Problem 157
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 ? (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
Verified Answer
The bond dissociation energy of \( \mathrm{X}_{2} \) is 400 kJ/mol. (b)
1Step 1: Understand the Relationship between Bond Energies
The bond dissociation energy of XY, X2, and Y2 are in the ratio 1:1:0.5. Let's denote the bond energies as B_XY, B_X2, and B_Y2 respectively, such that B_XY = B_X2 = 2*B_Y2. Given the ratio, if we express B_Y2 as B, then B_XY and B_X2 both equal 2B.
2Step 2: Use the Enthalpy of Formation
The enthalpy of formation \( \Delta H_f \) of compound XY is provided as \(-200 \text{ kJ/mol} \). According to the equation for the formation of XY from elements X and Y: \( \mathrm{X}_{2} + \mathrm{Y}_{2} \rightarrow 2\mathrm{XY} \).
3Step 3: Apply Hess's Law for the Enthalpy Change
Using Hess's Law, the enthalpy change for the reaction can be expressed as: \( \Delta H_f = \text{Sum of bond energies of reactants} - \text{Sum of bond energies of products} \). That is, \( \Delta H_f = B_{X2} + B_{Y2} - 2 \times B_{XY} \).
4Step 4: Substitute and Simplify the Equation
Substitute the known values: \(-200 = 2B + B - 2 \times 2B \). Simplifying the equation: \(-200 = 3B - 4B \). Further simplifying gives: \(-200 = -B \).
5Step 5: Solve for the Bond Energy of X2
From \(-B = -200\), we find \( B = 200 \text{ kJ/mol} \). Since \( B_{X2} = 2B \), it means the bond energy of X2 is \( 2 \times 200 = 400 \text{ kJ/mol} \).
Key Concepts
Enthalpy of FormationHess's LawChemical Thermodynamics
Enthalpy of Formation
The enthalpy of formation, symbolized as \( \Delta H_f \), is a concept in chemical thermodynamics used to understand the energy changes when a compound is formed from its elements. It specifically refers to the heat change that occurs during the formation of 1 mole of a substance from its constituent elements, under standard conditions (usually at 1 atm and 25°C or 298 K).
This concept helps in determining whether a reaction is endothermic or exothermic. In an exothermic reaction, \( \Delta H_f \) is negative, indicating that heat is released. Conversely, a positive \( \Delta H_f \) signifies an endothermic reaction, where heat is absorbed.
In the problem with the compound \( \mathrm{XY} \), a given \( \Delta H_f \) of \(-200 \text{kJ/mol}\) tells us that heat isreleased when \( \mathrm{XY} \) is formed. This insight facilitates calculations involving the energy required to form or break bonds within the substance, fundamental in understanding reaction energetics.
This concept helps in determining whether a reaction is endothermic or exothermic. In an exothermic reaction, \( \Delta H_f \) is negative, indicating that heat is released. Conversely, a positive \( \Delta H_f \) signifies an endothermic reaction, where heat is absorbed.
In the problem with the compound \( \mathrm{XY} \), a given \( \Delta H_f \) of \(-200 \text{kJ/mol}\) tells us that heat isreleased when \( \mathrm{XY} \) is formed. This insight facilitates calculations involving the energy required to form or break bonds within the substance, fundamental in understanding reaction energetics.
Hess's Law
Hess's Law is a powerful principle in chemical thermodynamics that states the total enthalpy change of a reaction is the same regardless of the pathway taken, provided the initial and final conditions are identical.
In simpler terms, Hess's Law allows us to break down a complex reaction into simpler steps, calculate the enthalpy changes for each of these steps, and then combine them to find the overall change. This is immensely helpful when direct measurement of enthalpy change for a reaction is challenging or impossible.
In our problem, Hess's Law is used to determine the bond dissociation energies. By leveraging the given enthalpy of formation \(( \Delta H_f \)) and the known bond energies from the reactants and products, we can arrive at the desired value by manipulating individual bond energies, a testament to the versatility and practical utility of Hess’s Law.
In simpler terms, Hess's Law allows us to break down a complex reaction into simpler steps, calculate the enthalpy changes for each of these steps, and then combine them to find the overall change. This is immensely helpful when direct measurement of enthalpy change for a reaction is challenging or impossible.
In our problem, Hess's Law is used to determine the bond dissociation energies. By leveraging the given enthalpy of formation \(( \Delta H_f \)) and the known bond energies from the reactants and products, we can arrive at the desired value by manipulating individual bond energies, a testament to the versatility and practical utility of Hess’s Law.
Chemical Thermodynamics
Chemical thermodynamics is the study of heat and work in chemical processes. This branch of chemistry examines how energy transitions and transfers occur during chemical reactions and how these transformations affect the physical and chemical properties of substances.
Core principles of chemical thermodynamics include understanding enthalpy, entropy, and Gibbs free energy, which help describe the spontaneity and feasibility of chemical reactions.
Core principles of chemical thermodynamics include understanding enthalpy, entropy, and Gibbs free energy, which help describe the spontaneity and feasibility of chemical reactions.
- **Enthalpy (H):** It denotes the heat content of a system at constant pressure.
- **Entropy (S):** This measures the degree of randomness or disorder within a system.
- **Gibbs Free Energy (G):** This determines the spontaneity of processes. A negative Gibbs free energy indicates a spontaneous reaction.
Other exercises in this chapter
Problem 155
An ideal gas expands in volume from \(1 \times 10^{-3} \mathrm{~m}^{3}\) to 1 \(\times 10^{-2} \mathrm{~m}^{3}\) at \(300 \mathrm{~K}\) against a constant press
View solution Problem 156
The enthalpies of combustion of carbon and carbon monoxide are \(-393.5\) and \(-283 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy of formation o
View solution Problem 158
Consider the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) carried out at constant temperature and pressure. If \(\Delta \ma
View solution Problem 161
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If \(\mathrm{T}_{\mathrm{i}}\) is the initial temperature and \(\mathr
View solution