To find the enthalpy change for the reaction \(\Delta H_{rxn}\), we use the enthalpies of formation (\(\Delta H_f\)) of the reactants and products involved in the chemical reaction. The enthalpy change for the reaction can be calculated using the formula:
[tex]\[
\Delta H_{rxn} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})
\][/tex]
Given:
- \(\Delta H_f \text{ (NH}_3\text{(g))} = -46.19 \, \text{kJ/mol}\)
- \(\Delta H_f \text{ (HCl(g))} = -92.30 \, \text{kJ/mol}\)
- \(\Delta H_f \text{ (NH}_4\text{Cl(s))} = -314.4 \, \text{kJ/mol}\)
In the given reaction:
[tex]\[
NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)
\][/tex]
First, identify the products and reactants:
- Reactants: \(NH_3(g)\) and \(HCl(g)\)
- Product: \(NH_4Cl(s)\)
Next, sum the enthalpies of formation for the products and reactants:
- For the product \(NH_4Cl(s)\):
[tex]\[
\sum \Delta H_f(\text{products}) = \Delta H_f \text{ (NH}_4\text{Cl(s))} = -314.4 \, \text{kJ/mol}
\][/tex]
- For the reactants \(NH_3(g)\) and \(HCl(g)\):
[tex]\[
\sum \Delta H_f(\text{reactants}) = \Delta H_f \text{ (NH}_3\text{(g))} + \Delta H_f \text{ (HCl(g))} = -46.19 \, \text{kJ/mol} + -92.30 \, \text{kJ/mol} = -138.49 \, \text{kJ/mol}
\][/tex]
Now, apply the reaction enthalpy formula:
[tex]\[
\Delta H_{rxn} = (-314.4 \, \text{kJ/mol}) - (-138.49 \, \text{kJ/mol})
\][/tex]
Simplify the expression:
[tex]\[
\Delta H_{rxn} = -314.4 \, \text{kJ/mol} + 138.49 \, \text{kJ/mol} = -175.90999999999997 \, \text{kJ/mol}
\][/tex]
Thus, the enthalpy change for the reaction, \(\Delta H_{rxn}\), is:
[tex]\[
\boxed{-175.91 \, \text{kJ}}
\][/tex]
