How are they related?

Energy Encountered Daily

What is Energy?  Defined as the ability to do work or create heat.  Many types of energy  Thermal  Light  Gravitational  Kinetic  Potential

Light Energy Review  How is light energy produced?  Electrons release light energy when they fall from a high energy level to a lower energy.  We’re now going to talk about energy released or used in a chemical reaction. Heat energy

Thermochemistry  The study of heat used or released in a chemical reaction.  Let’s investigate heat as it compares to temperature Let’s investigate heat as it compares to temperature

What is Heat? Form of energy because it can move things - E.g: Makes a hot air balloon rise. - Steam engines Measured in JOULES (J)

UNITS for HEAT ENERGY Heat energy is usually measured in either Joules, given by the unit (J), and kilojoules (kJ) or in calories, written shorthand as (cal), and kilocalories (kcal). 1 cal = J NOTE: This conversion correlates to the specific heat of water which is 1 cal/g o C or J/g o C.

Heat Vs Temperature The temperature of an object tells us how HOT it is Measured in degrees Celsius - °C It is NOT the same as heat energy although the two quantities are related. e.g. a beaker of water at 60 °C is hotter than a bath of water at 40 °C BUT the bath contains more joules of heat energy

Heating and Cooling cont… Heat energy always moves from: HOT object COOLER object e.g. Cup of water at 20 °C in a room at 30°C - gains heat energy and heats up – its temperature rises Cup of water at 20 °C in a room at 10°C loses heat energy and cools down – its temperature will fall.

Specific Heat Calculations  q = mCΔT q = heat (J or cal or Cal) cal = 1 Joule 1000 cal = 1 Cal (dietary calorie) m = mass (g) o C = specific heat (J/g o C or cal/g o C) ΔT = change in temperature ( o C or K) = T f - T i

Specific Heat  Specific heat of water = 1 cal /g o C or = J / g o C  Specific heat of most metals = < 1 J / g o C  Do metals heat slowly or quickly compared to water?  Do metals stay warm longer or shorter than water?

Practice Problem  How much energy is required to heat g of water from 2.0 o C to 24.0 o C? q = mCΔT m= g C = J/g o C ΔT= (24.0 – 2.0) o C = 22.0 o C q = (120.0g)(4.184 J/g o C)(22.0 o C) =

Practice Problem  How much heat (in kJ) is given off when 85.0 g of lead cools from o C to 10.0 o C? (Specific heat of lead = J/g o C) q = mCΔT m = 85.0 g C = J/g o C ΔT = (10.0 – 200.0) o C = o C q = (85.0 g)(0.129 J/g o C)( o C) = -

CHEMICAL THERMODYMANICS The first law of thermodynamics: Energy and matter can be neither created nor destroyed; only transformed from one form to another. The energy and matter of the universe is constant. The second law of thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe. The entropy is increasing. The third law of thermodynamics: The entropy of a perfect crystal at 0 K is zero. There is no molecular motion at absolute 0 K.

How Do Chemical Reactions Create Heat energy?  Consider the combustion of gasoline (octane) 2 C 8 H O 2  16 CO H 2 O  Potential Energy: Stored energy  Potential energy is stored in the bonds of the reactant s and the products  When bonds are broken, the energy is available  When product bonds form, some energy is used in these bonds  The excess energy is released as heat

Kinetic Energy  Directly related to temperature

Is Heat Used or Released?  Endothermic reactions used heat from the surroundings  Sweating  Refrigeration  Exothermic heat releases heat to the surroundings  Hot hands  Combustion  Exercise

Endothermic Reactions  Decrease in kinetic energy  decrease in temperature  heat will transfer from the environment to the system resulting in a cooler environment  Absorbs heat from its surrounding.  The system gains heat  Positive value for q   H = q =  0  H products  H reactants

Exothermic Reactions  Increase in kinetic energy  increase in temperature of system  heat released to the environment resulting in a hotter environment  Releases heat to its surroundings  The system loses heat  Negative value for q   H = q =  0  H products  H reactants

Enthalpy  Heat content for systems at constant pressure  Symbol is H  Terms heat and enthalpy are used interchangeably for this course   H = q = m C  T  Heat moves from ________ to ___________.

Law of Conservation of Energy  Energy is not lost or gained in a chemical reaction  In a chemical reaction potential energy is transferred to kinetic energy

Thermochemical Equations  An equation that includes the heat change  Example: write the thermochemical equation for this reaction  CaO(s) + H 2 O(l)  Ca(OH) 2 (s)  H = kJ CaO(s) + H 2 O(l)  Ca(OH) 2 (s) kJ

Stoichiometry and Thermochemistry Tin metal can be extracted from its oxide according to the following reaction: SnO 2 (s) + 4NO 2 (g) + 2H 2 O(l) kJ  Sn(s) + 4HNO 3 (aq) How much energy will be required to extract 59.5 grams of tin?

