Q9.76CP - Chemistry: Molecular Nature Of Matter And Change

Question

Even though so much energy is required to form a metal cation with a 2+ charge, the alkaline earth metals form halides with general formula 

 ${\mathbf{MX}}_2$, rather than MX. 

(a) Use the following data to calculate the  of MgCl: 

Mg(s)                      Mg(g)  = 148 kJ 

${\mathbf{Cl}}_2$  (g)          2Cl(g)  -243 kJ 

Mg(g)            (g) + e^-  =738 kJ 

Cl(g) +        (g)  = -349 kJ

   of MgCl = 783.5 kJ/mol.

(b) Is MgCl favoured energetically relative to Mg and  ? Explain. 

(c) Use Hess’s law to calculate ∆H° for the conversion of MgCl to   and Mg ( of   = -641.6 kJ/mol). 

(d) Is MgCl favoured energetically relative to  ? Explain.

Step-by-Step Solution

Answer

1. The ${\text{ΔH°}}_{\text{f}}$ of  MgCl is -125kJ.
2.  MgCl is energetically favoured than Mg and  because the enthalpy of formation  of is negative. The  $\text{ΔH°}$ for the conversion of  $\text{MgCl}$ to ${\text{MgCl}}_{\text{2}}$ and Mg is -392kJ.
3. ${\text{MgCl}}_{\text{2}}$ is energetically favoured because as the enthalpy of the formation of the ${\text{MgCl}}_{\text{2}}$ is more negative.

Enthalpy of formation of MgCl

1. The overall reaction of the above equation:

 

 $\mathbf{Mg}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\frac{1}{2}\mathrm{Cl}2\stackrel{\mathbf{\Delta H}{\mathbf{°}}_f}{\to }\mathbf{MgCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

According to Hess’s Law,

 

 

$\begin{array}{c}{\text{ΔH°}}_{\text{f}}\text{=(Enthalpy\hspace{0.17em}of\hspace{0.17em}Reactant)-(Enthalpy\hspace{0.17em}of\hspace{0.17em}Product)}\\ {\text{ΔH°}}_{\text{f}}\text{=148kJ+738kJ+}\frac{\text{243}}{\text{2}}\text{+(-349kJ)+(-783.5kJ)}\\ {\text{ΔH°}}_{\text{f}}\text{=-125kJ}\end{array}$

Greater stability of MgCl as compared to Mg and Cl 2

MgCl is energetically favoured than Mg and Cl₂ because the enthalpy of the formation of molecule is negative and which favourable. The reaction is exothermic reaction which means there is the liberation of heat after the formation of product.

Enthalpy of conversion of MgCl    to    MgCl 2   and    Mg

The overall reaction of the above situation:

 $\mathbf{2}\mathbf{MgCl}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\stackrel{\mathbf{\Delta H}{\mathbf{°}}_f}{\to }\mathbf{MgCl}\mathbf{2}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{Mg}$

 

According to Hess’s law, the enthalpy of formation is:

 $\begin{array}{c}{\text{ΔH°}}_{\text{f}}\text{=(EnthalpyofReactant)-(EnthalpyofProduct)}\\ {\text{ΔH°}}_{\text{f}}\text{=2×125kJ+(-641.6kJ)}\\ {\text{ΔH°}}_{\text{f}}\text{=-392kJ}\end{array}$

reason for greater stability of MgCl 2

${\mathrm{MgCl}}_2$ is energetically favoured because as the enthalpy of the formation of the  is more negative as compared to the heat of formation of   ${\mathrm{MgCl}}_2$. The reaction is exothermic reaction which means there is the liberation of heat after the formation of product. ${\mathrm{MgCl}}_2$  is unstable than the  molecule.