CHEMISTRY 161 Chapter 6 chem.hawaii/Bil301/welcome.html
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Transcript of CHEMISTRY 161 Chapter 6 chem.hawaii/Bil301/welcome.html
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