Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction entropy of the following chemical reaction: ch₃oh(g) + co(g) → hch₃co₂(l) 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 are: $S^{\circ}(\text{CH}_3\text{OH}(g)) = 239.8\ \frac{\text{J}}{\text{mol}\cdot\text{K}}$, $S^{\circ}(\text{CO}(g)) = 197.7\ \frac{\text{J}}{\text{mol}\cdot\text{K}}$, $S^{\circ}(\text{HCH}_3\text{CO}_2(l))$ (acetic acid) $= 158.0\ \frac{\text{J}}{\text{mol}\cdot\text{K}}$)

Step3: Calculate the sum of entropies of reactants

Reactants: $\text{CH}_3\text{OH}(g)$ (coefficient 1) and $\text{CO}(g)$ (coefficient 1).
$\sum mS^{\circ}(\text{reactants})=1\times S^{\circ}(\text{CH}_3\text{OH}(g)) + 1\times S^{\circ}(\text{CO}(g))=239.8 + 197.7 = 437.5\ \frac{\text{J}}{\text{K}}$

Step4: Calculate the sum of entropies of products

Product: $\text{HCH}_3\text{CO}_2(l)$ (coefficient 1).
$\sum nS^{\circ}(\text{products})=1\times S^{\circ}(\text{HCH}_3\text{CO}_2(l)) = 158.0\ \frac{\text{J}}{\text{K}}$

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

$\Delta S^{\circ}=\sum nS^{\circ}(\text{products})-\sum mS^{\circ}(\text{reactants})=158.0 - 437.5=-279.5\ \frac{\text{J}}{\text{K}}$ (round to zero decimal places: $-280$)

Answer:

$-280$