Question

Calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction: $$ \mathrm{S}+3 / 2 \mathrm{O}_{2} \rightleftarrows \mathrm{SO}_{3} $$ The heat formation of \(\mathrm{SO}_{3}\) at \(25^{\circ} \mathrm{C}\) is \(-94.45 \mathrm{Kcal} / \mathrm{mole}\) and the standard molar entropy changes for \(\mathrm{S}, \mathrm{O}_{2}\), and \(\mathrm{SO}_{3}\) at \(25^{\circ} \mathrm{C}\) are \(7.62,49.0\) and \(61.24 \mathrm{cal} / \mathrm{mole}^{\circ} \mathrm{K}\), respectively.

Short answer

The equilibrium constant, \(K\), for the given reaction at \(25^{\circ} \mathrm{C}\) is approximately \(3.64 \times 10^{64}\).

Step-by-step solution

Write the given information in an organized manner

Heat formation of \(\mathrm{SO}_{3}\) at \(25^{\circ} \mathrm{C}\): \(\Delta H_f = -94.45 \, \mathrm{Kcal} / \mathrm{mole}\) Standard molar entropy changes at \(25^{\circ} \mathrm{C}\): \( S(\mathrm{S}) = 7.62 \, \mathrm{cal} / \mathrm{mole}^{\circ} \mathrm{K}\) \( S(\mathrm{O}_{2}) = 49.00 \, \mathrm{cal} / \mathrm{mole}^{\circ} \mathrm{K}\) \( S(\mathrm{SO}_{3}) = 61.24 \, \mathrm{cal} / \mathrm{mole}^{\circ} \mathrm{K} \)

Calculate the change in entropy, \(\Delta S\), for the reaction

For the given reaction: \(\mathrm{S}+3/2 \, \mathrm{O}_{2} \rightleftarrows \mathrm{SO}_{3}\) \(\Delta S_{reaction} = S(\mathrm{SO}_{3}) - (S(\mathrm{S}) + \frac{3}{2} \times S(\mathrm{O}_{2}))\) \(\Delta S_{reaction} = 61.24 - (7.62 + \frac{3}{2} \times 49.00) \, \mathrm{cal}/\mathrm{mole}^{\circ}\mathrm{K}\) \(\Delta S_{reaction} = 61.24 - 81.12 = -19.88 \, \mathrm{cal}/\mathrm{mole}^{\circ}\mathrm{K}\)

Calculate the change in Gibbs Free Energy, \(\Delta G\), for the reaction

First, we need to convert \(\Delta H_f\) to \(\mathrm{cal} / \mathrm{mole}\): \(\Delta H_f = -94.45 \times 1000 = -94450 \, \mathrm{cal} / \mathrm{mole}\) Now we can use the Gibbs Free Energy equation: \(\Delta G = \Delta H - T\Delta S\) At \(25^{\circ} \mathrm{C}\), the temperature, \(T = 273.15 + 25 = 298.15 \, \mathrm{K}\) \(\Delta G = -94450 - (298.15 \times (-19.88)) \, \mathrm{cal} / \mathrm{mole}\) \(\Delta G = -94450 + 5930.22 = -88519.78 \, \mathrm{cal} / \mathrm{mole}\)

Calculate the equilibrium constant, \(K\)

We can calculate the equilibrium constant using the relation: \(K = e^{-\frac{\Delta G}{RT}}\) Where \(R\) is the gas constant, \(R = 1.987 \, \mathrm{cal} / \mathrm{mole} \cdot \mathrm{K}\) \(K = e^{-\frac{-88519.78}{(1.987 \times 298.15)}}\) \(K = e^{149.1536} \approx 3.64 \times 10^{64}\) The equilibrium constant, \(K\), for the given reaction at \(25^{\circ} \mathrm{C}\) is approximately \(3.64 \times 10^{64}\).

Key concepts

Gibbs Free Energy

Gibbs Free Energy, often represented as \( \Delta G \), is a thermodynamic potential that helps mediate chemical reactions. It combines enthalpy, which is heat content, and entropy, a measure of disorder or randomness. This energy concept predicts whether a reaction is spontaneous—that is, whether it occurs naturally without external assistance.

The important relationship to remember is the equation:

• \( \Delta G = \Delta H - T\Delta S \)

Here, \( \Delta H \) represents the change in enthalpy, \( T \) is the absolute temperature, and \( \Delta S \) is the change in entropy. If \( \Delta G \) is negative, the reaction tends to be spontaneous. Our exercise demonstrated this with \( \Delta G = -88,519.78 \) cal/mole, indicating the reaction is indeed spontaneous at 25°C.

Entropy Change

Entropy, signified as \( S \), measures how much disorder or randomness is in a system. Changes in entropy reveal how a reaction affects this disorder, and it's a crucial component of Gibbs Free Energy calculations.

To calculate the change in entropy for the reaction \( \mathrm{S} + \frac{3}{2} \mathrm{O_2} \rightleftharpoons \mathrm{SO_3} \), follow this method:

• \( \Delta S_{reaction} = S(\mathrm{SO_3}) - (S(\mathrm{S}) + \frac{3}{2} \times S(\mathrm{O_2})) \)

After substituting the known values, the result was \( \Delta S = -19.88 \) cal/mole°K. The negative value suggests less disorder in the products compared to the reactants, making it orderly.

Heat Formation

Heat formation, also known as enthalpy change (\( \Delta H_f \)), measures the heat absorbed or released during a chemical reaction. For our exercise, the heat formation of \( \mathrm{SO_3} \) is given as \(-94.45 \) Kcal/mol at 25°C.

This refers to the energy released when sulfur and oxygen combine to form sulfur trioxide. If the value is negative, as in this exercise, the process is exothermic, meaning it gives off heat.

In calculating Gibbs Free Energy, \( \Delta H_f \) must be converted to consistent units (cal/mole) for combining with entropy for Gibbs energy calculations. This conversion ensures an accurate interpretation of the thermodynamic properties.

Thermodynamics

Thermodynamics is an overarching principle guiding the behavior of energy and its transformation between forms in chemical reactions. This discipline provides vital insights into whether a reaction will occur and what energy changes accompany it.

In the exercise, thermodynamic calculations involved entropy, heat formation, and Gibbs Free Energy to derive the equilibrium constant (\( K \)). The constant \( K \) gives a sense of the reaction's favorability under specified conditions. The considerably large value \( 3.64 \times 10^{64} \) found for this reaction underlines its tendency to proceed towards the formation of \( \mathrm{SO_3} \) at room temperature.
Understanding these variables in concert exemplifies the real potency of thermodynamics as a predictive tool.