Question

From the values given for $\mathrm{\Delta }{H}^{\circ }$ and $\mathrm{\Delta }{S}^{\circ }$, calculate $\mathrm{\Delta }{G}^{\circ }$ 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? (a) $2\mathrm{PbS}(s)+3{\mathrm{O}}_2(g)⟶2\mathrm{PbO}(s)+2{\mathrm{SO}}_2(g)$ $\mathrm{\Delta }{H}^{\circ }=-844\mathrm{kJ};\mathrm{\Delta }{S}^{\circ }=-165\mathrm{J}/\mathrm{K}$ (b) $2{\mathrm{POCl}}_3(g)⟶2{\mathrm{PCl}}_3(g)+{\mathrm{O}}_2(g)$ $$\mathrm{\Delta }{H}^{\circ }=572\mathrm{kJ};\mathrm{\Delta }{S}^{\circ }=179\mathrm{J}/\mathrm{K}$$

Short answer

For reaction (a), the ΔG∘ at 298 K is -794.83 kJ, and the reaction is spontaneous under standard conditions. For reaction (b), the ΔG∘ at 298 K is 518.658 kJ and is not spontaneous under standard conditions. However, reaction (b) would become spontaneous at a temperature of approximately 3198 K.

Step-by-step solution

Formulate Gibbs free energy equation

Use the equation, ΔG∘ = ΔH∘ - TΔS∘, where T is the given temperature in Kelvin (298 K).

Calculate ΔG∘

Substitute the given values into the equation: ΔG∘ = -844 kJ - (298 K)(-165 J/K). Convert J to kJ by dividing by 1000: ΔG∘ = -844 kJ + (298 K)(0.165 kJ/K) = -844 kJ + 49.17 kJ = -794.83 kJ. The reaction is spontaneous at 298 K because ΔG∘ is negative. Since the reaction is spontaneous under standard conditions, we don't need to find the temperature at which it becomes spontaneous. (b) $2{\mathrm{POCl}}_3(g)⟶2{\mathrm{PCl}}_3(g)+{\mathrm{O}}_2(g)$ Given: $\mathrm{\Delta }{H}^{\circ }=572\mathrm{kJ}$; $\mathrm{\Delta }{S}^{\circ }=179\mathrm{J}/\mathrm{K}$ Step 1: Calculate ΔG∘ at 298 K

Formulate Gibbs free energy equation

Use the equation, ΔG∘ = ΔH∘ - TΔS∘, where T is the given temperature in Kelvin (298 K).

Calculate ΔG∘

Substitute the given values into the equation: ΔG∘ = 572 kJ - (298 K)(179 J/K). Convert J to kJ by dividing by 1000: ΔG∘ = 572 kJ - (298 K)(0.179 kJ/K) = 572 kJ - 53.342 kJ = 518.658 kJ. The reaction is not spontaneous at 298 K because ΔG∘ is positive. Step 2: Find the temperature when the reaction becomes spontaneous

Set ΔG∘ equal to zero

To find the temperature at which the reaction becomes spontaneous, set ΔG∘ = 0, then solve for T: 0 = ΔH∘ - TΔS∘.

Calculate the temperature

Rearrange the equation to solve for T: T = ΔH∘ / ΔS∘. Insert the given values: T = (572 kJ) / (0.179 kJ/K) = 3197.77 K. So, the reaction would become spontaneous at a temperature of approximately 3198 K.

Key concepts

Thermodynamic Spontaneity

Thermodynamic spontaneity refers to the tendency of a chemical reaction to occur without needing any external input of energy. It is primarily determined by the Gibbs Free Energy change ($\mathrm{\Delta }G$) of the reaction, which is calculated using the formula:

• $$\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }$$

In this equation, $\mathrm{\Delta }{H}^{\circ }$ is the change in enthalpy, $T$ is the temperature in Kelvin, and $\mathrm{\Delta }{S}^{\circ }$ is the change in entropy. If $\mathrm{\Delta }{G}^{\circ }$ is negative, it implies that the reaction is spontaneous and will proceed on its own once started.
If $\mathrm{\Delta }{G}^{\circ }$ is positive, the reaction is non-spontaneous and will require energy to proceed. At zero $\mathrm{\Delta }{G}^{\circ }$, the system is at equilibrium, meaning no net change occurs in the reaction. When assessing spontaneity, it's important to consider both the enthalpy and entropy changes, as these play a crucial role in determining the overall free energy change.

Enthalpy Change

Enthalpy change ($\mathrm{\Delta }{H}^{\circ }$) represents the heat absorbed or released by a system at constant pressure. It is expressed in kilojoules per mole (kJ/mol). A negative value of $\mathrm{\Delta }{H}^{\circ }$ suggests that the reaction is exothermic, meaning it releases heat to the surroundings.
Conversely, a positive $\mathrm{\Delta }{H}^{\circ }$ indicates an endothermic reaction, where heat is absorbed from the surroundings. For example, in reaction (a), the $\mathrm{\Delta }{H}^{\circ }=-844\text{ kJ}$ signifies an exothermic process, which contributes to the spontaneity of the reaction at 298 K. In contrast, reaction (b) has a $\mathrm{\Delta }{H}^{\circ }=572\text{ kJ}$, showing that it is endothermic and not spontaneously occurring under standard conditions. Enthalpy change affects the temperature dependency of a reaction's spontaneity. In general, highly exothermic reactions are more likely to be spontaneous, especially at lower temperatures.

Entropy Change

Entropy change ($\mathrm{\Delta }{S}^{\circ }$) describes the change in disorder or randomness of a system during a chemical reaction. It is measured in joules per Kelvin (J/K). A positive $\mathrm{\Delta }{S}^{\circ }$ indicates an increase in disorder—often favored in chemical processes.

• A good example is the transition from a solid phase to a gas, where molecules gain more freedom to move.

Conversely, a negative $\mathrm{\Delta }{S}^{\circ }$ means a decrease in entropy, resulting in increased order. In reaction (a), $\mathrm{\Delta }{S}^{\circ }=-165\text{ J/K}$ implies a reduction in randomness, yet the reaction remains spontaneous due to its highly negative $\mathrm{\Delta }{H}^{\circ }$. For reaction (b), $\mathrm{\Delta }{S}^{\circ }=179\text{ J/K}$ reveals that the disorder increases, which aids spontaneity at higher temperatures. Entropy plays a crucial role in determining reaction feasibility, particularly when competition between enthalpy and entropy exists, as seen by the temperature-dependent spontaneity of reaction (b). This highlights the delicacy in balancing system energy and entropy to achieve spontaneous reactions.