Thermochemistry

Your Job On a separate piece of paper, write what’s in green!

Haystack Fires: What’s Happening? Video http://youtu.be/Mb CydmSm7BU

ENERGY=ability to do work Unlike MATTER, Energy does not have physical traits such as odor, color, or taste. Only detectable due to its effects. Can’t be created, or destroyed, only transformed. Whatever goes in must come out. THERMOCHEMISTRY is the study of heat changes during chemical reactions and physical changes of state.

q = HEAT Heat is a form of energy. It is measured in temperature; Fahrenheit, Celsius, Kelvin. Heat always flows from a warmer object to a cooler object.

Endothermic and Exothermic reactions If the reaction or process requires or absorbs heat from the environment it is ENDOTHERMIC. (feels cool) ∆H Positive (gaining energy, heat) When heat is given off, the reaction is EXOTHERMIC. (feels warm) ∆H Negative (releasing energy), heat Enthalpy: ∆H Heat released or absorbed in a chemical or physical change,

Demonstration!

Phase Changes What happens to the temperature of water when ice is melted into liquid water, which is then boiled into steam?

Heat Capacity Heat capacity is the amount of heat required to change an objects temperature 1 degree C. The heat capacity of an object depends on both its mass and its chemical composition. (the more mass = higher heat capacity) A calorie is the amount of energy required to raise the temperature of 1g of pure water 1 degree C. A Joule is related to the calorie in that 1,000 calories = 4186 Joules or 1J = 0.239 cal.

J Specific Heat Heat (J) Specific Heat of a substance is the amount of heat it takes to raise the temperature of 1g of the substance 1°C Calculating specific heat Heat (J) Specific Heat = Mass(g) X Change in Temp or J ________________ C = m t

Calculating Specific Heat The temperature of a 95.4-g piece of copper increases from 25.0 degrees C to 48 degrees C when copper absorbs 849 J of heat. What is the specific heat? Heat (J) Specific Heat =Mass(g) X Change in Temp .39 J / (g degrees C)

Thermodynamics Lab!

Extra Slides The following slides contain the remaining content from chapter 17, which we won’t get to until later.

The Exothermic equations have the heat with the products ®. Exothermic reaction: Decrease in enthalpy means the total heat of the products is smaller than of the reactants, meaning the energy is lost to the environment during the reaction. The ∆H will be negative, indicating the chemical reaction is exothermic. When the balance equation of the reaction includes the heat change, it is called a thermochemical equation. Endothermic Reactions will have the heat located on the reactant side (L). The Exothermic equations have the heat with the products ®. Endothermic reaction: Increase in enthalpy means the total heat of the products is higher than of the reactants, meaning the energy is absorbed from the environment during the reaction. The ∆H will be negative, indicating the chemical reaction is endothermic.

q = H = m x C x T CALCULATING ENTHALPY Where q = the change in Heat which in turn is equal to the mass of water (m) times the specific heat capacity of water, times the change in temperature (T2-T1).

Calculating the Heat of Rxns 2NaHCO3(s) + 129 kJ-->Na2CO3(s) + H2O(g) + CO2(g) If you are given 2.24 moles of NaHCO3, how many kJ are required to decompose it? 2.24 mol NaHCO3 x 129 kJ = 145 kJ 2 mol NaHCO3

In Thermochemical Equations the state of matter must be addressed. H2O (l) +285.8 kJ ---> H2(g) + 1/2 O2 (g) H2O (g) +241.8 kJ---> H2 (g) + 1/2 O2 (g) Here, the vaporization of liquid water requires an extra 44 kJ. (285.8 - 241.8)

Heat and Changes of State When an object changes phase, it requires energy. The heat absorbed when changing from a solid to a liquid is called the Molar Heat of Fusion. (Hfus) Molar Heat of Solidification is the opposite. (Hsol) Additionally, there are Molar Heat of Vaporization (Hvap) and Molar Heat of Condensation (Hcond), and a Molar Heat of Solution (Hsoln)which deals with solubility.

Calculating the Energy of Phase Change How many grams of ice at 0 degrees C can be melted by 2.25 kJ of heat? (H fus = 6.01 kJ and H sol = -6.01 kJ) 2.25 kJ x 1 mol ice x 18 g ice = 6.74 g ice 6.01 kJ 1 mol ice

Hess’s Law of Heat Summation If two thermochemical equations can be combined to give one final equation, then the final heat change can also be calculated.

The End