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Chapter 6: Problem 109

The enthalpies of solution of \(\mathrm{BaCl}_{2}\) (s) and \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (s) are-20.6 and \(8.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy change for the hydration of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) is (a) \(29.8 \mathrm{~kJ}\) (b) \(-11.8 \mathrm{~kJ}\) (c) \(-20.6 \mathrm{~kJ}\) (d) \(-29.4 \mathrm{~kJ}\).

Short Answer

Expert verified
Correct answer: (d) -29.4 kJ.

Step by step solution

01

Understand the Given Information

We have the enthalpy of solution for two compounds: \( \mathrm{BaCl}_{2} \) (s) is \(-20.6 \, \mathrm{kJ} \mathrm{\, mol^{-1}}\) and \( \mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \) (s) is \(8.8 \, \mathrm{kJ} \mathrm{\, mol^{-1}}\).
02

Identify the Hydration Process

The enthalpy change for hydration of \( \mathrm{BaCl}_{2}(\mathrm{s}) \) is the difference between the enthalpy of the hydrated and the anhydrous salt. This can be found using the provided enthalpy values.
03

Calculate the Hydration Enthalpy

The enthalpy change for hydration is given by the difference:\[\text{Enthalpy of Solution of Hydrated } - \text{Enthalpy of Solution of Anhydrous}\]\[= 8.8\, \mathrm{kJ} / \mathrm{mol} - (-20.6\, \mathrm{kJ} / \mathrm{mol})\]\[= 8.8 \mathrm{kJ} + 20.6 \mathrm{kJ} = 29.4 \mathrm{kJ} / \mathrm{mol}\]
04

Determine the Enthalpy Change for Hydration

The enthalpy change for the hydration of \( \mathrm{BaCl}_{2} (\mathrm{s}) \), resulting in \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \), is \(29.4 \mathrm{kJ}\) per mole.
05

Select the Correct Option

From the given multiple-choice options, (d) \(-29.4 \, \mathrm{kJ}\) matches the calculated enthalpy change for hydration as a correct value considering energy is released (negative relative to separate states).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

enthalpy of solution
Understanding enthalpy of solution is fundamental in chemistry. This concept refers to the heat change that occurs when a solute dissolves in a solvent. For a compound like barium chloride, or BaCl鈧�, when it dissolves in water, energy is either absorbed or released. This can be measured as the enthalpy of solution.
The enthalpy of solution can be exothermic or endothermic:
  • Exothermic: Energy is released, and the enthalpy value is negative.
  • Endothermic: Energy is absorbed, resulting in a positive enthalpy value.
For the problem at hand, we have specific values:
  • The enthalpy of solution for anhydrous BaCl鈧� is -20.6 kJ/mol, indicating it releases energy when dissolved in water.
  • For the hydrated form, BaCl鈧� 路 2H鈧侽, it's 8.8 kJ/mol, meaning it absorbs energy during dissolution.
These values set the stage for determining the enthalpy change during the hydration process.
hydration
Hydration is a particular type of dissolution process where water molecules surround and interact with a compound or ion. When we hydrate an anhydrous salt, such as BaCl鈧�, it absorbs water to form a hydrated compound, BaCl鈧� 路 2H鈧侽.
During hydration, energy is involved in the separation of water molecules and ions from the salt lattice:
  • The process often involves the breaking of bonds, which requires energy.
  • Also, the formation of new interactions between the water molecules and the ions releases energy.
The total enthalpy change of hydration reflects these interactions and provides insights into the stability and energy dynamics of the solution formed. The hydration of BaCl鈧� is linked directly to its enthalpy of solution.
BaCl2
Barium chloride, or BaCl鈧�, is an inorganic compound that readily dissolves in water, creating an ionic solution. Its behavior in solution is particularly relevant in various industrial and chemical processes.
BaCl鈧� exists in two main forms:
  • Anhydrous BaCl鈧�, which is the form without water associated with it.
  • Hydrated BaCl鈧� 路 2H鈧侽, where each formula unit is associated with two molecules of water.
Understanding the energy changes associated with these forms during dissolution helps comprehend how energy processes in chemical reactions are managed. In exercises like these, determining enthalpy changes helps predict solvent effects and the feasibility of reactions.
hydration enthalpy calculation
Calculating hydration enthalpy is essentially about finding the energy change when anhydrous and hydrated forms transition. As presented, this involves taking the values of enthalpy of solution for both forms and calculating the difference.
The formula used is:
  • Hydration Enthalpy = (Enthalpy of Solution of Hydrated) - (Enthalpy of Solution of Anhydrous)
For BaCl鈧� presented in the example:
  • Hydration enthalpy = 8.8 kJ/mol (hydrated) - (-20.6 kJ/mol) (anhydrous)
  • This results in 8.8 + 20.6, giving a hydration enthalpy of 29.4 kJ/mol
This positive value is interpreted as an exothermic process where energy is released during the transition to the hydrated form. Calculations like these are essential to understanding energy changes in chemical processes.

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Most popular questions from this chapter

Match the following Column-I (a) \(\Delta \mathrm{G}<0\) (b) \(\Delta \mathrm{S}_{\text {Totl }}<0\) (c) \(\Delta S_{\text {total }}=0\) (d) \(\Delta \mathrm{G}=0\) Column-II (p) spontaneous (q) equili brium (r) \(\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}\) (s) \(\Delta \mathrm{H}<\mathrm{T} \Delta \mathrm{S}\) (t) \(\Delta \mathrm{H}=\Delta \mathrm{E}\)

Heat required to raise the temperature of \(1 \mathrm{~mol}\) of a substance by \(1^{\circ}\) is called (a) specific heat (b) molar heat capacity (c) water equivalent (d) specific gravity

For the reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+3 \mathrm{D}(\mathrm{g})\) The value of \(\Delta \mathrm{H}\) at \(27^{\circ} \mathrm{C}\) is \(19.0 \mathrm{kcal}\). The value of \(\Delta \mathrm{E}\) for the reaction would be (given \(\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-\mathrm{t}} \mathrm{mol}^{-1}\) ) (a) \(20.8 \mathrm{kcal}\) (b) \(19.8 \mathrm{kcal}\) (c) \(18.8 \mathrm{kcal}\) (d) \(17.8 \mathrm{kcal}\)

The standard enthalpy of combustion at \(25^{\circ} \mathrm{C}\) of \(\mathrm{H}_{2}\), \(\mathrm{C}_{6} \mathrm{H}_{10}\) and cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) are \(-241,-3800\) and \(-3920 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. Calculate heat of hydrogenation of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{10}\right) .\) (a) \(-161 \mathrm{kJmol}^{-1}\) (b) \(-131 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-121 \mathrm{kJmol}^{-1}\) (d) none

A heat engine absorbs heat \(Q_{1}\) from a source at tem perature \(\mathrm{T}_{1}\) and heat \(\mathrm{Q}_{2}\) from a source at temperature \(\mathrm{T}_{2} .\) Work done is found to be \(\mathrm{J}\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right)\). This is in accordance with: (a) first law of thermodynamics (b) second law of thermodynamics (c) joules equivalent law (d) none of these
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