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 \(\mathrm{I}_{2}\) (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 ICl(g) is +16.8 kJ/mol (option c).

Step 1: Write the Reaction Equation

The reaction for which we need the standard enthalpy of formation is the formation of ICl(g) from its elements in their standard states: \[ \mathrm{\frac{1}{2}Cl_{2}(g) + \frac{1}{2}I_{2}(s) \rightarrow ICl(g)} \]

Step 2: Identify the Known Enthalpy Changes

You've been given the enthalpies of several processes. Let's break these down:1. Decomposition of \( \mathrm{Cl_2(g)} \) into \( 2 \mathrm{Cl(g)} \): \( 242.3 \ \mathrm{kJ/mol} \)2. Decomposition of \( \mathrm{I_2(g)} \) into \( 2 \mathrm{I(g)} \): \( 151.0 \ \mathrm{kJ/mol} \)3. Decomposition of \( \mathrm{ICl(g)} \) into \( \mathrm{I(g)} + \mathrm{Cl(g)} \): \( 211.3 \ \mathrm{kJ/mol} \)4. Sublimation of \( \mathrm{I_2(s)} \) into \( \mathrm{I_2(g)} \): \( 62.76 \ \mathrm{kJ/mol} \)

Step 3: Apply Hess's Law to Find enthalpy of \( \mathrm{ICl(g)} \) formation

To find the enthalpy change for the formation of \( \mathrm{ICl(g)} \), we use Hess's law: 1. Sublimate \( \mathrm{\frac{1}{2}I_2(s)} \) to \( \mathrm{\frac{1}{2}I_2(g)} \): \[ \frac{62.76 \ \mathrm{kJ/mol}}{2} = 31.38 \ \mathrm{kJ/mol} \]2. Convert \( \mathrm{\frac{1}{2}Cl_{2}(g)} \) to \( \mathrm{Cl(g)} \): \[ \frac{242.3 \ \mathrm{kJ/mol}}{2} = 121.15 \ \mathrm{kJ/mol} \]3. Convert \( \mathrm{\frac{1}{2}I_{2}(g)} \) to \( \mathrm{I(g)} \): \[ \frac{151.0 \ \mathrm{kJ/mol}}{2} = 75.5 \ \mathrm{kJ/mol} \]4. Formation of \( \mathrm{ICl(g)} \) from \( \mathrm{I(g)} + \mathrm{Cl(g)} \): \[ -211.3 \ \mathrm{kJ/mol} \] Now combine these to find the enthalpy of formation of \( \mathrm{ICl(g)} \):\[ \Delta H_f = 31.38 + 121.15 + 75.5 - 211.3 = 16.73 \ \mathrm{kJ/mol} \]

Step 4: Select the Closest Given Option

Comparing our calculated result 16.73 kJ/mol with the given options, the closest value is +16.8 kJ/mol. Therefore, the standard enthalpy of formation for \( \mathrm{ICl} \) is given by option (c).

Key Concepts

Hess's Law

Hess's Law is an essential principle in chemistry that helps us determine the overall enthalpy change for a reaction. It states that the total enthalpy change for a chemical reaction is the same, no matter how many steps are involved in the process. This is because enthalpy is a state function, which means its change depends only on the initial and final states, not the path taken to get there.

When applying Hess's Law, we often use a series of known enthalpy changes from related reactions to calculate the \( \Delta H \) \( \text{(enthalpy change)} \) for an unknown reaction. In practice, this involves

• Adding up the enthalpy changes of the steps needed to reach the desired end reaction.
• Supplying the correct stoichiometric coefficients for each step.

In the original exercise, we applied Hess's Law to determine the standard enthalpy of formation for \( \text{ICl(g)} \). We added together the separately calculated enthalpy changes from known processes to find our final value, proving how powerful Hess’s law can be in solving such problems effectively.

Standard Enthalpy Changes

Standard enthalpy change is the enthalpy change when a chemical process occurs under standard conditions (usually 1 bar pressure and at a specified temperature, typically 298.15 K). This concept is vital because it allows chemists to compare enthalpy changes under common conditions, making data widely useful and consistent.

The standard enthalpy of formation, a specific type of standard enthalpy change, refers to the formation of one mole of a compound from its elements in their standard states. We utilize standard enthalpies of formation to calculate more complex reactions through Hess’s Law.

• It simplifies calculations by providing baseline values for reactions.
• Facilitates comparing the stability of different compounds.

In the exercise, the standard enthalpy of formation of \( \text{ICl(g)} \) was computed from known processes, highlighting its usefulness in quantifying the overall energy change in chemical formation.

Decomposition Reactions

Decomposition reactions are chemical reactions where a single compound breaks down into two or more smaller compounds or elements. This type of reaction is essential in understanding enthalpy changes because it often involves energy absorption to break bonds, illustrating important energy relationships in chemical processes.

In the original problem, various decomposition reactions formed part of the calculations:

• The breaking of \( \text{Cl}_2 \) into \( 2 \text{Cl} \ \text{atoms} \).
• The breakdown of \( \text{I}_2 \ \text{molecules} \) into individual \( \text{I} \ \text{atoms} \).
• Decomposition of \( \text{ICl(g)} \) into \( \text{I(g)} + \text{Cl(g)} \).

Each decomposition involved certain enthalpy changes, contributing to the calculation of the total enthalpy change for our target reaction. Understanding these decomposition processes helps to comprehend the energetic demands of breaking chemical bonds.

Sublimation

Sublimation is the process where a substance transitions directly from a solid to a gas without passing through the liquid state. This phase change requires energy input to overcome the forces holding the solid together, and it's expressed as an enthalpy change.

In our exercise, the sublimation of \( \text{I}_2 \ \text{(s)} \) to \( \text{I}_2 \ \text{(g)} \) was one of the steps. The enthalpy change for this process contributes to the overall calculation by representing the energy needed to convert solid iodine into a gaseous form, part of forming \( \text{ICl(g)} \).

• Sublimation is crucial in reactions where solid reactants need to be in the gaseous phase.
• The enthalpy of sublimation directly affects overall enthalpy changes in chemical processes involving solids turning to gas.

Understanding sublimation allows for better predictions and calculations of energy requirements in chemical reactions like those involving iodine in our given task.