Phase Changes - MCAT Chemical and Physical Foundations of Biological Systems
Card 1 of 231
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.
Which of the following best describes the Liquid-Solid Water Phase Change Reaction above?
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.
Which of the following best describes the Liquid-Solid Water Phase Change Reaction above?
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This is an example of a phase change physical reaction at the liquid-solid interface. There is no change in the organization of chemical bonds; therefore, it is not an example of a chemical reaction. There is also no change in the bonding partners of the constituent oxygen and hydrogen atoms of water. As a result, there is no change in oxidation numbers. Without a change in oxidation number values, it is not a redox reaction.
This is an example of a phase change physical reaction at the liquid-solid interface. There is no change in the organization of chemical bonds; therefore, it is not an example of a chemical reaction. There is also no change in the bonding partners of the constituent oxygen and hydrogen atoms of water. As a result, there is no change in oxidation numbers. Without a change in oxidation number values, it is not a redox reaction.
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Which best describes the significance of a "critical point" on a phase diagram?
Which best describes the significance of a "critical point" on a phase diagram?
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The critical point is defined by the highest temperature and highest pressure at which a substance exists in distinct liquid and gas phases. Beyond this point, there is no distinction between the liquid and gaseous phases.
The temperature and pressure at which a substance may exist in equilibrium between the solid, liquid, and gas phases is known as the triple-point.
The critical point is defined by the highest temperature and highest pressure at which a substance exists in distinct liquid and gas phases. Beyond this point, there is no distinction between the liquid and gaseous phases.
The temperature and pressure at which a substance may exist in equilibrium between the solid, liquid, and gas phases is known as the triple-point.
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What is the name of Point D?
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What is the name of Point D?
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Point D occurs at the intersection of the three segments, and is called the triple point. All three phases (solid, liquid, and gas) co-exist at the triple point.
Point B corresponds to the critical point. The freezing and boiling points are dependent on pressure, and are depicted as the lines between regions 1 and 2 (freezing) and between regions 2 and 3 (boiling).
Point D occurs at the intersection of the three segments, and is called the triple point. All three phases (solid, liquid, and gas) co-exist at the triple point.
Point B corresponds to the critical point. The freezing and boiling points are dependent on pressure, and are depicted as the lines between regions 1 and 2 (freezing) and between regions 2 and 3 (boiling).
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point D?
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point D?
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Point D is called the triple point. At the triple point, all three phases (solid, liquid, and gas) co-exist. The processes that are occurring are condensation and vaporization, sublimation and deposition, and freezing and melting.
Condensation and vaporization allow liquid and gas to co-exist. Sublimation and deposition allow solid and gas to co-exist. Freezing and melting allow solid and liquid to co-exist. In these terms, it makes sense that all six processes occur at the triple point.
Point D is called the triple point. At the triple point, all three phases (solid, liquid, and gas) co-exist. The processes that are occurring are condensation and vaporization, sublimation and deposition, and freezing and melting.
Condensation and vaporization allow liquid and gas to co-exist. Sublimation and deposition allow solid and gas to co-exist. Freezing and melting allow solid and liquid to co-exist. In these terms, it makes sense that all six processes occur at the triple point.
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
At temperatures and pressures above those at Point B, the material is called a .
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
At temperatures and pressures above those at Point B, the material is called a .
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Point B is known as the critical point. While identifying this point on the phase diagram is important, it is also important to know that at temperatures and pressures above the critical point, the solution is called a supercritical fluid, meaning that separate gas and liquid phases do not exist.
Point B is known as the critical point. While identifying this point on the phase diagram is important, it is also important to know that at temperatures and pressures above the critical point, the solution is called a supercritical fluid, meaning that separate gas and liquid phases do not exist.
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What are the phases in sections 1, 2, and 3, respectively?
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What are the phases in sections 1, 2, and 3, respectively?
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Section 1 is at high pressure and low temperature, meaning the solution is a solid. In section 2, the solution is at intermediate pressure and temperature, meaning it is a liquid. Section 3 is at high temperature and low pressure, meaning it is a gas.
Section 1 is at high pressure and low temperature, meaning the solution is a solid. In section 2, the solution is at intermediate pressure and temperature, meaning it is a liquid. Section 3 is at high temperature and low pressure, meaning it is a gas.
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point A?
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point A?
