Problem 162 - A Complete Resource Book in Chemistry for JEE Main

Question

The enthalpy changes for the following processes are listed below. \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{C} 1(\mathrm{~g}) ; 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~g})=21(\mathrm{~g}) ; 151.0 \mathrm{kJmol}^{-1}\) \(\mathrm{ICl}(\mathrm{g})=\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) ; 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~s})=\mathrm{I}_{2}(\mathrm{~g}) ; 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Given that the standard states for iodine and chlorine are \(\mathrm{I}_{2}(\mathrm{~s})\) and \(\mathrm{Cl},(\mathrm{g})\), the standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) is [2006] (a) \(-14.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+244.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) is \(+16.8\, \mathrm{kJ/mol}\), option (c).

Step 1: Write the Required Reaction

The standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) can be written as: \(\frac{1}{2}\mathrm{I}_2(\mathrm{s}) + \frac{1}{2}\mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{ICl}(\mathrm{g})\). This represents the formation of 1 mole of \(\mathrm{ICl}\) from its elements in their standard states.

Step 2: Analyze Given Reactions

Identify the relevant steps from the given data that eventually convert \(\frac{1}{2}\mathrm{I}_2(\mathrm{s}) + \frac{1}{2}\mathrm{Cl}_2(\mathrm{g})\) to \(\mathrm{ICl}(\mathrm{g})\). These include the sublimation of \(\mathrm{I}_2(\mathrm{s})\), the dissociation of \(\mathrm{Cl}_2(\mathrm{g})\), the dissociation of \(\mathrm{I}_2(\mathrm{g})\), and the recombination to form \(\mathrm{ICl}(\mathrm{g})\).

Step 3: Calculate Each Step of Enthalpy Changes

Calculate the enthalpy change for each required process:1. \(\mathrm{I}_2(\mathrm{s}) \rightarrow \frac{1}{2}\mathrm{I}_2(\mathrm{g})\): \(\frac{1}{2} \times 62.76 = 31.38\, \mathrm{kJ/mol}\).2. \(\frac{1}{2}\mathrm{I}_2(\mathrm{g}) \rightarrow \mathrm{I}(\mathrm{g})\): \(\frac{1}{2} \times 151.0 = 75.5\, \mathrm{kJ/mol}\).3. \(\frac{1}{2}\mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{Cl}(\mathrm{g})\): \(\frac{1}{2} \times 242.3 = 121.15\, \mathrm{kJ/mol}\).

Step 4: Calculate Enthalpy for Formation of ICl

Consider the reverse of the given reaction for \(\mathrm{ICl}(\mathrm{g})\): \(\mathrm{I}(\mathrm{g}) + \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{ICl}(\mathrm{g})\), which is \(-211.3\, \mathrm{kJ/mol}\) because it is the reverse of \(\mathrm{ICl}(\mathrm{g})\) dissociating.

Step 5: Sum All Enthalpy Changes

Add all individual changes to find the total enthalpy change:\(31.38 + 75.5 + 121.15 - 211.3 = 16.73\, \mathrm{kJ/mol}\). Round to two significant figures: \(16.8\, \mathrm{kJ/mol}\).

Step 6: Choose the Correct Option

Compare the calculated value with the options. The correct answer is option (c) \(+16.8\, \mathrm{kJ/mol}\).

Key Concepts

Thermodynamics

Thermodynamics is the study of energy, specifically how it moves and changes state within a system. This branch of physics helps us understand various phenomena, such as the energy transfers that occur in chemical reactions. In our context, when iodine and chlorine form the compound ICl, thermodynamics provides insights into how this energy exchange unfolds.
One of the key principles in thermodynamics is the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed. This principle is essential when calculating the enthalpy changes in reactions.

• Exothermic reactions release heat, usually leading to a decrease in enthalpy.
• Endothermic reactions, conversely, absorb heat, resulting in an increase in enthalpy.

By using thermodynamics, we can quantify the energy changes as iodine and chlorine transition from their individual states to forming ICl.

Chemical Reaction Enthalpy

Enthalpy of a reaction is a measure of the total heat content of the system. It corresponds to the energy absorbed or released during a chemical reaction. Understanding enthalpy changes is crucial for predicting how reactions occur.
In the formation of ICl, enthalpy changes allow us to determine whether the reaction is endothermic or exothermic. The reaction involves a series of steps, each with its own enthalpy change, calculated from standard thermodynamic data.

• Dissociation of I₂ and Cl₂ into atoms absorbs energy.
• Combining these atoms to form ICl releases energy.

The net enthalpy change provides insight into the stability of the produced ICl molecule and the feasibility of the reaction under specific conditions.

Standard State Conditions

Standard state conditions refer to a reference point used to calculate the properties of substances. It's a defined set of conditions—typically 1 atm pressure and 25°C (298 K)—under which the properties of substances are measured.
Knowing these conditions is important because they ensure consistency when comparing enthalpy values across different reactions. For instance, the enthalpy of formation for ICl is calculated using these standard conditions, ensuring that we can accurately compare it with other reactions.

• Elements are usually in their most stable form.
• For iodine, this is I₂(s), and for chlorine, it's Cl₂(g).

By adhering to these conditions, chemists can determine whether the reaction of interest is favored or not.

Hess's Law

Hess's Law is a principle that stems from the first law of thermodynamics. It states that the total enthalpy change of a reaction is the same whether it occurs in one or several steps. Therefore, regardless of the path taken, the energy change remains constant.
This concept is a powerful tool for determining reaction enthalpies that are not easily measurable directly. In the case of ICl formation, Hess's Law allows us to calculate the enthalpy change by summing the enthalpy changes of smaller, individual reactions. Each calculated step reflects a part of the overall process.

• Step-by-step changes are carefully added to obtain the final enthalpy change.
• This approach simplifies the calculation and interpretation of complex reactions.

As a result, even if individual steps cannot be directly measured, Hess's Law makes it possible to determine the enthalpy for a combined reaction accurately.