Chapter 8, Problem 66E

Use the following data (in kJ/mol) to estimate ∆E for the reaction S⁻(g)+e⁻ → S²⁻(g). Include an estimate of uncertainity. ∆E_sub is the energy of sublimation.

	Lattice Energy	I.E.of M	∆Esub of M
Na2S	–2203	495	109
K2S	–2052	419	90
Rb2S	–1949	409	82
Cs2S	–1850	382	78

$\begin{array}{llllll}\text{2Na}\left(s\right)\hfill & \text{+ S}\left(\text{s}\right)\hfill & \to {\text{Na}}_{\text{2}}\text{S}\left(s\right)\hfill & \text{Δ}E\text{=}\hfill & -\text{365}\text{kJ/mol}\hfill & \hfill \\ \text{2K}\left(s\right)\hfill & \text{+ S}\left(\text{s}\right)\hfill & \to {\text{K}}_{\text{2}}\text{S}\left(s\right)\hfill & \text{Δ}E\text{=}\hfill & -\text{381}\text{kJ/mol}\hfill & \hfill \\ \text{2Rb}\left(s\right)\hfill & \text{+ S}\left(\text{s}\right)\hfill & \to {\text{Rb}}_{\text{2}}\text{S}\left(s\right)\hfill & \text{Δ}E\text{=}\hfill & -\text{361}\text{kJ/mol}\hfill & \hfill \\ \text{2Cs}\left(s\right)\hfill & \text{+ S}\left(\text{s}\right)\hfill & \to {\text{Cs}}_{\text{2}}\text{S}\left(s\right)\hfill & \text{Δ}E\text{=}\hfill & -\text{360}\text{kJ/mol}\hfill & \hfill \\ \hfill & \text{S}\left(\text{s}\right)\hfill & \to \text{S}\left(g\right)\hfill & \text{Δ}E\text{=}\hfill & \text{277}\text{kJ/mol}\hfill & \hfill \\ \text{S}\left(\text{g}\right)\hfill & {\text{+e}}^{\text{-}}\hfill & \to {\text{S}}^{-}\left(g\right)\hfill & \text{Δ}E\text{=}\hfill & -\text{200}\text{kJ/mol}\hfill & \hfill \end{array}$

Assume that all values are known to ±1 kJ/mol.

Interpretation Introduction

Interpretation: The change in energy $\Delta E$ is to be calculated from the given data of the reaction. Also uncertainty in energy of sublimation is to be included.

Concept introduction: The change in energy of a reaction is defined as the sum of change in the internal energy of a system and the product of its absolute temperature and entropy.

To determine: The change in energy for the given reaction.

Answer to Problem 66E

Answer

The value of change in energy is $\underset{\_}{538kJ\pm 50k}\text{J}$.

Explanation of Solution

The given chemical reaction is

${\text{S}}^{-}\left(g\right)+{\text{e}}^{-}\to {\text{S}}^{2-}\left(g\right)$

• First step involved in the given reaction is sublimation of sodium $\text{Na}\left(s\right)$. The sublimation is the process in which change of phase occur from solid to gas. The energy change occur during this process is called sublimation energy. The reaction is,

$\text{2Na}\left(s\right)\to 2\text{Na}\left(g\right)$

The energy of sublimation of sodium is,

$\begin{array}{c}\Delta H_{\text{sub}}=2\times 109\text{kJ}\\ =218\text{kJ}\end{array}$

The energy of sublimation of ${\text{K}}_{\text{2}}\text{S}$ is,

$\begin{array}{c}\Delta H_{\text{sub}}=2\times 90\text{kJ}\\ =180\text{kJ}\end{array}$

The energy of sublimation of ${\text{Rb}}_{\text{2}}\text{S}$ is,

$\begin{array}{c}\Delta H_{\text{sub}}=2\times 82\text{kJ}\\ =164\text{kJ}\end{array}$

The energy of sublimation of ${\text{Cs}}_{\text{2}}\text{S}$ is,

$\begin{array}{c}\Delta H_{\text{sub}}=2\times 78\text{kJ}\\ =156\text{kJ}\end{array}$

