Chapter 6: Thermochemistry

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Chapter 6: Thermochemistry AP Chemistry Exam Part 2 Chapter 6: Thermochemistry final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? Energy is the capacity to do work (or to produce heat) Work is a force acting over a distance Heat is actually a form of energy Kinetic energy: energy due to the motion of an object KE = ½mv2 KE measured in joules final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? What is the kinetic energy of a 2.25 kg baseball moving at 113 m/s? Try below final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? KE=1/2mv2 (1/2)225*1132 1,436,512 joules final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? Energy is the flow of heat Endothermic reaction = reaction gain heat Exothermic reaction = reaction releases heat Energy is E, change in energy equation below ∆E =q +w q= heat q is positive in endothermic reactions q is negative in exothermic reactions b. w = work w is negative if the system does work w is positive if work is done on the system final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? w = -P∆V a. by a gas (through expansion) = ∆V is positive, w is negative b. to a gas (by compression) = ∆V is negative w is positive final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? final exam chapter 6

6.1 Nature of energy Concept 1: What is energy? final exam chapter 6

6.2 Enthalpy and Calorimetry Concept 2: Calculating Enthalpy and Calorimetry Calorimetry - science of measuring heat energy equations = q = mcΔT  where  q = heat energy  m = mass  c = specific heat  ΔT = change in temperature  Specific Heat Capacity = 1calorie - Energy required to raise the temp of 1 gram of a substance by 1OC final exam chapter 6

6.2 Enthalpy and Calorimetry Concept 2: Calculating Enthalpy and Calorimetry What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in calories?  specific heat of water = 4.18 J/g·°C  Try below final exam chapter 6

6.2 Enthalpy and Calorimetry Concept 2: Calculating Enthalpy and Calorimetry q = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]  q = (25 g)x(4.18 J/g·°C)x(100 °C)  q = 10450 J  4.18 J = 1 calorie  x calories = 10450 J x (1 cal/4.18 J)  x calories = 10450/4.18 calories  x calories = 2500 calories  Answer:  10450 J or 2500 calories of heat energy are required to raise the temperature of 25 grams of water from 0 °C to 100 °C. final exam chapter 6

Change in enthalpy (∆H) or change in heat of system Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy Change in enthalpy (∆H) or change in heat of system N2 (g) + 2O2 (g) 2NO2 (g) ∆H = +68 Endothermic 2NO2 (g)  N2 (g) + 2O2 (g) ∆H = - 68kJ Exothermic final exam chapter 6

Calculate ∆H for N2 (g) + 2O2 (g) 2NO2 (g) Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy Calculate ∆H for N2 (g) + 2O2 (g) 2NO2 (g) N2 (g) + O2 (g)  2NO (g) ∆H = 180kJ 2NO2  2NO (g) + O2 (g) ∆H = 112kJ See how you need to reverse the second equation and then change from endothermic to exothermic Add the two together to get final answer ∆H = +68 final exam chapter 6

Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy final exam chapter 6

Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy final exam chapter 6

Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy final exam chapter 6

Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy final exam chapter 6

Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy final exam chapter 6

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