Problem 78 - Chemistry

Question

The molar heats of fusion and sublimation of molecular iodine are \(15.27 \mathrm{~kJ} / \mathrm{mol}\) and \(62.30 \mathrm{~kJ} / \mathrm{mol}\), respectively. Estimate the molar heat of vaporization of liquid iodine.

Step-by-Step Solution

Answer

The molar heat of vaporization of iodine is approximately \(47.03 \, \mathrm{kJ/mol}\).

Step 1: Understand Hess's Law

Hess's Law of Constant Heat Summation states when a reaction can be expressed as the sum of a series of steps, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual steps.

Step 2: Apply Hess's Law to the problem

The problem can be visualized as: \nSolid iodine melts (fusion) into liquid iodine and then boils (vaporizes) into gaseous iodine. Alternatively, solid iodine can directly sublimate into gaseous iodine.\nThe molar heat of fusion is the heat change when one mole of solid iodine melts into liquid iodine, and the molar heat of sublimation is the heat change when one mole of solid iodine transforms directly into gaseous iodine.\nThe molar heat of vaporization (which we aim to calculate) is the heat change when one mole of liquid iodine vaporizes into gaseous iodine.

Step 3: Calculate the molar heat of vaporisation

According to Hess's Law, the molar heat of sublimation is equal to the sum of the molar heat of fusion and vaporization: \n\n\[\Delta H_{sublimation}= \Delta H_{fusion} + \Delta H_{vaporization}\]\n\nWe can rearrange the equation to calculate the molar heat of vaporization as follows:\n\n\[\Delta H_{vaporization}= \Delta H_{sublimation} − \Delta H_{fusion}\]\n\nThen substituting the given values:\n\n\[\Delta H_{vaporization} = 62.30 kJ/mol - 15.27 kJ/mol = 47.03 kJ/mol\]

Key Concepts

Molar Heat of Vaporization

Molar heat of vaporization refers to the amount of heat required to transform a mole of a substance from a liquid to a gaseous state. This process is crucial in understanding how substances behave when they change from liquid to gas. The concept is often used in thermodynamics and physical chemistry. When heat is added to a liquid, its molecules gain energy and begin to move more vigorously. At a certain point, they overcome the forces holding them together in the liquid, and the substance transitions to a gas. This is where the molar heat of vaporization comes into play. It's an expression of the energy required for breaking the intermolecular forces in the liquid state:

• A higher molar heat of vaporization indicates stronger intermolecular forces requiring more energy to convert the liquid to gas.
• For iodine, it helps to bridge the understandings of various states and their transitions from solid to liquid and to gas.

Applying Hess's Law in calculations allows us to link the molar heat of sublimation, fusion, and vaporization efficiently to find the required values.

Enthalpy Change

Enthalpy change, represented as \( \Delta H \), is a fundamental concept in chemistry that describes the heat content change in a system during a transformation. It's a key concept when studying heat exchanges in chemical reactions.There are different forms of enthalpy changes depending on the kind of reaction or state change:

• Fusion: Heat intake when a solid turns into a liquid.
• Vaporization: Heat intake when a liquid turns into a gas.
• Sublimation: Direct transition from solid to gas without passing through the liquid phase.

In the problem of iodine's phase changes, we use enthalpy changes to connect the amounts of heat involved during different phase transitions.Hess's Law states that the total enthalpy change is independent of the path taken. Therefore, it allows us to calculate the molar heat of vaporization by summing or subtracting known enthalpy changes.

Molecular Iodine

Molecular iodine (\( I_2 \)) is a diatomic molecule, meaning it consists of two iodine atoms bonded together. This element has special characteristics which make it interesting in terms of phase transitions and thermodynamic studies.Iodine is a solid at room temperature, with a shiny, metallic appearance. It can transition from solid to liquid (fusion), liquid to gas (vaporization), or directly from solid to gas (sublimation).Understanding molecular iodine's behavior is vital because:

• It showcases a clear example where sublimation is significant, by allowing transition without first forming a liquid state.
• Being a halogen, iodine reacts uniquely compared to other elements, often serving as an excellent subject to observe enthalpy changes.

In chemical experimentation and applications, molecular iodine's straightforward phase changes and predictable calculations make it a go-to example for exploring concepts like the heat of vaporization and general thermodynamic laws.