Question

$\Delta_{ f } H ^{\circ}$ of hypothetical $MX$ is $-250\, kJ \,mol ^{-1}$ and for $MX _{2}$ is $-600\, kJ \,mol ^{-1}$. The enthalpy of disproportionation of $MX$ is $-100 x \,kJ \,mol ^{-1}$. Find the value of x

Answer 1

$2 MX \rightarrow M + MX _{2} ; \Delta H ^{\circ}=$ ?

$M ( s )+\frac{1}{2} \rightarrow 2 \times MX ; \Delta_{ f } H _{1}^{\circ}=-250 \,kJ\, mol ^{-1}$

$M ( s )+ X _{2} \rightarrow MX _{2} ; \Delta_{ f } H _{2}^{-}=-600 \,kJ \,mol ^{-1}$

$\therefore \Delta H ^{\circ}=\Delta_{ f } H _{2}^{\circ}-2 \Delta_{ f } H _{1}^{\circ}$

$=-600-(2 \times 250)=-100 \,kJ \,mol ^{-1}$

$\therefore -100 x =-100$

$\therefore x =1$