Problem 157 - A Complete Resource Book in Chemistry for JEE Main

Question

If the bond dissociation energies of \(\mathrm{XY}, \mathrm{X}_{2}\) and \(\mathrm{Y}_{2}\) are in the ratio of \(1: 1: 0.5\) and \(\Delta \mathrm{H}_{f}\) for the formation of \(\mathrm{XY}\) is \(-200 \mathrm{~kJ} / \mathrm{mole}\). The bond dissociation energy of \(\mathrm{X}_{2}\) will be ? [2005] (a) \(100 \mathrm{~kJ} / \mathrm{mole}\) (b) \(400 \mathrm{~kJ} / \mathrm{mole}\) (c) \(600 \mathrm{~kJ} / \mathrm{mole}\) (d) \(800 \mathrm{~kJ} / \mathrm{mole}\)

Step-by-Step Solution

Answer

The bond dissociation energy of \(\mathrm{X}_{2}\) is 400 kJ/mol.

Step 1: Identify the given ratio

We are given the bond dissociation energies of \(\mathrm{XY}, \mathrm{X}_{2}, \) and \(\mathrm{Y}_{2}\) in the ratio of \(1:1:0.5\). This indicates that if the bond dissociation energy of \(\mathrm{XY}\) is \(x\), then \(\mathrm{X}_{2}\) is also \(x\) and \(\mathrm{Y}_{2}\) is \(0.5x\).

Step 2: Understand the reaction

The reaction to form \(\mathrm{XY}\) from \(\mathrm{X}_{2}\) and \(\mathrm{Y}_{2}\) is half of \(\mathrm{X}_{2} + \mathrm{Y}_{2} \rightarrow 2 \mathrm{XY}\).

Step 3: Use the formation enthalpy

The enthalpy change for the formation of \(\mathrm{XY}\) is given as \(-200 \mathrm{~kJ} / \mathrm{mole}\). This is for the reaction \(\frac{1}{2} \mathrm{X}_{2} + \frac{1}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}\).

Step 4: Write down the enthalpy equation

For the formation \(\frac{1}{2} \mathrm{X}_{2} + \frac{1}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}\), the change in enthalpy \(\Delta \mathrm{H}\) is the difference in dissociation energies: \[\text{Bond energy of } \mathrm{XY} - \frac{1}{2} \text{Bond energy of } \mathrm{X}_{2} - \frac{1}{2} \text{Bond energy of } \mathrm{Y}_{2} = -200 \mathrm{~kJ} / \mathrm{mole}\].

Step 5: Equate and solve using the ratios

Substitute the bond dissociation energy ratios: \[x - \frac{x}{2} - \frac{0.5x}{2} = -200 \]. Simplify to get \[x - \frac{0.5x}{2} - \frac{0.25x}{2} = -200 \]. Solve the simplified equation: \[x = 400 \mathrm{~kJ} / \mathrm{mole}\].

Step 6: Conclusion

From the calculated \(x\), the bond dissociation energy of \(\mathrm{X}_{2}\) is \(400 \mathrm{~kJ} / \mathrm{mole}\).

Key Concepts

Enthalpy of Formation

The enthalpy of formation provides insight into how much energy is involved when forming a compound from its elemental constituents.
This concept is central to understanding chemical reactions at an energetic level. For the compound XY, the given enthalpy of formation is equal to a change of (-200 \( ext{kJ/mol} \). This means that when the elements X and Y react to form one mole of XY, 200 kilojoules of energy are released.
This release indicates that the formation process is exothermic, with energy being released to the surroundings as XY bonds are created. This energetically favorable process allows us to understand the driving force behind the formation of different chemical substances. An important aspect to remember about enthalpy of formation is that it is reported under standard conditions, usually at 1 bar pressure and 298 K (25°C). These conditions provide a basis for comparing different reactions and their heat changes.
When tackling problems involving enthalpy, always consider if you are given standard enthalpy values or if the conditions differ, impacting the calculation.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry involves understanding the proportions and relationships in chemical reactions. It allows us to quantify how much of each substance participates in and results from a reaction.
This concept is essential to solve problems related to energy changes and bond breaking or formation.
For instance, when forming XY, we use the balanced reaction equation:\( \frac{1}{2} \text{X}_{2} + \frac{1}{2} \text{Y}_{2} \rightarrow \text{XY} \).The coefficients (\( \frac{1}{2} \)) indicate the number of moles needed. That is, to produce one mole of XY, we need half a mole of \( \text{X}_2 \) and half a mole of \( \text{Y}_2 \).
Such stoichiometric relationships are crucial for converting atomic level phenomena to a macroscopic scale, as it allows chemists to calculate the necessary or produced amounts based on resulting substances in a reaction.

• This concept aids in quantifying reactants and products using balanced chemical equations.
• It ensures that both mass and charge are conserved in reactions, a principle known as the conservation of mass.

By adhering to stoichiometry, chemists can predict the outcomes of chemical reactions better and optimize them for industrial and laboratory Process.

Thermodynamic Calculations

Thermodynamic calculations revolve around analyzing energy changes during chemical processes, making it possible to predict reaction feasibility and efficiency.One essential aspect of thermodynamics is bond dissociation energy, which measures the strength of a bond. Studying these energies helps assess whether breaking or forming bonds will require or release energy.
To calculate enthalpy changes (\( \Delta H \), we proceeded through several steps:1. **Identify Given Energies and Ratios:** Recognize the bond energies' ratios of XY, \( \text{X}_{2}, \) and \( \text{Y}_{2} \) as 1:1:0.5.2. **Derive Energy Equation:** Use the ratio (x for XY and \( \frac{x}{2} \) for \( \text{X}_{2} \) and \( \text{Y}_{2} \))in the energy equation:\[ x - \frac{0.5x}{2} - \frac{0.25x}{2} = -200 \].3. **Solve the Equation:** Solve to find that \( x = 400 \text{kJ/mol} \), determining that \( \text{X}_{2} \)'s bond energy is 400 \( \text{kJ/mol} \).

• Using balanced equations like this helps calculate unknowns in a chemical reaction.
• Such calculations also help predict if a reaction is exothermic or endothermic, impacting how it's utilized in different contexts.

Understanding these calculations enhances our grasp of reaction energetics and their potential applications.