Question

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

Short answer

The temperature at which white and red phosphorus are in equilibrium is approximately 962 K.

Step-by-step solution

Calculate the overall enthalpy and entropy changes

We need to find the difference in enthalpy and entropy between the reactants (white phosphorus) and the products (red phosphorus). To do this, we subtract the values for the reactants from the values for the products. $$\Delta H^\circ = \Delta H_{red}^\circ - \Delta H_{white}^\circ = -17.6\,\text{kJ/mol} - 0.00\,\text{kJ/mol} = -17.6\,\text{kJ/mol}$$ $$\Delta S^\circ = S_{red}^\circ - S_{white}^\circ = 22.80\,\text{J/mol}\cdot\text{K}-41.09\,\text{J/mol}\cdot\text{K} = -18.29\,\text{J/mol}\cdot\text{K}$$

Set up the standard free energy change equation

Now that we have the enthalpy and entropy changes, we can set up the equation for the standard free energy change: $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ Since we're trying to find the temperature \(T\) at which the system reaches equilibrium, we want \(\Delta G^\circ = 0\): $$0 = -17.6\,\text{kJ/mol} - T(-18.29\,\text{J/mol}\cdot\text{K})$$

Solve for the equilibrium temperature

To solve for \(T\), we need to first convert the enthalpy change to J/mol to match the units of the entropy change: $$0 = -17,600\,\text{J/mol} + 18.29\,\text{J/mol}\cdot\text{K}\cdot T$$ Next, divide by the entropy change to isolate \(T\): $$T = \frac{17,600\,\text{J/mol}}{18.29\,\text{J/mol}\cdot\text{K}}$$ Finally, calculate the temperature: $$T \approx 962\,\text{K}$$ Therefore, the temperature at which white and red phosphorus are in equilibrium is approximately 962 K.