Question

${\Delta }_f{H}^{\ominus }$ of hypothetical $MgCl$ is $-125kJmo{l}^{-1}$ and for $MgC{l}_2$ is $-642kJmo{l}^{-1}$. The enthalpy of disproportionation of $MgCl$ is $-49x$. Find the value of x.

• A. $x=8$
• B. $x=8.5$
• C. $x=16$
• D. None of these

Answer 1

The correct option is A $x=8$

The reaction for formation of MgCl is as shown.
$Mg\left(s\right)+1/2C{l}_2\left(g\right)⟶MgCl$......(i) $\Delta H_1=-125kJmo{l}^{-1}$
The reaction for formation of $MgC{l}_2$ is as shown
$Mg\left(s\right)+C{l}_2\left(g\right)⟶MgC{l}_2$......(ii) $\Delta H_2=-642kJmo{l}^{-1}$
The disproportionation reaction is as shown.
$2MgCl⟶Mg+MgC{l}_2$ ......(iii) $\Delta H=?$
The reaction (i) is multiplied by 2 and subtracted from reaction (ii) to obtain reaction (iii)
The enthalpy of disproportionation of MgCl is
$\Delta H=\Delta H_2-2\Delta H_1=-642-(2\times -125)=-392kJmo{l}^{-1}$
$\therefore -49x=-392$
$x=8$