91影视

(a)

First, write the ionic equation of the given reactants.

${\text{KOH and HNO}}_{\text{3}}$

${\text{K}}^{\text{+}}{\text{+ OH}}^{\text{-}}{\text{+H}}^{\text{+}}{\text{+ NO}}_{\text{3}}^{\text{-}}鈫�{\text{K}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O + NO}}_{\text{3}}^{\text{-}}$

Cancel out the spectator ions and write the net equation.

${\text{OH}}^{\text{-}}{\text{+H}}^{\text{+}}鈫�{\text{H}}_{\text{2}}\text{O}$

Then, from Appendix B, calculate the$鈭�H_{r脳n}^{鈭�}$.

role="math" localid="1663412574540" $$\begin{gathered} 鈭�H_{r脳n}^{鈭�}=鈭�{\text{苍鈭咹}}_f^{鈭�}\text{products -}鈭�\text{m}鈭�H_f^{鈭�}\text{reactants} \\ =鈭�H_f^{鈭�},{\text{H}}_{\text{2}}\text{O(l) -[}鈭�H_f^{鈭�},{\text{OH}}^{\text{-}}\text{(aq) +}鈭�H_f^{鈭�},{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)]} \\ =-285.840kJ/mol-[\text{( - 229.94kJ/mol)+0]}鈭�H_{r脳n}^{鈭�} \\ =\text{- 55.9kJ/mol} \end{gathered}$$

$\text{NaOH and HCl}$

${\text{Na}}^{\text{+}}{\text{+ OH}}^{\text{-}}{\text{+H}}^{\text{+}}{\text{+ Cl}}^{\text{-}}鈫�{\text{Na}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O + Cl}}^{\text{-}}$

Cancel out the spectator ions and write the net equation.

${\text{OH}}^{\text{-}}{\text{+H}}^{\text{+}}鈫�{\text{H}}_{\text{2}}\text{O}$

Since it has the same net equation as the previous reaction, the$鈭�H_{r脳n}^{鈭�}$will be the same. Therefore, for the reaction of NaOH and HCI,role="math" localid="1663412715397" $鈭�H_{r脳n}^{鈭�}=-55.9kJ/mol$

(b)

The reactants in the reactions in the previous problem are all strong acids and bases. The reaction between a strong acid and strong base is a neutratization reaction and they will all have the same net equation.

$$\begin{gathered} {\text{K}}^{\text{+}}{\text{+ OH}}^{\text{-}}{\text{+H}}^{\text{+}}{\text{+ Cl}}^{\text{-}}鈫�{\text{K}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{O + Cl}}^{\text{-}} \\ {\text{OH}}^{\text{-}}{\text{+H}}^{\text{+}}鈫�{\text{H}}_{\text{2}}\text{O} \end{gathered}$$

Hence the $鈭�H_{r脳n}^{鈭�}=-55.9kJ/mol$