Question #1 Born-Haber Cycle H = 31 H (g) H Question #1 Born-Haber Cycle Use Born-Haber cycles calculations to show why formation of the salt CaBr2 is favoured over CaBr. Given the following information for magnesium, oxygen, and magnesium oxide calculate the second electron gain Applying Hess,s law, we get `Delta_("lattice")H^(Θ)= 411.2 + 108.4 + 121 + 496 - 348.Here are 10 results for born haber cycle questions:Ĭhem 1711 Born-Haber Cycle, Practice Problems Chem 1711 Born-Haber Cycle, Practice Problems 1. The importance of the cycle is that, the sum of the enthalpy changes round a cycle is zero. The electron gain enthalpy, `Delta_(eg)H^(Θ) = -348.6 kJ mol^(-1)` (v) `Na^(+)(g)+Cl^(-)(g) rarr Na^(+)Cl^(-)(s)` The sequence of steps is shown in given figure and is shown as Born-Haber cycle. Let us now calculate the enthalpy of `Na^(+)Cl^(-)` (s) by following steps given below (i) `Na^(+)(s) rarr Na(g)`, Sublimation of sodium metal, `Delta_(sub)H^(Θ) = 108.4 kJ mol^(-1)` (ii) `Na(g) rarr Na^(+)(g)+e^(-)(g)`, The ionisation of sodium atoms, ionisation enthalpy `Delta_(i)H^(Θ) = 496 kJ mol^(-1)` (iii) `(1)/(2) Cl_(2)(g) rarr Cl(g)`, The dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy `(1)/(2)Delta_("bond") H^(Θ) = 121 kJ mol^(-1)` (iv) `Cl(g)+e^(-)(g) rarr Cl^(-)(g)`, electron gained by chlorine atoms. For the reaction `Na^(+)Cl^(-)(S) rarr Na^(+)(g)+Cl^(-)(g),Delta_("lattice")H^(Θ) = +788 kJ mol^(-1)` Since, it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber cycle. Solution :The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.
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