1

Constant-Volume Calorimetry

No heat enters or leaves!

qsys = qcal+ qrxn

qsys = 0

qrxn = - qcal

qcal = m x s x Dt

qcal = Ccal x Dt

Reaction at Constant V

DH ~ qrxn

DH = qrxn

2

A quantity of 1.435 g of naphthalene (C10H8), a pungent smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently the temperature of the water rose from 20.28ºC to 25.95ºC. If the heat capacity of the bomb plus water was 10.17 kJ/ºC, calculate the heat of combustion of naphthalene on a molar basis; that is find the molar heat of combustion.

Ccal= 10.17 kJ/oC

Dt = tfinal – tinitial = 25.95oC – 20.28oC = 5.67oC

qcal = Ccal Dt

= 10.17 kJ/oC x 5.67oC = 57.66 kJqcal = Ccal Dt

Dhcombqrxn

moles= -57.66 kJ

1.435 g=

128.2 g

molex = -5.151 x 103 kJ/mol

Problem 6.37

A 0.1375 g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity of 3024 J/ºC. The temperature increases by 1.126ºC. Calculate the heat given off by the burning Mg in kJ/mol.

3

4

Constant-Pressure Calorimetry

No heat enters or leaves!

qsys = qcal + qrxn

qsys = 0

qrxn = - (qcal)

qcal = m x s x Dt

qcal = Ccal x Dt

Reaction at Constant PDH = qrxn

5

A lead (Pb) pellet having a mass of 26.47 g at 89.98ºC was placed in a constant pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50ºC to 23.17ºC. What is the specific heat of the lead pellet?

qPb + qwater = 0

qPb = - qwater

qwater = ms∆t

Dt = tfinal – tinitial = 23.17oC – 22.50oC = 0.67oC

qwater = ms∆t = 100.0g x 4.184 J/goC x 0.67oC = 280.3 J

6

The heat lost be the pellet is equal to the heat gained by the water, so

qPb = -280.3 J

Dt = tfinal – tinitial = 23.17oC – 89.98oC = -66.81oC

qPb = ms∆t

-280.3 J= 26.47g x s x -66.81oC

s = 0.158 J/goC

Problem 6.82A 44.0 g sample of an unknown metal at 99.0ºC was placed in a constant pressure calorimeter containing 80.0 g of water at 24.0ºC. The final temperature of the system was found to be 28.4ºC. Calculate the specific heat capacity of the metal. (the specific heat of water is 4.184 J/gºC)

7

8

Heating Curve for Water

1. Ice

2. Ice + water

3. Water

4. Water + steam

ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = 4.184 J/g˚C

http://www.kentchemistry.com/links/Matter/HeatingCurve.htm9

Calculate the energy to heat 17.9g of ice from 0.00˚C to liquid water at 100.˚C.

(ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = 4.184 J/g˚C)

10

11

Chemistry in Action:

Fuel Values of Foods and Other Substances

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J

Substance DHcombustion (kJ/g)

Apple -2

Beef -8

Beer -1.5

Gasoline -34

12

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions.

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.

DH0 (O2) = 0f

DH0 (O3) = 142 kJ/molf

DH0 (C, graphite) = 0f

DH0 (C, diamond) = 1.90 kJ/molf

13

14

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

rxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn nDH0 (products)f= S mDH0 (reactants)fS-

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.)

15

C (graphite) + 1/2O2 (g) CO (g)

CO (g) + 1/2O2 (g) CO2 (g)

C (graphite) + O2 (g) CO2 (g)

16

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

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

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

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

1. Write the enthalpy of formation reaction for CS2

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

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

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ/mol

2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1 kJ/mol x 2rxn

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

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

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

17

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.

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

DH0rxn nDH0 (products)f= S mDH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

18

Chemistry in Action: Bombardier Beetle Defense

C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) DH0 = ?

C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol

H2O2 (aq) H2O (l) + ½O2 (g) DH0 = -94.6 kJ/mol

H2 (g) + ½ O2 (g) H2O (l) DH0 = -286 kJ/mol

DH0 = 177 - 94.6 – 286 = -204 kJ/mol

Exothermic!

Problem 6.62From the following data

C(graphite) + O2(g) CO2(g)

∆Hrxnº=-393.5 kJ/mol

H2(g) + ½O2(g) H2O(l)

∆Hrxnº=-285.8 kJ/mol

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)

∆Hrxnº=-3119.6 kJ/mol

calculate the enthalpy change for the reaction

2C(graphite) + 3H2(g) C2H6(g)19

20

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?

21

The Solution Process for NaCl

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

22

∆Hsoln = U + ∆Hhyd

+

DH0 = 788 + (-784) = 4 kJ/molrxn

The energy required to completely separate one mole of a solid ionic compound into gaseous ions is called lattice energy (U).

The energy change associated with the hydration process is called the heat of hydration, ∆Hhyd

)()()( gClgNasNaClenergy

)()()()( 2 aqClaqNagClgNa OH

)()()( gClgNasNaCl U = 788 kJ/mol

)()()()( 2 aqClaqNagClgNa OH ∆Hhydr = -784 kJ/mol

)()()( aqClaqNasNaCl ∆Hsoln = 4 kJ/mol

Problem 6.133

From the following data, calculate the heat of solution for KI:

23

NaCl NaI KCl KI

Lattice energy (kJ/mol)

788 686 699 632

Heat of Solution(kJ/mol)

4.0 -5.1 17.2 ?