The heat of neutralization for 2 moles of te | vedclass

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(A) The reaction is $2 	ext{LiOH} + H_2SO_4 ightarrow Li_2SO_4 + 2H_2O$. Since $	ext{LiOH}$ is a weak base,the heat of neutralization is the sum o

Vedclass · Accepted Answer

$22.5 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The reaction is $2 	ext{LiOH} + H_2SO_4 ightarrow Li_2SO_4 + 2H_2O$. Since $	ext{LiOH}$ is a weak base,the heat of neutralization is the sum of the heat of formation of water from $H^+$ and $OH^-$ $(-57.3 \ kJ \ mol^{-1})$ and the heat of ionization of the weak base. For $2 \ moles$ of $	ext{LiOH}$,the total heat of neutralization is given as $-69.6 \ kJ$. The heat of neutralization for $2 \ moles$ of a strong base would be $2 	imes (-57.3) = -114.6 \ kJ$. The difference is due to the heat of ionization of $2 \ moles$ of $	ext{LiOH}$: $\Delta H_{ionization} = \Delta H_{observed} - \Delta H_{strong} = -69.6 - (-114.6) = 45 \ kJ$. Thus,the heat of ionization for $1 \ mole$ of $	ext{LiOH}$ is $45 / 2 = 22.5 \ kJ \ mol^{-1}$.

The heat of neutralization for $2 \ moles$ of $\text{LiOH}$ and $1 \ mole$ of $H_2SO_4$ at $25^{\circ} C$ is $-69.6 \ kJ$. The heat of ionisation of $\text{LiOH}$ will be nearly $:-$

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