Exercises: Entropy and spontaneity

For the synthesis reaction of sulfur trioxide:

$$2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{SO_3(g)}$$

the standard molar entropy values at $298\,\text{K}$ are available: $S^\circ[\mathrm{SO_2(g)}] = 248.2\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$, $S^\circ[\mathrm{O_2(g)}] = 205.1\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$ and $S^\circ[\mathrm{SO_3(g)}] = 256.8\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

1. a) Justify qualitatively the expected sign of the standard entropy change of the system, based on the dispersal of matter and the number of available microstates.
2. b) Calculate the numerical value of $\Delta S^\circ$ for the reaction at $298\,\text{K}$ and confirm the previous prediction.

Consider the industrial synthesis reaction of ammonia from its elements at $298\,\text{K}$:

$$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$$

For this transformation the following standard thermodynamic data at that temperature are known: reaction enthalpy $\Delta H^\circ = -92.2\,\text{kJ}\times\text{mol}^{-1}$ and entropy change $\Delta S^\circ = -198.7\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

1. a) Calculate the standard Gibbs free energy change ($\Delta G^\circ$) of the reaction at $298\,\text{K}$, paying particular attention to the consistency of the units used.
2. b) State, justifying your answer using the appropriate thermodynamic criterion, whether the process is spontaneous under standard conditions at that temperature.

At $298\,\text{K}$ and $1\,\text{bar}$, the allotropic transformation of diamond carbon into graphite carbon has the following thermodynamic data:

$$\mathrm{C(s, diamante)} \rightarrow \mathrm{C(s, grafito)} \quad \Delta G^\circ_{298} = -2.90\,\text{kJ}\times\text{mol}^{-1}$$

However, diamond jewelry remains intact for millennia without appreciably transforming into graphite.

1. a) Justify whether the transformation is thermodynamically spontaneous under standard conditions at $298\,\text{K}$.
2. b) Explain the apparent discrepancy between the negative sign of $\Delta G^\circ$ and the empirical observation, clearly distinguishing between the thermodynamic concept of spontaneity and the rate of the process (kinetic control).

For a hypothetical endothermic gas-phase reaction, the standard Gibbs free energy change ($\Delta G^\circ$) has been plotted as a function of the absolute temperature $T$. In the accompanying graph

it can be seen that the line has a y-intercept of $+40\,\text{kJ}\times\text{mol}^{-1}$ and a slope of $-0.10\,\text{kJ}\times\text{mol}^{-1}\times\text{K}^{-1}$ over the interval analyzed ($0$ to $800\,\text{K}$).

1. a) Using the parameters of the line and the equation $\Delta G^\circ = \Delta H^\circ - T\times\Delta S^\circ$, determine the value of the threshold temperature (marked with "?" at the intersection with the horizontal axis) above which the process is favorable under standard conditions.
2. b) Discuss the limitations of the usual approximation that treats $\Delta H^\circ$ and $\Delta S^\circ$ as independent of temperature over wide intervals such as the one shown.

The thermal decomposition of calcium carbonate is a reaction of great importance in the manufacture of lime:

$$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$$

At $298\,\text{K}$, the standard values are $\Delta H^\circ = +178.3\,\text{kJ}\times\text{mol}^{-1}$ and $\Delta S^\circ = +160.5\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

1. a) Analyze qualitatively the sign of $\Delta H^\circ$ and $\Delta S^\circ$ using the table of the four possible cases of the Gibbs energy, and deduce how an increase in temperature affects the spontaneity of the process.
2. b) Calculate the minimum temperature (in degrees Celsius), assuming that $\Delta H^\circ$ and $\Delta S^\circ$ do not vary with temperature, for the decomposition to be spontaneous under standard conditions ($p_{\mathrm{CO_2}} = 1\,\text{bar}$).

