Question

The enthalpy of formation of $A{{l}_{2}}{{O}_{3}}$ and $C{{r}_{2}}{{O}_{3}}$ are -1596kJ and -1134kJ respectively. $\Delta H$ for the reaction,
\[2Al+C{{r}_{2}}{{O}_{3}}\to 2Cr+A{{l}_{2}}{{O}_{3}}\] is:
(A) -2730kJ
(B) -462kJ
(C) -1365kJ
(D) +2730kJ

Answer 1

Hint: $\Delta H$ is the change in enthalpy for the given reaction. Using the formula, \[\text{ }\!\!\Delta\!\!\text{ H=}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{products} \right)\text{-}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{reactants} \right)\], calculate the change in enthalpy for the reaction.

Complete step by step answer:
-The enthalpy of a system is the sum of internal energy of the system and the energy that results due to its pressure and volume.
-The enthalpy of formation of a compound is the energy required to form that compound from its elements or molecules present in their stable reference states.
-The enthalpy change is the difference between sum of enthalpy of products by enthalpy of reactants and is given by the formula, \[\text{ }\!\!\Delta\!\!\text{ H=}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{products} \right)\text{-}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{reactants} \right)\]
-According to the given data, the enthalpy of formation of $A{{l}_{2}}{{O}_{3}}$ is -1596kJ and enthalpy of formation for $C{{r}_{2}}{{O}_{3}}$ is -1134kJ. Also, for elements in their elemental state, enthalpy of formation is zero.
From the reaction, $2Al+C{{r}_{2}}{{O}_{3}}\to 2Cr+A{{l}_{2}}{{O}_{3}}$, let us substitute the values and find the enthalpy change.
\[\text{ }\!\!\Delta\!\!\text{ H=}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{products} \right)\text{-}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{reactants} \right)\]
\[\begin{align}
  & \text{ }\!\!\Delta\!\!\text{ H=}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{products} \right)\text{-}\sum{{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}}\left( \text{reactants} \right) \\
 & =2\times {{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}\left( Cr \right)+{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}\left( A{{l}_{2}}{{O}_{3}} \right)-2\times {{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}\left( Al \right)-{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}\text{H}\left( C{{r}_{2}}{{O}_{3}} \right) \\
 & =(2\times 0)+(-1596kJ)-(2\times 0)-(-1134kJ) \\
 & =-1596kJ+1134kJ \\
 & =-462kJ
\end{align}\]
\[\therefore \Delta H=-462kJ\]
Therefore, the change in enthalpy, $\Delta H$ for the reaction, $2Al+C{{r}_{2}}{{O}_{3}}\to 2Cr+A{{l}_{2}}{{O}_{3}}$ is -462kJ.
So, the correct answer is “Option B”.

Note:Remember for all metals in their elemental state, the enthalpy of formation is zero. Also, for gases like oxygen, hydrogen, nitrogen, the enthalpy of formation is zero.