Chapter 6, Problem 6A.4P

Interpretation Introduction

Interpretation:

Thermodynamically, the most stable solid at $190\text{K}$ has to be predicted.

Concept introduction:

The Gibbs free energy of the reaction is the key factor of the identification of the direction of spontaneity of the reaction.  For the reaction to be spontaneous, the Gibbs free energy of the reaction must be negative.

Answer to Problem 6A.4P

The most stable solid at $190\text{K}$ is ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$.

Explanation of Solution

For the formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$:

The formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is shown by the following chemical reaction.

    ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{s}\right){\Delta }_{\text{r}}{G}^{\Theta }=-23.6\text{kJ}{\text{mol}}^{\text{-1}}$

The equilibrium constant for the formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is calculated by the following formula.

    $K_1=\frac{p_{{\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)}}{p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}}$                                                                                                 (1)

Where,

• $p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}$ is the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ at $190\text{K}$.
• $p_{{\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)}$ is the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$.

The partial pressure of a substance in the solid state is $1\text{bar}$.  Therefore, the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$ is $1\text{bar}$.

It is given that the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $0.13\text{μbar}$.

The conversion of $\text{μbar}$ to $\text{bar}$ is done as,

    $1\text{μbar}=1\times {10}^{-6}\text{bar}$

Therefore, the conversion of $\text{0}\text{.13}\text{μbar}$ to $\text{bar}$ is done as,

    $\begin{array}{c}0.13\text{μbar}=0.13\times {10}^{-6}\text{bar}\\ =\text{1}\text{.3}\times {10}^{-7}\text{bar}\end{array}$

Therefore, the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $\text{1}\text{.3}\times {10}^{-7}\text{bar}$.

Substitute the value of the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$ in equation (1).

    $\begin{array}{c}{K}_1=\frac{1}{\text{1}\text{.3}\times {10}^{-7}}\\ =7.69\times {10}^6\end{array}$

Therefore, the equilibrium constant for the formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $7.69\times {10}^6$.

For the formation of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$:

The formation of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is shown by the following chemical reaction.

    ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)+{\text{HNO}}_{\text{3}}\left(\text{g}\right)\to {\text{HNO}}_{\text{3}}{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right){\Delta }_{\text{r}}{G}^{\Theta }=-57.2\text{kJ}{\text{mol}}^{\text{-1}}$

The equilibrium constant for the formation of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is calculated by the following formula.

    $K_2=\frac{p_{{\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)}}{p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}{p}_{{\text{HNO}}_3\left(g\right)}}$                                                                                      (2)

Where,

• $p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}$ is the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ at $190\text{K}$.
• $p_{{\text{HNO}}_3\left(g\right)}$ is the partial pressure of ${\text{HNO}}_3\left(\text{g}\right)$ at $190\text{K}$.
• $p_{{\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)}$ is the partial pressure of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$.

The partial pressure of a substance in the solid state is $1\text{bar}$.  Therefore, the partial pressure of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$ is $1\text{bar}$.

It is given that the partial pressure of ${\text{HNO}}_3\left(\text{g}\right)$ is $0.41\text{nbar}$.

The conversion of $\text{nbar}$ to $\text{bar}$ is done as,

    $1\text{nbar}=1\times {10}^{-9}\text{bar}$

Therefore, the conversion of $\text{0}\text{.41}\text{nbar}$ to $\text{bar}$ is done as,

    $\begin{array}{c}0.41\text{nbar}=0.41\times {10}^{-9}\text{bar}\\ =\text{4}\text{.1}\times {10}^{-10}\text{bar}\end{array}$

Therefore, the partial pressure of ${\text{HNO}}_3\left(\text{g}\right)$ is $\text{4}\text{.1}\times {10}^{-10}\text{bar}$.

The partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $\text{1}\text{.3}\times {10}^{-7}\text{bar}$.

Substitute the value of the partial pressure of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$, ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ and ${\text{HNO}}_3\left(\text{g}\right)$ in equation (2).

    $\begin{array}{c}{K}_2=\frac{1}{\left(\text{1}\text{.3}\times {10}^{-7}\right)\left(\text{4}\text{.1}\times {10}^{-10}\text{bar}\right)}\\ =\frac{1}{5.33\times {10}^{-17}}\\ =1.87\times {10}^{16}\end{array}$

Therefore, the equilibrium constant for the formation of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $1.87\times {10}^{16}$.

For the formation of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$:

The formation of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is shown by the following chemical reaction.

    ${\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)+{\text{HNO}}_{\text{3}}\left(\text{g}\right)\to {\text{HNO}}_{\text{3}}{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right){\Delta }_{\text{r}}{G}^{\Theta }=-85.6\text{kJ}{\text{mol}}^{\text{-1}}$

The equilibrium constant for the formation of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is calculated by the following formula.

