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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}