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First, it is important to identify what phases are occurring on each side of the line on which Point A rests. In section 1, the pressure is high and the temperature low, meaning the solution is a solid. In section two, both the pressure and temperature are intermediate, meaning the solution is a liquid. In other words, the segment that Point A is on is the equilibrium line between solid and liquid, thus, melting and freezing are occurring at Point A.
First, it is important to identify what phases are occurring on each side of the line on which Point A rests. In section 1, the pressure is high and the temperature low, meaning the solution is a solid. In section two, both the pressure and temperature are intermediate, meaning the solution is a liquid. In other words, the segment that Point A is on is the equilibrium line between solid and liquid, thus, melting and freezing are occurring at Point A.
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point C?
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point C?
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First, it is important to identify what phases are occurring on each side of the line on which Point C rests. In section two, both the pressure and temperature are intermediate, meaning the solution is a liquid. In section three, both the pressure and temperature are high, meaning the solution is a gas. In other words, the segment that Point C is on is the equilibrium line between liquid and gas, thus, vaporization and condensation are occurring at Point C.
First, it is important to identify what phases are occurring on each side of the line on which Point C rests. In section two, both the pressure and temperature are intermediate, meaning the solution is a liquid. In section three, both the pressure and temperature are high, meaning the solution is a gas. In other words, the segment that Point C is on is the equilibrium line between liquid and gas, thus, vaporization and condensation are occurring at Point C.
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point E?
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What processes are occurring at Point E?
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First, it is important to identify what phases are occurring on each side of the line on which Point E rests. In section 1, the pressure is high and the temperature low, meaning the solution is a solid. In section three, both the pressure and temperature are high, meaning the solution is a gas. In other words, the segment that Point E is on is the equilibrium line between solid and gas, thus, deposition and sublimation are occurring at Point E.
First, it is important to identify what phases are occurring on each side of the line on which Point E rests. In section 1, the pressure is high and the temperature low, meaning the solution is a solid. In section three, both the pressure and temperature are high, meaning the solution is a gas. In other words, the segment that Point E is on is the equilibrium line between solid and gas, thus, deposition and sublimation are occurring at Point E.
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Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What molecule has a phase diagram similar to the one presented?
Phase diagrams are used to depict changes in the properties of a solution at different temperatures and pressures. Below is a phase diagram of a polar solution.
What molecule has a phase diagram similar to the one presented?
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While this may seem like an obscure question, the MCAT specifically requires you to know the shape of the water phase diagram. Unique to only a few molecules, the solid phase in our diagram is less dense than the liquid phase, meaning that the solid/liquid phase line has a negative slope (this can be seen as segment AD in the above image). Water is one of the select few compounds with this characteristic.
While this may seem like an obscure question, the MCAT specifically requires you to know the shape of the water phase diagram. Unique to only a few molecules, the solid phase in our diagram is less dense than the liquid phase, meaning that the solid/liquid phase line has a negative slope (this can be seen as segment AD in the above image). Water is one of the select few compounds with this characteristic.
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The phase diagram for water (H2O) shows that it can go through all three phases at a pressure of 1atm. The phase diagram for carbon dioxide (CO2) shows that it can only be a solid or a gas at 1atm.
Based on this information, which statement is true?
The phase diagram for water (H2O) shows that it can go through all three phases at a pressure of 1atm. The phase diagram for carbon dioxide (CO2) shows that it can only be a solid or a gas at 1atm.
Based on this information, which statement is true?
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The phase diagram for carbon dioxide (CO2) shows that it will sublimate from a solid to a gas as temperature is increased at one atmosphere of pressure. If a solid immediately goes to a gas, we can conclude that the pressure is too low to allow the substance to first go through the liquid phase. As a result, we can conclude that the point in which CO2 is in all three phases (the triple point) will take place at a higher pressure than 1atm. Because water (H2O) is able to go through all of its three phases at a pressure of 1atm, we know that the triple point pressure is less than 1atm.
The phase diagram for carbon dioxide (CO2) shows that it will sublimate from a solid to a gas as temperature is increased at one atmosphere of pressure. If a solid immediately goes to a gas, we can conclude that the pressure is too low to allow the substance to first go through the liquid phase. As a result, we can conclude that the point in which CO2 is in all three phases (the triple point) will take place at a higher pressure than 1atm. Because water (H2O) is able to go through all of its three phases at a pressure of 1atm, we know that the triple point pressure is less than 1atm.
