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 δh° for the reaction.
compound δhf° (kj/mol) s° (j/mol·k)
so₃ (g) -396 256
so₂ (g) -297 248
nh₃ (g) -46 192.5
no (g) 90.4 210.6
h₂o (g) -241.8 188.7

Explanation:

To calculate the standard enthalpy change (\(\Delta H^\circ\)) for a reaction, we use the formula:

\[
\Delta H^\circ = \sum n \Delta H_f^\circ (\text{products}) - \sum m \Delta H_f^\circ (\text{reactants})
\]

where \(n\) and \(m\) are the stoichiometric coefficients of the products and reactants, respectively, and \(\Delta H_f^\circ\) is the standard enthalpy of formation.

Step 1: Identify the stoichiometric coefficients and \(\Delta H_f^\circ\) values for products and reactants.

Products:

• \(2 \, \text{mol}\) of \( \text{NO}(g) \) with \(\Delta H_f^\circ = 90.4 \, \text{kJ/mol}\)
• \(5 \, \text{mol}\) of \( \text{SO}_2(g) \) with \(\Delta H_f^\circ = -297 \, \text{kJ/mol}\)
• \(3 \, \text{mol}\) of \( \text{H}_2\text{O}(g) \) with \(\Delta H_f^\circ = -241.8 \, \text{kJ/mol}\)

Reactants:

• \(5 \, \text{mol}\) of \( \text{SO}_3(g) \) with \(\Delta H_f^\circ = -396 \, \text{kJ/mol}\)
• \(2 \, \text{mol}\) of \( \text{NH}_3(g) \) with \(\Delta H_f^\circ = -46 \, \text{kJ/mol}\)

Step 2: Calculate the sum of \(\Delta H_f^\circ\) for products.

\[

$$\begin{align*} \sum n \Delta H_f^\circ (\text{products}) &= (2 \times 90.4) + (5 \times -297) + (3 \times -241.8) \\ &= 180.8 + (-1485) + (-725.4) \\ &= 180.8 - 1485 - 725.4 \\ &= 180.8 - 2210.4 \\ &= -2029.6 \, \text{kJ} \end{align*}$$

\]

Step 3: Calculate the sum of \(\Delta H_f^\circ\) for reactants.

\[

$$\begin{align*} \sum m \Delta H_f^\circ (\text{reactants}) &= (5 \times -396) + (2 \times -46) \\ &= -1980 + (-92) \\ &= -1980 - 92 \\ &= -2072 \, \text{kJ} \end{align*}$$

\]

Step 4: Calculate \(\Delta H^\circ\) for the reaction.

\[

$$\begin{align*} \Delta H^\circ &= \sum n \Delta H_f^\circ (\text{products}) - \sum m \Delta H_f^\circ (\text{reactants}) \\ &= (-2029.6) - (-2072) \\ &= -2029.6 + 2072 \\ &= 42.4 \, \text{kJ} \end{align*}$$

\]

Answer:

The standard enthalpy change (\(\Delta H^\circ\)) for the reaction is \(\boxed{42.4 \, \text{kJ}}\).