To determine whether each of the reactions is endothermic or exothermic, we need to examine the sign of the enthalpy change ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) for each reaction:
1. Reaction 1:
[tex]\[
NaOH (s) \rightarrow Na^{+}(aq) + OH^{-}(aq) \quad \Delta H_{\text{rxn}} = -44.5 \, \text{kJ}
\][/tex]
The enthalpy change [tex]\(\Delta H_{\text{rxn}}\)[/tex] is given as [tex]\(-44.5\)[/tex] kJ. Because [tex]\(\Delta H_{\text{rxn}}\)[/tex] is negative, this indicates that the reaction releases heat to the surroundings. Therefore, this reaction is exothermic.
2. Reaction 2:
[tex]\[
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \quad \Delta H_{\text{rxn}} < 0
\][/tex]
The enthalpy change [tex]\(\Delta H_{\text{rxn}}\)[/tex] is indicated to be less than zero ([tex]\(\Delta H_{\text{rxn}} < 0\)[/tex]). When [tex]\(\Delta H_{\text{rxn}}\)[/tex] is less than zero, it means the reaction releases heat to the surroundings. As a result, this reaction is also exothermic.
3. Reaction 3:
[tex]\[
CH_3COOH(aq) + NaHCO_3(s) \rightarrow CO_2(g) + H_2O(l) + Na^{+}(aq) + CH_3COO^{-}(aq) \quad \Delta H_{\text{rxn}} > 0
\][/tex]
The enthalpy change [tex]\(\Delta H_{\text{rxn}}\)[/tex] is given as greater than zero ([tex]\(\Delta H_{\text{rxn}} > 0\)[/tex]). When [tex]\(\Delta H_{\text{rxn}}\)[/tex] is positive, it means the reaction absorbs heat from the surroundings. Thus, this reaction is endothermic.
Summarizing the results:
1. [tex]\[
NaOH (s) \rightarrow Na^{+}(aq) + OH^{-}(aq) \quad \Delta H_{\text{rxn}} = -44.5 \, \text{kJ}
\][/tex]
Exothermic
2. [tex]\[
CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(g) \quad \Delta H_{\text{rxn}} < 0
\][/tex]
Exothermic
3. [tex]\[
CH_3COOH(aq) + NaHCO_3(s) \rightarrow CO_2(g) + H_2O(l) + Na^{+}(aq) + CH_3COO^{-}(aq) \quad \Delta H_{\text{rxn}} > 0
\][/tex]
Endothermic
