CHEMISTRY 161

Chapter 6

www.chem.hawaii.edu/Bil301/welcome.html

Thermochemistry

Enthalpy of Reaction

heat released or absorbed by the system at a constant pressure

H = HH = Hproductsproducts - H - HreactantsreactantsH = HH = Hproductsproducts - H - Hreactantsreactants

Hproducts > Hreactants : H > 0 ENDOTHERMICHproducts > Hreactants : H > 0 ENDOTHERMIC

Hproducts < Hreactants : H < 0 EXOTHERMICHproducts < Hreactants : H < 0 EXOTHERMIC

Measurement of Heat Changes

H = H = ΔΔQQ ∞ ∞ ΔΔTT H = H = ΔΔQQ ∞ ∞ ΔΔTT

heatheat

SurroundingSurrounding

System temperature

increase

(pressure is constant)

H = H = ΔΔQ Q ∞ ∞ ΔΔTT H = H = ΔΔQ Q ∞ ∞ ΔΔTT

H = H = ΔΔQ = Q = const const × × ΔΔTT H = H = ΔΔQ = Q = const const × × ΔΔTT

HHmm = =ΔΔQQmm = = ccmpmp ΔΔTT HHmm = =ΔΔQQmm = = ccmpmp ΔΔTT

molar enthalpy change

temperaturechange

molar specific heat capacity

HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT

molar specific heat capacity

capability of substances to store heat and energy

ccmpmp] = J mol] = J mol-1-1 K K-1-1ccmpmp] = J mol] = J mol-1-1 K K-1-1

the J necessary to increase the temperature of

1 mol of a compound by 1 K

ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1

HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT

1.prepare two styrofoam cups

2. carry out chemical reactionin a compound with known cmp

ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1

3. measure temperature change

4. determine ΔHm

calorimeter

HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT

The Real Experiment

H = H = ΔΔQ = (Q = (ccmpmp × n) × n) ΔΔTT H = H = ΔΔQ = (Q = (ccmpmp × n) × n) ΔΔTT

/ × n

H = H = ΔΔQ = (Q = (ccmpmp × n + c × n + ccupcup) ) ΔΔTT H = H = ΔΔQ = (Q = (ccmpmp × n + c × n + ccupcup) ) ΔΔTT

100 ml of 0.5 M HCl is mixed with 100 ml 0.5 M NaOH in a constant pressure calorimeter (ccup = 335 J K-1). The initial

temperature of the HCl and NaOH solutions are 22.5C, and the final temperature of the solution is 24.9C. Calculate the molar heat of neutralization assuming the specific heat of

the solution is the same as for water.

ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1

1.Neutralization reactions2.Redox reactions

3. Precipitation reactions

Where does the ‘heat’ go?

Constant Volume Calorimeter

ΔΔQ = (Q = (ccmvmv × n + c × n + cbombbomb) ) ΔΔTT ΔΔQ = (Q = (ccmvmv × n + c × n + cbombbomb) ) ΔΔTT

ccmvmv(H(H22O) ≈ cO) ≈ cmpmp(H(H22O)O)ccmvmv(H(H22O) ≈ cO) ≈ cmpmp(H(H22O)O)

Constant Volume Calorimeter

H = H = ΔΔQ = Q = ccpp ΔΔTT H = H = ΔΔQ = Q = ccpp ΔΔTT

ΔΔE = E = ΔΔQ + Q + ΔΔWWΔΔE = E = ΔΔQ + Q + ΔΔWW

ΔΔQ = Q = ccvv ΔΔTT ΔΔQ = Q = ccvv ΔΔTT ΔΔW = - p W = - p ΔΔVV ΔΔW = - p W = - p ΔΔVV

E = ΔQ + Δw

CHANGING THE INTERNAL ENERGY

C8H18(l) + (25/2) O2(g)

8 CO2(g) + 9H2O(g) E lost as HEAT

and WORK

Einitial

Efinal

E lost as HEAT

EN

ER

GY

CHANGING THE INTERNAL ENERGY

different combinations of q and w

INT

ER

NA

L E

NE

RG

Y,

E INITIAL STATE

FINAL STATE

Q 0w = 0

Q 0w 0

E = ΔQ + Δw

SUMMARY

measurement of heat changes

HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT HHmm = = ΔΔQQmm = = ccmpmp ΔΔTT

ΔΔE = E = ΔΔQ + Q + ΔΔWWΔΔE = E = ΔΔQ + Q + ΔΔWW

ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1ccmpmp(H(H22O) = 75.3 J molO) = 75.3 J mol-1-1 K K-1-1

Homework

Chapter 6, p. 212-216problems