What is revealed in the reaction [tex]\( Ca(OH)_2(s) \rightarrow Ca^{2+}(aq) + 2OH^{-}(aq), \Delta H = -16.71 \)[/tex] ?

A. The reaction produces heat.
B. The reaction is endothermic.
C. The reaction occurs slowly.
D. Heat is one of the reactants.



Answer :

To answer this question, we need to analyze the given chemical reaction and the enthalpy change (ΔH) associated with it.

The reaction provided is:
[tex]\[ \text{Ca(OH)}_2(s) \rightarrow \text{Ca}^{2+}(aq) + 2 \text{OH}^-(aq) \][/tex]

The enthalpy change (ΔH) for this reaction is given as -16.71.

1. Understanding Enthalpy Change (ΔH):
- ΔH represents the change in enthalpy (heat content) during a reaction.
- A negative ΔH (-16.71) indicates that the reaction releases heat to its surroundings.

Next, let's evaluate each of the given options:

A. The reaction produces heat.
- Since ΔH is negative, this option is correct. The reaction releases heat, which means it produces heat.

B. The reaction is endothermic.
- Endothermic reactions absorb heat and have a positive ΔH. Since ΔH is negative here (-16.71), this reaction is exothermic and not endothermic. Thus, this option is incorrect.

C. The reaction occurs slowly.
- ΔH gives information about the heat change, not the rate of the reaction. Therefore, based on the given enthalpy change, we cannot conclude anything about the speed of the reaction. This option is incorrect.

D. Heat is one of the reactants.
- Since the reaction has a negative ΔH, heat is released as a product, not consumed as a reactant. Therefore, this option is incorrect.

Based on the analysis, the correct answer is:

A. The reaction produces heat.

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