# Ch 5 Thermochemistry - Mrs. Sedlock's Science Page

Ch 5 Thermochemistry CH4 + 2 O2 2 H2O + CO2 + heat Energy Energy Capacity heat

to do work or transfer A consequence of matter having mass, position, and motion Kinetic energy motion Potential energy - position Mass Energy E= mc2

Kinetic Energy Energy of motion of atoms and molecules KE = mv2 Potential Energy

Energy of position PE = mgh bonds stored in chemical Electrostatic Potential Energy

Like charges repel : smaller distance between like charges, higher the PE Large PE + +

Opposite charges attract: the smaller the distance between opposite charges, lower + the PE As separation approaches infinity, PE

approaches 0 Electrostatic Energy Potential Electrostatic Potential Energy E el = kQ1Q2

d Electric charge of particles (1.6 x 10-19 Coulombs) Constant distance State Functions Independent of path or number of steps it takes, only concerned with

final initial states. Usually indicated by a variable that is a capital letter State Functions Units of Energy Joule 1 J = kg m2 s2 See pg 162 for proof

of units Calorie Amount of energy required to raise the temp of 1 g of water 1 C 1 cal = 4.184 J (exactly) Nutrition Calorie = 1000 calories = 1 kcal

Transferring Energy Transferring Energy System part we are interested in Surroundings everything else

Transferring Energy Types of Systems Open matter and energy can be exchanged with surroundings (pot boiling water on stove) Closed can exchange energy with surroundings but not matter Isolated neither energy or matter is exchanged with surroundings

Exothermic and endothermic Exothermic energy is released CH4 + 2 O2 2 H2O + CO2 + heat Endothermic energy is absorbed

C2H6(g) + heat C2H2 + 2H2(g) Exothermic and endothermic Exothermic energy is released Endothermic energy is absorbed

Transferring Energy Energy can be transferred from the system to the surroundings (or vice versa) in two ways: Heat and work Work energy used to cause an object to move against a force w = f d Force = push or pull exerted on an object

(Fg = mass x force of gravity 9.8 m/s2) Transferring Energy Energy can be transferred from the system to the surroundings (or vice versa) in two ways: Heat and work Heat

energy that is transferred from a hot object to a cold one First Law of Thermodynamics Energy is conserved any energy lost by the system must be gained by the surroundings and vice versa First Law of Thermodynamics

Energy is conserved total internal energy of a system = sum of all energy quantitavitely: E = KE + PE This

of a system includes energy of subatomic particles hard to measure we calculate the change in the energy of the system EE Thermodynamic quantities Thermodynamic

quantities always have a number, a unit and a sign (positive or negative) E always indicates the final state initial state In chem initial state refers to reactants and final state refers to products

Heat and Work EE =q + w heat work Heat and Work

EE =q + w If heat is added (endothermic) or work is done on a system EE increases If heat is released (exothermic) or work is done by a system on its surroundings EE decreases

q + gained Heat andheat Work- Lost heat w + work done on system

- Work done by system EE + net gain - Net loss of energy by system energy by the system

Gases A(g) and B(g) are confined in a cylinderand-piston arrangement like that in Figure 5.4 and react to form a solid product C(s): A(g) As + B(g) C(s). the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves

downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? Solve Heat is transferred from the system to the surroundings, and work is done on the system by the surroundings, so

q is negative and w is positive: q = 1150 J and w = 480 kJ. ,E = q + w = (1150 J) + (480 J) = 670 P-V work Energy is an extensive property dependent on the amount of matter and dependent on temperature and

pressure P-V work When pressure is constant we can rewrite w = F d to w = -P EV change in volume pressure ( positive or 0)

P-V work w = -P EV if volume increases (expansion) then EV will be + , work is done by the system so w will be if volume decreases (compression) then EV will be - , work is done on the system by the surroundings so w will be + Enthalpy EH = Energy change at constant

pressure EH = EE + PEV EH = (qp + w) + ( -w) = qp (See text) Usually use enthalpy instead of internal energy b/c most chem rxns happen @ constant pressure

