Chapter 4, Problem 4.14E

Thermodynamic properties can also be determined for ions. Determine $\Delta H,\Delta S,\text{and}\Delta G$ for the following two reactions, which are simply reactions of dissolution:

$\begin{array}{c}{\text{NaHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}^{+}\left(\text{aq}\right)+{\text{HCO}}_3{}^{-}\left(\text{aq}\right)\\ {\text{Na2CO}}_{\text{3}}\left(\text{s}\right)\to 2{\text{Na}}^{+}\left(\text{aq}\right)+{\text{CO}}_3{}^{2-}\left(\text{aq}\right)\end{array}$

Assume standard conditions (standard concentration is $1\text{M}$ for ions in aqueous solution), and consult the table of thermodynamic properties in Appendix $2$. What similarities and differences are there?

Interpretation Introduction

Interpretation:

The values of $\Delta H,\Delta S,\text{and}\Delta G$ for the two reactions are to be calculated.

Concept introduction:

The standard Gibbs-energy gives the non-expansion work achieved from the system at a constant temperature and pressure for the reaction. The standard Gibbs-energy for the reaction is represented as $\Delta G°$.

Answer to Problem 4.14E

The values of $\Delta H,\Delta S,\text{and}\Delta G$ for the first reaction are $18.7\text{kJ}$, $48.6\text{J}/\text{K}$ and $2.27\text{kJ}$ respectively.

The values of $\Delta H,\Delta S,\text{and}\Delta G$ for the second reaction are $-25.77\text{kJ}$, $-73.69\text{J}/\text{K}$, and $-3.85\text{kJ}$ respectively. The values of $\Delta H,\Delta S,\text{and}\Delta G$ for the first reaction are positive; whereas, the values of $\Delta H,\Delta S,\text{and}\Delta G$ for the second reaction are negative.

Explanation of Solution

The first reaction is shown below.

${\text{NaHCO}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}^{+}\left(\text{aq}\right)+{\text{HCO}}_3{}^{-}\left(\text{aq}\right)$

The values of ${\Delta }_{\text{f}}H°,\Delta S°\text{and}{\Delta }_{\text{f}}G°$ for ${\text{NaHCO}}_{\text{3}}$ are $-950.81\text{kJ}/\text{mol}$, $101.7\text{J}/\text{mol}\cdot \text{K}$ and $-851.0\text{kJ}/\text{mol}$ respectively.

The values of ${\Delta }_{\text{f}}H°,\Delta S°\text{and}{\Delta }_{\text{f}}G°$ for ${\text{Na}}^{+}$ are $-240.12\text{kJ}/\text{mol}$, $59.1\text{J}/\text{mol}\cdot \text{K}$ and $-261.88\text{kJ}/\text{mol}$ respectively.

The values of ${\Delta }_{\text{f}}H°,\Delta S°\text{and}{\Delta }_{\text{f}}G°$ for ${\text{HCO}}_3{}^{-}$ are $-691.99\text{kJ}/\text{mol}$, $91.2\text{J}/\text{mol}\cdot \text{K}$ and $-586.85\text{kJ}/\text{mol}$ respectively.

The value of $\Delta H$ for the first reaction is calculated by the given formula.

$\Delta H={\Delta }_{\text{f}}H°\left({\text{Na}}^{+}\right)+{\Delta }_{\text{f}}H°\left({\text{HCO}}_3{}^{-}\right)-{\Delta }_{\text{f}}H°\left({\text{NaHCO}}_{\text{3}}\right)$ …(1)

Substitute the values of ${\Delta }_{\text{f}}H°$ for ${\text{NaHCO}}_{\text{3}}$, ${\text{Na}}^{+}$ and ${\text{HCO}}_3{}^{-}$ in equation (1).

$\begin{array}{rll}\Delta H& =& -240.12\text{kJ}/\text{mol}+\left(-691.99\text{kJ}/\text{mol}\right)-\left(-950.81\text{kJ}/\text{mol}\right)\\ & =& 18.7\text{kJ}\end{array}$

The value of $\Delta G$ for the first reaction is calculated by the given formula.

$\Delta G={\Delta }_{\text{f}}G°\left({\text{Na}}^{+}\right)+{\Delta }_{\text{f}}G°\left({\text{HCO}}_3{}^{-}\right)-{\Delta }_{\text{f}}G°\left({\text{NaHCO}}_{\text{3}}\right)$ …(2)

Substitute the values of ${\Delta }_{\text{f}}G°$ for ${\text{NaHCO}}_{\text{3}}$, ${\text{Na}}^{+}$ and ${\text{HCO}}_3{}^{-}$ in equation (2).

$\begin{array}{rll}\Delta G& =& -261.88\text{kJ}/\text{mol}+\left(-586.85\text{kJ}/\text{mol}\right)-\left(-851.0\text{kJ}/\text{mol}\right)\\ & =& 2.27\text{kJ}\end{array}$

The value of $\Delta S$ for the first reaction is calculated by the given formula.

