Question
What is the enthalpy change for the given reaction, if enthalpies of formation of $$A{l_2}{O_3}$$ and $$F{e_2}{O_3}$$ are $$ - 1670\,kJ\,mo{l^{ - 1}}$$ and $$ - 834\,kJ\,mo{l^{ - 1}}$$ respectively ?
$$F{e_2}{O_3} + 2Al \to A{l_2}{O_3} + 2Fe$$
A.
$$ - 836\,kJ\,mo{l^{ - 1}}$$
B.
$$ + 836\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 424\,kJ\,mo{l^{ - 1}}$$
D.
$$ + 424\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 836\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta H = \sum {\Delta {H_P} - \sum {\Delta {H_R}} } \cr
& \,\,\,\,\,\,\,\,\,\, = - 1670 - \left( { - 834} \right) \cr
& \,\,\,\,\,\,\,\,\,\, = - 836\,kJ\,mo{l^{ - 1}} \cr} $$