Enthalpy Enthalpy Indicate change : the sign of the enthalpy An ice cube melts absorbs energy

from surroundings , so endothermic and + EH 1 gram of butane (C4H10) is combusted to produce CO2 and H2O releases energy so exothermic and - EH In (a) the water that makes up the ice cube is the system. The ice cube absorbs heat

from the surroundings as it melts, so H is positive and the process is endothermic. In (b) the system is the 1 g of butane and the oxygen required to combust it. The combustion of butane in oxygen gives off heat, so H is negative and the process is exothermic. Ways to find Enthalpy

Experimentally Law) using calorimetry or use tabulated values (Hesss Calorimetry Experimentally

determine enthalpy when there is a temperature change measuring heat flow Heat capacity (c) - amount of heat needed to change the temp of an object by 1 Kelvin can use C (see sec 1.4) Molar

heat capacity cm heat per mole Specific heat capacity cs heat per gram Calorimetry Specific heat of a pure substance is a characteristic property of that substance q = mcET

Relating Heat, Temperature Change, and Heat Capacity a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/gK. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat? (

Answer: (a) 4.9 105 J, (b) 11 K decrease = 11 C decrease Calorimetry A calorimeter is used to determine the heat associated with a chemical reaction Calorimetry Reactants

and products are the system, water and calorimeter are surroundings If the calorimeter is sealed, the energy released by the reaction will be absorbed by the water Calorimetry Can use the change in temp of water

(knowing the heat capacity of water is 4.184 J/g C) to determine EH of a rxn, and 1st Law of Thermodynamics q solution mcET = -q

reaction = - (mcET) Calorimetery example A blacksmith drops a 1.50 kg piece of iron heated to 525C into 2.00 kg of water at 15. Given the molar heat capacity of iron is 25.1 J/mol and the molar heat capacity of water is 75.3 J/mol he final temperature of

the water. T= 53.1 Advanced: Measuring H Using a Coffee-Cup Calorimeter When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 C to 27.5

C. Calculate the enthalpy change for the reaction in kj/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/gK. Advanced: Measuring H Using a Coffee-Cup Calorimeter

Analyze Mixing solutions of HCl and NaOH results in an acidbase reaction: HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) We need to calculate the heat produced per mole of HCl, given the temperature increase of the solution, the number of moles of HCl and NaOH involved, and the density and specific heat of the solution. Plan The total heat produced can be calculated using Equation 5.23. The number of moles of HCl consumed in the reaction must be calculated from the volume and molarity of this

substance, and this amount is then used to determine the heat produced per mol HCl. Advanced: Measuring H Using a Coffee-Cup Calorimeter Solve Because the total volume of the solution is 100 mL,

mass = (100 mL)(1.0 g/mL) = 100 g T = 27.5 C 21.0 C = 6.5 C = 6.5 K qrxn = Cs m T = (4.18 J/gK)(100 g)(6.5 K) = 2.7 103 J = 2.7 kJ Calorimetry Constant volume

Combustion rxns Heat released is absorbed by water and calorimeter increasing the temp

Heat change corresponds to Enthalpy and phase changes Remember: phase changes require energy changes

But It how much energy? depends on the substance Phase Changes Requires energy change

When energy is addedmolecules move faster, overcome IMF and spread out When energy is released IMF pull the molecules together Changing States(Phases) of Matter sublimation endothermi

c Solid Melting endothermic Liquid freezing exothermi c

Vaporization endothermi c condensation exothermic deposition exothermi Gas

Examples of Sublimation and Deposition Freezer frost

Snow Moth crystals Solid air freshener Cirrus clouds Iodine to I2vapor Dry ice (CO2) Phase changes and changes in entropy Entropy- the measure of disorder Low entropy highly ordered ( solids)

High entropy highly disordered (gases) Heating and Cooling Curves Show energy and temperature changes Show where phase change occurs Melting Point & Freezing Point

M.P. /F.P. is the temperature at which a solid turns into a liquid or a liquid to a solid At the melting/ freezing point both solid and liquid can exist at the same time

At the boiling/condensatio n point both liquid and gas can exist at the same time Calculating Heat Change when there is a temperature change Change in temperature

Heat in calories q= mcE Mass in grams Specific heat (on chart) Use Q=mcT Temperature DOESNT

change while a substance is going through a phase change The energy thats being added or released is either moving particles apart or together If the temperature doesnt change during a phase change and energy is being added, what is

changing, the kinetic energy or potential energy? A phase change of Water http://mutuslab.cs.uwindsor.ca/sc hurko/animations/waterphases/sta tus_water.htm So To calculate the amount of heat required to go

through the phase change you need to know 1. the Heat of fusion (or vaporization, condensation, or solidification) for that substance 2. the mass of the substance Amount of heat energy Is called.. (calories) required to __ a gram of it at its ____