$\Delta S=\Delta S°\left({\text{Na}}^{+}\right)+\Delta S°\left({\text{HCO}}_3{}^{-}\right)-\Delta S°\left({\text{NaHCO}}_{\text{3}}\right)$ …(3)

Substitute the values of $\Delta S°$ for ${\text{NaHCO}}_{\text{3}}$, ${\text{Na}}^{+}$ and ${\text{HCO}}_3{}^{-}$ in equation (3).

$\begin{array}{rll}\Delta S& =& 59.1\text{J}/\text{mol}\cdot \text{K}+91.2\text{J}/\text{mol}\cdot \text{K}-101.7\text{J}/\text{mol}\cdot \text{K}\\ & =& 48.6\text{J}/\text{K}\end{array}$

The second reaction is shown below.

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{s}\right)\to 2{\text{Na}}^{+}\left(\text{aq}\right)+{\text{CO}}_3{}^{2-}\left(\text{aq}\right)$

The values of ${\Delta }_{\text{f}}H°,\Delta S°\text{and}{\Delta }_{\text{f}}G°$ for ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ are $-1130.77\text{kJ}/\text{mol}$, $138.79\text{J}/\text{mol}\cdot \text{K}$ and $-1048.01\text{kJ}/\text{mol}$ respectively.

The values of ${\Delta }_{\text{f}}H°,\Delta S°\text{and}{\Delta }_{\text{f}}G°$ for ${\text{Na}}^{+}$ for $-240.12\text{kJ}/\text{mol}$, $59.1\text{J}/\text{mol}\cdot \text{K}$, and $-261.88\text{kJ}/\text{mol}$ respectively.

The values of ${\Delta }_{\text{f}}H°,\Delta S°\text{and}{\Delta }_{\text{f}}G°$ for ${\text{CO}}_3{}^{2-}$ are $-676.3\text{kJ}/\text{mol}$, $-53.1\text{J}/\text{mol}\cdot \text{K}$ and $-528.1\text{kJ}/\text{mol}$, respectively.

The value of $\Delta H$ for the second reaction is calculated by the given formula.

$\Delta H=2{\Delta }_{\text{f}}H°\left({\text{Na}}^{+}\right)+{\Delta }_{\text{f}}H°\left({\text{CO}}_3{}^{2-}\right)-{\Delta }_{\text{f}}H°\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ …(4)

Substitute the value of ${\Delta }_{\text{f}}H°$ for ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$, ${\text{Na}}^{+}$ and ${\text{CO}}_3{}^{2-}$ in equation (4).

$\begin{array}{rll}\Delta H& =& 2\times -240.12\text{kJ}/\text{mol}+\left(-676.3\text{kJ}/\text{mol}\right)-\left(-1130.77\text{kJ}/\text{mol}\right)\\ & =& -25.77\text{kJ}\end{array}$

The value of $\Delta G$ for the second reaction is calculated by the given formula.

$\Delta G=2{\Delta }_{\text{f}}G°\left({\text{Na}}^{+}\right)+{\Delta }_{\text{f}}G°\left({\text{CO}}_3{}^{2-}\right)-{\Delta }_{\text{f}}G°\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ …(5)

Substitute the value of ${\Delta }_{\text{f}}G°$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$, ${\text{Na}}^{+}$ and ${\text{CO}}_3{}^{2-}$ in equation (5).

$\begin{array}{rll}\Delta G& =& 2\times -261.88\text{kJ}/\text{mol}+\left(-528.1\text{kJ}/\text{mol}\right)-\left(-1048.01\text{kJ}/\text{mol}\right)\\ & =& -3.85\text{kJ}\end{array}$

The value of $\Delta S$ for the second reaction is calculated by the ggiven formula.

$\Delta S=2\Delta S°\left({\text{Na}}^{+}\right)+\Delta S°\left({\text{CO}}_3{}^{2-}\right)-\Delta S°\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)$ …(6)

Substitute the value of $\Delta S°$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$, ${\text{Na}}^{+}$ and ${\text{CO}}_3{}^{2-}$ in equation (6).

$\begin{array}{rll}\Delta S& =& 2\times 59.1\text{J}/\text{mol}\cdot \text{K}+\left(-53.1\text{J}/\text{mol}\cdot \text{K}\right)-\left(138.79\text{J}/\text{mol}\cdot \text{K}\right)\\ & =& -73.69\text{J}/\text{K}\end{array}$

The values of $\Delta H,\Delta S,\text{and}\Delta G$ for the first reaction are positive; whereas, the values of $\Delta H,\Delta S,\text{and}\Delta G$ for the second reaction are negative.

Conclusion

The values of $\Delta H,\Delta S,\text{and}\Delta G$ for the first reaction are $18.7\text{kJ}$, $48.6\text{J}/\text{K}$ and $2.27\text{kJ}$.

The values of $\Delta H,\Delta S,\text{and}\Delta G$ for the second reaction are $-25.77\text{kJ}$, $-73.69\text{J}/\text{K}$ and $-3.85\text{kJ}$. The values of $\Delta H,\Delta S,\text{and}\Delta G$ of the first reaction are positive; whereas, the values of $\Delta H,\Delta S,\text{and}\Delta G$ for the second reaction are negative.