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? $$ \begin{array}{l}{\text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)} \\ {\Delta H^{\circ}=-844 \mathrm{kk} ; \Delta S^{\circ}=-165 \mathrm{J} / \mathrm{K}} \\\ {\text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g)} \\ {\Delta H^{\circ}=572 \mathrm{kJ} ; \Delta S^{\circ}=179 \mathrm{J} / \mathrm{K}}\end{array} $$

Short answer

For reaction (a), the Gibbs free energy change at 298 K is \(\Delta G^{\circ} = -794.83\,\mathrm{kJ}\), indicating that the reaction is spontaneous at this temperature. For reaction (b), the Gibbs free energy change at 298 K is \(\Delta G^{\circ} = 518.66\,\mathrm{kJ}\), indicating that the reaction is not spontaneous at this temperature. The reaction becomes spontaneous at approximately 3197 K.

Step-by-step solution

Calculate \(\Delta G^{\circ}\) at 298 K

Using the given values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and the equation for Gibbs free energy change: \[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} = -844 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (-165\,\mathrm{J} /\mathrm{K})) \] Convert the entropy change in J/K to kJ/K before performing the calculation: \[ \Delta G^{\circ} = -844 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (-0.165\,\mathrm{kJ} /\mathrm{K})) = -844\,\mathrm{kJ} + 49.17\,\mathrm{kJ} = -794.83\,\mathrm{kJ} \]

Determine, if the reaction is spontaneous at 298 K

Since the Gibbs free energy change is negative (\(\Delta G^{\circ} = -794.83\,\mathrm{kJ}\)) at 298 K, the reaction is spontaneous at this temperature. ##Reaction (b)##

Calculate \(\Delta G^{\circ}\) at 298 K

Using the given values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and the Gibb's free energy change equation: \[ \Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}= 572 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (179\,\mathrm{J}/\mathrm{K})) \] Convert the entropy change in J/K to kJ/K before performing the calculation: \[ \Delta G^{\circ} = 572 \,\mathrm{kJ} - (298\,\mathrm{K} \cdot (0.179\,\mathrm{kJ}/\mathrm{K}))= 572\,\mathrm{kJ} - 53.34\,\mathrm{kJ} = 518.66\,\mathrm{kJ} \]

Determine if the reaction is spontaneous at 298 K

Since the Gibbs free energy change is positive (\(\Delta G^{\circ}= 518.66\,\mathrm{kJ}\)) at 298 K, the reaction is not spontaneous at this temperature.

Find the temperature at which the reaction becomes spontaneous

To find the temperature at which the reaction becomes spontaneous, we're looking for the temperature at which \(\Delta G^{\circ} = 0\). So, set the equation for Gibbs free energy change equal to 0 and solve for T: \[ 0 = \Delta H^{\circ} - T \Delta S^{\circ} \Longrightarrow T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{572\,\mathrm{kJ}}{0.179\,\mathrm{kJ}/\mathrm{K}} \approx 3197\,\mathrm{K} \] The reaction (b) will become spontaneous at approximately 3197 K.

Key concepts

Spontaneous Reactions

Spontaneous reactions are natural processes that occur without needing external energy input. Spontaneity is determined by the Gibbs Free Energy change (ΔG). When ΔG is negative, a reaction proceeds spontaneously; when positive, it does not. For example, the reaction of lead sulfide (PbS) with oxygen (O₂) to form lead oxide (PbO) and sulfur dioxide (SO₂) is spontaneous at 298 K because ΔG is negative.

• Spontaneous: Negative ΔG
• Non-Spontaneous: Positive ΔG
• Equilibrium: ΔG = 0

A reaction may not be spontaneous at a given temperature but could become spontaneous if the temperature changes. This is because temperature can affect the entropy term in the Gibbs Free Energy equation, potentially changing the sign of ΔG.

Enthalpy Change

Enthalpy change (ΔH) refers to the heat absorbed or released during a reaction at constant pressure. It is a vital component of the Gibbs Free Energy equation: \[ \Delta G = \Delta H - T \Delta S \]In the given exercises, enthalpy helps determine whether a reaction is exo- or endothermic.

• Exothermic reactions release heat, resulting in negative ΔH.
• Endothermic reactions absorb heat, leading to positive ΔH.

For example, reaction (a) has a ΔH of -844 kJ, indicating it is exothermic. Conversely, reaction (b) is endothermic, with a ΔH of 572 kJ. Understanding the direction of heat flow helps predict how temperature changes might affect reaction spontaneity.

Entropy Change

Entropy change (ΔS) measures the disorder or randomness in a system. It indicates how distributed energy or matter becomes during a reaction. In the Gibbs Free Energy equation, ΔS is multiplied by the temperature to form the entropy term that affects ΔG. A positive ΔS suggests increased disorder, while a negative ΔS indicates reduced disorder:

• Increased disorder: Positive ΔS
• Reduced disorder: Negative ΔS

In the original exercises, reaction (a) has a negative ΔS of -165 J/K, showing a decrease in disorder, whereas reaction (b) has a positive ΔS of 179 J/K, indicating an increase in disorder. The entropy change helps determine the temperature dependence of spontaneity; even reactions with unfavorable ΔH can become spontaneous if ΔS is large and positive, particularly at higher temperatures.