Melt - melting point Heat of fusion= H fus Boil - boiling point Heat of vaporization Hvap Condense -condensation point

Heat of condensation Hcon Freeze - freezing point Heat of solidificationsol Sublime - Heat of sublimationsub Deposit -

Heat of depositiondep Values are the same for Heat of fusion / solidification Heat of vaporization / condensation Heat of sublimation / deposition Use Heat of x mass

Calculating H for a substance may require both formulas Calculate for each section, and add together Heat of vaporization/condensation Q = mcT (Heat capacity of gas) Heat of fusion/solidification

Q = mcT ( Heat capacity of liquid) Q = mcT (Heat capacity of solid) For Water, . Specific heat of water = 1 cal/goC Specific heat of ice = .5 cal/goC Specific heat of steam = .4cal/goC Heat Heat

Heat Heat of of of of fusion = 80 cal/g Solidification = 80 cal/g Vaporization = 540 cal/g

condensation = 540cal./g What is the amount of energy needed to raise the temperature of 30g of water from 15oC to 5oC? Specific heat of water = 1 cal/goC Specific heat of ice = .5 cal/goC Specific heat of steam = .4cal/goC Heat Heat

Heat Heat of of of of fusion = 80 cal/g Solidification = 80 cal/g Vaporization = 540 cal/g

condensation = 540cal./g What is the amount of energy that must be released to lower the temperature of 20g of water from 115oC to 75oC? Specific heat of water = 1 cal/goC Specific heat of ice = .5 cal/goC Specific heat of steam = .4cal/goC Heat

Heat Heat Heat of of of of fusion = 80 cal/g Solidification = 80 cal/g

Vaporization = 540 cal/g condensation = 540cal./g Enthalpy of Reactions Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H2O2(l) 2 H2O(l) + O2(g) H = 196 kJ

Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure. Answer: 14.4 kJ Hesss Law Calculating Values

Enthalpy from Tabulated EH is a state function so EH = Hf - Hi So

for chem rxn EH = Hproducts H reactants Hesss Law If a reaction is carried out in a series of steps, EH for overall reaction will be the sum of enthalpies for the

individual steps Used for measuring energy changes for difficult reactions Hesss Law The enthalpy of a reaction is the sum of

the enthalpies of the combined reactionsknown as Hesss Law Hesss Law states that, whatever mathematical operations are performed on a chemical reaction, the same mathematical operations are applied also to the heat of reaction : Hesss Law

Note that the physical states of the reactants and products are indicated (s,l,g, etc) H2(g) + O2(g) H2O(g) H = - 241.8 kJH = - 241.8 kJ H2(g) + O2(g) H2O(l) H = - 241.8 kJH = -285.8 kJ Hesss Law How to solve: Write

down target reaction Enthalpy is extensive property so dependent on number of moles If you reverse the reaction reverse EH sign

Check phase (solid, liquid, gas) *Hesss Law* 1. If the coefficients of a chemical reaction are all multiplied by a constant, EH rxn is multiplied by the same constant- once this is done, the reaction is no

longer standard, drop the Recall combustion of propane. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) EHrxn

= -2044 kJ when 1 mol of propane reacts with 5 mol of oxygen If we multiply the reaction by 2, 2C3H8(g) + 10 O2(g) 6CO2(g) + 8H2O(g) EHrxn = -4088 kJ If

we multiply the reaction by , C3H8(g) + 5/2 O2(g) 3/2CO2(g) + 2H2O(g) EHrxn = -4088 kJ 2. If two or more reactions are added together to obtain an overall reaction, the heats of these reactions are also added to give the heat of the overall reaction

Hesss Law Ex: N2(g) + O2(g) 68 kJ 2NO2(g) EH = We could also break down the rxn into 2 steps.

