Question

The melting point of benzene is 5.5 degree C. Predict the signs of Delta H, Delta S, and Delta G for the melting of benzene at: a. 0.0 °C ΔH = ΔS = ΔG = b. 15.0 °C ΔH = ΔS = ΔG =

Answer 1

a. At 0.0°C, the equilibrium state of benzene's melting yields ΔH = ΔS = ΔG = 0.

b. At 15.0°C, ΔH > 0, ΔS > 0, and ΔG's sign depends on the relationship between ΔH and TΔS.

Step-by-step explanation:

The process of melting involves changes in enthalpy (ΔH) and entropy (ΔS). At 0.0°C, where the melting point of benzene is 5.5°C, the transition from solid to liquid occurs at equilibrium. Therefore, at this temperature, there is no net change in enthalpy, entropy, or Gibbs free energy (ΔG), all equaling zero. The system is in a state of dynamic equilibrium between solid and liquid phases.

However, at 15.0°C, which is higher than the melting point of benzene, the sign of ΔH is positive as energy is required to overcome intermolecular forces during melting. Additionally, ΔS is positive as the liquid state usually has higher entropy than the solid state due to increased molecular disorder. The sign of ΔG at 15.0°C depends on the relationship between ΔH and ΔS. If ΔH is greater than TΔS (where T is the temperature in Kelvin), ΔG will be positive, indicating a non-spontaneous process. Conversely, if ΔH is less than TΔS, ΔG will be negative, indicating a spontaneous process.

To calculate ΔG precisely at 15.0°C, one would use the equation ΔG = ΔH - TΔS, considering the temperature in Kelvin. This would determine whether the process is spontaneous or non-spontaneous based on the relative values of ΔH and ΔS, influenced by the temperature.

Answer 2

The signs of ΔH, ΔS, and ΔG for the melting of benzene at different temperatures are determined based on the relationship between temperature and the melting point of the substance.

Step-by-step explanation:

The signs of ΔH, ΔS, and ΔG for the melting of benzene at different temperatures can be determined using the following guidelines:

a) 0.0 °C:

Since the melting point of benzene is 5.5 °C, melting at 0.0 °C would require the absorption of heat energy. Hence, ΔH would be positive.

The melting process usually involves an increase in the randomness of the molecules, leading to an increase in entropy. Therefore, ΔS would also be positive.

Using the equation ΔG = ΔH - TΔS, where T is the temperature in kelvin, we can calculate ΔG. At 0.0 °C (273.15 K), ΔG would depend on the relative magnitudes of ΔH and ΔS.

b) 15.0 °C:

At 15.0 °C, which is below the melting point of benzene, the substance would still be in the solid state. Hence, ΔH, ΔS, and ΔG would be zero because no melting occurs.