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 \(179 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Step by step solution

01

Write the general equation to determine ΔG° of the reaction

To calculate the standard Gibbs free energy change (ΔG°) for the reaction, we use the following equation: \[ \Delta G^\circ = \Delta G_{products}^\circ - \Delta G_{reactants}^\circ \] Where ΔG° is the Gibbs free energy change for the reaction, and ΔG°(products) and ΔG°(reactants) are the standard Gibbs free energies of the products and reactants, respectively.

02

Determine the ΔG° of the products and reactants

The standard Gibbs free energy of formation (ΔGf°) values for POCl3(g) and PCl3(g) are given as -502 kJ/mol and -270 kJ/mol, respectively. Use these values to determine the standard Gibbs free energy of the products and reactants: For reactants (2 mol POCl3): \( \Delta G_{reactants}^\circ = 2 \times (-502 \, kJ/mol) = -1004\, kJ/mol \) For products (2 mol PCl3 and 1 mol O2): \( \Delta G_{products}^\circ = 2 \times (-270\, kJ/mol) + 0 = -540\, kJ/mol \) (Note: The ΔGf° of elemental O2 in its standard state is 0.)

03

Calculate ΔG° for the reaction

Now, substitute the values of ΔG°(products) and ΔG°(reactants) into the general equation: \[ \Delta G^\circ = -540\, kJ/mol - (-1004\, kJ/mol) = 464\, kJ/mol \] So, ΔG° for this reaction is +464 kJ/mol. #b. Determine if the reaction is spontaneous under standard conditions at 298 K#

04

Check the sign of ΔG°

If the sign of ΔG° is negative, the reaction is spontaneous, and if the sign is positive, it is non-spontaneous. In this case, ΔG° is +464 kJ/mol, which means the reaction is not spontaneous under standard conditions at 298 K. #c. Find the temperature at which the reaction becomes spontaneous#

05

Write the Gibbs free energy equation that includes enthalpy and entropy

To determine the temperature at which the reaction becomes spontaneous, we will use the Gibbs free energy equation that includes enthalpy (ΔH°) and entropy (ΔS°): \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] Where ΔH° is the standard enthalpy change, T is the temperature, and ΔS° is the standard entropy change.

06

Rearrange the equation to solve for the temperature

We want to know the temperature at which the reaction becomes spontaneous, which means ΔG° < 0. Rearrange the equation to solve for T: \[ T > \frac{\Delta H^\circ}{\Delta S^\circ} \]

07

Calculate ΔH° using ΔG° and ΔS°

We are given ΔS° = 179 J/(K·mol) for the reaction. We also know ΔG° = +464 kJ/mol from part a. Now, we need to calculate ΔH°: Use the equation: \( \Delta H^\circ = \Delta G^\circ + T \Delta S^\circ \) For the temperature (T), we will use the standard condition temperature of 298 K: \(\Delta H^\circ = 464\, kJ/mol + 298\, K \times (179\, J/(K \cdot mol)) \) Convert entropy into kJ to match units with ΔG and ΔH: \(\Delta H^\circ = 464\, kJ/mol + 298\, K \times (0.179\, kJ/(K \cdot mol)) \) \(\Delta H^\circ = 464\, kJ/mol + 53.342\, kJ/mol \) \(\Delta H^\circ = 517.34\, kJ/mol \)

08

Calculate the temperature at which the reaction becomes spontaneous

Now that we have the values of ΔH° and ΔS°, we can use the equation we derived earlier to find the temperature at which the reaction becomes spontaneous: \[ T > \frac{\Delta H^\circ}{\Delta S^\circ} \] \( T > \frac{517.34\, kJ/mol}{0.179\, kJ/(K \cdot mol)} \) \( T > 2890\, K \) Therefore, the reaction becomes spontaneous under standard conditions at temperatures above 2890 K.

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