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A chemist measures the energy change [tex]$\Delta H$[/tex] during the following reaction:
[tex]\[
2 \text{NO}_2(g) \rightarrow \text{N}_2\text{O}_4(g) \quad \Delta H = -55.3 \text{kJ}
\][/tex]

Use the information to answer the following questions.

1. This reaction is...
- A. endothermic.
- B. exothermic.

2. Suppose 53.6 g of [tex]$\text{NO}_2$[/tex] react. Will any heat be released or absorbed?
- A. Yes, absorbed.
- B. Yes, released.

3. If you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed. Round your answer to 3 significant digits.

Sagot :

GinnyAnsweridn
Certainly! Let's go through the problem step-by-step to address each part of the question:

### Step 1: Determine the Nature of the Reaction
The reaction given is:
[tex]\[2 \, NO_2(g) \rightarrow N_2O_4(g) \quad \Delta H = -55.3 \, \text{kJ}\][/tex]

The enthalpy change ([tex]\(\Delta H\)[/tex]) for this reaction is negative, which indicates that the reaction releases heat. Therefore, the reaction is exothermic.

### Answer:
This reaction is... [tex]\(\boxed{\text{exothermic}}\)[/tex].

### Step 2: Determine if Heat is Released or Absorbed
Since the reaction is exothermic, heat will be released during the reaction.

### Answer:
Will any heat be released or absorbed? [tex]\(\boxed{\text{Yes, released.}}\)[/tex]

### Step 3: Calculate the Amount of Heat Released
#### Given Information:
- Molar mass of [tex]\(NO_2 = 46.0055 \, \text{g/mol}\)[/tex]
- Mass of [tex]\(NO_2\)[/tex] reacting = 53.6 g
- [tex]\(\Delta H = -55.3 \, \text{kJ}\)[/tex] for 2 moles of [tex]\(NO_2\)[/tex]

#### Step-by-Step Calculation:
1. Calculate the moles of [tex]\(NO_2\)[/tex] reacting:
[tex]\[ \text{Moles of } NO_2 = \frac{\text{Mass of } NO_2}{\text{Molar mass of } NO_2} = \frac{53.6 \, \text{g}}{46.0055 \, \text{g/mol}} = 1.1650780884894198 \, \text{mol} \][/tex]

2. Calculate the energy change per mole of [tex]\(NO_2\)[/tex]:
Since [tex]\(\Delta H = -55.3 \, \text{kJ}\)[/tex] is for 2 moles of [tex]\(NO_2\)[/tex],
[tex]\[ \text{Energy change per mole of } NO_2 = \frac{-55.3 \, \text{kJ}}{2} = -27.65 \, \text{kJ/mol} \][/tex]

3. Calculate the total heat released for 53.6 g of [tex]\(NO_2\)[/tex]:
[tex]\[ \text{Heat released} = \text{Energy change per mole of } NO_2 \times \text{Moles of } NO_2 = -27.65 \, \text{kJ/mol} \times 1.1650780884894198 \, \text{mol} = -32.214 \, \text{kJ} \][/tex]

4. Round the answer to 3 significant digits:
[tex]\[ \text{Heat released} = -32.214 \, \text{kJ} \approx -32.2 \, \text{kJ} \][/tex]

### Answer:
The amount of heat released: [tex]\(\boxed{32.2 \, \text{kJ}}\)[/tex]

In summary:
1. The reaction is exothermic.
2. Yes, heat will be released.
3. The amount of heat released when 53.6 g of [tex]\(NO_2\)[/tex] react is 32.2 kJ.
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