Animated Solution for Chemistry - Chemical Thermodynamics: Given : (i) $\text{C(graphite)}+{\text{O}}_2(g)\to {\text{CO}}_2(g)$; ${\Delta }_r{H}^{\circ }=x{\text{ kJ mol}}^{-1}$ (ii) $\text{C(graphite)}+\frac{1}{2}{\text{O}}_2(g)\to {\text{CO}}_2(g)$; ${\Delta }_r{H}^{\circ }=y{\text{ kJ mol}}^{-1}$ (iii) $\text{CO}(g)+\frac{1}{2}{\text{O}}_2(g)\to {\text{CO}}_2(g)$; ${\Delta }_r{H}^{\circ }=z{\text{ kJ mol}}^{-1}$ Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?

Imagine you are planning a road trip from City A to City C. You can either take a direct highway, or you can stop at City B for lunch before continuing to City C. Regardless of the route you choose, your total change in elevation from the start of your trip to the end will be exactly the same. This intuitive idea is the heart of Hess's Law of Constant Heat Summation in thermodynamics.

In this problem, we are given three thermochemical equations. However, before we dive into the math, we need to put on our detective glasses. There is a subtle typo in the second equation provided in the question:

If you look closely, the oxygen atoms are not balanced. One oxygen atom on the reactant side cannot magically become two on the product side! Furthermore, the first equation already tells us that forming one mole of ${\text{CO}}_2$ requires a full mole of ${\text{O}}_2$. Therefore, the correct second equation must represent the incomplete combustion of carbon to form carbon monoxide:

Now that we have corrected the typo, let's map out our chemical journey. Our ultimate destination (the direct route) is the complete combustion of graphite to form carbon dioxide. This is represented by the first equation:

This direct path has an enthalpy change of $x$.

Alternatively, we can take the two-step route.

Step 1: We partially burn the graphite to form carbon monoxide.

Step 2: We take that carbon monoxide and burn it with the remaining oxygen to finally reach carbon dioxide.

According to Hess's Law, if we add the chemical equations of the intermediate steps, their enthalpy changes must also add up to give the enthalpy change of the overall reaction. Let's verify this by adding Step 1 and Step 2:

Notice that $\text{CO}(g)$ appears on both the reactant and product sides, so it cancels out. The two half-moles of ${\text{O}}_2$ combine to form one full mole. We are left with:

This perfectly matches our direct reaction! Therefore, the enthalpy of the direct reaction ($x$) must equal the sum of the enthalpies of the two steps ($y$ and $z$).

This elegant relationship showcases the power of state functions in thermodynamics. The energy change depends only on where you start and where you finish, not how you get there.