Question

consider the following reaction at 25 °c:
5 so₃(g) + 2 nh₃(g) → 2 no(g) + 5 so₂(g) + 3 h₂o(g)
given the information in the table, calculate δs° for the reaction.

Explanation:

To calculate the standard entropy change ($\Delta S^\circ$) for a reaction, we use the formula:

$$\Delta S^\circ = \sum nS^\circ(\text{products}) - \sum mS^\circ(\text{reactants})$$

where $n$ and $m$ are the stoichiometric coefficients of the products and reactants, respectively, and $S^\circ$ is the standard molar entropy of each substance.

Step 1: Identify the standard molar entropies (from the table, which is not provided here, but we'll assume typical values or use the general method)

Let's assume the following standard molar entropies (in $\text{J/mol·K}$) for the substances involved (these are approximate values; actual values should be taken from the given table):

• $S^\circ(\text{SO}_3(\text{g})) \approx 256.8 \, \text{J/mol·K}$
• $S^\circ(\text{NH}_3(\text{g})) \approx 192.8 \, \text{J/mol·K}$
• $S^\circ(\text{NO}(\text{g})) \approx 210.8 \, \text{J/mol·K}$
• $S^\circ(\text{SO}_2(\text{g})) \approx 248.2 \, \text{J/mol·K}$
• $S^\circ(\text{H}_2\text{O}(\text{g})) \approx 188.8 \, \text{J/mol·K}$

Step 2: Calculate the total entropy of the products

For the products:

• $\text{NO}(\text{g})$: coefficient = 2, so $2 \times 210.8 = 421.6 \, \text{J/K}$
• $\text{SO}_2(\text{g})$: coefficient = 5, so $5 \times 248.2 = 1241.0 \, \text{J/K}$
• $\text{H}_2\text{O}(\text{g})$: coefficient = 3, so $3 \times 188.8 = 566.4 \, \text{J/K}$

Total entropy of products:
$$\sum nS^\circ(\text{products}) = 421.6 + 1241.0 + 566.4 = 2229.0 \, \text{J/K}$$

Step 3: Calculate the total entropy of the reactants

For the reactants:

• $\text{SO}_3(\text{g})$: coefficient = 5, so $5 \times 256.8 = 1284.0 \, \text{J/K}$
• $\text{NH}_3(\text{g})$: coefficient = 2, so $2 \times 192.8 = 385.6 \, \text{J/K}$

Total entropy of reactants:
$$\sum mS^\circ(\text{reactants}) = 1284.0 + 385.6 = 1669.6 \, \text{J/K}$$

Step 4: Calculate $\Delta S^\circ$

Using the formula:
$$\Delta S^\circ = 2229.0 - 1669.6 = 559.4 \, \text{J/K}$$

(Note: If the actual values from the table are different, the calculation will change accordingly. For example, if the table gives:

• $S^\circ(\text{SO}_3(\text{g})) = 256.8 \, \text{J/mol·K}$
• $S^\circ(\text{NH}_3(\text{g})) = 192.8 \, \text{J/mol·K}$
• $S^\circ(\text{NO}(\text{g})) = 210.8 \, \text{J/mol·K}$
• $S^\circ(\text{SO}_2(\text{g})) = 248.2 \, \text{J/mol·K}$
• $S^\circ(\text{H}_2\text{O}(\text{g})) = 188.8 \, \text{J/mol·K}$

the calculation remains as above. If the table has different values, substitute them in Steps 2 and 3.)

Final Answer

Assuming the standard molar entropies as above, $\Delta S^\circ \approx \boxed{559.4 \, \text{J/K}}$ (or the value calculated with the actual table values).

Answer:

To calculate the standard entropy change ($\Delta S^\circ$) for a reaction, we use the formula:

$$\Delta S^\circ = \sum nS^\circ(\text{products}) - \sum mS^\circ(\text{reactants})$$

where $n$ and $m$ are the stoichiometric coefficients of the products and reactants, respectively, and $S^\circ$ is the standard molar entropy of each substance.

Step 1: Identify the standard molar entropies (from the table, which is not provided here, but we'll assume typical values or use the general method)

Let's assume the following standard molar entropies (in $\text{J/mol·K}$) for the substances involved (these are approximate values; actual values should be taken from the given table):

• $S^\circ(\text{SO}_3(\text{g})) \approx 256.8 \, \text{J/mol·K}$
• $S^\circ(\text{NH}_3(\text{g})) \approx 192.8 \, \text{J/mol·K}$
• $S^\circ(\text{NO}(\text{g})) \approx 210.8 \, \text{J/mol·K}$
• $S^\circ(\text{SO}_2(\text{g})) \approx 248.2 \, \text{J/mol·K}$
• $S^\circ(\text{H}_2\text{O}(\text{g})) \approx 188.8 \, \text{J/mol·K}$

Step 2: Calculate the total entropy of the products

For the products:

• $\text{NO}(\text{g})$: coefficient = 2, so $2 \times 210.8 = 421.6 \, \text{J/K}$
• $\text{SO}_2(\text{g})$: coefficient = 5, so $5 \times 248.2 = 1241.0 \, \text{J/K}$
• $\text{H}_2\text{O}(\text{g})$: coefficient = 3, so $3 \times 188.8 = 566.4 \, \text{J/K}$

Total entropy of products:
$$\sum nS^\circ(\text{products}) = 421.6 + 1241.0 + 566.4 = 2229.0 \, \text{J/K}$$

Step 3: Calculate the total entropy of the reactants

For the reactants:

• $\text{SO}_3(\text{g})$: coefficient = 5, so $5 \times 256.8 = 1284.0 \, \text{J/K}$
• $\text{NH}_3(\text{g})$: coefficient = 2, so $2 \times 192.8 = 385.6 \, \text{J/K}$

Total entropy of reactants:
$$\sum mS^\circ(\text{reactants}) = 1284.0 + 385.6 = 1669.6 \, \text{J/K}$$

Step 4: Calculate $\Delta S^\circ$

Using the formula:
$$\Delta S^\circ = 2229.0 - 1669.6 = 559.4 \, \text{J/K}$$

(Note: If the actual values from the table are different, the calculation will change accordingly. For example, if the table gives:

• $S^\circ(\text{SO}_3(\text{g})) = 256.8 \, \text{J/mol·K}$
• $S^\circ(\text{NH}_3(\text{g})) = 192.8 \, \text{J/mol·K}$
• $S^\circ(\text{NO}(\text{g})) = 210.8 \, \text{J/mol·K}$
• $S^\circ(\text{SO}_2(\text{g})) = 248.2 \, \text{J/mol·K}$
• $S^\circ(\text{H}_2\text{O}(\text{g})) = 188.8 \, \text{J/mol·K}$

the calculation remains as above. If the table has different values, substitute them in Steps 2 and 3.)

Final Answer

Assuming the standard molar entropies as above, $\Delta S^\circ \approx \boxed{559.4 \, \text{J/K}}$ (or the value calculated with the actual table values).