    $K_3=\frac{p_{{\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)}}{{\left(p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}\right)}^2{p}_{{\text{HNO}}_3\left(g\right)}}$                                                                                 (3)

Where,

• $p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}$ is the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ at $190\text{K}$.
• $p_{{\text{HNO}}_3\left(g\right)}$ is the partial pressure of ${\text{HNO}}_3\left(\text{g}\right)$ at $190\text{K}$.
• $p_{{\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)}$ is the partial pressure of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$.

The partial pressure of a substance in the solid state is $1\text{bar}$.  Therefore, the partial pressure of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$ is $1\text{bar}$.

The partial pressure of ${\text{HNO}}_3\left(\text{g}\right)$ is $\text{4}\text{.1}\times {10}^{-10}\text{bar}$.

The partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $\text{1}\text{.3}\times {10}^{-7}\text{bar}$.

Substitute the value of the partial pressure of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$, ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ and ${\text{HNO}}_3\left(\text{g}\right)$ in equation (3).

    $\begin{array}{c}{K}_3=\frac{1}{{\left(\text{1}\text{.3}\times {10}^{-7}\right)}^2\left(\text{4}\text{.1}\times {10}^{-10}\right)}\\ =\frac{1}{1.69\times {10}^{-14}\times \text{4}\text{.1}\times {10}^{-10}}\\ =\frac{1}{6.92\times {10}^{-24}}\\ =1.44\times {10}^{23}\end{array}$

Therefore, the equilibrium constant for the formation of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $1.44\times {10}^{23}$.

For the formation of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$:

The formation of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is shown by the following chemical reaction.

    ${\text{3H}}_{\text{2}}\text{O}\left(\text{g}\right)+{\text{HNO}}_{\text{3}}\left(\text{g}\right)\to {\text{HNO}}_{\text{3}}{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right){\Delta }_{\text{r}}{G}^{\Theta }=-112.8\text{kJ}{\text{mol}}^{\text{-1}}$

The equilibrium constant for the formation of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is calculated by the following formula.

    $K_4=\frac{p_{{\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)}}{{\left(p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}\right)}^3{p}_{{\text{HNO}}_3\left(g\right)}}$                                                                                 (4)

Where,

• $p_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}$ is the partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ at $190\text{K}$.
• $p_{{\text{HNO}}_3\left(g\right)}$ is the partial pressure of ${\text{HNO}}_3\left(\text{g}\right)$ at $190\text{K}$.
• $p_{{\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)}$ is the partial pressure of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$.

The partial pressure of a substance in the solid state is $1\text{bar}$.  Therefore, the partial pressure of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ is $1\text{bar}$.

The partial pressure of ${\text{HNO}}_3\left(\text{g}\right)$ is $\text{4}\text{.1}\times {10}^{-10}\text{bar}$.

The partial pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ is $\text{1}\text{.3}\times {10}^{-7}\text{bar}$.

Substitute the value of the partial pressure of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$, ${\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$ and ${\text{HNO}}_3\left(\text{g}\right)$ in equation (4).

    $\begin{array}{c}{K}_4=\frac{1}{{\left(\text{1}\text{.3}\times {10}^{-7}\right)}^3\left(\text{4}\text{.1}\times {10}^{-10}\right)}\\ =\frac{1}{2.197\times {10}^{-21}\times \text{4}\text{.1}\times {10}^{-10}}\\ =\frac{1}{9.00\times {10}^{-31}}\\ =1.11\times {10}^{30}\end{array}$

Therefore, the equilibrium constant for the formation of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $1.11\times {10}^{30}$.

The relation between the equilibrium constant and the standard Gibbs energy of the reaction is shown below.

    ${\Delta }_{\text{r}}{G}^{\Theta }=-RT\ln K$                                                                                         (5)

Where,

• ${\Delta }_{\text{r}}{G}^{\Theta }$ is the standard Gibbs energy of the reaction.
• $R$ is the gas constant.
• $T$ is the temperature.
• $K$ is the equilibrium constant for the reaction.

The equilibrium constant for the formation of ${\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $7.69\times {10}^6$.

The equilibrium constant for the formation of ${\text{HNO}}_3{\text{.H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $1.87\times {10}^{16}$.

The equilibrium constant for the formation of ${\text{HNO}}_3{\text{.2H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $1.44\times {10}^{23}$.

The equilibrium constant for the formation of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ at $190\text{K}$ is $1.11\times {10}^{30}$.

Equation (5) indicates that higher the value of equilibrium constant, higher will be the negative value of the standard Gibbs energy of the reaction.  Higher the negative value of the standard Gibbs energy of the reaction, higher will be the spontaneity of the reaction or the reaction will be thermodynamically stable.

The value of the equilibrium constant for the formation of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ is highest.  Therefore, the reaction for the formation of ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ will give the highest negative value of the standard Gibbs energy or ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ will be the most stable.

Hence, ${\text{HNO}}_3{\text{.3H}}_{\text{2}}\text{O}\left(\text{s}\right)$ is the most stable solid at $190\text{K}$.