To calculate the standard enthalpy change (∆H°) for the reaction

$\text{N}_2\text{O}(g) + \text{NO}_2(g) \rightarrow 3 \text{NO}(g),$

we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps, regardless of the pathway taken.

Given Reactions:

1. $\text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g)$ $\Delta H° = 180.7 \, \text{kJ}$
2. $2 \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \text{NO}_2(g)$ $\Delta H° = -113.1 \, \text{kJ}$
3. $2 \text{N}_2\text{O}(g) \rightarrow 2 \text{N}_2(g) + \text{O}_2(g)$ $\Delta H° = -163.2 \, \text{kJ}$

Step 1: Rearranging Reactions

1. To form $\text{N}_2O(g)$, we reverse reaction 3:

   $2 \text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{N}_2\text{O}(g) \quad \Delta H° = +163.2 \, \text{kJ}$

   Since we need only 1 mole of $\text{N}_2O$, we divide the entire reaction by 2:

   $\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{N}_2\text{O}(g) \quad \Delta H° = \frac{163.2}{2} = 81.6 \, \text{kJ}$

2. Use reaction 2 directly, but we need only 1 mole of $\text{NO}_2$:

   $\text{NO}_2(g) \rightarrow \text{NO}(g) + \frac{1}{2}\text{O}_2(g) \quad \Delta H° = \frac{-113.1}{2} = -56.55 \, \text{kJ}$

3. We will use reaction 1 directly for the production of 3 moles of $\text{NO}$:

   $\text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g) \quad \Delta H° = 180.7 \, \text{kJ}$

   To form 3 moles of $\text{NO}$, we adjust:

   $\text{N}_2(g) + \text{O}_2(g) \rightarrow 3 \text{NO}(g) \quad \Delta H° = \frac{180.7}{2} + 180.7 \approx 271.05 \, \text{kJ}$

Step 2: Summing the Adjusted Reactions

Now, we combine these three reactions:

1. From $\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{N}_2\text{O}(g)$ $\Delta H° = 81.6 \, \text{kJ}$
2. From $\text{NO}_2(g) \rightarrow \text{NO}(g) + \frac{1}{2}\text{O}_2(g)$ $\Delta H° = -56.55 \, \text{kJ}$
3. From $\text{N}_2(g) + \text{O}_2(g) \rightarrow 3 \text{NO}(g)$ $\Delta H° = 271.05 \, \text{kJ}$

Final Calculation:

Summing these enthalpies:

$\Delta H° = 81.6 - 56.55 + 271.05 = 296.1 \, \text{kJ}$

Thus, the enthalpy change for the reaction

$\text{N}_2\text{O}(g) + \text{NO}_2(g) \rightarrow 3 \text{NO}(g)$

is

$\Delta H° = 296.1 \, \text{kJ}.$

Summary

The calculated enthalpy change for the reaction is $296.1 \, \text{kJ}$.

Do you want more details or have any questions? Here are some related questions:

1. How does Hess's Law apply to other types of reactions?
2. What are the limitations of Hess's Law?
3. Can you explain the significance of the signs of ∆H° values?
4. How can we determine if a reaction is exothermic or endothermic using enthalpy changes?
5. What are some real-world applications of Hess's Law?

Tip: Always ensure the reactions are balanced before applying Hess’s Law to avoid errors in calculations.