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Chapter 6: Problem 66

Why is the lattice energy of a solid always a positive quantity? Why is the hydration of ions always a negative quantity?

Short Answer

Expert verified
The lattice energy of a solid is always a positive quantity because separating a solid into gaseous ions requires the input of energy, which is an endothermic process. The hydration of ions is a negative quantity because ions forming bonds with water molecules and getting hydrated releases energy, which is an exothermic process.

Step by step solution

01

Understanding Lattice Energy

Lattice energy refers to the energy required to break apart an ionic solid and convert all its atoms into gaseous ions. In other words, it's the energy needed to separate ions while they are in their crystal lattice form.
02

Why is Lattice Energy Positive?

As lattice energy is about breaking apart the ionic solid and converting it to gaseous ions, it requires the input of energy which is an endothermic process (takes in energy). And since on the energy scale, anything that needs energy is usually considered as positive, hence, the lattice energy of a solid is always a positive quantity.
03

Understanding Ion Hydration

Ionic hydration or ion hydration refers to the process of surrounding an ion with water molecules. This happens when ionic solids are put into water. As ions have a charge, they attract the opposite charge on the water molecule and end up being surrounded by water molecules.
04

Why is Hydration Energy Negative?

As water molecules surround the ion, it forms bonds. The formation of bonds is an exothermic process (releases energy). As releasing energy moves down on the energy scale and we traditionally regard it as a negative change, the hydration of ions is always a negative quantity.

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Most popular questions from this chapter

From the enthalpy of formation for \(\mathrm{CO}_{2}\) and the following information, calculate the standard enthalpy of formation for carbon monoxide (CO). $$ \begin{aligned} \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H^{\circ} &=-283.0 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$Why can't we obtain it directly by measuring the enthalpy of the following reaction?$$\mathrm{C}(\text { graphite })+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g)$$

Suggest ways (with appropriate equations) that would enable you to measure the \(\Delta H_{\mathrm{f}}^{\circ}\) values of \(\mathrm{Ag}_{2} \mathrm{O}(s)\) and \(\mathrm{CaCl}_{2}(s)\) from their elements. No calculations are necessary.

Describe two exothermic processes and two endothermic processes.

Calculate the standard enthalpy change for the reaction $$2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)$$ given that$$\begin{array}{l}2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-822.2 \mathrm{~kJ} / \mathrm{mol}\end{array}$$

Explain the meaning of this thermochemical equation: $$\begin{array}{r}4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\\\Delta H=-904 \mathrm{~kJ} / \mathrm{mol}\end{array}$$
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