Problem 66 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The reaction (a) with values \(\Delta H^{\circ} = -844,000 \mathrm{~J}\) and \(\Delta S^{\circ} = -165 \mathrm{~J} / \mathrm{K}\) is spontaneous at 298 K, with a \(\Delta G^{*} = -794,830 \mathrm{~J}\). The reaction (b) with values \(\Delta H^{\circ} = 572,000 \mathrm{~J}\) and \(\Delta S^{\circ} = 179 \mathrm{~J} / \mathrm{K}\) is non-spontaneous at 298 K, with a \(\Delta G^{*} = 518,658 \mathrm{~J}\). However, it becomes spontaneous at approximately 3197 K.

Step 1: Calculate the final value of \(\Delta G^{*}\)

\[ \Delta G^{*} = -794,830 \mathrm{~J} \] Since the value of \(\Delta G^{*}\) is negative, the reaction is spontaneous at 298 K. (b) Reaction: \(2 \mathrm{POCl}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{O}_{\text {(a }}(\mathrm{a})\) Given: \(\Delta H^{\circ} = 572 \mathrm{~kJ}\) and \(\Delta S^{\circ} = 179 \mathrm{~J} / \mathrm{K}\) #Step 1: Convert to appropriate units# Convert the enthalpy value to J/mol. \(\Delta H^{\circ} = 572,000 \mathrm{~J}\) #Step 2: Calculate \(\Delta G^{*}\) at 298 K# Using the Gibbs free energy equation, calculate \(\Delta G^{*}\): \[ \Delta G^{*} = \Delta H^{\circ} - T\Delta S^{\circ} \] \[ \Delta G^{*} = 572,000 \mathrm{~J} - (298 \mathrm{~K})(179 \mathrm{~J} / \mathrm{K}) \] \[ \Delta G^{*} = 572,000 \mathrm{~J} - 53,342 \mathrm{~J} \]

Step 2: Calculate the final value of \(\Delta G^{*}\)

\[ \Delta G^{*} = 518,658 \mathrm{~J} \] Since \(\Delta G^{*}\) is positive, the reaction is non-spontaneous at 298 K. #Step 3: Determine the temperature for spontaneous reaction# To determine the temperature at which the reaction becomes spontaneous, set \(\Delta G^{*} = 0\) and solve for \(T\): \[ 0 = \Delta H^{\circ} - T\Delta S^{\circ} \] \[ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} \]

Step 3: Calculate the temperature at which the reaction becomes spontaneous

\[ T = \frac{572,000 \mathrm{~J}}{179 \mathrm{~J} / \mathrm{K}} \] \[ T = 3196.65 \mathrm{~K} \] At a temperature of approximately 3197 K, the reaction becomes spontaneous.

Key Concepts

Enthalpy

Enthalpy is an essential concept when discussing Gibbs free energy. It is often represented by the symbol \( \Delta H \). Enthalpy captures the total heat content or energy change in a chemical system during a reaction.

• **Enthalpy Change**: This reflects whether a reaction absorbs or releases heat. It is crucial to determining reaction spontaneity.
• **Exothermic Reactions**: When \( \Delta H \) is negative, the reaction releases energy, often making it spontaneous.
• **Endothermic Reactions**: A positive \( \Delta H \) indicates energy absorption, which might result in a non-spontaneous reaction under standard conditions.
  

In the provided exercise, for reactions (a) and (b), \( \Delta H \) was noted to be \(-844 \mathrm{kJ}\) and \(572 \mathrm{kJ}\), respectively, revealing their exothermic and endothermic nature.

Entropy

Entropy, denoted as \( \Delta S \), describes the randomness or disorder within a system. It's a key player in determining the spontaneity of a reaction.

• **Increasing Entropy (Positive \( \Delta S \))**: Often favors spontaneity since systems naturally tend toward greater disorder.
• **Decreasing Entropy (Negative \( \Delta S \))**: May lead to non-spontaneity under standard conditions due to decreased disorder.
  

In the initial exercise, the reactions exhibited \( \Delta S \) values of \(-165 \mathrm{~J}/\mathrm{K}\) for reaction (a) and \(179 \mathrm{~J}/\mathrm{K}\) for reaction (b). This indicated decreasing entropy in reaction (a) and increasing entropy in reaction (b), impacting their spontaneous behavior significantly.

Spontaneity

Spontaneity in chemical reactions indicates whether a reaction can proceed on its own without additional energy input. It's primarily driven by Gibbs Free Energy (\( \Delta G^{*} \)).

• **Spontaneous Reaction**: Occurs when \( \Delta G^{*} \) is negative, meaning the system's total energy has decreased sufficiently.
• **Non-Spontaneous Reaction**: Happens when \( \Delta G^{*} \) is positive, suggesting external energy is needed for the reaction to proceed.
  

The equation \( \Delta G^{*} = \Delta H - T\Delta S \) is used to calculate the Gibbs Free Energy. If \( \Delta G^{*} \) is negative, a reaction is spontaneous, just as with reaction (a) in the exercise example at \(298 \mathrm{~K}\). However, if it's positive, like reaction (b), the reaction remains non-spontaneous at the given temperature.
As learned, adjusting temperature can affect spontaneity, exemplified by reaction (b) becoming spontaneous at \(3197 \mathrm{~K}\).