Question

Calculate$∆{\mathrm{H}}^{°}$ for each of the following reactions using the data in Appendix 4:

$$\begin{gathered} \mathbf{4}\mathbf{Na}\left(s\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}{\mathbf{Na}}_{\mathbf{2}}\mathbf{O}\left(s\right) \\ \mathbf{2}\mathbf{Na}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{\to }\mathbf{2}\mathbf{NaOH}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right) \\ \mathbf{2}\mathbf{Na}\left(s\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{Na}}_{\mathbf{2}}\mathbf{O}\left(s\right)\mathbf{+}\mathbf{CO}\left(g\right) \end{gathered}$$

Explain why a water or carbon dioxide fire extinguisher might not be effective in putting out a sodium fire.

Short answer

The$∆\mathrm{H}$ for$4\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)$ is $-832\mathrm{kJ}$.

The$∆\mathrm{H}$ for$2\mathrm{Na}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$ is $-368\mathrm{kJ}$.

The$∆\mathrm{H}$ for$2\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\to {\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)+\mathrm{CO}\left(\mathrm{g}\right)$ is $-133\mathrm{kJ}$.

When sodium metal combines with water or carbon dioxide as the "extinguishing agent” according to reactions (2) and (3) respectively. These generate the flammable gases, hydrogen and carbon monoxide. Hence, the fire extinguisher containing water or carbon dioxide might not be effective in putting out a sodium fire.

Step-by-step solution

Calculate the enthalpy change for the given reactions.

The value of$∆\mathrm{H}$ for the reaction$4\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)$ can be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)\right)\right]-\left[4\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Na}\left(\mathrm{s}\right)\right)+1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)\right]\\ & =& \left[2\mathrm{mol}\times -416\mathrm{kJ}/\mathrm{mol}\right]-\left[4\mathrm{mol}\times 0+1\mathrm{mol}\times 0\right]\\ & =& -832\mathrm{kJ}\end{array}$

Hence, the$∆\mathrm{H}$ for$4\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)$ is $-832\mathrm{kJ}$.

The value of$∆\mathrm{H}$ for the reaction$2\mathrm{Na}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$ can be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{NaOH}\left(\mathrm{s}\right)\right)+1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)\right]-\left[2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Na}\left(\mathrm{s}\right)\right)+2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)\right]\\ & =& \left[2\mathrm{mol}\times -470\mathrm{kJ}/\mathrm{mol}+2\mathrm{mol}\times 0\right]-\left[2\mathrm{mol}\times 0+2\mathrm{mol}\times -286\mathrm{kJ}/\mathrm{mol}\right]\\ & =& -368\mathrm{kJ}\end{array}$

Hence, the$∆\mathrm{H}$ for$2\mathrm{Na}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$ is $-368\mathrm{kJ}$.

The value of$∆\mathrm{H}$ for the reaction$2\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\to {\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)+\mathrm{CO}\left(\mathrm{g}\right)$ can be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)\right)+1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{CO}\left(\mathrm{g}\right)\right)\right]-\left[2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left(\mathrm{Na}\left(\mathrm{s}\right)\right)+1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)\right]\\ & =& \left[1\mathrm{mol}\times -416\mathrm{kJ}/\mathrm{mol}+1\mathrm{mol}\times -110.5\mathrm{kJ}/\mathrm{mol}\right]-\left[2\mathrm{mol}\times 0+1\mathrm{mol}\times -393.5\mathrm{kJ}/\mathrm{mol}\right]\\ & =& -133\mathrm{kJ}\end{array}$

Therefore, the$∆\mathrm{H}$ for $2\mathrm{Na}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\to {\mathrm{Na}}_2\mathrm{O}\left(\mathrm{s}\right)+\mathrm{CO}\left(\mathrm{g}\right)$is $-133\mathrm{kJ}$.

The reason.

When sodium metal combines with water or carbon dioxide as the "extinguishing agent” according to reactions (2) and (3) respectively. These generate the flammable gases, hydrogen and carbon monoxide. Hence, the fire extinguisher containing water or carbon dioxide might not be effective in putting out a sodium fire.