Nitrogen dioxide and its dimer form the following equilibrium in the gas phase at $298.15\,\text{K}$:

$$\mathrm{N_2O_4(g)} \rightleftharpoons 2\,\mathrm{NO_2(g)}$$

For this system at that temperature, the standard Gibbs free energy change is known to be $\Delta G^\circ = +4.73\,\text{kJ}\times\text{mol}^{-1}$. Data: $R = 8,314\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

1. a) Clearly distinguish the physical meaning of $\Delta G$ and $\Delta G^\circ$. What value does $\Delta G$ take when the system reaches chemical equilibrium?
2. b) Calculate the value of the standard thermodynamic equilibrium constant $K_p$ at $298.15\,\text{K}$ and explain in which direction a mixture with reaction quotient $Q=1$ would initially evolve. Do not assume that the value of K alone determines all the mole fractions. Use $\Delta G^\circ=-RT\ln K$ and $\Delta G=\Delta G^\circ+RT\ln Q$, with gaseous activities $p_i/p^\circ$ and $p^\circ=1\,\text{bar}$.

In iron metallurgy, a classic stage of indirect reduction in the blast furnace is:

$$\mathrm{Fe_2O_3(s)} + 3\,\mathrm{CO(g)} \rightarrow 2\,\mathrm{Fe(s)} + 3\,\mathrm{CO_2(g)}$$

The following standard thermodynamic data at $298\,\text{K}$ are available:

$\Delta H_f^\circ[\mathrm{Fe_2O_3(s)}] = -824.2\,\text{kJ}\times\text{mol}^{-1}$; $\Delta H_f^\circ[\mathrm{CO(g)}] = -110.5\,\text{kJ}\times\text{mol}^{-1}$; $\Delta H_f^\circ[\mathrm{CO_2(g)}] = -393.5\,\text{kJ}\times\text{mol}^{-1}$; $S^\circ[\mathrm{Fe_2O_3(s)}] = 87.4\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$; $S^\circ[\mathrm{CO(g)}] = 197.7\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$; $S^\circ[\mathrm{Fe(s)}] = 27.3\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$; $S^\circ[\mathrm{CO_2(g)}] = 213.7\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$.

1. a) Calculate $\Delta H^\circ$ and $\Delta S^\circ$ for the reaction at $298\,\text{K}$. Interpret the numerical value obtained for $\Delta S^\circ$ in terms of the number of microstates and the balance of gaseous species.
2. b) Calculate $\Delta G^\circ$ at $298\,\text{K}$ and at $1000\,\text{K}$, assuming that $\Delta H^\circ$ and $\Delta S^\circ$ remain constant. State, with reasoning, whether the increase in temperature thermodynamically favors or disfavors the standard spontaneity of the reaction.

The vaporization equilibrium of water at a pressure of $1\,\text{atm}$ is described by:

$$\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2O(g)}$$

The normal boiling point is $373.15\,\text{K}$ ($100\,^\circ\text{C}$), with a standard enthalpy of vaporization of $\Delta H^\circ_{\text{eb}} = +40.66\,\text{kJ}\times\text{mol}^{-1}$.

1. a) Calculate the standard entropy change of vaporization $\Delta S^\circ_{\text{eb}}$ at $373.15\,\text{K}$. Justify its sign in terms of energy dispersal and the increase in molecular microstates.
2. b) Calculate $\Delta G^\circ$ of vaporization at $298.15\,\text{K}$, assuming that $\Delta H^\circ$ and $\Delta S^\circ$ do not depend on temperature. In light of this result, explain why liquid water placed in an open dish at $25\,^\circ\text{C}$ can evaporate spontaneously if the surroundings are unsaturated and the air is renewed, even though the value of $\Delta G^\circ$ obtained is positive. In this exercise the gaseous reference state is explicitly set at 1 atm (not 1 bar); the symbol $\circ$ refers to that convention. To discuss the dish, distinguish between the total pressure of the air and the partial pressure of the water vapor.