Identify I, II, III in the given Born-Haber c | vedclass

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

Question

(B) The process $I$ represents the direct dissolution of the solid $AB(s)$ into its aqueous ions,which is the enthalpy of solution,$\Delta_{sol}H^o$.T

Vedclass · Accepted Answer

$\Delta_{sol}H^o, \Delta_{lattice}H^o, \Delta_{hyd}H^o$

Vedclass · Answer

(B) The process $I$ represents the direct dissolution of the solid $AB(s)$ into its aqueous ions,which is the enthalpy of solution,$\Delta_{sol}H^o$.The process $II$ represents the breaking of the ionic lattice into gaseous ions,which is the lattice enthalpy,$\Delta_{lattice}H^o$.The process $III$ represents the hydration of gaseous ions to form aqueous ions,which is the enthalpy of hydration,$\Delta_{hyd}H^o$.According to Hess's Law,$\Delta_{sol}H^o = \Delta_{lattice}H^o + \Delta_{hyd}H^o$.Comparing this with the diagram:$I = \Delta_{sol}H^o$$II = \Delta_{lattice}H^o$$III = \Delta_{hyd}H^o$Therefore,the correct sequence is $\Delta_{sol}H^o, \Delta_{lattice}H^o, \Delta_{hyd}H^o$.

Identify $I, II, III$ in the given Born-Haber cycle diagram for the dissolution of an ionic solid $AB(s)$:
$AB(s) \xrightarrow{I} A^+(aq) + B^-(aq)$
$AB(s) \xrightarrow{II} A^+(g) + B^-(g)$
$A^+(g) + B^-(g) \xrightarrow{III} A^+(aq) + B^-(aq)$

Similar Questions

The bond dissociation enthalpies of $H_2$, $Cl_2$, and $HCl$ are $434 \text{ kJ mol}^{-1}$, $242 \text{ kJ mol}^{-1}$, and $431 \text{ kJ mol}^{-1}$ respectively. Calculate the enthalpy of formation of $HCl$.

Enthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:

Explain the bond enthalpy.

The standard enthalpy of combustion at $25\,^{\circ}C$ of hydrogen,cyclohexene $(C_6H_{10})$ and cyclohexane $(C_6H_{12})$ are $-241$,$-3800$ and $-3920\, kJ/mol$ respectively. Calculate the heat of hydrogenation of cyclohexene in $kJ/mol$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Identify $I, II, III$ in the given Born-Haber cycle diagram for the dissolution of an ionic solid $AB(s)$:$AB(s) \xrightarrow{I} A^+(aq) + B^-(aq)$$AB(s) \xrightarrow{II} A^+(g) + B^-(g)$$A^+(g) + B^-(g) \xrightarrow{III} A^+(aq) + B^-(aq)$