Question

What is the enthalpy change for the given reaction, if the enthalpies of formation of $A{l_2}{O_3}$ and $F{e_2}{O_3}$ are $- 1670kJmo{l^{ - 1}}$ and $- 834kJmo{l^{ - 1}}$ respectively?
$F{e_2}{O_3} + 2Al \to A{l_2}{O_3} + 2Fe$
(A) $- 836kJmo{l^{ - 1}}$
(B) $+ 836kJmo{l^{ - 1}}$
(C) $- 424kJmo{l^{ - 1}}$
(D) $+ 424kJmo{l^{ - 1}}$

Answer 1

The enthalpy change for the reaction is given by the change in the enthalpies of the product minus the change in the enthalpies of the reactants. The enthalpy formation of metals is zero which is constant. In the given reaction aluminum and iron are metals which have the enthalpy formation of zero.

Complete Step By Step Answer:
Enthalpy is defined as the amount of heat absorbed or released when the number of moles of reactants was completely reacted to give products.
Given reaction is $F{e_2}{O_3} + 2Al \to A{l_2}{O_3} + 2Fe$
In the above reaction, ferric oxide which has the molecular formula of $F{e_2}{O_3}$ reacts with aluminium metal to form aluminium oxide and iron metal.
Chemical reactions are of different types based on the reaction mechanism. Here in the above reaction, the single displacement was undergone.
The enthalpy change for a given reaction will be equal to the enthalpy formation of products minus the enthalpy formation of reactants.
Thus, by substituting the values of enthalpy formation in the below formula,
$\Delta H = \Delta {H_p} - \Delta {H_R}$
$\Delta H = \left( { - 1670 + 0} \right) - \left( { - 834 + 0} \right)$
By simplification, the enthalpy change value will be $\Delta H = - 836kJmo{l^{ - 1}}$
Thus, when the enthalpies of formation of $A{l_2}{O_3}$ and $F{e_2}{O_3}$ are $- 1670kJmo{l^{ - 1}}$ and $- 834kJmo{l^{ - 1}}$ then the enthalpy change for a given reaction will be $- 836kJmo{l^{ - 1}}$ .
Option A is the correct one.

Note:
The enthalpy formation of only one reactant and one product were given, whereas the enthalpy formation of aluminum and iron was not given. They should be taken as zero. As the enthalpy formation of elements in the metallic state is zero.