Problem 64 - Chemistry: The Central Science in SI Units, Global Edition

Question

From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at \(298 \mathrm{~K}\). If the reaction is not spontaneous under standard conditions at \(298 \mathrm{~K}\), at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} $$

Step-by-Step Solution

Answer

For reaction (a), the calculated value of \(\Delta G^{\circ} = -764.82 \, \mathrm{kJ}\) indicates that the reaction is spontaneous at \(298 \mathrm{~K}\). For reaction (b), the calculated value of \(\Delta G^{\circ} = 161.662 \, \mathrm{kJ}\) indicates that the reaction is not spontaneous at \(298 \mathrm{~K}\), but it becomes spontaneous at a temperature of approximately \(3196 \mathrm{~K}\).

Step 1: (a) Calculate \(\Delta G^{\circ}\) for the reaction

Using the given values for the reaction: $$ \Delta H^{\circ} = -844 \mathrm{~kJ} $$ $$ \Delta S^{\circ} = -165 \mathrm{~J} / \mathrm{K} $$ We will plug these values into the formula for \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\): $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = -844\times10^3 \, \mathrm{J} - (298 \, \mathrm{K})(-165 \, \mathrm{J/K}) $$ Now we'll calculate the value of delta G: $$ \Delta G^{\circ} = -764820 \, \mathrm{J} = -764.82 \, \mathrm{kJ} $$

Step 2: (a) Analyze the spontaneity of the reaction

Since the value of \(\Delta G^{\circ} < 0\), the reaction is spontaneous at \(298 \mathrm{~K}\). Therefore, there is no need to calculate the temperature at which the reaction becomes spontaneous.

Step 3: (b) Calculate \(\Delta G^{\circ}\) for the reaction

Using the given values for the reaction: $$ \Delta H^{\circ} = 572 \mathrm{~kJ} $$ $$ \Delta S^{\circ} = 179 \mathrm{~J} / \mathrm{K} $$ We will plug these values into the formula for \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\): $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = 572\times10^3 \, \mathrm{J} - (298 \, \mathrm{K})(179 \, \mathrm{J/K}) $$ Calculating the value of delta G: $$ \Delta G^{\circ} = 161662 \, \mathrm{J} = 161.662 \, \mathrm{kJ} $$

Step 4: (b) Analyze the spontaneity of the reaction

Since the value of \(\Delta G^{\circ} > 0\), the reaction is not spontaneous at \(298 \mathrm{~K}\). We need to find the temperature at which the reaction becomes spontaneous.

Step 5: (b) Find the temperature for spontaneous reaction

To find the temperature at which \(\Delta G^{\circ} = 0\), we will plug the value of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) into the formula, and solve for T: $$ 0 = 572\times10^3 \, \mathrm{J} - T(179 \, \mathrm{J/K}) $$ Now solve for T: $$ T = \frac{572\times10^3 \, \mathrm{J}}{179 \, \mathrm{J/K}} \approx 3196 \, \mathrm{K} $$ So, the reaction becomes spontaneous at a temperature of approximately \(3196 \mathrm{~K}\).

Key Concepts

Spontaneity of Reactions

In chemistry, the spontaneity of reactions refers to whether a chemical reaction can occur without any external influence, like added energy. The key indicator for this is the Gibbs Free Energy change, denoted as \( \Delta G \). A reaction is considered spontaneous if \( \Delta G < 0 \). This means that the process can happen on its own under given conditions, releasing free energy.
For non-spontaneous reactions, where \( \Delta G > 0 \), external energy would be required to initiate the reaction. For instance, in the exercise we reviewed, reaction (a), where \( \Delta G = -764.82 \text{ kJ} \), is spontaneous, suggesting it will occur naturally without extra input at \( 298 \text{ K} \).
Meanwhile, reaction (b) had \( \Delta G = 161.662 \text{ kJ} \), indicating it is non-spontaneous at standard temperature, requiring either a change in temperature or conditions to become spontaneous.

Thermodynamic Calculations

Thermodynamic calculations form the core of determining reaction spontaneity and involve using the Gibbs Free Energy equation:

\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]

This formula incorporates three variables: enthalpy change (\( \Delta H \)), temperature (\( T \)), and entropy change (\( \Delta S \)). Since enthalpy and entropy often have units that differ (kJ for \( \Delta H \) and J/K for \( \Delta S \)), you need to convert them to matching units, typically Joules, for accurate calculations.
In our example problem, for each reaction at \( 298 \text{ K} \), we calculated \( \Delta G^{\circ} \) by ensuring unit consistency and applying given values to the equation. The calculations give insight into whether energy is absorbed or released in the process, helping us understand readiness to occur spontaneously.
Solving thermodynamic problems accurately requires handling these conversions with care and precision to understand the behavior of chemical reactions under specific conditions.

Enthalpy and Entropy

Enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) are thermodynamic properties fundamental to understanding chemical reactions.
Enthalpy involves the heat content in a system. A negative \( \Delta H \) (exothermic reaction) releases heat, lowering system energy, more likely leading to spontaneity. For example, reaction (a) with \( \Delta H = -844 \text{ kJ} \) is spontaneous at standard temperature, suggesting that it naturally releases free energy.
On the other hand, entropy reflects the degree of disorder or randomness in a system. Reactions with increased entropy (positive \( \Delta S \)) tend to be spontaneous since systems naturally progress towards more disorder. In our exercise, reaction (b)'s \( \Delta S = 179 \text{ J/K} \) indicates increased entropy, but the high enthalpy prevents spontaneity at \( 298 \text{ K} \).
Hence, analyzing both \( \Delta H \) and \( \Delta S \) helps predict reaction behavior. When balancing endothermic reactions with significant entropy increases and exothermic reactions despite entropy changes, these concepts crucially inform thermodynamic spontaneity and feasibility.