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# What is the enthalpy change in the following reaction, if enthalpies of formation of $\,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,$ and $\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ are $-1670\text{kJmo}{{\text{l}}^{\text{-1}}}$ and $-834\text{kJmo}{{\text{l}}^{\text{-1}}}$ respectively.$\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\,\text{+}\,\,\text{2Al}\,\,\to \,\,\text{A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}\,\,\text{+}\,\text{2Fe}$ (A) $-836\text{kJmo}{{\text{l}}^{\text{-1}}}$(B) $+836\text{kJmo}{{\text{l}}^{\text{-1}}}$(C) $-42\text{4kJmo}{{\text{l}}^{\text{-1}}}$(D) $+424\text{kJmo}{{\text{l}}^{\text{-1}}}$

Last updated date: 09th Sep 2024
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Hint: Enthalpy is the amount of heat evolved or absorbed when the number of moles of the reactant according to the balanced chemical reaction has completely reacted. Enthalpy change of a reaction at constant pressure is represented by $\vartriangle \text{H}$.

Complete Solution :
Heat of formation is the enthalpy change when one mole of a substance is formed from its elements in their most abundant naturally occurring form or in their standard and stable state form. For example, the standard state of oxygen, carbon and sulphur are ${{O}_{2}}$ gas, ${{\text{C}}_{\text{graphite}}}$ and ${{\text{S}}_{\text{rhombic}}}$ respectively. Heat of formation of a reaction is represented by $\vartriangle {{\text{H}}_{\text{f}}}$. The standard heat of formation is represented by $\vartriangle \text{H}_{\text{f}}^{\text{o}}$. Standard heat of formation of all the elements is in stable state and is taken to be zero.
So the change in enthalpy of the reaction -
$\Delta {{\text{H}}_{\text{reaction}}}\,\,=\,\,\sum{\Delta }{{\text{H}}_{\text{f(product)}}}\,+\,\,\sum{\Delta {{\text{H}}_{\text{f(reactant)}}}}.....(1)$
So to calculate the change in the reaction given are-
$\sum{\Delta }{{\text{H}}_{\text{f(product)}}}\,\,=$ $\text{-1670kJmo}{{\text{l}}^{\text{-1}}}$
$\sum{\Delta {{\text{H}}_{\text{f(reactant)}}}}\,=\,$ $\text{-834kJmo}{{\text{l}}^{\text{-1}}}$
Where $\sum{\Delta }{{\text{H}}_{\text{f(product)}}}\,\,=$ enthalpy of formation of product
$\sum{\Delta {{\text{H}}_{\text{f(reactant)}}}}\,=\,$ Enthalpy of formation of product
So on putting the above values in the equation …1
$\Delta {{\text{H}}_{\text{reaction}}}\,\,=\,\,\sum{\Delta }{{\text{H}}_{\text{f(product)}}}\,+\,\,\sum{\Delta {{\text{H}}_{\text{f(reactant)}}}}.....(1)$
\begin{align} & \Delta {{\text{H}}_{\text{reaction}}}\,\,=-1670\,\,-\,-\,834 \\ & \Delta {{\text{H}}_{\text{reaction}}}\,\,=\,\,-1670\,\,+\,\,834 \\ & \Delta {{\text{H}}_{\text{reaction}}}\,\,\,=\,\,-836\text{kJmo}{{\text{l}}^{\text{-1}}} \\ \end{align}
So, the correct answer is “Option A”.

Note: The enthalpy of formation will be positive for endothermic reaction, while negative for exothermic reaction. If a thermodynamic reaction is multiplied by a coefficient then, enthalpy change of reaction is also multiplied by coefficient.
- More stable species or compounds have lower enthalpy of formation.
- Heat of combustion of a reaction is always exothermic.