Question

What is the enthalpy of the disproportion of MgCl if the enthalpy of formation of hypothetical MgCl is -125 kJ/mol and the $MgC{l}_2$ is -642 kJ mol${}^{-1}$?

• A. -767 kJ mol${}^{-1}$.
• B. 767 kJ mol${}^{-1}$.
• C. -392 kJ mol${}^{-1}$.
• D. 392 kJ mol${}^{-1}$.

Answer 1

The correct option is C -392 kJ mol${}^{-1}$.

$Mg\left(s\right)+C{l}_2\left(g\right)⟶MgC{l}_2\left(s\right)$

$△H_1$ = - 642 kJ mol${}^{-1}$.

$Mg\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)⟶MgCl\left(s\right)$

$△H_1$ = - 125 kJ mol${}^{-1}$.

$2MgCl⟶MgC{l}_2+Mg$

$\Rightarrow \Delta H=\Delta H_1-2\Delta H_2$

= -642 - 2 $\times$ (-125) = -392 kJ mol${}^{-1}$.

Hence, the correct option is $C$