ap chemistry exam part 2 chapter 6: thermochemistry 1final exam chapter 6

final exam chapter 6 1

AP Chemistry Exam Part 2

Chapter 6: Thermochemistry


6.1 Nature of energyConcept 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



• What is the kinetic energy of a 2.25 kg baseball moving at 113 m/s?

• Try below



• KE=1/2mv2

• (1/2)225*1132

• 1,436,512 joules



• 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

•



• w = -P∆Va. by a gas (through expansion) = ∆V is positive, w is negativeb. to a gas (by compression) = ∆V is negativew is positive


6.2 Enthalpy and CalorimetryConcept 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

http://chemistry.about.com/od/workedchemistryproblems/a/Heat-Capacity-Phase-Change-Example-Problem.htm

http://chemistry.about.com/od/workedchemistryproblems/a/Specific-Heat-Example-Problem.htm



• 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



• 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.


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


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


Chapter 6.3 Hess’s Law Concept 3: Hess’s Law and calculating change in change in enthalpy

ap chemistry exam part 2 chapter 6: thermochemistry 1final exam chapter 6

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