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[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

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...

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system

Calculate the enthalpy change for a chemical reaction involving 1 mole of $\mathrm{H_2S}$ (gas) and $\mathrm{H_2O}$ (liquid) and $\mathrm{SO_2}$ (gas) from their standard formation enthalpies, using Hess's Law which states:

$$
\Delta H=\Sigma \Delta H_{\text {(products) }}-\Sigma \Delta H_{\text {(reactants) }},
$$

where $\Sigma \Delta H_{\text {(products) }}$ and $\Sigma \Delta H_{\text {(reactants) }}$ are the sums of the standard formation enthalpies of the products and reactants of the reaction, respectively.

Given values:

$\Delta H_{\text{products}} = -286.0 \text{ kJ} - 297.0 \text{ kJ}$

$\Delta H_{\text{reactants}} = -20.17 \text{ kJ}$

Calculate $\Delta H$ and determine the nature of the reaction (whether it is exothermic or not).

Present the answer in LaTex format: \boxed{Your answer}
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