#Phase Changes & Energy: Your Ultimate AP Chem Review 🚀

Hey there, future AP Chem master! Let's break down phase changes and energy, making sure you're totally prepped for anything the exam throws your way. We'll go through heating/cooling curves, phase diagrams, and tackle some practice problems. Let's get started!

#Heating Curves: The Endothermic Journey 📈

Heating curves show how temperature changes as you add energy to a substance. It's all about that endothermic life!

###### Image Courtesy of Schoenherr & Diamantopoulos Chemistry Videos

• X-axis: Time or energy added
• Y-axis: Temperature 🌡️
• As you add energy, you move from solid → liquid → gas.

#Plateaus: The Phase Change Zones 🧊🔥

• Melting: Solid → Liquid. Energy is the heat of fusion (H_f).
• Vaporizing: Liquid → Gas. Energy is the heat of vaporization (H_v).

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#Example Calculation: Ice to Steam 🧊→ 😤

Let's say we want to know how much energy is needed to change 30.0g of ice at -20°C to steam at 140°C. Here's the breakdown:

• Given values:

  • Specific heat of ice (c_ice) = 2.108 J/g°C
  • Specific heat of water (c_water) = 4.18 J/g°C
  • Specific heat of steam (c_steam) = 2.010 J/g°C
  • H_f of H₂O = 334 J/g
  • H_v of H₂O = 2260 J/g

• Steps:

  1. 🧊Heating ice: q = mcΔT = (30.0 g)(2.108 J/g°C)(20°C) = 1264.8 J
  2. Melting ice: q = H_f(m) = (334 J/g)(30.0 g) = 10020 J
  3. 💧Heating water: q = mcΔT = (30.0 g)(4.18 J/g°C)(100°C) = 12540 J
  4. Vaporizing water: q = H_v(m) = (2260 J/g)(30.0 g) = 67800 J
  5. 😤Heating steam: q = mcΔT = (30.0 g)(2.010 J/g°C)(40°C) = 2412 J
  6. Total: 1264.8 + 10020 + 12540 + 67800 + 2412 = 94036.8 J = 94.0 kJ

#Cooling Curves: The Exothermic Return Trip 📉

Cooling curves show exothermic processes, the opposite of heating curves. Energy is released as a substance cools.

###### Image Courtesy of Schoenherr & Diamantopoulos Chemistry Videos

• The curve is the same shape as a heating curve, but the energy is released instead of absorbed.
• Condensation: Gas → Liquid. Heat of condensation = -H_v
• Freezing: Liquid → Solid. Heat of freezing = -H_f

#Phase Diagrams: Mapping States of Matter 🗺️

Phase diagrams show the relationship between pressure and temperature in determining the state of matter.

###### Image Courtesy of Aakash Shah

• X-axis: Temperature (°C)
• Y-axis: Pressure (atm)
• Solid, Liquid, Gas: These are the three main phases, each occupying a specific region of the diagram.

#Key Points on the Diagram

• Triple Point: The unique point where all three phases coexist. 🤯
• Critical Point: The point beyond which a distinct liquid phase no longer exists; you get a supercritical fluid or gas.

###### Image Courtesy of UCSC Physics

#Practice FRQ: Putting It All Together 💪

Let's tackle a real AP-style question that combines multiple concepts!

#Question

Propane (C₃H₈) is a common fuel. Let's break it down:

1. (a) Write a balanced equation for the complete combustion of propane gas, yielding CO₂(g) and H₂O(l).
2. (b) Calculate the volume of air (21% O₂ by volume) at 30°C and 1.00 atm needed to burn 10.0 g of propane.
3. (c) The heat of combustion of propane is -2220.1 kJ/mol. Calculate the heat of formation (ΔH_f°) of propane, given ΔH_f° of H₂O(l) = -285.3 kJ/mol and ΔH_f° of CO₂(g) = -395.3 kJ/mol.
4. (d) If all the heat from burning 30.0 g of propane is transferred to 8.00 kg of water, calculate the increase in water temperature. (c_water = 4.18 J/g·K)

#Solution

1. (a) C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

2. (b)

   • Moles of propane = 10.0 g / 44.1 g/mol = 0.227 mol
   • Moles of O₂ = 0.227 mol C₃H₈ * (5 mol O₂ / 1 mol C₃H₈) = 1.135 mol
   • Volume of O₂ = (1.135 mol * 0.0821 L·atm/mol·K * 303 K) / 1.00 atm = 28.2 L
   • Volume of air = 28.2 L / 0.21 = 134.3 L

3. (c)

   • ΔH_comb = ΣnΔH_f°(products) - ΣnΔH_f°(reactants)
   • -2220.1 kJ/mol = [3(-395.3) + 4(-285.3)] - [ΔH_f°(C₃H₈)]
   • ΔH_f°(C₃H₈) = -104.6 kJ/mol

4. (d)

   • Moles of propane = 30.0 g / 44.1 g/mol = 0.680 mol
   • Heat released = 0.680 mol * 2220.1 kJ/mol = 1509.7 kJ = 1509700 J
   • q = mcΔT
   • 1509700 J = (8000 g)(4.18 J/g·K)ΔT
   • ΔT = 45.1 K

#Final Exam Focus 🎯

• Heating and Cooling Curves: Know how to use q=mcΔT and H_f(m)/H_v(m). Understand what's happening on the slopes vs. plateaus.
• Phase Diagrams: Be able to identify the triple point, critical point, and different phases.
• Calorimetry: Practice problems involving heat transfer and specific heat.
• Thermochemistry: Master Hess's Law and heat of formation calculations.
• Unit Conversions: Always double-check units! J/g, J/mol, kJ/g, etc.

#Last-Minute Tips ⏰

• Time Management: Don't spend too long on one question. Move on and come back if you have time.
• Show Your Work: Even if you don't get the final answer, you can earn partial credit for setting up the problem correctly.
• Stay Calm: You've got this! Take deep breaths and trust your preparation.

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Practice Question

Practice Questions

#Multiple Choice Questions

1. The heat of vaporization of a substance is always (A) less than the heat of fusion (B) equal to the heat of fusion (C) greater than the heat of fusion (D) zero

2. At the triple point of a substance, which of the following is true? (A) Only the solid phase is present (B) Only the liquid phase is present (C) Only the gas phase is present (D) All three phases are present in equilibrium

3. If 100 g of water at 20°C is heated to 80°C, how much heat is absorbed by the water? (Specific heat of water = 4.18 J/g°C) (A) 4180 J (B) 25080 J (C) 250800 J (D) 41800 J

#Free Response Question

Consider the following heating curve for a 100g sample of a substance:

[Imagine a heating curve with three distinct slopes and two plateaus]

1. (a) Identify the phase(s) present during segment II.
2. (b) Calculate the amount of heat needed to melt the substance. (Given H_f = 50 J/g)
3. (c) Calculate the specific heat capacity of the substance in the liquid phase, given that the temperature increases from 50°C to 100°C when 1000 J of heat is added to the liquid phase.
4. (d) Explain why the heat of vaporization is greater than the heat of fusion for most substances.

#Solution to FRQ

1. (a) Solid and liquid
2. (b) q = mH_f = (100 g)(50 J/g) = 5000 J
3. (c) q = mcΔT; 1000 J = (100 g)c(50°C); c = 0.2 J/g°C
4. (d) During vaporization, all intermolecular forces must be overcome, whereas only some are broken during melting. This requires more energy, hence a higher heat of vaporization.

You've got this! Go rock that AP Chem exam! 💪