Given the data,

N2(g)  +  O2(g)  -->  2 NO(g)                           ∆H° = 180.7 kJ

2 NO(g)  +  O2(g)  -->  2 NO2(g)                     ∆H° = -113.1 kJ

2 N2O(g)  -->  2 N2(g)  +  O2(g)                      ∆H° = -163.2 kJ

use Hess’s Law to calculate ∆H° for the following reaction.

N2O(g)  +  NO2(g)    3 NO(g)

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.

use Hess’s Law to calculate ∆H° for the reaction N2O(g) + NO2(g) → 3 NO(g)

Explore how to calculate the standard enthalpy change (∆H°) for the reaction N2O(g) + NO2(g) → 3 NO(g) using Hess's Law. This solution details the rearrangement of given reactions and the step-by-step process, resulting in ∆H° = 296.1 kJ.

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