This presentation is the property of its rightful owner.
1 / 30

# Chapter 5: PowerPoint PPT Presentation

Chapter 5:. Thermochemistry. Thermochemistry:. Energy Kinetic & Potential First Law of Thermo internal energy, heat & work endothermic & exothermic processes state functions Enthalpy Enthalpies of Reaction. Calorimetry heat capacity and specific heat constant-pressure calorimetry

## Related searches for Chapter 5:

Chapter 5:

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 5:

Thermochemistry

### Thermochemistry:

• Energy

• Kinetic & Potential

• First Law of Thermo

• internal energy, heat & work

• endothermic & exothermic processes

• state functions

• Enthalpy

• Enthalpies of Reaction

• Calorimetry

• heat capacity and specific heat

• constant-pressure calorimetry

• bomb calorimetry (constant-volume calorimetry)

• Hess’s Law

• Enthalpies of Formation

• for calculation of enthalpies of reaction

• Foods and Fuels

Energy

• work is a form of energy w = F x d

• energy is the capacity to do work or transfer heat

• Kinetic Energy

• energy of motion E = ½ mv2

• potential energy

• energy of position

• applies to electrostatic energy

• applies to chemical energy (energy of bonds)

• energy units

• one joule = energy of a 2 kg mass moving at 1 m/s

• E = ½ mv2 (½)(2 kg) (m/s)2 = kg m2/s2 = 1 J

• 1 cal = 4.184 J 1 kcal = 1 food calorie (Cal)

Systems & Surroundings

• system -- chemicals in the reaction

• surroundings -- container & all outside environment

• closed system can exchange energy (but not matter) with its surroundings

Closed System

energy(as heat or work)

2H2(g) + O2(g)

¯

2H2O(l)+ energy

(system)

no exchg of matterwith surroundings

First Law of Thermo.

• Energy is always conserved

• any energy lost by system, must be gained by surroundings

• Internal Energy -- total energy of system

• combination of all potential and kinetic energy of system

• incl. motions & interactions of of all components

• we measure the changes in energy E = Efinal - Einitial

• + D E = Efinal > Einitial system has gained E from surroundings

• - D E = Efinal < Einitial system has lost E to surroundings

• Relating D E to heat and work

• D E = q + w q is positive if heat goes from surroundings to system w is positive if work is done on system by surroundings

+q +w

- q - w

surroundings

surroundings

system

system

heat

heat

work

work

Endothermic

• system absorbs heat or heat flows into the system

Exothermic

• system gives off heat or heat flows out of the system

State Function

• a property of a system that is determined by specifying its condition or state (T, P, etc.)

• internal energy is a state function, \DE depends only on Efinal & Einitial

Enthalpy

• for most reactions, most of the energy exchanged is in the form of heat, that heat transfer is called enthalpy, H

• enthalpy is a state function

• like internal energy, we can only measure the change in enthalpy, DH

• DH = qp when the process occurs under constant pressureDH = Hfinal - Hinitial = qp

• - DH Þ exothermic process

• +DH Þ endothermic process

system

DH > 0

surroundings

system

surroundings

DH < 0

Enthalpies of Reaction

• D Hrxn = Hprod - Hreact

• enthalpy is an extensive property

• magnitude of D H depends directly on the amount of reactant

• C(s) + 2H2(g)® CH4(g)D H = -74.8 kJ/mol

• 2C(s) + 4H2(g)® 2CH4(g)D H = -149.6 kJ/2mol

• enthalpy change for forward rxn is equal in magnitude but opposite in sign for the reverse rxn

• CH4(g) ® C(s) + 2H2(g)D H = +74.8 kJ/mol

• C(s) + 2H2(g)® CH4(g) D H = - 74.8 kJ/mol

• enthalpy change for a reaction depends on the state of the reactants and products

• C(g) + 2H2(g)® CH4(g)D H = -793.2 kJ/mol

• 2H2(g) + O2(g)® 2H2O(g)D H = -486.6 kJ/mol

• 2H2(g)+ O2(g)® 2H2O(l)D H = -571.7 kJ/mol

DH = Hfinal - Hinitial

H2O(g)

-241.8 kJ

44 kJ

-

+

Enthalpy

H2O(l)

-285.8 kJ

Practice Ex. 5.2:

• Hydrogen peroxide can decompose to water and oxygen . Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.

