Question

The thermite reaction is spectacular and exothermic.  Iron(III) oxide, Fe2O3, and metallic aluminum produce molten iron and aluminum oxide in a few seconds, according to the equation(/4):

 Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe

 (a)  Given the following, calculate ΔH for the thermite reaction.  Show your work.                        2 Al(s) + O2 (g) →Al2O3(s)         ΔH = -1670 kJ

                 2 Fe(s) + O2 (g) → Fe2O3(s)      ΔH = - 822 kJ

Answer 1

Gives

$Fe_2o_3++2Al\rightarrow Al_2o_3+2Fe$

$2Al+o_2\rightarrow Al_2o_3$ ∆H=-1670KJ

$2Fe+o_2\rightarrow fe_2o_3$ ∆H=-822kJ

Now we can written as

$$∆H_{rxn}=1mol\times(-822kJ)-(1670)\times1+2\times0$$

$$∆H_{rxn}=-822+1670KJ=848KJ$$