The enthalpies of formation of `Al_2O_3` and `Cr_2O_3` are -1596 kJ and -1134 KJ, respectively . `DeltaH` for reaction `2Al+Cr_2O_3 to 2Cr +Al_2O_3` is _____

Options

A. `-1365` kJ
B. 2730 kJ
C. `-2730` kJ
D. `-462` kJ

Answer

D

Explanation

`2Al+Cr_2O_3 to 2Cr + Al_2O_3`
`DeltaH_r^0=[2DeltaH_f (Cr)+DeltaH_f (Al_2O_3) ]-[2DeltaH_f (Al) + DeltaH_f (Cr_2O_3)]`
`DeltaH_r^0=[0+(-1596 kJ ) ]-[0+(-1134)]`
`DeltaH_r^0`=-1596 kJ + 1134 kJ
`DeltaH_r^0`=-462 kJ