• 2H2O2(l)® 2H2O(l) + O2(g)D H = -196 kJ

• 5.00 g H2O2(l) x 1 mol = 0.147 mol H2O2(l) 34.0 g H2O2(l)

• 0.147 mol H2O2(l) x -196 kJ H2O2(l) = -14.4 kJ 2 mol

Calorimetry

• experimental determination of D H using heat flow

heat capacity

• measures the energy absorbed using temperature change

• the amount of heat required to raise its temp. by 1 K

• molar heat capacity -- heat capacity of 1 mol of substance

specific heat

• heat energy required to raise some mass of a substance to some different temp.

• specific heat = quantity of heat trans. (g substance) (temp. change)

• = q . m DT

• S.H. = joule g K

• q = (S.H.) (g substance) (D T)

remember: this is change in temp.

Practice Ex. 5.3:

• Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temp. increases by 12.0 °C if the specific heat of the rocks is 0.82 J/gK.

• S.H. x g x DT = joules

• What unit should be in the solution?

• joules -- quantity of heat

• 0.82 J x 50.0 x 103 g x 12.0 K = 4.9 x 105 Jg K

Constant-Pressure Calorimetry

• D H = qp at constant pressure as in coffee cup calorimeter

• heat gained by solution = qsoln

• \qsoln = (S.H.soln)(gsoln)(DT)

• heat gained by solution must that which is given off by reaction

• \qrxn = - qsoln = - (S.H.soln)(gsoln)(DT)

must be opposite in sign

if DT is positive then qrxn is exothermic

Practice Ex. 5.4:

• When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from 22.30°C to 23.11°C. Calculate D H for this reaction, assuming that the combined solution has a mass of 100.0 g and a S.H. = 4.18 J/g °C.

• AgNO3(aq) + HCl(aq) ® AgCl(s) + HNO3(aq)

• qsoln = 4.18 J x 100.0 g soln x 0.81°C = 3.39 x 102 J g °C

• qrxn = - qsoln = - 3.39 x 102 J = - 68,000 J or 0.00500 mol - 68 kJ/mol

insulating cup

rxn

q

soln

Bomb Calorimetry (Constant-Volume)

• bomb calorimeter has a pre-determined heat capacity

• sample is combusted in the calorimeter and D T is used to determine the heat change of the reaction

• qrxn = - Ccalorimeter x D T

heat capacity of calorimeter

because rxn is exothermic

thermometer

insulation

water

rxn

Practice Ex. 5.5:

• A 0.5865 g sample of lactic acid, HC3H5O3, is burned in a calorimeter with C = 4.812 kJ/°C. Temp. increases from 23.10°C to 24.95°C. Calculate heat of combustion per gram and per mole. D T = +1.85°C

• qrxn = - (4.812 kJ/°C) (1.85°C) = - 8.90 kJ per 0.5865 g lactic acid-8.90 kJ = - 15.2 kJ/g 0.5865 g

• - 15.2 kJ x 90 .1 g = - 1370 kJ/mol 1 g 1 mol

Hess’s Law

• rxns in one step or multiple steps are additive because they are state functions

• eg.

• CH4(g) + 2O2(g)® CO2(g) + 2H2O(g)D H = - 802 kJ

• CH4(g) + 2O2(g)® CO2(g) + 2H2O(l)D H = - 890 kJ

2H2O(g)® 2H2O(l)D H = - 88 kJ

Practice Ex. 5.6:

• Calculate D H for the conversion of graphite to diamond:

• Cgraphite® Cdiamond

• Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ

• Cdiamond + O2(g)® CO2(g)D H = -395.4 kJ

• Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ

• CO2(g)® Cdiamond + O2(g)D H = 395.4 kJ

Cgraphite® CdiamondD H = + 1.9 kJ

Enthalpies of Formation

• enthalpies are tabulated for many processes

• vaporization, fusion, formation, etc.

• enthalpy of formation describes the change in heat when a compound is formed from its constituent elements, DHf

• standard enthalpy of formation, DHfo, are values for a rxn that forms 1 mol of the compound from its elements under standard conditions, 298 K, 1 atm

• For elemental forms:

• eg. C(s) graphite, Ag(s) , H2(g) , O2(g) , etc.

• DHfo, for any element is = 0

• used for calculation of enthalpies of reaction, DHrxn

• DHrxn = S DHfo prod - S DHfo react

Practice Ex. 5.9:

• Given this standard enthalpy of reaction, use the standard enthalpies of formation to calculate the standard enthalpy of formation of CuO(s)(

• CuO(a) + H2(g)® Cu(s) + H2O(l)DHo = -130.6 kJ

• DHrxn = S DH f o prod - S DH f o react

• -130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)]

• DHfo CuO = -155.2 kJ/mol