Question

1. calculate the value of \\( \delta h_{\mathrm{rxn}}^{\circ} \\) (in \\( \mathrm{kj} \\)) for the following reaction:

\\( 2 \mathrm{h}_{2} \mathrm{~s}(g)+3 \mathrm{o}_{2}(g) \
ightarrow 2 \mathrm{h}_{2} \mathrm{o}(l)+2 \mathrm{so}_{2}(g) \\)
\

$$\begin{tabular}{|c|c|} \\hline substance & \\( \\delta h_{\\mathrm{f}}^{\\circ} \\mathrm{kj} / \\mathrm{mol} \\) \\\\ \\hline \\( \\mathrm{h}_{2} \\mathrm{~s}(g) \\) & \\( -20.15 \\) \\\\ \\( \\mathrm{h}_{2} \\mathrm{o}(l) \\) & \\( -285.8 \\) \\\\ \\( \\mathrm{so}_{2}(g) \\) & \\( -296.4 \\) \\\\ \\hline \\end{tabular}$$

a. \\( -561.8 \mathrm{~kj} \\)
b. \\( -1124 \mathrm{~kj} \\)
c. \\( -1143 \mathrm{~kj} \\)
d. \\( -1204 \mathrm{~kj} \\)
e. \\( -732.3 \mathrm{~kj} \\)

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 the sum of $\Delta H^{\circ}_{f}$ for products

For products: $n_1 = 2$ (for $H_2O(l)$) and $n_2=2$ (for $SO_2(g)$).
$\sum n\Delta H^{\circ}_{f}(\text{products})=2\times(- 285.8)+2\times(-296.4)$
$=-571.6-592.8=-1164.4\ \text{kJ/mol}$

Step3: Calculate the sum of $\Delta H^{\circ}_{f}$ for reactants

For reactants: $m_1 = 2$ (for $H_2S(g)$) and $m_2 = 3$ (for $O_2(g)$, and $\Delta H^{\circ}_{f}(O_2(g)) = 0$).
$\sum m\Delta H^{\circ}_{f}(\text{reactants})=2\times(-20.15)+3\times0=-40.3\ \text{kJ/mol}$

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

$\Delta H^{\circ}_{rxn}=-1164.4-(-40.3)=-1164.4 + 40.3=-1124.1\approx - 1124\ \text{kJ/mol}$

Answer:

B. -1124 kJ