Question

calculating enthalpy of reaction
ammonia nh₃(g), δhf = - 46.19 kj/mol reacts with hydrogen chloride hcl(g), δhf = - 92.30 kj/mol to form ammonium chloride nh₄cl(s), δhf = - 314.4 kj/mol according to this equation:
nh₃(g) + hcl(g) → nh₄cl(s)
what is δhrxn for this reaction?

kj
is this reaction endothermic or exothermic?

Explanation:

Step1: Recall enthalpy - reaction formula

The enthalpy of a reaction $\Delta H_{rxn}$ is given by $\Delta H_{rxn}=\sum\Delta H_{f(products)}-\sum\Delta H_{f(reactants)}$.

Step2: Identify reactants and products enthalpies

For the reaction $NH_3(g)+HCl(g)
ightarrow NH_4Cl(s)$, $\Delta H_{f}(NH_3)= - 46.19\ kJ/mol$, $\Delta H_{f}(HCl)=-92.30\ kJ/mol$ and $\Delta H_{f}(NH_4Cl)=-314.4\ kJ/mol$.

Step3: Calculate $\Delta H_{rxn}$

$\Delta H_{rxn}=\Delta H_{f}(NH_4Cl)-[\Delta H_{f}(NH_3)+\Delta H_{f}(HCl)]$
$=-314.4-(-46.19 - 92.30)$
$=-314.4-(-138.49)$
$=-314.4 + 138.49=-175.91\ kJ$.

Step4: Determine reaction type

Since $\Delta H_{rxn}=-175.91\ kJ<0$, the reaction is exothermic.

Answer:

-175.91 kJ; Exothermic