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Given the following information: Heat of sublimation of \(\operatorname{Li}(s)=166 \mathrm{~kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{HCl}=427 \mathrm{~kJ} / \mathrm{mol}\) Ionization energy of \(\operatorname{Li}(g)=520 . \mathrm{kJ} / \mathrm{mol}\) Electron affinity of \(\mathrm{Cl}(g)=-349 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy of \(\operatorname{LiCl}(s)=-829 \mathrm{~kJ} / \mathrm{mol}\) Bond energy of \(\mathrm{H}_{2}=432 \mathrm{~kJ} / \mathrm{mol}\) Calculate the net change in energy for the following reaction: $$ 2 \mathrm{Li}(s)+2 \mathrm{HCl}(g) \longrightarrow 2 \mathrm{LiCl}(s)+\mathrm{H}_{2}(g) $$

Short Answer

Expert verified
The net change in energy for the given reaction is \(3276\text{ kJ}\).

Step by step solution

01

Identify the energy values involved in the reaction

We will first list all the values provided in the exercise for ease of reference. - Heat of sublimation of Li(s): 166 kJ/mol - Bond energy of HCl: 427 kJ/mol - Ionization energy of Li(g): 520 kJ/mol - Electron affinity of Cl(g): -349 kJ/mol - Lattice energy of LiCl(s): -829 kJ/mol - Bond energy of H₂: 432 kJ/mol
02

Determine energy changes for bond breaking

In the given reaction, bonds in solid lithium and HCl need to be broken. Let's calculate the energy required to break these bonds: Energy to break 2 Li(s) bonds: \(2 \times 166\text{ kJ/mol} = 332\text{ kJ}\) Energy to break 2 HCl bonds: \(2 \times 427\text{ kJ/mol} = 854\text{ kJ}\) Total energy required for bond breaking: \(332\text{ kJ} + 854\text{ kJ} = 1186\text{ kJ}\)
03

Determine energy changes for bond formation

In the given reaction, new bonds are formed in LiCl and H₂. Let's calculate the energy released during the formation of these bonds: Energy released in the formation of 2 LiCl bonds: \(2 \times (- 829\text{ kJ/mol}) = - 1658\text{ kJ}\) Energy released in the formation of 1 H₂ bond: \(-432\text{ kJ/mol}\) Total energy released for bond formation: \(-1658\text{ kJ} - 432\text{ kJ} = -2090\text{ kJ}\)
04

Calculate the net change in energy for the reaction

Now, we will calculate the net change in energy by subtracting the energy released for bond formation from the energy required for bond breaking: Net change in energy = Energy required for bond breaking - Energy released for bond formation \(= 1186\text{ kJ} - (- 2090\text{ kJ}) = 1186\text{ kJ} + 2090\text{ kJ} = 3276\text{ kJ}\) So, the net change in energy for the given reaction is \(3276\text{ kJ}\).

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Most popular questions from this chapter

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for potassium chloride. $$ \mathrm{K}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s) $$ $$ \begin{array}{lr} \text { Lattice energy } & -690 . \mathrm{kJ} / \mathrm{mol} \\ \text { Ionization energy for } \mathrm{K} & 419 \mathrm{~kJ} / \mathrm{mol} \\\ \text { Electron affinity of Cl } & -349 \mathrm{~kJ} / \mathrm{mol} \\ \text { Bond energy of } \mathrm{Cl}_{2} & 239 \mathrm{~kJ} / \mathrm{mol} \\ \text { Enthalpy of sublimation for } \mathrm{K} & 64 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Rank the following bonds in order of increasing ionic character: \(\mathrm{N}-\mathrm{O}, \mathrm{Ca}-\mathrm{O}, \mathrm{C}-\mathrm{F}, \mathrm{Br}-\mathrm{Br}, \mathrm{K}-\mathrm{F}\).

An alternative definition of electronegativity is Electronegativity \(=\) constant (I.E. \(-\) E.A.) where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 7 to calculate the (I.E. \(-\) E.A.) term for \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\), and \(\mathrm{I}\). Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are \(1678,1255,1138\), and \(1007 \mathrm{~kJ} / \mathrm{mol}\), respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals 4.0. Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

Write electron configurations for the most stable ion formed by each of the elements Te, \(\mathrm{Cl}, \mathrm{Sr}\), and \(\mathrm{Li}\) (when in stable ionic compounds).

Write Lewis structures and predict the molecular structures of the following. (See Exercises 105 and 106 .) a. \(\mathrm{OCl}_{2}, \mathrm{KrF}_{2}, \mathrm{BeH}_{2}, \mathrm{SO}_{2}\) b. \(\mathrm{SO}_{3}, \mathrm{NF}_{3}, \mathrm{IF}_{3}\) c. \(\mathrm{CF}_{4}, \mathrm{SeF}_{4}, \mathrm{KrF}_{4}\) d. \(\mathrm{IF}_{5}, \mathrm{AsF}_{5}\) Which of these compounds are polar?

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