Problem 49 - Chemistry: Principles and Reactions

Question

Red phosphorus is formed by heating white phosphorus. Calculate the temperature at which the two forms are at equilibrium, given $$ \text { white } \text { P: } \Delta H_{\mathrm{f}}^{\circ}=0.00 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ}=41.09 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} $$ red \(\mathrm{P}: \Delta H_{\mathrm{f}}^{\circ}=-17.6 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{S}^{\circ}=22.80 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\).

Step-by-Step Solution

Answer

Answer: White phosphorus and red phosphorus are in equilibrium at approximately 962 K.

Step 1: 1. Write down the Gibbs free energy equation for both forms of phosphorus

Since at equilibrium, the change in Gibbs free energy, \(\Delta G\) is zero, we can write the following equations for both white phosphorus (W) and red phosphorus (R): $$ \Delta G_W = \Delta H_W - T\Delta S_W \\ \Delta G_R = \Delta H_R - T\Delta S_R $$

Step 2: 2. Write down the Gibbs free energy for the reaction between white phosphorus and red phosphorus

We can express the reaction as White P \(\rightleftharpoons\) Red P. The change in Gibbs free energy for the reaction, \(\Delta G_{reaction}\), is given by the difference between the \(\Delta G\) of products and reactants. In this case, it is: $$ \Delta G_{reaction} = \Delta G_R - \Delta G_W $$

Step 3: 3. Set up the expression for the equilibrium condition

Since \(\Delta G_{reaction} = 0\) at equilibrium, we can set up the expression: $$ \Delta G_R - \Delta G_W = 0 $$

Step 4: 4. Substitute the Gibbs free energy equations into the equilibrium expression

Substituting the expressions for \(\Delta G_W\) and \(\Delta G_R\) from step 1, the equilibrium expression becomes: $$ (\Delta H_R - T\Delta S_R) - (\Delta H_W - T\Delta S_W) = 0 $$

Step 5: 5. Simplify the equation and solve for the temperature (T)

Rearrange the equation to solve for T: $$ T(\Delta S_R - \Delta S_W) = \Delta H_R - \Delta H_W $$ Now plug in the given values for \(\Delta H_W\), \(\Delta H_R\), \(S_W\), and \(S_R\): $$ T(22.80\,\text{J/mol·K} - 41.09\,\text{J/mol·K}) = (-17.6\,\text{kJ/mol} - 0\,\text{kJ/mol}) $$ $$ T(-18.29\,\text{J/mol·K}) = -17.6 \times 10^3\,\text{J/mol} $$ Divide both sides by -18.29 J/mol·K to obtain the equilibrium temperature: $$ T = \frac{-17.6 \times 10^3\,\text{J/mol}}{-18.29\,\text{J/mol·K}} = 962\,\text{K} $$ Hence, the temperature at which white phosphorus and red phosphorus are at equilibrium is approximately 962 K.

Key Concepts

Gibbs Free Energy

Gibbs free energy, symbolized by the letter G, is a thermodynamic quantity that represents the maximum amount of work that a system can perform at constant temperature and pressure. It's a crucial concept for understanding chemical equilibrium because it helps determine whether a reaction will proceed spontaneously. The change in Gibbs free energy, denoted by \( \Delta G \), for a process is given by the equation:\[ \Delta G = \Delta H - T\Delta S \]

Here, \( \Delta H \) is the change in enthalpy, \( T \) is the absolute temperature in Kelvin, and \( \Delta S \) is the change in entropy. When \( \Delta G \) is negative, the process occurs spontaneously; when it's positive, the process is non-spontaneous, and when it's zero, the system is at equilibrium, meaning there is no net change in the composition of the reaction mixture over time.

Enthalpy Change

Enthalpy change, represented by \( \Delta H \) in chemical equations, is a measure of the heat absorbed or released during a reaction at constant pressure. It's an important concept in determining the energy changes within a chemical system. Endothermic reactions absorb heat (\(\Delta H > 0\)), making the surroundings cooler, while exothermic reactions release heat (\(\Delta H < 0\)), heating up the surroundings.

For the transformation of white phosphorus to red phosphorus, the enthalpy change helps us understand part of the energy dynamics during the equilibrium. Calculating the difference in enthalpy between the two forms of phosphorus is one step in determining the equilibrium temperature, as seen in the original exercise.

Entropy

Entropy, symbolized as S, is a measure of the disorder or randomness in a system. The greater the disorder, the higher the entropy. Entropy change, indicated by \( \Delta S \), occurs when a system transitions from one state to another. An increase in entropy (\(\Delta S > 0\)) means that the disorder of the system is increasing, often associated with the dispersal of energy and matter.

Within the context of the exercise, we compare the entropy of white phosphorus and red phosphorus to help understand how the system's disorder changes during the equilibrium process. The equilibrium condition reflects a balance between the enthalpy and entropy changes at a certain temperature, contributing to the overall Gibbs free energy change of zero.

Equilibrium Temperature

Equilibrium temperature is the temperature at which a system reaches a state of equilibrium, where there is no net change in the concentrations of reactants and products over time. This concept is closely tied to the concepts of Gibbs free energy, enthalpy, and entropy, as the equilibrium temperature is a condition where the total free energy change of a reaction mixture is zero.

In the step-by-step solution provided, the equilibrium temperature for the transition between white phosphorus and red phosphorus is calculated. By setting the Gibbs free energy change for the reaction to zero (which occurs at equilibrium) and solving for temperature, we determine that equilibrium will occur at approximately 962 K. This calculation is vital for predicting the conditions under which the two forms of phosphorus can coexist without further conversion to one another.