Question

Using the following data, calculate the standard heat of formation of ICl \((g)\) in \(\mathrm{kJ} / \mathrm{mol} :\) $$\begin{array}{ll}{\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)} & {\Delta H^{\circ}=242.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{I}(g)} & {\Delta H^{\circ}=151.0 \mathrm{kJ}} \\ {\mathrm{ICl}(g) \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g)} & {\Delta H^{\circ}=211.3 \mathrm{kJ}} \\ {\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)} & {\Delta H^{\circ}=62.8 \mathrm{kJ}}\end{array}$$

Short answer

The standard heat of formation of ICl(g) is 48.15 kJ/mol.

Step-by-step solution

Write the desired reaction for the formation of ICl(g)

We want to find the standard heat of formation for ICl(g), which means we are looking for the heat change when ICl is formed from its elements in their standard states. The desired reaction is: \[ \frac{1}{2}\mathrm{I}_{2}(s) + \frac{1}{2}\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ICl}(g) \]

Manipulate the given reactions to match the desired reaction

From the given reactions, we need to rearrange and manipulate the equations to achieve the desired reaction. 1. Multiply the first reaction by \(\frac{1}{2}\) to have one Cl(g) produced: \[\frac{1}{2}\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g); \quad \Delta H^{\circ} = \frac{1}{2}\cdot 242.3\,\mathrm{kJ} = 121.15\,\mathrm{kJ}\] 2. Multiply the second reaction by \(\frac{1}{2}\) to have one I(g) produced: \[\frac{1}{2}\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}(g); \quad \Delta H^{\circ} = \frac{1}{2}\cdot 151.0\,\mathrm{kJ} = 75.5\,\mathrm{kJ}\] 3. Reverse the third given reaction: \[ \mathrm{I}(g) + \mathrm{Cl}(g) \longrightarrow \mathrm{ICl}(g); \quad \Delta H^{\circ} = -211.3\,\mathrm{kJ}\]

Sum the manipulated reactions

Now, we will sum the manipulated reactions as well as their heats to find the heat of formation of ICl(g): Reaction 1: \[\frac{1}{2}\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{Cl}(g); \quad \Delta H^{\circ} = 121.15\,\mathrm{kJ}\] Sublimation reaction: \[\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g); \quad \Delta H^{\circ} = 62.8\,\mathrm{kJ} \] Reaction 2: \[\frac{1}{2}\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}(g); \quad \Delta H^{\circ} = 75.5\,\mathrm{kJ}\] Reaction 3 (reversed): \[ \mathrm{I}(g) + \mathrm{Cl}(g) \longrightarrow \mathrm{ICl}(g); \quad \Delta H^{\circ} = -211.3\,\mathrm{kJ}\] Summing these reactions, we get the desired reaction: \[\frac{1}{2}\mathrm{I}_{2}(s) + \frac{1}{2}\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ICl}(g)\]

Calculate the standard heat of formation of ICl(g)

Summing the corresponding heats of reactions to calculate the standard heat of formation of ICl(g): \[\Delta H_{f}^{\circ}(\mathrm{ICl}) = 121.15\,\mathrm{kJ} + 62.8\,\mathrm{kJ} + 75.5\,\mathrm{kJ} - 211.3\,\mathrm{kJ} = 48.15\,\mathrm{kJ/mol}\] The standard heat of formation of ICl(g) is 48.15 kJ/mol.

Key concepts

Enthalpy Change

Enthalpy change, symbolized by \( \Delta H \), is the difference between the energy of the products and the reactants in a chemical reaction. It indicates whether a reaction consumes or releases energy. In our exercise, enthalpy changes are given for various reactions involving chlorine and iodine in different states.
These changes tell us how much energy is required or released when these substances undergo formation or decomposition.
If the enthalpy change is negative (\( \Delta H < 0 \)), the reaction is exothermic, meaning it releases energy. Conversely, if \( \Delta H > 0 \), the reaction is endothermic, meaning it absorbs energy from the surroundings.

• For instance, the reaction \( \mathrm{I}(g) + \mathrm{Cl}(g) \rightarrow \mathrm{ICl}(g) \), has an enthalpy change of \(-211.3 \text{ kJ} \), indicating it's exothermic and releases energy when forming ICl(g).
• Manipulating these enthalpy values was crucial in calculating the standard heat of formation for ICl(g).

Understanding enthalpy change is critical because it helps predict reaction behavior, such as feasibility and energy requirements.

Chemical Thermodynamics

Chemical thermodynamics involves the study of energy transformations in chemical processes. It's a branch of science concerned with understanding how heat and energy affect chemical reactions.
This exercise is a typical example of applying chemical thermodynamics to predict and calculate energy changes during the formation of ICl(g) from its constituent elements.
Thermodynamics allows us to determine the spontaneity and equilibrium of reactions. For example, using Hess's Law, one of the fundamental principles of thermodynamics, we can add or subtract known enthalpy changes to find an unknown enthalpy change, like in the calculation of the standard heat of formation as given in the solution.

• In this context, we are dealing with standard conditions, usually defined at 25°C and 1 atm, allowing for consistency across thermodynamic calculations.
• By understanding chemical thermodynamics, we can design reactions with desired energy profiles or optimize existing processes for energy efficiency.

Manipulating Reactions

Manipulating reactions is a crucial step in arriving at the desired chemical equation, especially when calculating enthalpy changes for formations. This involves using mathematical techniques to rearrange and simplify given chemical equations to match the target equation.
In our exercise, we needed to derive the enthalpy change for the formation of \( \mathrm{ICl}(g) \) by aligning the given reactions to our desired reaction pathway.
The steps included scaling reactions' stoichiometry and reversing reactions, which involves changing the direction of the reaction and so inverting the sign of the enthalpy change.

• For example, reversing the reaction \( \mathrm{ICl}(g) \rightarrow \mathrm{I}(g) + \mathrm{Cl}(g) \) yields the desired reaction with \( \Delta H^{\circ} = -211.3 \text{ kJ} \).
• By adding or subtracting these manipulated reactions, we systematically build the overall reaction pathway and derive its enthalpy change, utilizing the concept of Hess's Law.
• Thus, manipulating reactions is a core skill in chemical thermodynamics, allowing precise calculations under given constraints.

Understanding how to manipulate reactions ensures accurate predictions regarding energy changes and reactions, ultimately assisting in efficient energy management.