Problem 55 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

For an ionic solid \(\mathrm{MX}_{2}\), where \(\mathrm{X}\) is monovalent, the enthalpy of formation of the solid from \(\mathrm{M}(\mathrm{s})\) and \(\mathrm{X}_{2}(\mathrm{~g})\) is \(1.5\) times the electron gain enthalpy of \(\mathrm{X}(\mathrm{g})\). The first and second ionization enthalpies of the metal (M) are \(1.2\) and \(2.8\) times of the enthalpy of sublimation of \(\mathrm{M}(\mathrm{s})\). The bond dissociation enthalpy of \(\mathrm{X}_{2}(\mathrm{~g})\) is \(0.8\) times the first ionization enthalpy of metal and it is also equal to one-fifth of the magnitude of lattice enthalpy of \(\mathrm{MX}_{2}\). If the electron gain enthalpy of \(\mathrm{X}(\mathrm{g})\) is \(-96 \mathrm{kcal} / \mathrm{mol}\), then what is the enthalpy of sublimation (in \(\mathrm{kcal} / \mathrm{mol}\) ) of the metal (M)? (a) \(41.38\) (b) \(52.5\) (c) \(48.0\) (d) \(38.27\)

Step-by-Step Solution

Answer

The enthalpy of sublimation of the metal (M) is 48.0 kcal/mol.

Step 1: Understand the Problem

The problem states that the enthalpy of formation of an ionic solid MX2 is given in terms of the electron gain enthalpy of X(g), and relates other enthalpies like ionization enthalpies of M, enthalpy of sublimation of M, bond dissociation enthalpy of X2(g) and the lattice enthalpy of MX2. We are given the electron gain enthalpy of X(g) and we need to find the enthalpy of sublimation of M(s).

Step 2: Relate Given Enthalpies

It is given that the bond dissociation enthalpy of X2(g) is 0.8 times the first ionization enthalpy of M. This dissociation enthalpy is also one-fifth of the lattice enthalpy of MX2. Using these relationships, we can set up a proportion to find the enthalpy of sublimation of M(s).

Step 3: Express Enthalpies of M in Terms of Sublimation Enthalpy

According to the given problem, the first and second ionization enthalpies of M are 1.2 and 2.8 times the enthalpy of sublimation of M(s), respectively. Let the enthalpy of sublimation of M be 'S'. Then, the first ionization enthalpy of M is 1.2S and the second ionization enthalpy is 2.8S.

Step 4: Set Up Equation for Bond Dissociation Enthalpy of X2(g)

The bond dissociation enthalpy of X2(g) is given to be 0.8 times the first ionization enthalpy of M, which is 0.8 * 1.2S. This bond dissociation enthalpy is also one-fifth of the lattice enthalpy of MX2. Therefore, 0.8 * 1.2S = one-fifth of the lattice enthalpy of MX2.

Step 5: Use Electron Gain Enthalpy of X to Find Lattice Enthalpy

The problem provides the electron gain enthalpy of X(g) as -96 kcal/mol and states that the enthalpy of formation of MX2 is 1.5 times this value. We can write the lattice enthalpy of MX2 in terms of the electron gain enthalpy and use this to find the enthalpy of sublimation.

Step 6: Solve for Enthalpy of Sublimation (S)

Using the previous steps, we have two expressions involving the lattice enthalpy of MX2. By equating these, we can solve for the enthalpy of sublimation S and find the answer.

Key Concepts

Enthalpy of Formation

When we discuss the enthalpy of formation, we refer to the total energy change that occurs when one mole of a compound is formed from its elements in their standard states. For an ionic solid like MX2, this involves the energy changes from the sublimation of the metal (M) from solid to gas, dissociation of the diatomic molecule X2, and the subsequent formation of the ionic lattice. Understanding this concept is crucial, because it sets the foundation for calculating and comparing the stabilities of different compounds. It is important to note that the standard enthalpy of formation for a pure element in its standard state is zero.

Ionization Enthalpies

Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom or ion. The first ionization enthalpy is the energy needed to remove the first electron, and the second ionization enthalpy is for the second electron. These values are indicative of the tightness of the electronic hold within the atom and generally increase across a period in the periodic table. In our problem, the first and second ionization enthalpies are given as multiples of the enthalpy of sublimation of the metal, linking these two types of enthalpy. A higher ionization enthalpy means it is more difficult to remove an electron from the atom, which can affect the compound's properties.

Bond Dissociation Enthalpy

The bond dissociation enthalpy refers to the energy required to break a bond in a molecule and separate the atoms completely. For diatomic molecules like X2, it's the energy needed to break the bond between the two X atoms. In our exercise, this value plays a pivotal role in establishing the relationship between different enthalpies and is expressed as a fraction of the first ionization enthalpy of metal M. This concept helps us understand the stability of chemical bonds. Remember that the bond dissociation energy for a single bond in a polyatomic molecule can be different from that in a diatomic molecule.

Lattice Enthalpy

Lattice enthalpy is the measure of the strength of the forces between the ions in an ionic solid. It's the amount of energy released when gaseous ions combine to form an ionic solid. The concept of lattice enthalpy is vital as it provides insight into the stability of an ionic structure. A more negative lattice enthalpy indicates a more stable lattice. In terms of calculations, lattice enthalpy can be used to predict physical properties such as melting points and solubilities. The problem we are analyzing uses the bond dissociation enthalpy and the electron gain enthalpy to form a chain of proportional relationships, all leading back to the lattice enthalpy.

Electron Gain Enthalpy

Electron gain enthalpy, sometimes simply called electron affinity, is the energy change when an electron is added to a neutral atom in the gaseous state to form a negative ion. If energy is released during this process, the electron gain enthalpy is negative, indicating that the atom has a tendency to gain electrons. In our exercise, the electron gain enthalpy of X(g) is a given constant, and since the formation enthalpy of MX2 is related to it, applying this concept helps us decipher the relative stabilities and reactivities of different elements and ions. This understanding is key when predicting how substances will interact in a chemical reaction.

Physical Chemistry Problems

When solving physical chemistry problems, it's important to systematically approach the exercise by thoroughly understanding the relationships between different thermodynamic quantities. Step-by-step problem-solving includes identifying known quantities, understanding how they relate to one another, and applying relevant equations or principles. For instance, in the given problem, interrelations between different enthalpies are identified and used to set up equations that ultimately allow us to find the enthalpy of sublimation of metal M. These types of problems enhance your understanding of fundamental chemistry concepts and are key in illustrating how theory is applied to practical chemistry.