Question

Find${\mathbf{\Delta S}}^{\mathbf{o}}$for the combustion of ammonia to nitrogen dioxide and water vapor. Is the sign ofrole="math" localid="1663359837067" ${\mathbf{\Delta S}}^{\mathbf{o}}$as expected?

Short answer

The combustion of ammonia to nitrogen dioxide and water vapor is obtained as: ${\mathrm{\Delta S}}^{\circ }=-115.08\mathrm{J}/\mathrm{K}$.

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${\mathrm{NH}}_3$combustion reaction and balance the equation.

$4{\mathrm{NH}}_3\left(\mathrm{g}\right)+7{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{NO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

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

$\begin{array}{rcl}\mathrm{S}°\left({\mathrm{NH}}_3\left(\mathrm{g}\right)\right)& =& 193\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)& =& 205.0\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{NO}}_2\left(\mathrm{g}\right)\right)& =& 239.9\mathrm{J}/\mathrm{mol}\mathrm{K}\\ \mathrm{S}°\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)& =& 188.72\mathrm{J}/\mathrm{mol}\mathrm{K}\end{array}$

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

$\begin{array}{c}{\mathrm{\Delta S}}^{\circ }=\sum {\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{\circ }\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{\circ }\left(\mathrm{reactant}\right)...........................\left(1\right)\\ =\left[{\mathrm{nS}}^{\circ }\left({\mathrm{NO}}_2\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right)\right]-\left[{\mathrm{nS}}^{\circ }\left({\mathrm{NH}}_3\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)\right]....................\left(2\right)\\ =\left(4\right(239.9)+6(188.72\left)\right]-\left[4\right(193)+7(205.0\left)\right])\mathrm{J}/\mathrm{mol}\times \mathrm{K}....................\left(3\right)\\ {\mathrm{\Delta S}}^{\circ }=-115.08\mathrm{J}/\mathrm{K}.....................\left(4\right)\end{array}$

The omen is not good. All of the chemical species found are gaseous compounds. The total number of moles on the product side is smaller than the total number of moles on the reactant side, as can be seen:$\mathrm{\Delta n}=-1$. As a result, fewer gaseous molecules imply less unpredictability and hence reduced entropy.

Therefore, the value is: ${\mathrm{\Delta S}}^{\circ }=-115.08\mathrm{J}/\mathrm{K}$.