Calculate \(\Delta_{\mathrm{f}} H^{\circ}\) for aqueous chloride ion from the following data: \(\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}), \quad \Delta_{\mathrm{f}} H^{\mathrm{o}}\) \(=-92.4 \mathrm{~kJ}\) \(\mathrm{HCl}(\mathrm{g})+n \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) \(\Delta H^{\circ}=-74.8 \mathrm{~kJ}\) \(\Delta_{\mathrm{f}} H^{\circ}\left(\mathrm{H}^{+}\right.\), aq. \()=0.0 \mathrm{~kJ}\) (a) \(0.0\) (b) \(+83.6 \mathrm{~kJ}\) (c) \(+167.2 \mathrm{~kJ}\) (d) \(-167.2 \mathrm{~kJ}\)

Short Answer

Expert verified
\(\Delta_{\mathrm{f}} H^{\circ} \text{ for } \mathrm{Cl}^{-}(\mathrm{aq}) = -92.4 \mathrm{kJ} + (-74.8 \mathrm{kJ}) = -167.2 \mathrm{kJ}\)

Step by step solution

01

Write the Hess's Law Statement

According to Hess's Law, if a reaction is the sum of two or more other reactions, the heat change for the total reaction is the sum of the heat changes for the individual reactions.
02

Write the Formation Reaction for Chloride Ion

Write the thermochemical equation for the formation of a chloride ion in water. It must be the combination of the given reactions which includes the formation of gaseous HCl and its dissolution in water to create hydrochloric acid.
03

Express the Formation Reaction

Combine the two provided reactions to form the reaction: \(\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) + n H_2O(l) \rightarrow H^+(aq) + Cl^-(aq)\).
04

Calculate the Standard Enthalpy of Formation for Chloride Ion

Sum the enthalpies of the given reactions, keeping in mind that the standard enthalpy of formation for the hydrogen ion in aqueous solution is zero, to find the standard enthalpy of formation for the chloride ion.

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Key Concepts

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

Understanding Hess's Law
In chemistry, Hess's Law is a principle that relates to the total enthalpy change in a chemical reaction. It states that the total enthalpy change, regardless of the number of steps or intermediate stages of the reaction, is the same as if the reaction were to happen in a single step. This is because enthalpy is a state function, which means its value is determined only by the current state of the system and not by the path the system took to reach that state.

Now, let's take a real-world analogy to simplify this concept: imagine building a house. The total cost to build it will be the same whether you pay for it in one lump sum or in smaller payments over time. Similarly, when it comes to chemical reactions, the total heat change (like the total cost of the house) doesn't care about the

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Most popular questions from this chapter

The word standard in molar enthalpy change implies (a) temperature \(298 \mathrm{~K}\) and pressure 1 atm (b) any temperature and pressure 1 atm (c) any temperature and pressure 1 bar (d) any temperature and pressure

The factor of \(\Delta G\) values is important in metallurgy. The \(\Delta G\) values for the following reactions at \(800^{\circ} \mathrm{C}\) are given as: \(\mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) ; \Delta G=-544 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \rightarrow 2 \mathrm{ZnS}(\mathrm{s}) ; \Delta G=-293 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s}) ; \Delta G=-480 \mathrm{~kJ}\) The \(\Delta G\) for the reaction: \(2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g})\) \(\rightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})\) will be (a) \(-357 \mathrm{~kJ}\) (b) \(-731 \mathrm{~kJ}\) (c) \(-773 \mathrm{~kJ}\) (d) \(-229 \mathrm{~kJ}\)

Study the following thermodynamic data given by E. H. P. Cordfunke, A. S. Booji and M. Y. Furkalionk. (i) \(\mathrm{DyCl}_{3}(\mathrm{~s}) \rightarrow \mathrm{DyCl}_{3}(\) aq., in \(4.0 \mathrm{M}-\mathrm{HCl}) ;\) \(\Delta H^{0}=-180.06 \mathrm{kJmol}^{-1}\) (ii) \(\mathrm{Dy}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}, 4.0 \mathrm{M}) \rightarrow \mathrm{DyCl}_{3}\) \((\mathrm{aq}\), in \(4.0 \mathrm{M}-\mathrm{HCl})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})\) \(\Delta H^{\circ}=-699.43 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (iii) \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{HCl}_{\text { }}(\mathrm{aq},\), \(4.0 \mathrm{M}) ; \Delta H^{\circ}=-158.31 \mathrm{~kJ} \mathrm{~mol}^{-1}\) What is \(\Delta H_{\mathrm{f}}^{0}\) of \(\mathrm{DyCl}_{3}(\mathrm{~s})\) from these data? (a) \(-248.58 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-994.30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-3977.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-1469.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The standard heat of combustion of propane is \(-2220.1 \mathrm{~kJ} / \mathrm{mol}\). The standard heat of vaporization of liquid water is \(44 \mathrm{~kJ} / \mathrm{mol}\). What is the \(\Delta H^{\text {o }}\) of the reaction: \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ?\) (a) \(-2220.1 \mathrm{~kJ}\) (b) \(-2044.1 \mathrm{~kJ}\) (c) \(-2396.1 \mathrm{~kJ}\) (d) \(-2176.1 \mathrm{~kJ}\)

The lattice energy of solid \(\mathrm{NaCl}\) is 180 kcal/mol. The dissolution of the solid in water, in the form of ions is endothermic to the extent of \(1 \mathrm{kcal} / \mathrm{mol}\). If the solvation energies of \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions are in the ratio \(6: 5\), what is the enthalpy of hydration of sodium ion? (a) \(-85.6 \mathrm{kcal} / \mathrm{mol}\) (b) \(-97.6 \mathrm{kcal} / \mathrm{mol}\) (c) \(82.6 \mathrm{kcal} / \mathrm{mol}\) (d) \(100 \mathrm{kcal} / \mathrm{mol}\)

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