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MCAT Physical › Physical Chemistry
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.
The same scientist in the passage measures the variables of another reaction in the lab. He knows that this reaction is spontaneous under standard conditions, with a standard free energy change of –43 kJ/mol. Using laboratory-calculated variables, he determines that the Gibbs Free Energy has a value of 0 kJ/mol. What can we say about this reaction?
The reaction had reached equilibrium when the scientist made his measurements.
It indicates that the reaction will produce 43 kJ of energy for every mole that is allowed to react spontaneously under current conditions.
43 kJ of energy must be put into the system to have the reaction proceed spontaneously.
The scientist made an error, all spontaneous reactions have a negative Gibbs Free Energy value.
The reaction will always produce 43 kJ of heat when it reacts to completion.
Explanation
A reaction with a zero free energy change is at equilibrium. At standard conditions, not at equilibrium, this reaction would have a Gibbs free energy change of –43 kJ/mol, would be spontaneous, and would be able to produce 43 kJ of useful work for every mole that reacts.
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.
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.
Consider the following overall reaction:
The rate law for this reaction is determined to be:
What can we conclude about the reaction, based on the rate law?
The slow step involves two nitrogen dioxide molecules
The fast step involves two nitrogen dioxide molecules
The overall reaction must be done twice
Carbon monoxide is a solid, so it is not included in the rate law
Explanation
Since the rate law does not match the overall reaction, we can assume that the reaction has multiple steps.
In a multistep reaction, the slowest step will determine the rate law. As a result, we can conclude that the overall reaction has a slow step, in which two nitrogen dioxide molecules react. The coefficient becomes the exponent. Note that this is only the case because we are working with the identified slow step, which can only be determined experimentally.
Carbon monoxide only enters into the equation in the faster step, and is not included in the overall rate law.
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.
The same scientist in the passage measures the variables of another reaction in the lab. He knows that this reaction is spontaneous under standard conditions, with a standard free energy change of –43 kJ/mol. Using laboratory-calculated variables, he determines that the Gibbs Free Energy has a value of 0 kJ/mol. What can we say about this reaction?
The reaction had reached equilibrium when the scientist made his measurements.
It indicates that the reaction will produce 43 kJ of energy for every mole that is allowed to react spontaneously under current conditions.
43 kJ of energy must be put into the system to have the reaction proceed spontaneously.
The scientist made an error, all spontaneous reactions have a negative Gibbs Free Energy value.
The reaction will always produce 43 kJ of heat when it reacts to completion.
Explanation
A reaction with a zero free energy change is at equilibrium. At standard conditions, not at equilibrium, this reaction would have a Gibbs free energy change of –43 kJ/mol, would be spontaneous, and would be able to produce 43 kJ of useful work for every mole that reacts.
Consider the following overall reaction:
The rate law for this reaction is determined to be:
What can we conclude about the reaction, based on the rate law?
The slow step involves two nitrogen dioxide molecules
The fast step involves two nitrogen dioxide molecules
The overall reaction must be done twice
Carbon monoxide is a solid, so it is not included in the rate law
Explanation
Since the rate law does not match the overall reaction, we can assume that the reaction has multiple steps.
In a multistep reaction, the slowest step will determine the rate law. As a result, we can conclude that the overall reaction has a slow step, in which two nitrogen dioxide molecules react. The coefficient becomes the exponent. Note that this is only the case because we are working with the identified slow step, which can only be determined experimentally.
Carbon monoxide only enters into the equation in the faster step, and is not included in the overall rate law.
You are charging your cell phone battery with your cell phone charger. What can you conclude about this process?
The cell phone battery is acting as an electrolytic cell because it is acquiring voltage
The cell phone battery is acting as a galvanic cell because it is acquiring voltage
The cell phone battery is acting as a galvanic cell because it is releasing voltage
The cell phone battery is acting as an electrolytic cell because it is releasing voltage
Explanation
A single battery can act as both a galvanic and an electrolytic cell. When a battery is discharging it is considered to be a galvanic cell because it is undergoing a spontaneous redox reaction and is losing voltage. On the other hand, when a battery is charging, it is acquiring voltage (from the phone charger that is connected to an outlet) and is considered an electrolytic cell.
Recall that electrolytic cells facilitate nonspontaneous reactions and require energy to carry out these unfavorable reactions. Charging a battery is a nonspontaneous process (because the reaction involved is the reverse of the reaction that occurs when the battery is discharging) and requires energy in the form of voltage input.
You are charging your cell phone battery with your cell phone charger. What can you conclude about this process?
The cell phone battery is acting as an electrolytic cell because it is acquiring voltage
The cell phone battery is acting as a galvanic cell because it is acquiring voltage
The cell phone battery is acting as a galvanic cell because it is releasing voltage
The cell phone battery is acting as an electrolytic cell because it is releasing voltage
Explanation
A single battery can act as both a galvanic and an electrolytic cell. When a battery is discharging it is considered to be a galvanic cell because it is undergoing a spontaneous redox reaction and is losing voltage. On the other hand, when a battery is charging, it is acquiring voltage (from the phone charger that is connected to an outlet) and is considered an electrolytic cell.
Recall that electrolytic cells facilitate nonspontaneous reactions and require energy to carry out these unfavorable reactions. Charging a battery is a nonspontaneous process (because the reaction involved is the reverse of the reaction that occurs when the battery is discharging) and requires energy in the form of voltage input.
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.
Boiling point is the temperature a liquid needs to achieve in order to begin its transformation into a gaseous state. Campers and hikers who prepare food during their trips have to account for differences in atmospheric pressure as they ascend in elevation. During the ascent, the decrease in atmospheric pressure changes the temperature at which water boils.
Further complicating the matter is the observation that addition of a solute to a pure liquid also changes the boiling point. Raoult’s Law can be used to understand the changes in boiling point if a non-volatile solute is present, as expressed here.
In this law, 
A scientist is studying solution chemistry to better understand vapor pressure. He finds that, for one solution he creates, the beaker is cool to the touch after the solute is fully dissolved. Which of the following is true of this solution? (Note: The beaker, solute, and solvent are the system, the remainder of the universe is the surroundings)
It forms spontaneously only at high temperatures
It forms spontaneously only at low temperatures
It always forms spontaneously
It never forms spontaneously
It forms spontaneously only if dissolution decreases entropy of the system
Explanation
The act of dissolving a solute in a solvent is a local increase in entropy, converting a single molecule to multiple ions. The absorption of heat from the surroundings (cool beaker) indicates that this is an endothermic dissolution. We can look at the equation for Gibbs free energy to evaluate the possible answers.
In order to be spontaneous, the reaction must have a negative Gibbs free energy. To accomplish this, a reaction may have a negative enthalpy (exothermic) and positive entropy, however we know that our reaction has a positive enthalpy (endothermic) and positive entropy. A reaction will be spontaneous if it has a positive 
