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Ammonia [tex]$\left( NH_3(g), \Delta H_{f} = -46.19 \, \text{kJ/mol} \right)$[/tex] reacts with hydrogen chloride [tex]$\left( HCl(g), \Delta H_{f} = -92.30 \, \text{kJ/mol} \right)$[/tex] to form ammonium chloride [tex]$\left( NH_4Cl(s), \Delta H_{f} = -314.4 \, \text{kJ/mol} \right)$[/tex] according to this equation:

[tex]\[ NH_3(g) + HCl(g) \rightarrow NH_4Cl(s) \][/tex]

What is [tex]$\Delta H_{rxn}$[/tex] for this reaction?

[tex]\[\Delta H_{rxn} = \square \, \text{kJ}\][/tex]



Answer :

GinnyAnswer
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]

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