Animated Solution for Chemistry - Chemical Thermodynamics: The standard enthalpies of formation of ${\text{Al}}_2{\text{O}}_3$ and $\text{CaO}$ are $-1675{\text{ kJ mol}}^{-1}$ and $-635{\text{ kJ mol}}^{-1}$ respectively. For the reaction, $3\text{CaO}+2\text{Al}⟶3\text{Ca}+{\text{Al}}_2{\text{O}}_3$ the standard reaction enthalpy ${\Delta }_r{H}^{\circ }$ ......... kJ. (Round off to the nearest integer).

Let's embark on a journey through the energetic landscape of a fascinating chemical reaction. We are looking at the reduction of calcium oxide by aluminum, a process that can be beautifully visualized using a Hess's Law cycle.

Imagine a foundation built upon the elements in their most stable standard states: solid calcium, solid aluminum, and gaseous oxygen. From this foundation, we can construct our reactants, $3\text{CaO}$ and $2\text{Al}$, and our products, $3\text{Ca}$ and ${\text{Al}}_2{\text{O}}_3$. The energy required to build these compounds from the ground up is known as their standard enthalpy of formation.

Hess's Law

To find the standard reaction enthalpy, ${\Delta }_r{H}^{\circ }$, we rely on the elegant principle of Hess's Law. It tells us that the total enthalpy change of a reaction is independent of the pathway taken. Therefore, we can calculate it by taking the total enthalpy of formation of the products and subtracting the total enthalpy of formation of the reactants.

Mathematically, this is expressed as:

Let's expand this formula for our specific reaction. We must account for every single mole involved. For the products, we have one mole of ${\text{Al}}_2{\text{O}}_3$ and three moles of $\text{Ca}$. For the reactants, we have three moles of $\text{CaO}$ and two moles of $\text{Al}$.

Here is a crucial concept where many students stumble. By thermodynamic convention, the standard enthalpy of formation for any element in its most stable standard state is exactly zero.

Because solid calcium and solid aluminum are elements in their standard states, their formation enthalpies vanish from our equation.

This beautifully simplifies our master equation to:

Now, we carefully substitute the given values. The formation enthalpy of aluminum oxide is $-1675{\text{ kJ mol}}^{-1}$. For calcium oxide, it is $-635{\text{ kJ mol}}^{-1}$. We must not forget to multiply the calcium oxide value by its stoichiometric coefficient, three!

Multiplying $-3$ with $-635$ gives us a positive $1905$.

Finally, adding these values together yields exactly $230\text{ kJ}$.

This positive value is highly significant; it tells us the reaction is endothermic, meaning it absorbs heat from its surroundings to proceed. Understanding these energy flows is the true heart of chemical thermodynamics!