Thermochemistry Chapter 17

Thermochemistry Thermochemistry is the study of energy changes that occur during chemical reactions and changes in state of matter.

Thermochemistry The energy is stored in bonds between atoms and is called chemical energy. It is a form of potential energy.

Heat Heat (q) is energy transferred from one object to another because of the temperature difference between them. Heat always flows from a warmer object to a cooler object.

Types of Reactions In an endothermic process a reaction absorbs/requires/gains heat. – Melting, photosynthesis, cooking In an exothermic process a reaction releases/loses heat. – Freezing, cellular respiration, fire

Measuring Heat Heat flow is measured in two common units, the calorie (c) and the joule (J). A calorie is the amount of heat needed to raise the temperature of one gram of water 1°C. 1 J = cal J = 1 cal

Measuring Heat The specific heat of a substance is the amount of heat it takes to raise the temperature of 1 g of a substance 1°C Specific heat (C) is calculated with the following equation: C = q m x ΔT Where q is heat, m is mass, and ΔT is change in temperature

Practice specific heat problem The temperature of a 98.4 g piece of copper increases from 25°C to 48°C when copper absorbs 849J of heat. What is the specific heat of copper?

Practice specific heat problem The temperature of a 98.4 g piece of copper increases from 25°C to 48°C when copper absorbs 849J of heat. What is the specific heat of copper? C = q m x ΔT

Practice specific heat problem The temperature of a 98.4 g piece of copper increase from 25°C to 48°C when copper absorbs 849J of heat. What is the specific heat of copper? C = 849J (98.4g) x (23°C)

Practice specific heat problem The temperature of a 98.4 g piece of copper increase from 25°C to 48°C when copper absorbs 849J of heat. What is the specific heat of copper? C = 849J (98.4g) x (23°C) C =

Practice specific heat problem The temperature of a 98.4 g piece of copper increase from 25°C to 48°C when copper absorbs 849J of heat. What is the specific heat of copper? C = 849J (98.4g) x (23°C) C = = 0.38 J/g  ° C

One more practice problem When 435 J of heat is added to 3.4 g of olive oil at 21°C, the temperature increases to 85°C. What is the specific heat of the olive oil? C = q m x ΔT

One more practice problem When 435 J of heat is added to 3.4 g of olive oil at 21°C, the temperature increases to 85°C. What is the specific heat of the olive oil? C = g x 64°C

One more practice problem When 435 J of heat is added to 3.4 g of olive oil at 21°C, the temperature increases to 85°C. What is the specific heat of the olive oil? C = g x 64°C = = 2.0 J/g  ° C

Enthalpy The heat constant of a system is called enthalpy (H), measured in Joules (J) The heat absorbed or released by a system or reaction is called the change in enthalpy or ΔH You can calculate enthalpy with this equation: ΔH=-m x C x ΔT m= mass, C = specific heat, ΔT = change in temp

Practice Enthalpy Problem 50.0g of water is heated from 25°C to 32°C. The specific heat of water is 4.18 J/g°C. Calculate the enthalpy change in this reaction.

Practice Enthalpy Problem 50.0g of water is heated from 25°C to 32°C. The specific heat of water is 4.18 J/g°C. Calculate the enthalpy change in this reaction. ΔH=-m x C x ΔT

Practice Enthalpy Problem 50.0g of water is heated from 25°C to 32°C. The specific heat of water is 4.18 J/g°C. Calculate the enthalpy change in this reaction. ΔH=-(50.0g) x (4.18) x (32-25)

Practice Enthalpy Problem 50.0g of water is heated from 25°C to 32°C. The specific heat of water is 4.18 J/g°C. Calculate the enthalpy change in this reaction. ΔH=-(50.0g) x (4.18) x (32-25) =-1463 J = J

Another Practice Calculate the change in enthalpy when you increase the temperature of 250g of water from 20 to 46C. The specific heat of water is 4.18 J/g°C.

Another Practice Calculate the change in enthalpy when you increase the temperature of 250g of water from 20 to 46C. The specific heat of water is 4.18 J/g°C. ΔH=-m x C x ΔT

Another Practice Calculate the change in enthalpy when you increase the temperature of 250g of water from 20 to 46C. The specific heat of water is 4.18 J/g°C. ΔH=-(250) x (4.18) x (46-20)

Another Practice Calculate the change in enthalpy when you increase the temperature of 250g of water from 20 to 46C. The specific heat of water is 4.18 J/g°C. ΔH=-(250) x (4.18) x (46-20)=

Another Practice Calculate the change in enthalpy when you increase the temperature of 250g of water from 20 to 46C. The specific heat of water is 4.18 J/g°C. ΔH=-(250) x (4.18) x (46-20)= =-30000J

Math Practice Complete the practice problems for specific heat and enthalpy.

Phase Changes: Freezing and Melting When a substance melts, it absorbs heat, called the heat of fusion.

Phase Changes: Freezing and Melting When a substance freezes, it loses heat, called the heat of solidification. The quantity of heat absorbed by one mole of a melting solid is exactly the same as the quantity of heat released when one mole of the liquid freezes.

Phase changes: Vaporization and Condensation When liquid evaporates, it absorbs heat, called heat of vaporization.

Phase changes: Vaporization and Condensation When gas condenses to liquid, it releases heat, called heat of condensation. The quantity of heat absorbed by a vaporizing liquid is exactly the same as the quantity of heat released when the vapor condenses.

Specific Heat of Metals Lab 1. Read and annotate the lab procedure. 2. Complete the pre-lab. 3. This is a stand-alone lab report! You will either handwrite or type and print it – it does not go in your notebooks! 4. Be ready to conduct the lab next class period.