Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate \(\Delta H_{f}^{\circ}\) for chloride ion from the following data: \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HCl}(\mathrm{g})\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-92.4 \mathrm{~kJ}\) \(\mathrm{HCl}(\mathrm{g})+\mathrm{nH}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}\) (aq) \(\Delta \mathrm{H}_{\mathrm{Hyd}}=-74.8 \mathrm{~kJ}\) \(\Delta \mathrm{H}_{f}\left[\mathrm{H}^{+}\right]=0.0 \mathrm{~kJ}\) (a) \(-189 \mathrm{~kJ}\) (b) \(-167 \mathrm{~kJ}\) (c) \(+167 \mathrm{~kJ}\) (d) \(-191 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

(b) -167 kJ

Step 1: List the Given Reactions and Data

We are given two reactions: 1. Formation of HCl gas: \( \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-92.4 \mathrm{~kJ} \)2. Dissolution of HCl: \( \mathrm{HCl}(\mathrm{g})+\mathrm{nH}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}), \Delta \mathrm{H}_{\mathrm{Hyd}}=-74.8 \mathrm{~kJ} \) Additionally, the standard enthalpy of formation of \(\mathrm{H}^{+}(\mathrm{aq})\) is given as 0 kJ/mol.

Step 2: Calculate Enthalpy Change for Cl⁻

The objective is to find \( \Delta H_{f}^{\circ} \) for \( \mathrm{Cl}^{-}(\mathrm{aq}) \). Use Hess's law:\[ \Delta H_{f}^{\circ}(\mathrm{Cl}^{-}(\mathrm{aq})) = \Delta H_{f}^{\circ}(\mathrm{HCl}(\mathrm{g})) + \Delta H_{\mathrm{Hyd}}(\mathrm{HCl}) - \Delta H_{f}(\mathrm{H}^{+}) \]Substitute the known values:\[ \Delta H_{f}^{\circ}(\mathrm{Cl}^{-}(\mathrm{aq})) = (-92.4 \mathrm{~kJ}) + (-74.8 \mathrm{~kJ}) - (0.0 \mathrm{~kJ}) \] Calculate the result:\[ \Delta H_{f}^{\circ}(\mathrm{Cl}^{-}(\mathrm{aq})) = -167.2 \mathrm{~kJ} \]

Step 3: Select the Closest Option

The calculated \( \Delta H_{f}^{\circ}(\mathrm{Cl}^{-}(\mathrm{aq})) \) is \(-167.2 \mathrm{~kJ}\), which matches closely with option (b).

Key Concepts

Hess's law

Hess's law is an incredibly useful principle in chemistry that helps us calculate the enthalpy change of a reaction, even if the reaction doesn't occur directly. It is based on the idea that enthalpy is a state function. This means the change in enthalpy is independent of the path taken between the initial and final states.

Using Hess's law, if multiple reactions add up to the overall reaction, the total enthalpy change is simply the sum of the enthalpy changes of each step. This becomes very useful when direct measurement is difficult or impossible.

• The law is rooted in the conservation of energy, implying that energy cannot be created or destroyed, just transformed.

In our exercise, we used Hess's law to determine the enthalpy of the chloride ion by summing the standard enthalpy of formation of hydrogen chloride gas and its enthalpy of hydration after dissolving in water. This makes it easier to calculate enthalpy changes in multi-step chemical processes.

standard enthalpy of formation

The standard enthalpy of formation refers to the heat change when one mole of a compound forms from its elements in their standard states under standard conditions, usually 298K and 1 bar pressure. It is a key concept in thermodynamics and aids in tabulating energies of various substances.

• When the standard enthalpy of formation is negative, it indicates an exothermic reaction.
• This means the formation releases heat.
• A zero value, like that for hydrogen ions (\( \Delta H_{f}[\mathrm{H}^{+}] = 0.0 \mathrm{~kJ} \)), implies that it's the reference point by convention.

In this exercise, we calculated the enthalpy for the formation of \( \mathrm{Cl}^{-}(\mathrm{aq}) \). We started by considering the formation of HCl gas, using its known enthalpy of formation value of \(-92.4 \mathrm{~kJ}\).

Through this process, we learn how the standard enthalpy of formation values can be used to describe the energy changes in chemical equations accurately. This is essential for building intuition about how various chemical reactions behave energetically.

enthalpy of hydration

The enthalpy of hydration is the energy change when ions dissolve in water to form a solution. This process can either be endothermic or exothermic, although hydration typically releases heat.

In this case, the dissolution of HCl gas in water is given an enthalpy change, known as the enthalpy of hydration, of \(-74.8 \mathrm{kJ}\).

• The process can be seen as HCl gas being dissociated into hydrogen ions and chloride ions in an aqueous solution.
• This shows how water’s unique properties cause energy absorption or release when solutes dissolve.

To find the enthalpy of the chloride ion, the enthalpy of hydration was combined with the formation enthalpy of HCl gas. In practical terms, it explains the solvation process during the dissolution, an essential step for understanding aqueous chemistry interactions.

Using these insights, you can predict whether reactions involving ions will absorb or release energy, which informs not only theoretical calculations but also practical applications in areas like material science and industrial processes.