Question

what is \\( \delta h^{\circ}_{rxn} \\) for the following reaction?
\\( \mathrm{sio}_{2}(s)+4 \mathrm{hcl}(g) \
ightarrow \mathrm{sicl}_{4}(g)+2 \mathrm{h}_{2} \mathrm{o}(g) \\)
\\( \

$$\begin{array}{lc}\\text { substance } & \\delta h^{\\circ}{ }_{f}(\\mathrm{~kj} / \\mathrm{mol}) \\\\ \\mathrm{sio}_{2}(s) & -859.3 \\\\ \\mathrm{sicl}_{4}(g) & -662.8 \\\\ \\mathrm{hcl}(g) & -92.3 \\\\ \\mathrm{h}_{2} \\mathrm{o}(g) & -241.8\\end{array}$$

\\)
\\( \bigcirc-1856.2 \mathrm{~kj} / \mathrm{mol} \\)
\\( \bigcirc-1372.6 \mathrm{~kj} / \mathrm{mol} \\)
\\( \bigcirc-47.0 \mathrm{~kj} / \mathrm{mol} \\)
\\( \bigcirc 82.1 \mathrm{~kj} / \mathrm{mol} \\)
\\( \bigcirc 530.6 \mathrm{~kj} / \mathrm{mol} \\)

Explanation:

Step1: Recall the formula for $\Delta H^{\circ}_{rxn}$

The formula is $\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(\text{products})-\sum m\Delta H^{\circ}_{f}(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients.

Step2: Calculate $\sum n\Delta H^{\circ}_{f}(\text{products})$

For products: $\text{SiCl}_{4}(g)$ and $2\text{H}_{2}\text{O}(g)$.
$\sum n\Delta H^{\circ}_{f}(\text{products})=\Delta H^{\circ}_{f}(\text{SiCl}_{4}(g)) + 2\Delta H^{\circ}_{f}(\text{H}_{2}\text{O}(g))$
Substitute the values: $(- 662.8)+2\times(-241.8)=-662.8 - 483.6=-1146.4$ kJ/mol.

Step3: Calculate $\sum m\Delta H^{\circ}_{f}(\text{reactants})$

For reactants: $\text{SiO}_{2}(s)$ and $4\text{HCl}(g)$.
$\sum m\Delta H^{\circ}_{f}(\text{reactants})=\Delta H^{\circ}_{f}(\text{SiO}_{2}(s))+4\Delta H^{\circ}_{f}(\text{HCl}(g))$
Substitute the values: $(-859.3)+4\times(-92.3)=-859.3-369.2 = - 1228.5$ kJ/mol.

Step4: Calculate $\Delta H^{\circ}_{rxn}$

$\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(\text{products})-\sum m\Delta H^{\circ}_{f}(\text{reactants})$
$\Delta H^{\circ}_{rxn}=-1146.4-(-1228.5)=-1146.4 + 1228.5 = 82.1$ kJ/mol.

Answer:

$82.1$ kJ/mol