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

$\Delta H$ 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.
$\text{ }\\!\\!\Delta\\!\\!\text{ H=}\sum{{{\text{ }\\!\\!\Delta\\!\\!\text{ }}_{\text{f}}}\text{H}}\left( \text{products} \right)\text{-}\sum{{{\text{ }\\!\\!\Delta\\!\\!\text{ }}_{\text{f}}}\text{H}}\left( \text{reactants} \right)$

& \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{aligned}$$ $$\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.