Animated Solution for Chemistry - Chemical Thermodynamics: Comprehension Passage

Imagine you are standing at a crossroads of temperatures and phases. We have two distinct phases, $\alpha$ and $\beta$, and we need to uncover their entropy and enthalpy differences at a cool $300\text{ K}$. However, the data provided in the graph is centered around a much hotter $600\text{ K}$.

How do we bridge this gap? The beauty of thermodynamics lies in state functions. Because entropy ($S$) and enthalpy ($H$) depend only on the current state of the system and not the path taken to get there, we can construct a thermodynamic cycle to connect these states across the two temperatures.

First, let's decode the graph at $600\text{ K}$. The y-axis gives us the value of $S_T-S_0$. For the $\beta$ phase, this value is $6$, and for the $\alpha$ phase, it is $5$.

We are given that the entropy of both phases is identical at absolute zero ($0\text{ K}$), meaning $S_{0(\beta )}=S_{0(\alpha )}$. Therefore, the entropy change $\Delta S$ at $600\text{ K}$ is simply the difference between these two values:

Furthermore, $600\text{ K}$ is the phase transition temperature. At this temperature, the two phases coexist in perfect equilibrium, which thermodynamically means the change in Gibbs free energy, $\Delta G$, is exactly zero. Using the relation $\Delta G=\Delta H-T\Delta S$, we can easily find the enthalpy change at $600\text{ K}$:

Now, we need to relate the entropy at $600\text{ K}$ to the entropy at $300\text{ K}$. We know that entropy changes with temperature according to the formula $\Delta S=\int \frac{C_P}{T}dT$, which integrates to $C_P\ln \left(\frac{T_2}{T_1}\right)$ for a constant heat capacity.

Let's apply this to our cycle. There are two paths to go from the $\alpha$ phase at $300\text{ K}$ to the $\beta$ phase at $600\text{ K}$. We can either undergo the phase transition at $300\text{ K}$ and then heat the $\beta$ phase, or we can heat the $\alpha$ phase to $600\text{ K}$ and then undergo the phase transition. Equating the total entropy change for both paths gives:

Substituting the known values ($\Delta S_{600}=1$, $\Delta C_P=1$, and $\ln 2=0.69$), we get:

To find the enthalpy change at $300\text{ K}$, we use Kirchhoff's equation, which is essentially the enthalpy version of the cycle we just built. It relates the enthalpy change of a reaction at two different temperatures using the difference in heat capacities:

Plugging in our temperatures and the constant $\Delta C_P$:

By leveraging the path independence of state functions, we successfully navigated from $600\text{ K}$ down to $300\text{ K}$. We found that $\Delta S_{300}=0.31{\text{ J K}}^{-1}{\text{ mol}}^{-1}$ and $\Delta H_{300}=300{\text{ J mol}}^{-1}$. This problem is a masterclass in combining graphical interpretation with fundamental thermodynamic cycles!