MCAT Physical - Thermodynamics of Phase Changes

A scientist prepares an experiment to demonstrate the second law of thermodynamics for a chemistry class. In order to conduct the experiment, the scientist brings the class outside in January and gathers a cup of water and a portable stove.

The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:

Gibbs Free Energy Formula:

ΔG = ΔH – TΔS

Liquid-Solid Water Phase Change Reaction:

H2O(l) ⇌ H2O(s) + X

The scientist prepares two scenarios.

Scenario 1:

The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.

Scenario 2:

The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.

After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment.

Scenario 3:

Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.

Scenario 4:

The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.

In scenario 1, the Gibbs Free Energy and Keq of the Liquid-Solid Water Phase Change Reaction, as the reaction begins, is best characterized as __________.

negative and increasing, respectively

positive and increasing, respectively

zero and decreasing, respectively

undefined and decreasing, respectively

negative and decreasing, respectively

Explanation

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.

The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:

Gibbs Free Energy Formula:

ΔG = ΔH – TΔS

Liquid-Solid Water Phase Change Reaction:

H2O(l) ⇌ H2O(s) + X

The scientist prepares two scenarios.

Scenario 1:

The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.

Scenario 2:

The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.

After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment.

Scenario 3:

Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.

Scenario 4:

The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.

When the scientist moves the frozen water to the stove, the water melts. If the temperature of the stove is kept such that the reaction is at equilibrium, and the water is almost entirely melted, which of the following is a possible Keq value for the Liquid-Solid Water Phase Change Reaction as it is written?

$2.3*10^{-15}$

$\dpi{100} \small 4.2\times 10^{3}$

$\dpi{100} \small 2.9\times 10^{15}$

Explanation

Once the water is almost entirely melted, the reaction is heavily skewed to the left side of the reaction. Notice the reaction is written with liquid water on the left. The equilibrium constant equation is written with the product (right-side) concentrations over the reactant concentrations; therefore, if there is a relative abundance of the reactants to products, K will be less than 1, but not all the way down to zero.

The molar heats of fusion and vaporization for water are given below:

$\Delta H_{vap} = 40.7 \frac{kJ}{mol}$

$\Delta H_{fus} = 6.01 \frac{kJ}{mol}$

of water solidify and of water vaporize within a closed system. What is the change in energy of the surroundings?

of energy is absorbed from the surroundings

of energy is released to the surroundings

The system experiences no overall change in energy

Explanation

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.

$70 kg * \frac{1000g}{1kg}*\frac{1 mol}{18 g}*\frac{-6.01 kJ}{1 mol} = -20,000 kJ$

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:

$15 kg * \frac{1000g}{1kg}*\frac{1 mol}{18 g}*\frac{40.7 kJ}{1 mol} = 30,000 kJ$

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).

$\small 100J$ of energy is added to a $\small 1g$ sample, which undergoes a phase change from liquid to gas as a result. The temperature remains constant throughout the process. What does the $\small 100J$ quantity represent?

Specific heat of vaporization

Heat of vaporization

Enthalpy of vaporization

Specific heat capacity

Explanation

The energy required to cause a liquid-gas phase change is called the heat of vaporization, and is usually given in the units of kilojoules per mole. Enthalpy of vaporization and heat of vaporization are the same quantity.

When the heat of vaporization is given for a single gram of substance, with the units of Joules per gram, it is termed the specific heat of vaporization. This is the quantity given in the question: the energy required to vaporize exactly one gram of the sample.

Specific heat capacity refers to the energy needed to raise the temperature of one gram of a sample by one degree Celsius.

The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:

Gibbs Free Energy Formula:

ΔG = ΔH – TΔS

Liquid-Solid Water Phase Change Reaction:

H2O(l) ⇌ H2O(s) + X

The scientist prepares two scenarios.

Scenario 1:

The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.

Scenario 2:

The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.

After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment.

Scenario 3:

Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.

Scenario 4:

The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.

Which of the following are possible quantities for the “X” present on the right side of the Liquid-Solid Water Phase Change Reaction?

I. Entropy

II. Heat

III. Water Vapor

II only

I only

I and II

I and III only

I, II, and III

Explanation

As water freezes, it releases heat into the universe to balance the local loss of entropy encountered by ordering atoms into a rigid solid structure. Entropy, in contrast, is not a commodity that can be released by a reaction in the same way as heat. Also, water vapor would not be released on only one side of the equation. Water vapor would exist with in equilibrium with both solid and liquid, but would not appear as an additional product with the solid phase in this reaction.

The following diagram shows the temperature of a substance as constant heat is supplied. Suppose the substance began in a solid state.

Vaporization_graph2

What is the freezing point of the substance?

Explanation

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.

The enthalpy of fusion for water is $80\frac{cal}{g}$ .

The specific heat capacity for ice is $0.5 \frac{cal}{g*\ ^oC}$ .

The specific heat capacity for water is $1.0\frac{cal}{g*\ ^oC}$ .

How much heat is necessary to raise of water from to ?

Explanation

Solving this problem means solving for three steps.

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

2. Melting the ice (changing phases).

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

Steps 1 and 3 are both solved by the equation $q = mc\Delta T$ .

Step 2 is solved using the enthalpy of fusion, and is multiplied by the number of grams being melted: $q = (2g)(80\frac{cal}{g})$ .

Combining the steps, we get the following expression.

$q=mc_{ice}(T_2-T_1)+m\Delta H_f+mc_{water}(T_3-T_2)$

The molar heats of fusion and vaporization for water are given below:

$\Delta H_{vap} = 40.7 \frac{kJ}{mol}$

$\Delta H_{fus} = 6.01 \frac{kJ}{mol}$

Given the following information, which substance would be expected to have an intermolecular force strength similar to that of water?

Substance	ΔHfus	ΔHvap
Acetone	5.72 kJ/mol	29.1 kJ/mol
Ethanol	4.60 kJ/mol	43.5 kJ/mol
Hydrogen	0.12 kJ/mol	0.90 kJ/mol
Ammonia	5.65 kJ/mol	23.4 kJ/mol

Ethanol

Ethanol and ammonia

Acetone and ammonia

Hydrogen

Explanation

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.

The following diagram shows the temperature of a substance as constant heat is supplied. Suppose the substance began in a solid state.

Vaporization_graph2

The slope of the graph is zero at intervals B and D because __________.

the heat that is supplied is being used to overcome intermolecular forces

no phase changes are taking place at these intervals

no energy is being supplied to the system

the substance is undergoing fusion at these intervals

Explanation

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.

The molar heats of fusion and vaporization for water are given below:

$\Delta H_{vap} = 40.7 \frac{kJ}{mol}$

$\Delta H_{fus} = 6.01 \frac{kJ}{mol}$

A certain amount of snow is formed in the atmosphere, and $5.2 * 10^{6} kJ$ of energy is released. How much water was solidified?

$3.4 * 10^{4}\ lb$

$1.6 * 10^{7}\ lb$

$5.0 * 10^{3}\ lb$

$2.3 * 10^{6}\ lb$

Explanation

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.

$Q=n(-\Delta H_{fus})$