Question

Given the following data for sodium $$ \begin{array}{l} \mathrm{Na}(s): S^{\circ}=51.2 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ \mathrm{Na}(g): S^{\circ}=153.6 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \quad \Delta H_{\mathrm{f}}^{\circ}=108.7 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ estimate the temperature at which sodium sublimes at 1 atm. $$ \mathrm{Na}(s) \rightleftharpoons \mathrm{Na}(g) $$

Short answer

Answer: The sublimation temperature of sodium at 1 atm pressure is approximately 1061.52 K.

Step-by-step solution

Calculate the standard enthalpy of sublimation

Since we have the standard enthalpy of formation for gaseous sodium \((\Delta H^{\circ}_{\mathrm{f}})\), we can use it to find the standard enthalpy of sublimation \((\Delta H^{\circ}_{\mathrm{sub}})\) as follows: $$ \Delta H^{\circ}_{\mathrm{sub}} = \Delta H^{\circ}_{\mathrm{f}} (\mathrm{Na} (g)) - \Delta H^{\circ}_{\mathrm{f}} (\mathrm{Na} (s)) $$ Given that the standard enthalpy of formation for a pure element in its standard state is zero, we have: $$ \Delta H^{\circ}_{\mathrm{sub}} = 108.7 \mathrm{~kJ} / \mathrm{mol} $$

Calculate the standard entropy change for sublimation

To calculate the standard entropy change during sublimation \((\Delta S^{\circ}_{\mathrm{sub}})\), we can use the standard entropy values for solid and gaseous sodium as follows: $$ \Delta S^{\circ}_{\mathrm{sub}} = S^{\circ}_{\mathrm{g}} - S^{\circ}_{\mathrm{s}} $$ Substituting the given values: $$ \Delta S^{\circ}_{\mathrm{sub}} = (153.6 - 51.2) \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} = 102.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} $$

Apply the Gibbs Free Energy equation

The reaction is at equilibrium when the Gibbs free energy change \((\Delta G^{\circ}_{\mathrm{sub}})\) is equal to zero, which gives us the following equation: $$ 0 = \Delta H^{\circ}_{\mathrm{sub}} - T \Delta S^{\circ}_{\mathrm{sub}} $$ Rearranging the equation to solve for the temperature, we have: $$ T = \frac{\Delta H^{\circ}_{\mathrm{sub}}}{\Delta S^{\circ}_{\mathrm{sub}}} $$

Calculate the sublimation temperature

Using the values calculated in Steps 1 and 2: $$ T = \frac{108.7 \times 10^3 \mathrm{~J} / \mathrm{mol}}{102.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}} = 1061.52 \mathrm{~K} $$ The sublimation temperature of sodium at 1 atm is approximately 1061.52 K.