Question

Nitrogen dioxide, NO2(g) (Delta.Hf = 33.84 kJ/mol), is decomposed according to the following reaction:

2 upper N upper O subscript 2 (g) right arrow upper N subscript 2 (g) plus 2 upper O subscript 2 (g).

What is the enthalpy change when 2.50 mol of nitrogen dioxide decomposes?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
13.5 kJ of energy released
13.5 kJ of energy absorbed
84.6 kJ of energy released
84.6 kJ of energy absorbed

Answer 1

To calculate the enthalpy change (\( \Delta H \)) for the decomposition of nitrogen dioxide (NO\(_2\)), we can use the formula you provided:

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

### Reaction
The balanced equation for the decomposition of nitrogen dioxide is:

\[
2 \text{NO}_2(g) \rightarrow \text{N}_2(g) + 2 \text{O}_2(g)
\]

### Standard Enthalpy of Formation (\( \Delta H_f \))
Given:
- \( \Delta H_f \) for NO\(_2\) is \( 33.84 \, \text{kJ/mol} \)
- \( \Delta H_f \) for N\(_2(g)\) and O\(_2(g)\) is \( 0 \, \text{kJ/mol} \) since these are elements in their standard state.

### Calculation
1. **Reactants:**
- For 2 moles of NO\(_2\):
\[
\Delta H_f(\text{reactants}) = 2 \times 33.84 \, \text{kJ/mol} = 67.68 \, \text{kJ}
\]

2. **Products:**
- For 1 mole of N\(_2\) and 2 moles of O\(_2\):
\[
\Delta H_f(\text{products}) = 1 \times 0 \, \text{kJ/mol} + 2 \times 0 \, \text{kJ/mol} = 0 \, \text{kJ}
\]

3. **Substituting into the equation:**
\[
\Delta H = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants})
\]
\[
\Delta H = 0 \, \text{kJ} - 67.68 \, \text{kJ} = -67.68 \, \text{kJ}
\]

### Enthalpy Change for 2.5 moles of NO\(_2\)
Since the reaction is for 2 moles of NO\(_2\), we need to find the enthalpy change per mole:
- The enthalpy change for 2 moles is \(-67.68 \, \text{kJ}\).
- Thus, for 1 mole:
\[
\Delta H_{\text{per mole}} = \frac{-67.68 \, \text{kJ}}{2} = -33.84 \, \text{kJ}
\]

Now, for 2.5 moles:
\[
\Delta H = -33.84 \, \text{kJ/mol} \times 2.5 \, \text{mol} = -84.6 \, \text{kJ}
\]

### Conclusion
The enthalpy change when 2.50 mol of nitrogen dioxide decomposes is:
\[
\text{84.6 kJ of energy released}
\]

So the correct answer is **84.6 kJ of energy released**.