Q20.37 P - Chemistry: Molecular Nature Of Matter And Change

Question

Find ${\mathbf{\Delta S}}^{\mathbf{o}}$ for the reaction of nitrogen monoxide with hydrogen to form ammonia and water vapor. Is the sign of  ${\mathbf{\Delta S}}^{\mathbf{o}}$ as expected?

Step-by-Step Solution

Answer

The combustion of nitrogen monoxide with hydrogen to form ammonia and water vapor is obtained as: ${\mathrm{\Delta S}}^{\circ }=-310.86\mathrm{J}/\mathrm{K}$.

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 combustion reaction of NO with ${\mathrm{H}}_2$ and balance the equation.

$2\mathrm{NO}\left(\mathrm{g}\right)+5{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

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

$\begin{array}{rcl}\mathrm{S}°\left(\mathrm{NO}\left(\mathrm{g}\right)\right)& =& 210.65\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)& =& 130.6\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left({\mathrm{NH}}_3\left(\mathrm{g}\right)\right)& =& 193\mathrm{J}/\mathrm{mol} \mathrm{K}\\ \mathrm{S}°\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)& =& 188.72\mathrm{J}/\mathrm{mol} \mathrm{K}\end{array}$

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

$\begin{array}{rcl}{\mathrm{\Delta S}}^{\circ }& =& \sum {\mathrm{n}}_{\mathrm{p}}{\mathrm{S}}^{\circ }\left(\mathrm{product}\right)-\sum {\mathrm{n}}_{\mathrm{r}}{\mathrm{S}}^{\circ }\left(\mathrm{reactant}\right)...........................\left(1\right)\\ & =& \left[{\mathrm{nS}}^{\circ }\left({\mathrm{NH}}_3\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right)\right]-\left[{\mathrm{nS}}^{\circ }\left(\mathrm{NO}\left(\mathrm{g}\right)\right)+{\mathrm{nS}}^{\circ }\left({\mathrm{H}}_2\left(\mathrm{g}\right)\right)\right]....................\left(2\right)\\ & & =\left(2\right(193)+2(188.72\left)\right]-\left[2\right(210.65)+5(130.6\left)\right])\mathrm{J}/\mathrm{mol}\times \mathrm{K}....................\left(3\right)\\ & =& -310.86\mathrm{J}/\mathrm{K}.....................\left(4\right)\end{array}$

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: $\mathrm{\Delta n}=-3$. As a result, fewer gaseous molecules imply less unpredictability and hence reduced entropy.

Therefore, the value is: ${\mathrm{\Delta S}}^{\circ }= -310.86\mathrm{J}/\mathrm{K}$.