N2(g) + O2(g) 2NO(g) EH2 = 180 kJ 2NO(g) + O2(g) 2NO2(g) + EH3 = -112 kJ kJ EH2 + EH3 = 68

3. Ex: *Hesss Law* EHforward rxn = - EHreverse rxn C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) EHrxn = -2044 kJ

when 1 mol of propane reacts with 5 mol of oxygen In this ex, burning propane produced a large amount of heat. It would be almost impossible to experimentally calculate the reverse reaction. But the Hesss Law states hat if we reverse the reaction, we can reverse the sign of EH So the energy needed to produce the reverse reaction would be + 2044 kJ Sample Exercise Using Hesss Law to Calculate H

The enthalpy of reaction for the combustion of C to CO 2 is 393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is 283.0 kJ/mol C: 1. COUsing these data, calculate the enthalpy for the combustion of C to : 3. Solution

Analyze: We are given two thermochemical equations, and our goal is to combine them in such a way as to obtain the third equation and its enthalpy change. Plan: We will use Hesss law. In doing so, we first note the numbers of moles of substances among the reactants and products in the target equation, (3).We then manipulate equations (1) and (2) to give the same number of moles of these substances, so that when the resulting

equations are added, we obtain the target equation. At the same time, we keep track of the enthalpy changes, which we add. Sample Exercise Using Hesss Law to Calculate H Continued Practice Exercise

Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is 393.5 kJ/mol, and that of diamond is 395.4 kJ/mol: C(graphite) + O2(g) CO2(g) H = 393.5 kJ C(diamond) + O2(g) CO2(g) H = 395.4 kJ Calculate H for the conversion of graphite to diamond: C(graphite) C(diamond)

H = ? Answer: +1.9 kJ Sample Exercise Using Three Equations with Hesss Law to Calculate H Calculate H for the reaction 2 C(s) + H2(g) C2H2(g)

given the following chemical equations and their respective enthalpy changes: Solution Analyze We are given a chemical equation and asked to calculate its H using three chemical equations and their associated enthalpy changes. Plan We will use Hesss law, summing the three equations or their reverses and multiplying each by an appropriate

coefficient so that they add to give the net equation for the reaction of interest. At the same time, we keep track of the H values, reversing their signs if the reactions are reversed and multiplying them by whatever coefficient is employed in the equation. Sample Exercise Using Three Equations with Hesss Law to Calculate H Continued Practice Exercise Calculate H for the reaction

NO(g) + O(g) given the following information Answer: 304.1 kJ NO2(g) Hesss Law Example Ex: Using

the enthalpies of combustion for graphite and diamond, calculate the EHrxn for the conversion of graphite to diamond C graphite(s) Cdiamond(s) Hesss Law example cont.

The combustion reactions are: C graphite(s) + O2(g) CO2(g) EH = -394 kJ C

diamond(s) + O2(g) CO2(g) EH = -396 kJ Hesss Law Example continued If we reverse the second reaction and add them together the compounds that are on both sides of the reaction cancel out

C graphite(s) + + O2(g) CO2(g) EH = -394kJ

CO2(g) Cdiamond(s) + O2(g) EH = -(-396kJ) Cgraphite(s) Cdiamond(s) EHrxn = +2kJ Hesss Law Another Example Given the following data:

4CuO(s) 2Cu2O(s) + O2(g) EH = 288 kJ Cu2O(s) Cu(s) + CuO(s) Calculate 2Cu

(s) EH = 11 kJ the EHrxn for the following reaction + O2(g) 2CuO(s) Hesss Law continued Standard

heat of reaction, EHrxn ( indicates standard conditions of 25C and 1 atm) is the heat produced when the specified number of moles in the balanced equation reacts Ex: C3H8(g) + EH

5O2(g) 3CO2(g) + 4H2O(g) of rxn = -2044 kJ when 1 mol of propane reacts with 5 mol of oxygen Specific EH EHf = heat energy absorbed or released during synthesis of one mole of a compound from its elements at standard conditions

EHsol = heat energy absorbed or released when a substance dissolves in a solvent EHcomb = heat energy released when a substance reacts with oxygen to form CO2 and H2O

Hesss Law If you are given the EHf of compounds, you can use equation below to find EH rxn Enthalpy of Rxn = Hesss Law EH

f substance formed from its elements EH f - forms one mole of compound from its elements with everything in standard states Standard state - @ 1 atm pressure and 298 K Standard enthalpy change enthalpy change associated with reactants and products in standard states

Hesss Law If element exists in more than one form under standard conditions, use most stable form ex: O2 not O C(graphite) not C (diamond) EH f units are kJ/mole

EH f of an element is 0 (no formation rxn)

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