MCAT Physical : Physical Chemistry
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Example Questions
Example Question #22 : Thermochemistry And Energetics
Which of the following will result in heat transfer?
Enthalpy increases
A temperature gradient exists
Entropy increases
Internal energy is stable
A temperature gradient exists
Heat is a form of energy resulting from molecular velocity. Temperature is a scale to measure the amount of internal energy stored as heat, known as enthalpy. Entropy is the relative disorder of a system.
In order for a heat transfer to occur, the system must not be in equilibrium. In other words, there must be a difference in heat energy between two regions of the system, which will allow heat energy to travel from the region of high energy to the region of low energy. This condition is easily identified by a temperature gradient, in which heat will from the region of high temperature to the region of low temperature until the system is at an equilibrium temperature.
Example Question #1 : Laws Of Thermodynamics
Which of the following is an implication of the zeroth law of thermodynamics?
Entropy of the universe is always increasing
Energy in the universe cannot be created or destroyed
If systems A and B are in thermal equilibrium with system C, then systems A, B, and C are in thermal equilibrium with each other
A pure crystal kept at 
If systems A and B are in thermal equilibrium with system C, then systems A, B, and C are in thermal equilibrium with each other
This question asks about the zeroth law of thermodynamics which states if two systems are in thermal equilibrium with another system, then all three systems will be in thermal equilibrium with each other.
The other answer choices in this question talk about the other three laws of thermodynamics. First law of thermodynamics states that energy of the universe is constant and that energy cannot be created or destroyed. The second law states that the entropy of the universe is constantly increasing, even if local entropy decreases. The third law states that a pure crystal at absolute zero 
Example Question #9 : Laws Of Thermodynamics
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 5m/s and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 10m/s, slides into child 1, moving at 2m/s.
Collision 3:
The two children collide while traveling in opposite directions at 10m/s each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of +8m/s.
In collision 1, child 1 stops moving under the opposing force of friction. After she stops moving, the energy she lost to the lake as heat, via friction, is exactly equal to the amount of kinetic energy she lost from the moment she struck the lake surface to when she stopped. What law prevents this lost heat energy from ever moving the child again?
First Law of Thermodynamics
Zeroth Law of Thermodynamics
The principle that entropy of the universe always decreases
Second Law of Thermodynamics
Third Law of Thermodynamics
Second Law of Thermodynamics
The Second Law of Thermodynamics states that entropy of the universe always increases. The loss of heat energy to the lake due to friction leads to an increase in the entropy of the universe, and thus the return of that energy to move the child would lead to a decrease in the entropy of the universe.
Nevertheless, the hypothetical movement of the child with this energy is allowed by the First Law of Thermodynamics.
Example Question #1 : Thermodynamic Systems And Calorimetry
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 this situation described in the passage, which of the following is true?
The system is all of the universe, except the stove used to input energy.
The cup of water is the system, and the stove is the surroundings.
The cup of water is the system, and the snow is the surroundings.
The cup of water, snow, and stove comprise the system; the universe is the surroundings
The cup of water is the system, and the snow, stove, and remainder of the universe are the surroundings.
The cup of water is the system, and the snow, stove, and remainder of the universe are the surroundings.
The Gibbs Free Energy equation makes use of clearly delineated systems and surroundings. In this example, the water is freezing or melting depending on conditions. This is accompanied by thermal exchanges with other players, such as the snow and stove. Thus, the water is the system, and everything else (technically, everything else in the universe) comprises the surroundings.
Example Question #2 : Thermodynamic Systems And Calorimetry
The combustion of propane is given by the following formula.
If the heats of formation for CO, CO2, H2O, and C3H8 are -110.5 kJ/mol, -393.5 kJ/mol, -241.8 kJ/mol, and -103.85 kJ/mol, respectively, what is the heat of reaction for the combustion of propane?
641.95kJ
1477.9kJ
–1477.9kJ
–641.95kJ
–1477.9kJ
Given the fact that combustion reactions are exothermic, you should expect the heat of reaction to be negative (ruling out two answer choices). The heat of reaction is equal to the heat of formation of the products minus the heat of formation of the reactants. Be sure to refer to the balanced equation for the correct number of moles for each compound.
Oxygen is not included, as it is in elemental form and therefore has a heat of formation equal to zero.
Example Question #1 : Thermodynamic Systems And Calorimetry
The combustion of liquid hexane in air at 298K gives gaseous carbon dioxide and liquid water, as shown in this reaction.
Calculate the 
To calculate the 
Now we can plug in the given values and solve for the enthalpy of reaction.
Example Question #4 : Thermodynamic Systems And Calorimetry
How much energy must a stove transfer to completely transform 
To turn the water into steam, the stove must first raise the temperature of the water, and then provide energy to change the phase. This requires two distinct steps.
Find the energy to raise the temperature of the water using the equation 
Now, we need to find the energy needed to convert the water to a gas. We will use the equation 
Finally, we add the energy for the two steps to find the total energy required.
Example Question #5 : Thermodynamic Systems And Calorimetry
In the above reaction, how much heat will be released if 74.0g of sulfur reacts with excess oxygen? Round to the nearest 10kJ.
Basically, this is a unit conversion problem. Starting with grams of sulfur, convert to moles of sulfur, and finally to kJ of heat released. Note that the given enthalpy of reaction, 
Note that a negative enthalpy of reaction means the process is exothermic and releases heat, while a positive enthalpy of reaction would mean the process is endothermic and absorbs heat.
Example Question #6 : Thermodynamic Systems And Calorimetry
A 50g sample of an unknown substance is heated to 100oC in a tub of boiling water. It is then quickly removed and placed into an insulated jar holding 200mL of water, initially at 20oC. The final equilbrium temperature of the system is 30oC. Approximately, what is the specific heat of this unknown substance?
Specific heat of water is 4.187J/goC.
Density of water is 1g/mL.
Since the system is isolated, the amount of heat transferred away from the unknown substance must equal the heat transferred to the water. To calculate these heats, use
For the water, 
For the unknown substance,
Set these values equal to get cx = 2.39 J/goC, approximately 2.4 J/goC.
Example Question #7 : Thermodynamic Systems And Calorimetry
How much energy is needed to change a 50g ice cube at -30oC into 50g of water at 40oC? Use the above quantities as needed, and round the answer to the nearest kJ.
17kJ
28kJ
7kJ
11kJ
22kJ
28kJ
There are three distinct steps in this transformation: 1) heating the ice from -30oC to 0oC, 2) melting the ice at 0oC, and 3) heating the water from 0oC to 40oC.
When there is a temperature change, we use 
1)
2)
3)
Adding these three pieces together gives the total enthalpy change, which is equal to change in energy.
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