Explain Enthalpy of Solution $\Delta_{sol}H^{\theta}$.
(N/A) The enthalpy of solution of a substance is the enthalpy change when $1 \ mole$ of it dissolves in a specified amount of solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving the substance in an infinite amount of solvent when the interactions between the ions are negligible.
When an ionic compound dissolves in a solvent,the ions leave their ordered positions on the crystal lattice. These are now more free in solution. Simultaneously,solvation of these ions occurs. This is shown diagrammatically for an ionic compound,$AB_{(s)}$:
$\Delta_{sol}H^{\theta} = \Delta_{lattice}H^{\theta} + \Delta_{hyd}H^{\theta}$
For most ionic compounds,$\Delta_{sol}H^{\theta}$ is positive and the dissociation process is endothermic.
When an ionic compound dissolves in a solvent,the ions leave their ordered positions on the crystal lattice. These are now more free in solution. Simultaneously,solvation of these ions occurs. This is shown diagrammatically for an ionic compound,$AB_{(s)}$:
$\Delta_{sol}H^{\theta} = \Delta_{lattice}H^{\theta} + \Delta_{hyd}H^{\theta}$
For most ionic compounds,$\Delta_{sol}H^{\theta}$ is positive and the dissociation process is endothermic.
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DifficultMHT CET 2020
View SolutionThe heat of formation of $CO_{(g)}$ and $CO_{2(g)}$ are $-26.4 \ kcal$ and $-94.0 \ kcal$ respectively. The heat of combustion of carbon monoxide will be $... \ kcal$.
Medium
View SolutionComment on the thermodynamic stability of $NO_{(g)}$,given:
$\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{(g)}; \Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}; \Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$
$\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{(g)}; \Delta_r H^{\ominus} = 90 \, kJ \, mol^{-1}$
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow NO_{2(g)}; \Delta_r H^{\ominus} = -74 \, kJ \, mol^{-1}$
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View SolutionConsider the following reactions:
$(i)$ $H_{(aq)}^{+} + OH^{-}_{(aq)} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_1 \ kJ \ mol^{-1}$
$(ii)$ $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$
$(iii)$ $CO_{2_{(g)}} + H_{2_{(g)}} \longrightarrow CO_{(g)} + H_2O_{(l)}$,$\Delta H = -X_3 \ kJ \ mol^{-1}$
$(iv)$ $C_2H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + H_2O_{(l)}$,$\Delta H = -X_4 \ kJ \ mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is
$(i)$ $H_{(aq)}^{+} + OH^{-}_{(aq)} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_1 \ kJ \ mol^{-1}$
$(ii)$ $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$
$(iii)$ $CO_{2_{(g)}} + H_{2_{(g)}} \longrightarrow CO_{(g)} + H_2O_{(l)}$,$\Delta H = -X_3 \ kJ \ mol^{-1}$
$(iv)$ $C_2H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + H_2O_{(l)}$,$\Delta H = -X_4 \ kJ \ mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is
MediumAIPMT 2007
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