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Thermochemistry. 2H 2 ( g ) + O 2 ( g ) 2H 2 O ( l ) + energy. H 2 O ( g ) H 2 O ( l ) + energy. energy + 2HgO ( s ) 2Hg ( l ) + O 2 ( g ). energy + H 2 O ( s ) H 2 O ( l ).

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slide2

2H2(g) + O2(g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2(g)

energy + H2O (s) H2O (l)

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

slide3

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

DH = H (products) – H (reactants)

DH = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

Hproducts > Hreactants

DH < 0

DH > 0

slide4

H2O (s) H2O (l)

DH = 6.01 kJ

Thermochemical Equations

Is DH negative or positive?

System absorbs heat

Endothermic

DH > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

slide5

DH = -890.4 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O (l)

Thermochemical Equations

Is DH negative or positive?

System gives off heat

Exothermic

DH < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

slide6

2H2O (s) 2H2O (l)

H2O (s) H2O (l)

H2O (l) H2O (s)

DH =-6.01 kJ

DH = 6.01 kJ/mol ΔH = 6.01 kJ

DH = 2 mol x 6.01 kJ/mol= 12.0 kJ

Thermochemical Equations

  • The stoichiometric coefficients always refer to the number of moles of a substance
  • If you reverse a reaction, the sign of DH changes
  • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
slide7

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

x

H2O (l) H2O (g)

H2O (s) H2O (l)

3013 kJ

1 mol P4

x

DH = 6.01 kJ

DH = 44.0 kJ

1 mol P4

123.9 g P4

Thermochemical Equations

  • The physical states of all reactants and products must be specified in thermochemical equations.

P4(s) + 5O2(g) P4O10(s)DHreaction = -3013 kJ

= 6470 kJ

266 g P4

slide8

DH0 (O2) = 0

DH0 (O3) = 142 kJ/mol

DH0 (C, graphite) = 0

DH0 (C, diamond) = 1.90 kJ/mol

f

f

f

f

Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

slide10

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

-

[

+

]

[

+

]

=

DH0

DH0

rxn

rxn

dDH0 (D)

cDH0 (C)

aDH0 (A)

bDH0 (B)

f

f

f

f

-

DH0 (reactants)

S

DH0 (products)

S

=

f

f

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

6.6

slide11

2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l)

-

S

S

=

DH0

DH0

DH0

-

[

]

[

+

]

=

rxn

rxn

rxn

[ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ

=

12DH0 (CO2)

2DH0 (C6H6)

f

f

= - 3267 kJ/mol C6H6

6DH0 (H2O)

-6535 kJ

f

2 mol

DH0 (reactants)

DH0 (products)

f

f

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

6.6

slide12

C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ

rxn

S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ

rxn

CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ

rxn

2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ

C(graphite) + 2S(rhombic) CS2 (l)

C(graphite) + 2S(rhombic) CS2 (l)

rxn

rxn

C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ

+

CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ

rxn

DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ

rxn

Calculate the standard enthalpy of formation of CS2 (l) given that:

1. Write the enthalpy of formation reaction for CS2

2. Add the given rxns so that the result is the desired rxn.

slide13

The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

DHsoln = Hsoln - Hcomponents

Which substance(s) could be used for melting ice?

Which substance(s) could be used for a cold pack?

slide14

The Solution Process for NaCl

DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

energy diagrams
Energy Diagrams

Exothermic

Endothermic

  • Activation energy (Ea) for the forward reaction
  • Activation energy (Ea) for the reverse reaction
  • (c) Delta H
slide16

First Law of Thermodynamics

Energy can be converted from one form to another but energy cannot be created or destroyed.

slide17

The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = ms

Heat (q) absorbed or released:

q = msDt

q = CDt

Dt = tfinal - tinitial

6.5

slide18

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

s of Fe = 0.444 J/g •0C

Dt = tfinal – tinitial = 50C – 940C = -890C

q = msDt

= 869 g x 0.444 J/g •0C x –890C

= -34,000 J

slide19

Constant-Pressure Calorimetry

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcal)

qwater = msDt

qcal = CcalDt

Reaction at Constant P

DH = qrxn

No heat enters or leaves!

slide21

Phase Changes

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure.

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm.

11.8

slide22

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure.

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.

slide23

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Freezing

Melting

slide24

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid.

Sublimation

Deposition

DHsub = DHfus + DHvap

( Hess’s Law)

sample problem
Sample Problem
  • How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?

Step 1: Heat the ice Q=mcΔT

Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ

Step 2: Convert the solid to liquid ΔH fusion

Q = 2.0 mol x 6.01 kJ/mol = 12 kJ

Step 3: Heat the liquid Q=mcΔT

Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ

sample problem1
Sample Problem
  • How much heat is required to change 36 g of H2O from -8 deg C to 120 deg C?

Step 4: Convert the liquid to gas ΔH vaporization

Q = 2.0 mol x 44.01 kJ/mol = 88 kJ

Step 5: Heat the gas Q=mcΔT

Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ

Now, add all the steps together

0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