Alright, let's solve the problem using Hess's Law step-by-step. Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps that lead to the overall reaction.
Given the following reactions and their respective enthalpy changes ([tex]\(\Delta H\)[/tex]):
1. [tex]\(2 NO_2(g) \rightarrow N_2O_4(g)\)[/tex]
[tex]\(\Delta H = -57.0 \, kJ/mol\)[/tex]
2. [tex]\(N_2(g) + O_2(g) \rightarrow 2 NO(g)\)[/tex]
[tex]\(\Delta H = 180.6 \, kJ/mol\)[/tex]
3. [tex]\(2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)\)[/tex]
[tex]\(\Delta H = 114.4 \, kJ/mol\)[/tex]
We need to find the enthalpy change ([tex]\(\Delta H\)[/tex]) for the following reaction:
[tex]\(N_2(g) + 2 O_2(g) \rightarrow N_2O_4(g)\)[/tex]
### Step-by-Step Solution:
1. First, we need to understand the target reaction and the given reactions:
- Target Reaction: [tex]\(N_2(g) + 2 O_2(g) \rightarrow N_2O_4(g)\)[/tex]
- Reaction 1: [tex]\(2 NO_2(g) \rightarrow N_2O_4(g)\)[/tex]
- Reaction 2: [tex]\(N_2(g) + O_2(g) \rightarrow 2 NO(g)\)[/tex]
- Reaction 3: [tex]\(2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)\)[/tex]
2. Re-engineer the given reactions to achieve the target reaction:
- Reaction 1 can be used as it is:
[tex]\[
2 NO_2(g) \rightarrow N_2O_4(g) \quad (\Delta H = -57.0 \, kJ/mol)
\][/tex]
- Reaction 2 can also be used directly:
[tex]\[
N_2(g) + O_2(g) \rightarrow 2 NO(g) \quad (\Delta H = 180.6 \, kJ/mol)
\][/tex]
- Reaction 3 needs to be reversed to produce [tex]\(2 NO_2(g)\)[/tex] from [tex]\(2 NO(g) + O_2(g)\)[/tex], and the sign of [tex]\(\Delta H\)[/tex] will change:
[tex]\[
2 NO_2(g) \rightarrow 2 NO(g) + O_2(g) \quad (\Delta H = -114.4 \, kJ/mol)
\][/tex]
3. Adjust the stoichiometry if necessary and combine reactions:
Based on the stoichiometry, if we sum the given reactions in their proper forms, we'll get:
- Reaction 2:
[tex]\[
N_2(g) + O_2(g) \rightarrow 2 NO(g) \quad (\Delta H = 180.6 \, kJ/mol)
\][/tex]
- Reversed Reaction 3:
[tex]\[
2 NO_2(g) \rightarrow 2 NO(g) + O_2(g) \quad (\Delta H = -114.4 \, kJ/mol)
\][/tex]
- Reaction 1:
[tex]\[
2 NO_2(g) \rightarrow N_2O_4(g) \quad (\Delta H = -57.0 \, kJ/mol)
\][/tex]
4. Combine the enthalpy changes:
Since Reaction 3 is used in reverse, the enthalpy change of [tex]\(114.4 \, kJ/mol\)[/tex] will be negative. Add the enthalpy changes for these reactions:
[tex]\[
\Delta H_{\text{total}} = 180.6 \, kJ/mol + (-114.4 \, kJ/mol) + (-57.0 \, kJ/mol)
\][/tex]
[tex]\[
\Delta H_{\text{total}} = 180.6 \, kJ/mol - 114.4 \, kJ/mol - 57.0 \, kJ/mol
\][/tex]
[tex]\[
\Delta H_{\text{total}} = 66.4 \, kJ/mol
\][/tex]
Therefore, the overall enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction [tex]\(N_2(g) + 2 O_2(g) \rightarrow N_2O_4(g)\)[/tex] is [tex]\(66.4 \, kJ/mol\)[/tex].
