Problem 170 - A Complete Resource Book in Chemistry for JEE Main

Question

On the basis of the following thermochemical data: \(\left(\Delta \mathrm{G}^{0} \mathrm{H}+(\mathrm{aq})=0\right)\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) \(\Delta \mathrm{H}=57.32 \mathrm{~kJ}\) \(\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) \(\Delta \mathrm{H}=-286.20 \mathrm{~kJ}\) The value of enthalpy of formation of \(\mathrm{OH}^{-}\)ion at \(25^{\circ} \mathrm{C}\) is: (a) \(-228.88 \mathrm{~kJ}\) (b) \(+228.88 \mathrm{~kJ}\) (c) \(-343.52 \mathrm{~kJ}\) (d) \(-22.88 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The enthalpy of formation of \(\mathrm{OH}^-\) is \(-228.88 \mathrm{~kJ/mol}\).

Step 1: Understanding the Goal

The problem is asking us to find the enthalpy of formation for the hydroxide ion, \(\mathrm{OH}^-\), at \(25^{\circ} \mathrm{C}\). We have two main reactions given with their enthalpies, and we know the Gibbs free energy for \(\mathrm{H}^+\) is zero. We will use these to determine the enthalpy of formation for \(\mathrm{OH}^-\).

Step 2: Applying Hess's Law

According to Hess's Law, the total enthalpy change of a reaction is the sum of the enthalpy changes of each step of the reaction. First, consider the formation of water from its elements and the dissociation of water into \(\mathrm{H}^+\) and \(\mathrm{OH}^-\). We will combine these reactions to isolate \(\mathrm{OH}^-\).

Step 3: Writing the Reactions and Enthalpy Change

Given:1. \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq})\) with \(\Delta \mathrm{H} = 57.32 \mathrm{~kJ/mol}\).2. \(\mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) with \(\Delta \mathrm{H} = -286.20 \mathrm{~kJ/mol}\).

Step 4: Combining the Reactions

We rearrange the first equation to start from elements:\(\mathrm{H}_{2}(\mathrm{~g}) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq})\) combined enthalpy change is the sum of the reactions: \((-286.20 + 57.32) \mathrm{~kJ/mol} = -228.88 \mathrm{~kJ/mol}\).

Step 5: Result for Enthalpy of Formation

From the combined equation, the enthalpy change \(-228.88 \mathrm{~kJ/mol}\) corresponds to the formation of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) from \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\). Since the Gibbs free energy for \(\mathrm{H}^{+}(\mathrm{aq})\) is 0, we attribute this enthalpy value to the formation of \(\mathrm{OH}^-\). The enthalpy of formation for \(\mathrm{OH}^-\) is \(-228.88 \mathrm{~kJ/mol}\).

Key Concepts

Thermochemical Data

Thermochemical data provides essential information about the heat changes that occur during chemical reactions. It includes figures like enthalpy, entropy, and Gibbs free energy, which are all crucial in predicting the direction and extent of chemical processes. In our exercise, we are mainly dealing with enthalpy, which is the heat absorbed or released at constant pressure.

• Enthalpy (94H) measures the total heat content in a system.
• It can be positive (endothermic) or negative (exothermic).
• The exercise provides specific thermochemical data, detailing the enthalpy of formation for water and its dissociation into ions.

Understanding this data allows us to apply Hess's Law and find out the enthalpy for the formation of chemical species like hydroxide ion. Hence, mastering thermochemical data helps in accurately predicting how reactions will proceed and evaluating their feasibility.

Hess's Law

Hess's Law is a foundational principle in thermochemistry that states the total enthalpy change of a chemical reaction is the same, no matter how many steps it takes to get there. This means we can calculate the enthalpy change of a reaction even if it occurs through multiple steps by simply adding up the enthalpy changes of the individual reactions.

Using this principle:

• We can calculate enthalpies for complex reactions by breaking them into smaller, known reactions.
• In our problem, we utilized Hess's Law by combining given reactions to isolate the enthalpy change for the hydroxide ion formation.
• This approach made it possible to find the enthalpy of formation for 94H of 9F for 9F by summing the contributions of each part.

Hess's Law simplifies the process of obtaining hard-to-measure enthalpy changes, making it incredibly valuable for chemists working with thermochemical data.

Hydroxide Ion

The hydroxide ion, 9E9F, is a fundamental species in chemistry, particularly in the study of acids and bases. It is an essential part of water's dissociation and plays a significant role in various chemical reactions.

• The hydroxide ion is a negatively charged ion, consisting of oxygen and hydrogen bonded together.
• It contributes to the basicity of a solution and is crucial in neutralization reactions, where acids react with bases to form water.
• Understanding the formation enthalpy of 9E9F aids in predicting the behavior and energy changes in reactions involving bases and acids.

As demonstrated in our exercise, knowing the enthalpy change associated with the formation of hydroxide ions allows us to accurately determine important thermodynamic properties of reactions, giving insight into their spontaneity and thermal profile.