How to solve 1. Use your stoichiometry 2. Treat heat as a reactant or product SnO 2 (s) + 4NO 2 (g) + 2H 2 O(l) kJ  Sn(s) + 4HNO 3 (aq) 59.5 g Sn 1 mol Sn 192 kJ 1 g Sn 1 mol Sn

If an Object feels hot, it means it is undergoing a change with a  H that is: a. positive b. negative c. whether the object feels hot or not is unrelated to its  H d. I don’t know 

If the object feels hot, it means it is undergoing : a. an exothermic reaction b. an endothermic reaction c. whether it feels hot or not is unrelated to whether it is undergoing an exothermic or an endothermic change

How does ice melt?

Molar Heat of Fusion  Heat absorbed by one mole of a substance during melting  Constant temperature   H fus  H 2 O(s)  H 2 O(l)  H = 6.01 kJ/mol

Molar Heat of Solidification  Heat lost when 1 mole of a liquid solidifies  Temperature is constant   H solid   H fus = -  H solid  H 2 O(l)  H 2 O(s)  H = kJ/mol

Molar Heat of Vaporization  Heat needed to vaporize 1 mole of a liquid   H vap  H 2 O(l)  H 2 O(g)  H vap = 40.7 kJ/mol

Molar Heat of Condensation  Heat released when 1 mole of vapor condenses   H cond  H 2 O(g)  H 2 O(s)  H cond = kJ/mol   H vap = -  H cond

Themochemistry problems Problem #1: How many kJ are required to heat 45.0 g of H2O at 25.0 C and then boil it all away? Comment: We must do two calculations and then sum the answers. 1) The first calculation uses this equation: q = (mass) (Δt) (C p ) q = (45.0 g) (75.0 C) (4.184 J g¯1 C¯1) q = J = kJ Δt = 75.0 °C mass = 45.0 g C p = J g¯ 1 °C¯ 1 03/23/14

Problem #1: How many kJ are required to heat 45.0 g of H2O at 25.0 C and then boil it all away? q = (moles of water) (ΔH vap ) q = (45.0 g / 18.0 g mol¯1) (40.7 kJ/mol) q = kJ kJ kJ = 116 kJ (to three sig figs) 03/23/14

Equal masses of hot water and ice are mixed together. All of the ice melts and the final temperature of the mixture is 0 o C. If the ice was originally at 0 o C, what was the initial temperature of the hot water? Solution: Let us assume we have 18.0 g of ice and 18.0 g of hot water present. The key is to realize that the only thing the ice did is melt, it did not change its temperature. So, let us calculate the amount of heat needed to melt our 18.0 g (or, 1.00 mole) of ice: q = (6.02 kJ/mol) (1.00 mol) = 6.02 kJ The only source of heat is the hot water, which provides 6020 J (I converted the 6.02 kJ to J.) of heat. Let us calculate the temperature change of 18.0 g of hot water as it loses 6020 J of heat: 6020 J = (18.0 g) (x) (4.184 J g¯ 1 °C¯ 1 )x = 79.9 °C The hot water was at an initial temperature of 79.9 °C (since everything ended up at a final temperature of 0 °C. 03/23/14

Thermochemistry Problems How many kJ need to be removed from a g sample of water, initially at 25.0 o C, in order to freeze it at 0 o C? 1) q = (mass) (Δt) (C p ) q = (120.0 g) (25.0 C) (4.184 J g¯ 1 °C¯ 1 ) q = 12,552 J = kJ 2) q = (moles of water) (ΔH vap ) q = (120.0 g / 18.0 g mol¯ 1 ) (6.02 kJ/mol) q = kJ 3) Summing up the values from the two steps gives 52.8 kJ.

Phase Change Diagram for Water

Phase Change Diagram

The House that Heats Itself 

Calorimetry  Method used to determine the heat involved in a physical or chemical change.  Relies on the law of conservation of energy

Calorimeter

Simple Calorimeter

Calorimetry Math  Heat gained by the water = q  Heat lost by the system = -q mC  T = q  T = T f –T i, m = mass, C = specific heat q gained by water = q lost by system  q water = - q system  mC  T = -mC  T (mass H 2 O)(spec. heat H 2 O)(  T H 2 O) = - (mass sys)(spec. heat sys)(  T sys)

Standard Heat of Reaction  Heat change for the equation as it is written  H =  H f (products) -  H f (reactants) Standard Heats of Formation (  H f )  Change in enthalpy when 1 mole of the compound is formed from its elements in their standard states at 25 o C and kPa

Hess’s Law  A way to calculate the heat of a reaction that may be too slow or too fast to collect data from.  Add together several reactions that will result in the desired reaction. Add the ΔH for these reactions in the same way.   H total =  H products -  H reactants