Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction entropy of the following chemical reaction: \\(\ce{al_{2}o_{3}(s) + 3h_{2}(g) -> 2al(s) + 3h_{2}o(g)}\\) round your answer to zero decimal places.

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: $S^{\circ}(\text{Al}_2\text{O}_3(s)) = 50.9\ \frac{\text{J}}{\text{mol·K}}$, $S^{\circ}(\text{H}_2(g)) = 130.7\ \frac{\text{J}}{\text{mol·K}}$, $S^{\circ}(\text{Al}(s)) = 28.3\ \frac{\text{J}}{\text{mol·K}}$, $S^{\circ}(\text{H}_2\text{O}(g)) = 188.8\ \frac{\text{J}}{\text{mol·K}}$)

Step3: Calculate the entropy of products

For products: $2\ \text{mol of Al}(s)$ and $3\ \text{mol of H}_2\text{O}(g)$
$\sum nS^{\circ}(\text{products})=2\times S^{\circ}(\text{Al}(s)) + 3\times S^{\circ}(\text{H}_2\text{O}(g))$
$=2\times28.3 + 3\times188.8$
$= 56.6+566.4 = 623\ \frac{\text{J}}{\text{K}}$ (per mole of reaction)

Step4: Calculate the entropy of reactants

For reactants: $1\ \text{mol of Al}_2\text{O}_3(s)$ and $3\ \text{mol of H}_2(g)$
$\sum mS^{\circ}(\text{reactants})=1\times S^{\circ}(\text{Al}_2\text{O}_3(s)) + 3\times S^{\circ}(\text{H}_2(g))$
$= 50.9+3\times130.7$
$= 50.9 + 392.1=443\ \frac{\text{J}}{\text{K}}$ (per mole of reaction)

Step5: Calculate $\Delta S^{\circ}$

$\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})$
$= 623 - 443=180\ \frac{\text{J}}{\text{K}}$ (rounded to zero decimal places)

Answer:

180