Question

If standard enthalpies of formation of $ CaCl (s) $ (hypothetical) and that of $ CaCl_2(s) $ are $ − 188 \,J \,mol^{-1} $ and $ − 795\, kJ\, mol^{−1} $ respectively, calculate the value of standard heat of reaction for the following disproportionation reaction
$ 2CaCl(s) \longrightarrow CaCl2(s) +Ca(s)$

• A. $ − 607 \,kJ\, mol^{−1} $
• B. $ + 607 \,kJ\, mol^{−1} $
• C. $ − 419 \,kJ\, mol^{−1} $
• D. $ + 419 \,kJ\, mol^{−1} $

Answer 1

Answer: (C)

In disproportionation reaction same species (here $Ca ^{2+}$ ) is oxidised as well as reduced.



Standard heat of reaction, $\Delta H^{\circ}=\Delta H_{P}^{\circ}-\Delta H_{R}^{\circ}$

$=\Delta H_{F}^{\circ}\left( CaCl _{2}\right)+\Delta H_{F}^{\circ}( Ca )-2 \Delta H_{F}^{\circ}( CaCl )$

$=(-795+0)-(-2 \times 188)$

$=-419\, kJ\, mol ^{-1}$