Q20.36 P - Chemistry: Molecular Nature Of Matter And Change

Question

Find  ${\mathbf{\Delta S}}^{\mathbf{o}}$for the combustion of methane to carbon dioxide and liquid water. Is the sign of  ${\mathbf{\Delta S}}^{\mathbf{o}}$as expected?

Step-by-Step Solution

Answer

The combustion of methane to carbon dioxide and liquid water is obtained as: ${\mathrm{\Delta S}}^{\circ }=-242.52\mathrm{J}/\mathrm{K}$.

Step 1: Define Thermodynamics.

Thermodynamics is the study of the connections between heat, work, temperature, and energy. Thermodynamic principles specify how energy develops within a system and whether it is capable of having a positive impact on its surroundings.

Step 2: Evaluating the ΔS o .

First, write the ${\mathrm{CH}}_4$ combustion reaction and balance the equation.

${\mathrm{CH}}_4\left(\mathrm{g}\right)+2{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

After that, make a list of the ${\mathrm{S}}^{\mathrm{o}}$ values for each compound.

$\begin{array}{rcl}{\mathrm{S}}^{\circ }\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)& =& 205.0\mathrm{J}/\mathrm{mol} \mathrm{K}\\ {\mathrm{S}}^{\circ }\left({\mathrm{CH}}_4\left(\mathrm{g}\right)\right)& =& 186.1\mathrm{J}/\mathrm{mol} \mathrm{K}\\ {\mathrm{S}}^{\circ }\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)& =& 69.94\mathrm{J}/\mathrm{mol} \mathrm{K}\\ {\mathrm{S}}^{\circ }\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)& =& 213.7\mathrm{J}/\mathrm{mol} \mathrm{K}\end{array}$

Solve for the ${\mathrm{\Delta S}}^{\mathrm{o}}$ now as:

$\begin{array}{rcl}{\mathrm{\Delta S}}^{\circ } & =& \sum {\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{\circ }\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{\circ }\left(\mathrm{reactant}\right)...........................\left(1\right)\\ & =& \left[{\mathrm{nS}}^{\circ }\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right)\right]-\left[{\mathrm{nS}}^{\circ }\left({\mathrm{CH}}_4\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)\right]....................\left(2\right)\\ & & =\left(1\right(213.7)+2(69.94\left)\right]-\left[1\right(186.1)+2(205.0\left)\right])\mathrm{J}/\mathrm{mol}\times \mathrm{K}....................\left(3\right)\\ & =& -242.52\mathrm{J}/\mathrm{K}.....................\left(4\right)\end{array}$

The sign is not favorable. On the reactant side, all of the chemical species are gaseous compounds. Gaseous and liquid chemicals make up the chemical species found on the production side. Liquids, on the other hand, are more structured than gases. As a result, the reaction goes from being less structured to be more organized, with decreased entropy.

Therefore, the value is: ${\mathrm{\Delta S}}^{\circ } = -242.52\mathrm{J}/\mathrm{K}$.