• Second step involved is the ionization of sodium $\left(\text{Na}\right)$ in gas phase. It is the energy required to remove an electron from a neutral atom or ion. The reaction is,

$\text{2Na}\left(g\right)\to 2{\text{Na}}^{+}\left(g\right)+2{\text{e}}^{-}$

The energy of ionization for sodium is,

$\begin{array}{c}2I.E=2\times 495\text{kJ}\\ =990\text{kJ}\end{array}$

The energy of ionization of ${\text{K}}_{\text{2}}\text{S}$ is,

$\begin{array}{c}2I.E=2\times 419\text{kJ}\\ =838\text{kJ}\end{array}$

The energy of ionization of ${\text{Rb}}_{\text{2}}\text{S}$ is,

$\begin{array}{c}2I.E=2\times 409\text{kJ}\\ =818\text{kJ}\end{array}$

The energy of ionization of ${\text{Cs}}_{\text{2}}\text{S}$ is,

$\begin{array}{c}2I.E=2\times 382\text{kJ}\\ =764\text{kJ}\end{array}$

• Third step involved in the formation of sodium ion is sublimation of solid sulfur. The reaction is,

$\text{S}\left(s\right)\to \text{S}\left(g\right)$

The enthalpy of sublimation of sodium, ${\text{K}}_{\text{2}}\text{S}$ , ${\text{Rb}}_{\text{2}}\text{S}$ and ${\text{Cs}}_{\text{2}}\text{S}$ reaction is,

$\Delta H^{\text{'}}{}_{\text{sub}}=277\text{kJ}$

Fourth step involved the formation of sulfur ion in gas phase. The reaction is,

$\text{S}\left(g\right)+{\text{e}}^{-}\to {\text{S}}^{-}\left(g\right)$

The electron affinity of sodium, ${\text{K}}_{\text{2}}\text{S}$ , ${\text{Rb}}_{\text{2}}\text{S}$ and ${\text{Cs}}_{\text{2}}\text{S}$ reaction is,

$E{A}_{\text{1}}=-\text{200}\text{kJ}$

• Fifth step involved the formation of ${\text{S}}^{\text{2}-}$ ion. The reaction is,

${\text{S}}^{-}\left(g\right)+{\text{e}}^{-}\to {\text{S}}^{\text{2}-}\left(g\right)$

The electron affinity of this reaction is,

$E{A}_{\text{2}}=x$

Where,

• $x$ is unknown value.

• Sixth step involved the formation of solid sodium sulfide from gaseous phase. The reaction is,

${\text{2Na}}^{+}\left(g\right)+{\text{S}}^{\text{2}-}\left(g\right)\to {\text{Na}}_{\text{2}}\text{S}\left(s\right)$

The lattice energy for sodium reaction is,

$L.E=-\text{2203}\text{kJ}$

The lattice energy for ${\text{K}}_{\text{2}}\text{S}$ is,

$L.E=-\text{2052}\text{kJ}$

The lattice energy for ${\text{Rb}}_{\text{2}}\text{S}$ is,

$L.E=-\text{1949}\text{kJ}$

The lattice energy for ${\text{Cs}}_{\text{2}}\text{S}$ is,

$L.E=-\text{1850}\text{kJ}$

The overall reaction is obtained by adding the above processes,

$\text{2Na}\left(g\right)+{\text{S}}^{\text{2}-}\to {\text{Na}}_{\text{2}}\text{S}\left(s\right)$

The enthalpy of formation for sodium reaction is,

$\Delta H_{\text{f}}^{\text{o}}=-\text{365}\text{kJ}$

The enthalpy of formation for ${\text{K}}_{\text{2}}\text{S}$ reaction is,

$\Delta H_{\text{f}}^{\text{o}}=-\text{381}\text{kJ}$

The enthalpy of formation for ${\text{Rb}}_{\text{2}}\text{S}$ reaction is,

$\Delta H_{\text{f}}^{\text{o}}=-\text{361}\text{kJ}$

The enthalpy of formation for ${\text{Cs}}_{\text{2}}\text{S}$ reaction is,

$\Delta H_{\text{f}}^{\text{o}}=-\text{360}\text{kJ}$

Formula

The formula of enthalpy of formation is,

$\Delta H_{\text{f}}^{\text{o}}=\Delta H_{\text{sub}}+E.A_1+E.A_2+2I.E+\Delta H^{\text{'}}{}_{\text{sub}}$ (1)

Substitute the values of $\Delta H_{\text{f}}^{\text{o}},\Delta H_{\text{sub}},E.A_1,2I.E$ and $\Delta H^{\text{'}}{}_{\text{sub}}$ for the reaction of ${\text{Na}}_{\text{2}}\text{S}$ in above expression.

