AP Exam Discussion and Thermodynamics Lab Review Notes

AP Exam and Essay Discussion

• Discussion about the DBQ (Document-Based Question) and LAQ (Long Essay Question) on the AP exam.
• Some students found the DBQ prompt hard to understand, while others thought it was easy.
• A student spent too much time on the DBQ and had only twenty minutes left for the LAQ.
• The student wrote about Confucianism in the Song Dynasty due to lack of knowledge about Unit 9 topics like penicillin and the polio vaccine.
• AP exams are graded in July by teachers who fly to a central location like Kansas City and grade essays in a gym.
• The grading process is methodical and based on a system that outlines specific points to hit for different scores.

Thermodynamics Lab Review

• Labs will be graded based on the quality of answers, showing work, and units.
• The lab is worth about 10 points total.
• Students should aim for one good trial with an answer close to 334 or within 10% of that value.
• The biggest source of error in the lab comes from the melting or not melting of the ice.
• The mass of water resulting from the ice melting is the most finicky ingredient.
• A relatively large change in temperature (around 20 degrees) makes small errors less significant.
• Waiting too long or not long enough, or not melting enough ice can sway results by grams.
• The critical component of the experiment is the ice.

Energy Equations and Change of State

• The instructor does not like grading bad labs because it takes longer.
• Students should fix errors and ensure they include all work and units.
• Answers should be rounded to an appropriate number of significant digits (two or three).
• The instructor provided clues about potential errors related to the ice that could cause results to vary.
• A guaranteed test question will focus on the change of state without a change in temperature.
• During a change of state, there is no temperature change; the energy is used for the phase transition.

Heat of Fusion and Vaporization

• Fusion refers to melting or freezing.
• Heat of fusion is positive when melting and negative when freezing because energy is lost when changing to a solid.
• Vaporization is positive 2260 when changing to a gas, and condensation is negative 2260.

Hess's Law and Thermodynamics

• Hess's Law refers to the total energy required to go from point A to point B in a chemical reaction.
• Using second and third columns from thermodynamics thermochemical data sheets to find Gibbs free energy.
• The goal is to understand Gibbs free energy and its relationship to enthalpy and entropy.
• Changes in entropy and Gibbs free energy are calculated similarly to enthalpy by subtracting the reactants from the products.

Calculating Changes in Thermodynamic Properties

• \Delta H = \sum H{products} - \sum H{reactants}
• \Delta S = \sum S{products} - \sum S{reactants}
• \Delta G = \sum G{products} - \sum G{reactants}

Spontaneity and Gibbs Free Energy

• A reaction is spontaneous if it occurs on its own without needing external energy.
• Exothermic reactions tend to be spontaneous because they release energy.
• Endothermic reactions require energy and are less likely to be spontaneous.
• Negative \Delta H (exothermic) favors spontaneity.

Entropy and Disorder

• Entropy relates to chaos and disorder.
• The general trend in the universe is towards chaos.
• An increase in entropy is favorable for spontaneity.
• Entropy is directly related to temperature: solids have less entropy than liquids, and gases have the most.

Temperature and Spontaneity

• A positive change in entropy favors spontaneity.
• A negative change in enthalpy (exothermic) favors spontaneity.
• Negative \Delta G indicates a spontaneous reaction.
• Positive \Delta G indicates a non-spontaneous reaction.

Conditions for Spontaneity

• If \Delta H is negative and \Delta S is positive, the reaction is always spontaneous.
• If \Delta H is positive and \Delta S is negative, the reaction is never spontaneous.
• If \Delta H and \Delta S are both negative or both positive, temperature determines spontaneity.

Influence of Entropy and Enthalpy

• The magnitude of \Delta H and \Delta S influences the reaction.
• If \Delta S is large enough, a slightly endothermic reaction can still be spontaneous.
• Units must be consistent when using the equation \Delta G = \Delta H - T\Delta S (Joules vs Kilojoules).

Temperature's Role in Spontaneity

• Non-spontaneous reactions require energy input to occur and increasing temperature can provide this energy.
• Increasing temperature will increase the entropy.
• Warmer temperatures help in making a reaction spontaneous.
• The body maintains a constant temperature to facilitate spontaneous reactions needed for survival.

Practice Problems

• Calculations can be done using tabulated values or the equation \Delta G = \Delta H - T\Delta S.
• Practice both methods to ensure understanding and accuracy.
• Pure elements have Gibbs free energies of zero.

Understanding spontaneity through an example of hydrogenation of ethane gas

• The reaction has a negative \Delta S of -0.1207 so it does not favor spontaneity and is exothermic so it favors spontaneous reaction.

Gibbs free energy equation

• \Delta G = \Delta H - T\Delta S

Microwave Radiation Problem

• Microwave radiation is absorbed by water and converted into heat.
• Wavelength of the radiation is 12.5 centimeters.
• Goal: Calculate the number of photons required to increase the temperature of 1 \times 10^2 milliliters of water from 20 degrees Celsius to 200 degrees Celsius.

Equations and Constants

• c = \lambda \nu (c = speed of light, \lambda = wavelength, \nu = frequency)
• E = h\nu (E = energy, h = Planck's constant)
• c = 3.0 \times 10^8 \frac{m}{s}
• h = 6.626 \times 10^{-34} J\cdot s

Combining Equations

• E = \frac{hc}{\lambda}

Calculations

• Convert wavelength from centimeters to meters: 12.5 cm = 0.125 m
• E = \frac{(6.626 \times 10^{-34} J\cdot s)(3.0 \times 10^8 \frac{m}{s})}{0.125 m}
• E = 1.59 \times 10^{-24} J

Relating to Specific Heat

• q = mc\Delta T (q = heat, m = mass, c = specific heat, \Delta T = change in temperature)
• Volume to mass conversion: 100 mL of water = 100 g
• \Delta T = 200 - 20 = 180 ^\circ C
• c = 4.18 \frac{J}{g \cdot ^\circ C}
• q = (100 g)(180 ^\circ C)(4.18 \frac{J}{g \cdot ^\circ C}) = 75240 J

Number of Photons

• Number of photons = \frac{Total Energy}{Energy per Photon}
• Number of photons = \frac{75240 J}{1.59 \times 10^{-24} J/photon}