How do I solve this AP Chemistry problem on melting point and phase change?

Question

The melting point of a fictional X is 225°C at 10.0 atm. If the density of the solid phase of X is 2.67 g/cm^3 at 10.0 atm. And the density of the liquid phase is 2.78 g/cm^3 at 10.0 atm. Predict whether the normal melting point of X will be less than equal to or greater than 225°C.

Answer 1

To predict whether the normal melting point of substance X will be less than, equal to, or greater than 225°C, we can use the Clausius-Clapeyron equation. The Clausius-Clapeyron equation relates the vapor pressure of a substance at one temperature and pressure to its vapor pressure at another temperature and pressure when it undergoes a phase change (in this case, from solid to liquid).

The equation is as follows:

\[ \ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Where:

- \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \), respectively.

- \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization.

- \( R \) is the gas constant.

- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin.

In this problem, we have the melting point of X at 225°C, which we need to convert to Kelvin:

\[ T_1 = 225°C + 273.15 = 498.15 \, \text{K} \]

We also know that the densities of the solid and liquid phases are given at 10.0 atm. This means that the vapor pressures of the solid and liquid phases of X at 10.0 atm are equal to the pressure, which is 10.0 atm.

Therefore, we can set \( P_1 = P_2 = 10.0 \, \text{atm} \) in the Clausius-Clapeyron equation. Rearranging the equation, we get:

\[ \ln(1) = \frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Since \( \ln(1) = 0 \), we have:

\[ 0 = \frac{\Delta H_{\text{vap}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Now, we want to predict whether the normal melting point (\( T_2 \)) will be less than, equal to, or greater than 225°C (\( T_1 \)). To do this, we need to consider the sign of the right-hand side of the equation.

1. If \( \frac{1}{T_1} - \frac{1}{T_2} \) is positive, then \( T_2 \) will be greater than \( T_1 \), meaning the normal melting point is greater than 225°C.

2. If \( \frac{1}{T_1} - \frac{1}{T_2} \) is negative, then \( T_2 \) will be less than \( T_1 \), meaning the normal melting point is less than 225°C.

3. If \( \frac{1}{T_1} - \frac{1}{T_2} \) is zero, then \( T_2 \) will be equal to \( T_1 \), meaning the normal melting point is equal to 225°C.

So, you need to calculate \( \frac{1}{T_1} - \frac{1}{T_2} \) and determine its sign to make your prediction.