Question

If standard enthalpies of formation of $CaCl\left(s\right)$ (hypothetical) and that of $Ca{Cl}_2\left(s\right)$ are $-188J{mol}^{-1}$ and $-795J{mol}^{-1}$ respectively. Calculate the value of standard heat of reaction for the following disproportionation reaction.

$2CaCl\left(s\right)⟶Ca{Cl}_2\left(s\right)+Ca\left(s\right)$

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

Answer 1

The correct option is C $-419kJ{mol}^{-1}$

$\underset{+1}{2Ca}Cl\left(s\right)\to \underset{+2}{Ca}C{l}_2\left(s\right)+\underset{0}{Ca}$

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

Standard heat of reaction, $\Delta H^o=\Delta H_P^o-\Delta H_R^o$

$=\Delta H_f^o\left(Ca{Cl}_2\right)+\Delta H_f^o\left(Ca\right)-2\Delta H_f^o\left(CaCl\right)=-419kJ{mol}^{-1}$

Hence, option $C$ is correct.