Question

The standard molar entropies of $\text{SO}_2(g)$, $\text{SO}_3(g)$, and $\text{O}_2(g)$ are:

• $S^\circ_{\text{SO}_2} = 250$ J/K·mol
• $S^\circ_{\text{SO}_3} = 257$ J/K·mol
• $S^\circ_{\text{O}_2} = 205$ J/K·mol

Calculate the standard molar entropy change for the reaction:

$$2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$$

The standard entropy change ($\Delta S^\circ$) is given by:

$$\Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}$$

• A. -198 J/K·mol
• B. -191 J/K·mol
• C. 198 J/K·mol
• D. 191 J/K·mol
• E. -1219 J/K·mol

Hint:

To calculate the standard entropy change for a reaction, subtract the sum of the standard entropies of the reactants from the sum of the standard entropies of the products.

Answer 1

The Correct Option is B

The standard entropy change $\Delta S^\circ$ for the reaction is given by: $$\Delta S^\circ = \sum \left( S^\circ_{{products}} \right) - \sum \left( S^\circ_{{reactants}} \right)$$
Step 1: Write the expression for entropy change: $$\Delta S^\circ = \left[ 2 \times S^\circ_{{SO}_3(g)} \right] - \left[ 2 \times S^\circ_{{SO}_2(g)} + S^\circ_{{O}_2(g)} \right]$$
Step 2: Substitute the given standard entropies: $$\Delta S^\circ = \left[ 2 \times 257 \right] - \left[ 2 \times 250 + 205 \right]$$ $$\Delta S^\circ = 514 - \left[ 500 + 205 \right]$$ $$\Delta S^\circ = 514 - 705$$ $$\Delta S^\circ = -191 \, {J/K·mol}$$
Thus, the standard molar entropy change for the reaction is $-191 \, {J/K·mol}$.