The enthalpy of neutralization of a strong ac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The neutralization reaction is: $H^{+}_{(aq)} + OH^{-}_{(aq)} \longrightarrow H_{2}O_{(l)}$Given $\Delta H_{neut}^{\circ} = -57.32 \ kJ/mol$.The e

Vedclass · Accepted Answer

$-228.52$

Vedclass · Answer

(C) The neutralization reaction is: $H^{+}_{(aq)} + OH^{-}_{(aq)} \longrightarrow H_{2}O_{(l)}$Given $\Delta H_{neut}^{\circ} = -57.32 \ kJ/mol$.The enthalpy of reaction is given by: $\Delta H_{neut}^{\circ} = \Delta H_{f}^{\circ}(H_{2}O_{(l)}) - [\Delta H_{f}^{\circ}(H^{+}_{(aq)}) + \Delta H_{f}^{\circ}(OH^{-}_{(aq)})]$By convention,$\Delta H_{f}^{\circ}(H^{+}_{(aq)}) = 0 \ kJ/mol$.Substituting the values: $-57.32 = -285.84 - [0 + \Delta H_{f}^{\circ}(OH^{-}_{(aq)})]$$\Delta H_{f}^{\circ}(OH^{-}_{(aq)}) = -285.84 + 57.32 = -228.52 \ kJ/mol$.

The enthalpy of neutralization of a strong acid by a strong base is $-57.32 \ kJ/mol$. The enthalpy of formation of water is $-285.84 \ kJ/mol$. The enthalpy of formation of hydroxyl ion is......$kJ/mol$. (Assume $\Delta H_{f}^{\circ}(H^{+}_{(aq)}) = 0 \ kJ/mol$)

Similar Questions

The bond energy (in $kcal \ mol^{-1}$) of a $C-C$ single bond is approximately:

The standard enthalpies of combustion of $C_6H_{6(l)}$,$C(graphite)$ and $H_{2(g)}$ are respectively $-3270 \ kJ \ mol^{-1}$,$-394 \ kJ \ mol^{-1}$ and $-286 \ kJ \ mol^{-1}$. What is the standard enthalpy of formation of $C_6H_{6(l)}$ in $kJ \ mol^{-1}$?

Given that the bond energy of $H-H$ bonds is $436 \ kJ/mol$,$O-H$ bonds is $464 \ kJ/mol$,and $O=O$ bonds is $496 \ kJ/mol$,what is the approximate heat of reaction for $2H_2 + O_2 \longrightarrow 2H_2O$ in $kJ/mol$?

Given: $2Fe + \frac{3}{2} O_2 \to Fe_2O_3$,$\Delta H = -193.4 \ kJ$;
$Mg + \frac{1}{2} O_2 \to MgO$,$\Delta H = -140.2 \ kJ$.
What is the $\Delta H$ of the reaction $3Mg + Fe_2O_3 \to 3MgO + 2Fe$ in $kJ$?

The heat of combustion of solid benzoic acid at constant volume is $-321.30 \ kJ$ at $27^{\circ} C$. The heat of combustion at constant pressure is $(-321.30 - x R) \ kJ$. The value of $x$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module