whether a reaction will be exothermic or endothermic. In our specific problem, we break half of \( \text{H}-\text{H} \) and \( \text{Cl}-\text{Cl} \) bonds, and form an \( \text{HCl} \) bond. Substituting values into our formula, \( \Delta H_{\text{reaction}} = \frac{1}{2}(430) + \frac{1}{2}(240) - E_{\text{HCl}} \), we can solve for \( E_{\text{HCl}} \) and find the bond enthalpy for \( \text{HCl} \). It's a systematic way of understanding the energy shifts in reactions.

Enthalpy of Formation

Enthalpy of formation, denoted as \( \Delta H_f \), is the heat change that results when one mole of a compound is formed from its elements in their standard states. It is usually reported in \( \text{kJ/mol} \). This value helps us understand the relative stability of compounds; a negative \( \Delta H_f \) suggests that the compound is more stable than its constituent elements. In our example, the \( \Delta H_f \) for \( \text{HCl} \) is \(-90 \text{ kJ/mol} \), indicating that forming \( \text{HCl} \) from hydrogen gas and chlorine gas releases energy. This information is critical for calculating bond enthalpies and understanding the spontaneity and environmental implications of chemical reactions.