Animated Solution for Chemistry - Chemical Thermodynamics: Comprehension Passage

Let's embark on a journey to extract fundamental thermodynamic properties from a simple straight-line graph. We are given a heterogeneous reaction:

The first step is to write down the equilibrium constant, $K$. Since $\mathbf{X}$ and $\mathbf{Y}$ are pure solids, their active masses are taken as unity. They gracefully exit the stage, leaving only the gaseous product $\mathbf{Z}$. Thus, the equilibrium constant is simply the partial pressure of $\mathbf{Z}$ divided by the standard pressure:

Taking the natural logarithm on both sides, we get $\ln K=\ln \left(\frac{p_z}{p^{\ominus }}\right)$. This is a crucial realization because it tells us that the y-axis of our given graph is exactly $\ln K$.

To connect this graph to enthalpy ($\Delta H^{\ominus }$) and entropy ($\Delta S^{\ominus }$), we must call upon the fundamental equations of thermodynamics. We know that the standard Gibbs free energy change is given by:

Simultaneously, it is related to the equilibrium constant by the isotherm equation:

Equating these two expressions gives us our master equation:

Let's rearrange this to isolate $\ln K$, matching the y-axis of our graph. Dividing the entire equation by $-RT$, we obtain:

Now, look at the x-axis of the graph. It is plotted as $\frac{10^4}{T}$. Let's define our variables: $y=\ln K$ and $x=\frac{10^4}{T}$. This means $\frac{1}{T}=\frac{x}{10^4}$. Substituting this into our rearranged equation yields:

This is the beautiful equation of a straight line, $y=mx+c$, where the slope $m=-\frac{\Delta H^{\ominus }}{10^{4}R}$ and the y-intercept $c=\frac{\Delta S^{\ominus }}{R}$.

To find the standard enthalpy change, we need the slope of the line. The graph provides two clear, unambiguous points: $(10,-3)$ and $(12,-7)$. Let's calculate the slope $m$:

Now, we equate our theoretical slope to this calculated value:

Solving for $\Delta H^{\ominus }$, we get:

Substituting the value of the gas constant $R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$:

Converting this to kilojoules, we arrive at our first answer: $\Delta H^{\ominus }=166.28{\text{ kJ mol}}^{-1}$.

For the second part of the problem, we need the standard entropy change, $\Delta S^{\ominus }$, which is hidden within the y-intercept $c$. We can find $c$ by substituting one of our known points into the line equation $y=mx+c$. Let's use the point $(10,-3)$:

From our theoretical equation, we know that the intercept $c=\frac{\Delta S^{\ominus }}{R}$. Therefore:

Calculating this product gives us $141.338{\text{ J K}}^{-1}{\text{ mol}}^{-1}$. Rounding to two decimal places, we get our final answer: $\Delta S^{\ominus }=141.34{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

It is fascinating how a simple straight-line plot of experimental data can reveal the deep thermodynamic secrets of a chemical reaction!