Animated Solution for Chemistry - Chemical Thermodynamics: The following reaction is performed at $298\text{ K}$ $2\text{NO}(g)+{\text{O}}_2(g)\rightleftharpoons 2{\text{NO}}_2(g)$ The standard free energy of formation of $\text{NO}(g)$ is $86.6\text{ kJ/mol}$ at $298\text{ K}$. What is the standard free energy of formation of ${\text{NO}}_2(g)$ at $298\text{ K}$? ($K_p=1.6\times 10^{12}$)

Let's dive into a classic problem from Chemical Thermodynamics that beautifully connects the standard free energy of a reaction with its equilibrium constant. We are given the reaction:

We are provided with the standard free energy of formation for $\text{NO}(g)$, which is $86.6\text{ kJ/mol}$, and the equilibrium constant $K_p=1.6\times 10^{12}$ at a temperature of $298\text{ K}$. Our goal is to find the standard free energy of formation for the product, ${\text{NO}}_2(g)$.

To visualize this, imagine a thermodynamic cycle. At the base, we have our constituent elements in their standard states: ${\text{N}}_2(g)$ and ${\text{O}}_2(g)$. These elements can either form our reactants ($2\text{NO}+{\text{O}}_2$) or directly form our products ($2{\text{NO}}_2$). The difference in energy between these two pathways is exactly the standard free energy of the reaction, $\Delta G_r^{\circ }$.

There are two fundamental equations we need to bridge together. First, the standard free energy of the reaction can be calculated from the formation energies of the individual species:

Applying this to our specific reaction, and remembering to multiply by the stoichiometric coefficients, we get:

Second, we know that the standard free energy of a reaction is intimately linked to the equilibrium constant $K_p$ through the isotherm equation:

By equating these two expressions for $\Delta G_r^{\circ }$, we create a powerful equation that contains our unknown variable:

Now, let's substitute the known values. Here is where we must be extremely careful with units! The gas constant $R$ is typically expressed in Joules, but our formation energy for $\text{NO}$ is given in kilojoules ($86.6\text{ kJ/mol}$). We must convert this to $86600\text{ J/mol}$. Furthermore, remember that the standard free energy of formation for an element in its most stable standard state, like ${\text{O}}_2(g)$, is exactly zero.

Our final task is simple algebraic rearrangement. We want to isolate the term $2\Delta G_f^{\circ }({\text{NO}}_2)$. We do this by moving the reactant energy term to the left side of the equation:

Finally, to find the formation energy for a single mole of ${\text{NO}}_2$, we divide the entire right side by 2 (which is the same as multiplying by $0.5$):

This perfectly matches option (d). The beauty of this problem lies in the seamless integration of Hess's Law principles with chemical equilibrium, reminding us that thermodynamics is a perfectly consistent logical framework.