To determine what is revealed in the reaction [tex]\( \text{Ca(OH)}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq) \)[/tex] with [tex]\( \Delta H = -16.71 \)[/tex], we need to understand the significance of the enthalpy change ([tex]\( \Delta H \)[/tex]) value given.
Here are the steps to analyze the problem:
1. Understanding Enthalpy Change ([tex]\( \Delta H \)[/tex]):
- Enthalpy change ([tex]\( \Delta H \)[/tex]) is the heat change at constant pressure.
- If [tex]\( \Delta H \)[/tex] is negative, the reaction releases heat to the surroundings.
- If [tex]\( \Delta H \)[/tex] is positive, the reaction absorbs heat from the surroundings.
2. Interpreting the Given [tex]\( \Delta H \)[/tex] Value:
- The given reaction has [tex]\( \Delta H = -16.71 \)[/tex].
- The negative sign indicates that this reaction releases heat.
3. Determining the Nature of the Reaction:
- Since the reaction releases heat, it is classified as an exothermic reaction.
4. Conclusion Based on the Given Options:
- Option A ("The reaction is endothermic") is incorrect because the reaction is exothermic.
- Option B ("The reaction produces heat") is correct because an exothermic reaction releases heat.
- Option C ("Heat is one of the reactants") is incorrect because heat is released, not absorbed.
- Option D ("The reaction occurs slowly") is irrelevant to the information given, as [tex]\( \Delta H \)[/tex] does not provide information regarding the reaction rate.
Therefore, the correct option is:
B. The reaction produces heat.
