Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction entropy of the following chemical reaction:
$4\ce{pcl_{3}}(g) \
ightarrow \ce{p_{4}}(g) + 6\ce{cl_{2}}(g)$
round your answer to zero decimal places.
$\square \frac{\text{j}}{\text{k}}$

Explanation:

Step1: Recall the formula for standard reaction entropy

The formula for the standard reaction entropy ($\Delta S^{\circ}$) is $\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively, and $S^{\circ}$ is the standard molar entropy.

Step2: Find the standard molar entropies (from ALEKS Data tab, typical values: $S^{\circ}(\text{PCl}_3(g)) = 311.7\ \text{J/(mol·K)}$, $S^{\circ}(\text{P}_4(g)) = 280.0\ \text{J/(mol·K)}$, $S^{\circ}(\text{Cl}_2(g)) = 223.1\ \text{J/(mol·K)}$)

Step3: Calculate the entropy of products

For products: $\text{P}_4(g)$ has coefficient 1, $\text{Cl}_2(g)$ has coefficient 6.
$\sum nS^{\circ}(\text{products})=1\times S^{\circ}(\text{P}_4(g)) + 6\times S^{\circ}(\text{Cl}_2(g))=1\times280.0 + 6\times223.1=280.0+1338.6 = 1618.6\ \text{J/K}$ (per 4 moles of $\text{PCl}_3$? Wait, no, stoichiometry: reaction is $4\text{PCl}_3(g)
ightarrow\text{P}_4(g)+6\text{Cl}_2(g)$, so moles of reactant: 4 moles of $\text{PCl}_3$. So reactant entropy: $4\times S^{\circ}(\text{PCl}_3(g)) = 4\times311.7 = 1246.8\ \text{J/K}$

Step4: Calculate $\Delta S^{\circ}$

$\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})=1618.6 - 1246.8=371.8\ \text{J/K}$. Rounding to zero decimal places gives 372.

Answer:

372