Question

Find ${\mathbf{\Delta S}}^{\mathbf{o}}$for the combustion of ethane$\mathbf{\left(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\mathbf{\right)}$to carbon dioxide and gaseous water. Is the sign of${\mathbf{\Delta S}}^{\mathbf{o}}$as expected?

Short answer

The combustion of ethane to carbon dioxide and gaseous water is obtained as: ${\mathrm{\Delta S}}^{\circ }=94.4\mathrm{J}/\mathrm{K}$.

Step-by-step solution

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.

Evaluating the ΔSo.

First, write the${\mathrm{C}}_2{\mathrm{H}}_6$combustion reaction and balance the equation.

$2{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+7{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

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

$\begin{array}{rcl}\mathrm{S}°\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)& =& 205.0\mathrm{J}/\mathrm{molK}\\ \mathrm{S}°\left({\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)\right)& =& 229.5\mathrm{J}/\mathrm{molK}\\ \mathrm{S}°\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)& =& 188.72\mathrm{J}/\mathrm{molK}\\ \mathrm{S}°\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)& =& 213.7\mathrm{J}/\mathrm{molK}\end{array}$

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

$\begin{array}{rcl}{\mathrm{\Delta S}}^{\circ }& =& \sum {\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{\circ }\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{\circ }\left(\mathrm{reactant}\right).....................\left(1\right)\\ & =& \left[{\mathrm{nS}}^{\circ }\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)\right]-\left[{\mathrm{nS}}^{\circ }\left({\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)\right]....................\left(2\right)\\ & =& \left(\right(213.7)+6(188.75\left)\right]-\left[2\right(229.5)+7(205.0\left)\right])\mathrm{J}/\mathrm{mol}\times \mathrm{K}.....................\left(3\right)\\ & =& 94.4\mathrm{J}/\mathrm{mol}\times \mathrm{K}............................\left(4\right)\end{array}$

The omen is favourable. All of the chemical species found are gaseous compounds. The total number of moles on the product side is more than the total number of moles on the reactant side, as:$\mathrm{\Delta n}=+1$. As a result, having more gaseous molecules means having greater unpredictability and higher entropy.

Therefore, the value is: ${\mathrm{\Delta S}}^{\circ }=94.4\mathrm{J}/\mathrm{K}$.