Question

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

\\( 2 h_2 s(g)+3 o_2(g) \
ightarrow 2 h_2 o(l)+2 so_2(g) \\)

a. -561.8 kj
b. -1124 kj
c. -1143 kj
d. -1204 kj
e. -732.3 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 $\sum n\Delta H^{\circ}_{f}(\text{products})$

For products: $n_{H_2O} = 2$, $\Delta H^{\circ}_{f}(H_2O(l))=- 285.8\ \text{kJ/mol}$; $n_{SO_2}=2$, $\Delta H^{\circ}_{f}(SO_2(g))=-296.4\ \text{kJ/mol}$.

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

For reactants: $m_{H_2S}=2$, $\Delta H^{\circ}_{f}(H_2S(g))=-20.15\ \text{kJ/mol}$; $O_2(g)$ is in its standard state, so $\Delta H^{\circ}_{f}(O_2(g)) = 0\ \text{kJ/mol}$.

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

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Answer:

B. -1124 kJ