Question

Consider the reaction $$ 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) $$ a. Calculate \(\Delta G^{\circ}\) for this reaction. The \(\Delta G_{\mathrm{f}}^{\circ}\) values for \(\mathrm{POCl}_{3}(g)\) and \(\mathrm{PCl}_{3}(g)\) are \(-502 \mathrm{~kJ} / \mathrm{mol}\) and \(-270 . \mathrm{kJ} / \mathrm{mol}\), respectively. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{~K}\) ? c. The value of \(\Delta S^{\circ}\) for this reaction is \(179 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short answer

The change in standard Gibbs free energy (ΔG°) for the given reaction is 464 kJ/mol. The reaction is NOT spontaneous under standard conditions at 298 K. The reaction is spontaneous at standard conditions for temperatures greater than 2890 K.

Step-by-step solution

Calculate ΔGf° for all the reactants and products in the equation

We have the ΔGf° values for POCl3(g) and PCl3(g). Let's write them down: ΔGf°(POCl3) = -502 kJ/mol ΔGf°(PCl3) = -270 kJ/mol We can calculate ΔGf° for O2(g) as it is in its standard state (an element in its most stable form under standard conditions), and which has a ΔGf° value of 0. ΔGf°(O2) = 0 kJ/mol

Calculate ΔG° for the reaction

Using the stoichiometric coefficients in the balanced equation and the calculated ΔGf° values, we can calculate the ΔG° for the reaction: ΔG° = (2ΔG°(PCl3) + ΔG°(O2)) - (2ΔG°(POCl3)) Substitute the values: ΔG° = (2 × (-270 kJ/mol) + 0 kJ/mol) - (2 × (-502 kJ/mol)) ΔG° = -540 kJ/mol + 1004 kJ/mol ΔG° = 464 kJ/mol So the change in standard Gibbs free energy (ΔG°) for this reaction is 464 kJ/mol. #b. Is this reaction spontaneous under standard conditions at 298 K?#

Determine if the reaction is spontaneous at 298 K

A reaction is spontaneous under standard conditions if the ΔG° for the reaction is negative. In our case, ΔG° = 464 kJ/mol, which is positive. Therefore, the reaction is NOT spontaneous under standard conditions at 298 K. #c. At what temperatures is this reaction spontaneous at standard conditions?#

Calculate ΔH° for the reaction

Using the fact that ΔG° = ΔH° - TΔS°, we can solve for ΔH°. We know that ΔS° = 179 J/mol·K, and we already found ΔG° = 464 kJ/mol. Thus, we have: 464 kJ/mol = ΔH° - (298 K) × (179 J/mol·K) Note that we should convert ΔG° to J/mol to match the units of ΔS°: 464 kJ/mol × (1000 J/1 kJ) = 464000 J/mol Now, we solve for ΔH°: ΔH° = 464000 J/mol + (298 K × 179 J/mol·K) ΔH° = 464000 J/mol + 53342 J/mol ΔH° = 517342 J/mol So the change in standard enthalpy (ΔH°) for this reaction is 517342 J/mol.

Find the temperature range for spontaneity

Recall the equation ΔG° = ΔH° - TΔS°. A reaction is spontaneous when ΔG° < 0. Therefore, we need to find a temperature for which this inequality holds: ΔH° - TΔS° < 0 We already found that ΔH° = 517342 J/mol and ΔS° =179 J/mol·K. Substitute these values: 517342 J/mol - T × 179 J/mol·K < 0 Now, isolate T to find the temperature range: T > (517342 J/mol)/(179 J/mol·K) T > 2890 K Hence, the reaction is spontaneous at standard conditions for temperatures greater than 2890 K.

Key concepts

Enthalpy Change

Enthalpy change, denoted by \( \Delta H \), is the total heat content of a system. In chemical reactions, it represents the difference in energy between reactants and products. When a reaction results in the release of heat, \( \Delta H \) is negative, indicating an exothermic reaction. Conversely, when the reaction absorbs heat, \( \Delta H \) is positive, indicating an endothermic reaction.

In our scenario, for the reaction \( 2 \text{POCl}_3(g) \rightarrow 2 \text{PCl}_3(g) + \text{O}_2(g) \), calculating \( \Delta H \) helps understand the energy dynamics under standard conditions. It is calculated using Gibbs energy and entropy changes:

• Start by finding \( \Delta H \) using the formula: \( \Delta G = \Delta H - T \Delta S \).
• Here, \( \Delta G \) is the change in Gibbs free energy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.
• You adjust energy units accordingly to keep them consistent when plugging into equations.

This calculation yields insights into whether heat is being absorbed or released, a crucial factor for predicting spontaneity.

Entropy Change

Entropy change, represented as \( \Delta S \), measures the degree of disorder or randomness in a system. In chemical reactions, an increase in entropy occurs when products are more disordered than reactants. This is generally associated with greater molecular motion or separation.

For instance, in our reaction where \( 2 \text{POCl}_3(g) \) decomposes into \( 2 \text{PCl}_3(g) + \text{O}_2(g) \), \( \Delta S \) is given as \( 179 \text{ J/K} \cdot \text{mol} \). Here's the breakdown:

• Positive \( \Delta S \) suggests an increase in disorder — the reactants form additional gas, \( \text{O}_2(g) \), increasing randomness.
• Entropy changes affect how temperature influences a reaction's spontaneity. A positive \( \Delta S \) can favor spontaneity at higher temperatures.

Understanding \( \Delta S \) clarifies how particles' distribution affects the energy flow and reaction feasibility.

Spontaneity of Reaction

Whether a reaction is spontaneous depends on the sign of its Gibbs free energy change \( \Delta G \). A reaction is considered spontaneous if \( \Delta G < 0 \). This means that under given conditions, the reaction can occur without external input.

In our case, where we calculated \( \Delta G \) to be 464 kJ/mol, the positive sign indicates non-spontaneity at 298 K. To explore spontaneity further:

• Adjusting temperature can influence \( \Delta G \). A higher temperature can make reactions with positive entropy changes spontaneous.
• The equation \( \Delta G = \Delta H - T \Delta S \) shows how both enthalpy and entropy play roles, with temperature being a critical modifier.
• For this reaction, a temperature greater than 2890 K leads to spontaneity due to a favorable balance of \( \Delta H \) and \( T \Delta S \).

Conclusively, understanding spontaneity helps predict reaction viability under different conditions.

Standard Conditions

Standard conditions refer to a set benchmark that allows the consistent study of reactions across different experiments. Defined as 298 K (25°C) and 1 atmosphere of pressure, these conditions are symbolized with a "°" in thermodynamic quantities like \( \Delta G^\circ \), \( \Delta H^\circ \), and \( \Delta S^\circ \).

These conditions are essential for:

• Ensuring that comparisons between different reactions are reliable, as all reactions are assessed under the same baseline conditions.
• Helping predict reaction behavior consistently in theoretical settings, simplifying practical applications like drug design or material synthesis.
• Providing a reference point which, when altered, aids in studying how variables such as temperature or pressure influence specific reactions.

Using standard conditions simplifies complex energetic calculations, making thermodynamic properties more usable and predictions more accurate.