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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Similar Questions
The enthalpies of formation of $CO_{2(g)}$,$H_2O_{(g)}$,and $C_2H_{4(g)}$ are $-393.7$,$-241.8$,and $52.3 \ kJ \ mol^{-1}$ respectively. What will be the enthalpy of combustion of ethylene at $298 \ K$ and $1 \ atm$ pressure in $kJ \ mol^{-1}$?
Difficult
View SolutionChoose the reaction$(s)$ from the following options,for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$
MediumIIT 2019
View SolutionFrom the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:
$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ $\Delta H = 42.09 \ kJ \ mol^{-1}$ $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ $\Delta H = -242 \ kJ \ mol^{-1}$ $H_{2(g)} \rightarrow 2 H_{(g)}$ $\Delta H = 436 \ kJ \ mol^{-1}$ $O_{2(g)} \rightarrow 2 O_{(g)}$ $\Delta H = 496 \ kJ \ mol^{-1}$
| $1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ | $\Delta H = 42.09 \ kJ \ mol^{-1}$ |
| $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ | $\Delta H = -242 \ kJ \ mol^{-1}$ |
| $H_{2(g)} \rightarrow 2 H_{(g)}$ | $\Delta H = 436 \ kJ \ mol^{-1}$ |
| $O_{2(g)} \rightarrow 2 O_{(g)}$ | $\Delta H = 496 \ kJ \ mol^{-1}$ |
Calculate the heat of combustion (in $kJ$) of methane from the following data:
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$
DifficultAP EAMCET 2002
View SolutionAt $300 \, K$,the standard enthalpies of formation of $C_6H_5COOH_{(s)}$,$CO_{2_{(g)}}$,and $H_2O_{(l)}$ are $-408$,$-393$,and $-286 \, kJ \, mol^{-1}$ respectively. What is the heat of combustion of benzoic acid at constant volume in $kJ$ (in $.75$)? $(R = 8.31 \, J \, mol^{-1} \, K^{-1})$
Difficult
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