Multiple Correct Answer type .
The standard enthalpy change of formation of Al2O3(s) is −1669 kJ mol−1 and the standard enthalpy change of formation of Fe2O3(s) is −822 kJ mol−1.
(a) ∆Hο for the following reaction. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s) is -847kJ
(b) the reaction in (a) is exothermic
(c) the magnitude of the entropy change for this reaction will be minimal
(d) No significant change in positional or phase disorder
19initially , i was thinking about all,,,but i refrained because i found out the sign of ΔH0 to be the other way round....may be i have done some calculation errors !
106delta H rkn = delta H products - delta H reactants
= -1669 - (-822) = -847 kJ
as delta H<0 so reaction is exothermic
Entropy change is minimal bcz all species involved are solids.. so no appreciable change in disorder... further moles of reactants=moles of products
there is no phase disorder bcz all the species are solids.