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A hockey player cannot skate on dry ice due to its .
A hockey player cannot skate on dry ice due to its .
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Because dry ice is frozen carbon dioxide, it does not have the same liquid-solid equilibrium line as water and traditional ice. Its liquid-solid equilibrium line is positive, which means that increased pressure will only cause the dry ice to remain solid. If it were negative, then increased pressure would cause melting as seen with water. Skating on solids is like skating on dirt or rocks. The other choices concerning solid and gas does not apply since no gases are involved.
Because dry ice is frozen carbon dioxide, it does not have the same liquid-solid equilibrium line as water and traditional ice. Its liquid-solid equilibrium line is positive, which means that increased pressure will only cause the dry ice to remain solid. If it were negative, then increased pressure would cause melting as seen with water. Skating on solids is like skating on dirt or rocks. The other choices concerning solid and gas does not apply since no gases are involved.
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A brass disk with an outer diameter of 
has a 
diameter hole cut through its center. If the disk is initially at 
and is then placed into a freezer at 
, what is the final area of the hole?
The linear expansion coefficient for brass is: 
A brass disk with an outer diameter of
The linear expansion coefficient for brass is:
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First, find the original area of the hole, in square meters:
We will also convert the temperatures to Kelvin.
The linear thermal expansion equation is:
Similarly, the thermal expansion of an area is given by the equation:
We are given the value of the constant, we know the original area, and we have the change in temperature. Using these values, we can solve for the change in area.
Find the final area of the hole by adding the change in area to the original area.
First, find the original area of the hole, in square meters:
We will also convert the temperatures to Kelvin.
The linear thermal expansion equation is:
Similarly, the thermal expansion of an area is given by the equation:
We are given the value of the constant, we know the original area, and we have the change in temperature. Using these values, we can solve for the change in area.
Find the final area of the hole by adding the change in area to the original area.
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What conditions best favor the gaseous state?
What conditions best favor the gaseous state?
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Low pressure ensures that the molecules are not confined to a more organized state such as liquid or solid, while high temperature means higher kinetic energy, which means that the molecules have more energy to move away from one another and into the gaseous state. Low temperature would mean molecules have less energy to move away from one another, and high pressure will force molecules to be in a more organized state.
Low pressure ensures that the molecules are not confined to a more organized state such as liquid or solid, while high temperature means higher kinetic energy, which means that the molecules have more energy to move away from one another and into the gaseous state. Low temperature would mean molecules have less energy to move away from one another, and high pressure will force molecules to be in a more organized state.
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Raining is an example of what type of phase change?
Raining is an example of what type of phase change?
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Raining occurs when water vapors become liquid water in the sky. Deposition is the phase change from gas to solid, sublimation is the phase change form solid to gas, and evaporation is the phase change from liquid to gas.
Raining occurs when water vapors become liquid water in the sky. Deposition is the phase change from gas to solid, sublimation is the phase change form solid to gas, and evaporation is the phase change from liquid to gas.
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What is the phase change at the contact point between an ice hockey player's skate and the ice?
What is the phase change at the contact point between an ice hockey player's skate and the ice?
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Melting occurs because the skates apply pressure to the ice. Due to the negative slope of solid-liquid equilibrium line for water, increased pressure results in a change from solid to liquid. None of the other answer choices do not have to do with the equilibrium between solid and liquid.
Melting occurs because the skates apply pressure to the ice. Due to the negative slope of solid-liquid equilibrium line for water, increased pressure results in a change from solid to liquid. None of the other answer choices do not have to do with the equilibrium between solid and liquid.
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Which of the following describes the concept of sweating?
Which of the following describes the concept of sweating?
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Sweating reduces skin temperature because the sweat that is secreted to the skin evaporates, which is an endothermic process. Thus, heat is absorbed from the body and skin to drive the evaporation of the sweat on the skin.
Sweating reduces skin temperature because the sweat that is secreted to the skin evaporates, which is an endothermic process. Thus, heat is absorbed from the body and skin to drive the evaporation of the sweat on the skin.
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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.
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?
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.
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?
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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.
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.
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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.
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
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.
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
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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.
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.
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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 .
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 .
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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.
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.
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