Freezing point is always a temperature, meaning it must correspond to a single point on the y-axis. During the freezing/melting process, heat is absorbed/released from the system without any change in temperature. This results in the temperature plateau at the freezing point.

While the substance is freezing during interval B, the freezing point is the temperature at which it freezes, X.

During intervals B and D, fusion and vaporization are taking place, respectively. Instead of raising the temperature of the substance, the energy that is added during these phase changes is used to overcome intermolecular forces. The energy breaks the attraction between particles, allowing them to separate and gain the properties of a higher energy phase (liquid or gas).

When the slope is not zero, the phase is steady and the added heat energy is used to increase the molecular kinetic energy of the particles, resulting in a temperature increase.

Use the given molar heat of fusion to calculate the amount of snow formed. Remember that heats of fusion and solidification are opposite processes, so the magnitudes of molar heats of fusion and solidification are the same and signs are opposite. This indicates that fusion (melting) is endothermic, while solidification is exothermic.

The given change in energy will be negative, since the question states that it is released from the system into the atmosphere.

Convert moles to grams, and grams to pounds.

We must use the given heats of fusion and vaporization to calculate the energy change involved in these two processes.

First, calculate the energy change when  of water solidifies. Convert the mass to moles and multiply by the heat released during freezing. This value will be equal in magnitude, but opposite in sign to the heat of fusion.

From this calculation we find that  of heat is released into the surroundings (a negative sign denotes an exothermic process). 

Next, find the energy change associated with the vaporization of of water, using the given heat of vaporization:

We find that  of energy is absorbed when this quantity of water is vaporized.  

Adding the two together we find a total of .

Since this is a positive number, that means that the energy is absorbed from the surroundings (endothermic).

Solving this problem means solving for three steps.

1. The heat needed to raise the temperature from –20^oC to 0^oC.

2. Melting the ice (changing phases).

3. Raising the water temperature from 0^oC to 50^oC.

Steps 1 and 3 are both solved by the equation .

Step 2 is solved using the enthalpy of fusion, and is multiplied by the number of grams being melted: .

Combining the steps, we get the following expression.

Vaporization and condensation refer to the transition from liquid to gas and from gas to liquid, respectively. Fusion and freezing, in contrast, refer to the transition from solid to liquid and from liquid to solid, respectively. Vaporization (boiling) and fusion (melting) each require an input of energy, making them endothermic processes with positive changes in enthalpy. Condensation and freezing result in a decrease in energy and an output of enthalpy, making them exothermic.

The given enthalpy of fusion tells us that are consumed for every mole of water. If two moles of water are present, then are consumed during melting.

Enthalpy of vaporization will be equal and opposite the enthalpy of condensation. Since condensation is exothermic, heat will be released and the change in enthalpy must be negative (not positive).

Heat of condensation:

Intermolecular forces play a key role in determining the energy required for phase changes. Strong intermolecular forces result in more resistance to changes that result in greater distance between molecules (greater entropy), as the forces cause the molecules to "stick" to one another. When a liquid is vaporized, the strength of the intermolecular force is overcome; similar heats of vaporization indicate similar intermolecular forces. A similar concept governs the transition from solid to liquid. If two compounds share similar enthalpies of fusion and vaporization, then they likely have similar intermolecular forces.

In the given table, ethanol enthalpy values are most similar to those of water, meaning it likely has similar intermolecular forces.

Plateaus in the graph represent phase changes: period B shows the transitions between solid and liquid, and period D shows the transitions between liquid and gas. When the substance transitions through period D, it undergoes either vaporization (C to E transition) or condensation (E to C transition). If the substances starts out at interval E and ends at interval C, then it has undergone condensation.

Condensation involves transition from a high energy gas to a lower energy liquid, and has a net decrease in heat energy and temperature. This heat release is known as an exothermic process.

Notice that most of these are true statements (the only incorrect statement is that fusion is an exothermic process). The key is finding the statement that best explains why the heat of vaporization is greater than the heat of fusion. Essentially, we are looking for the reason why a transition from liquid to gas requires more energy than a transition from solid to liquid.

It is true that more heat is required to vaporize a given quantity of water, but the reason for this can be found on a molecular level. Gas molecules have a relatively high kinetic energy. The large difference in kinetic energies between the states accounts for the difference in heats of fusion and vaporization. Solids are the lowest energy state, followed by liquids, and then gases. The difference in energy between equal amounts of solid and liquid is given by the heat of fusion, while the difference in energy between equal amounts of liquid and gas is given by the heat of vaporization. The discrepancy in energy is reflected in the difference between these two heat quantities.

As scenario 1 begins, the reaction is spontaneous as written, and so the Gibbs Free Energy is negative. Additionally, the Van't Hoff equation proves that Keq increases with decreasing temperature in exothermic reactions. As this reaction is exothermic and placed in low-temperature conditions, the relative abundance of the products will become the prevailing state.  Keq, therefore, increases.