$\begin{array}{c}\Delta H_{\text{f}}^{\text{o}}=\Delta H_{\text{sub}}+E.A_1+E.A_2+2I.E+\Delta H^{\text{'}}{}_{\text{sub}}\\ -365\text{kJ}=218\text{kJ}+990\text{kJ}-2203\text{kJ}-200\text{kJ}+E.A_2+277\text{kJ}\\ E.A_2=553\text{kJ}\end{array}$

Substitute the values of $\Delta H_{\text{f}}^{\text{o}},\Delta H_{\text{sub}},E.A_1,2I.E$ and $\Delta H^{\text{'}}{}_{\text{sub}}$  for the reaction of ${\text{Rb}}_{\text{2}}\text{S}$ in equation (1).

$\begin{array}{c}\Delta H_{\text{f}}^{\text{o}}=\Delta H_{\text{sub}}+E.A_1+E.A_2+2I.E+\Delta H^{\text{'}}{}_{\text{sub}}\\ -361\text{kJ}=164\text{kJ}+818\text{kJ}+277\text{kJ}-200\text{kJ}+E.A_2-1949\text{kJ}\\ E.A_2=529\text{kJ}\end{array}$

Substitute the values of $\Delta H_{\text{f}}^{\text{o}},\Delta H_{\text{sub}},E.A_1,2I.E$ and $\Delta H^{\text{'}}{}_{\text{sub}}$  for the reaction of ${\text{K}}_{\text{2}}\text{S}$ in equation (1).

$\begin{array}{c}\Delta H_{\text{f}}^{\text{o}}=\Delta H_{\text{sub}}+E.A_1+E.A_2+2I.E+\Delta H^{\text{'}}{}_{\text{sub}}\\ -381\text{kJ}=180\text{kJ}+838\text{kJ}+277\text{kJ}-200\text{kJ}+E.A_2-2052\text{kJ}\\ E.A_2=576\text{kJ}\end{array}$

Substitute the values of $\Delta H_{\text{f}}^{\text{o}},\Delta H_{\text{sub}},E.A_1,2I.E$ and $\Delta H^{\text{'}}{}_{\text{sub}}$  for the reaction of ${\text{Cs}}_{\text{2}}\text{S}$ in equation (1).

$\begin{array}{c}\Delta H_{\text{f}}^{\text{o}}=\Delta H_{\text{sub}}+E.A_1+E.A_2+2I.E+\Delta H^{\text{'}}{}_{\text{sub}}\\ -360\text{kJ}=2\left(78\right)\text{kJ}+2\left(382\right)\text{kJ}+277\text{kJ}-200\text{kJ}+E.A_2-1850\text{kJ}\\ E.A_2=493\text{kJ}\end{array}$

The average value of electron affinity is calculated using the formula,

$\overline{E.A_2}=\frac{\sum E.A_2}{n}$

Where,

• $\overline{E.A_2}$ is the average of electron affinity.
• $n$ is number of compounds $\left(\text{4}\right)$.
• $E.A_2$ is electron affinity of each compound

Substitute the values of $E.A_2$ and $n$ in the above equation.

$\begin{array}{c}\overline{E.A_2}=\frac{\sum E.A_2}{n}\\ =\frac{\left(553+576+529+493\right)\text{kJ}}{4}\\ =538\text{kJ}\end{array}$

The uncertainty in energy of sublimation is $\text{50}\text{kJ}$. Hence, the change in enthalpy including uncertainty in sublimation is $\underset{\_}{538kJ\pm 50k}\text{J}$.

Conclusion

The change in energy is equal to sum of change in energy in reactant and change in product.  The steps involved in calculation of energy change are sublimation, ionization, dissociation and formation of compound.