Download Ch. 5: Thermochemistry

First Law of Thermodynamics
(Law of Conservation of Energy)




The combined amount of matter and energy in
the universe is constant.
Potential energy
Kinetic energy
ΔE = q + W
What are the symbols?




Internal Energy: ΔE
Heat transferred (into or out of system): q
Work done (on or by the system):
w
Enthalpy = heat gained or lost by system: ΔH

ΔH < 0: Exothermic process
Mullis
1
q and ΔH: Signs
Sign is determined by the experience of the
system:
Heat out
Heat in
System
(The reaction
occurs in here!)
System
(The reaction
occurs in here!)
ΔH > 0
q>0
Both q & Δ H are positive.
Endothermic Rxn
ΔH < 0
q<0
Both q & Δ H are negative.
Exothermic Rxn
Mullis
2
Relationships: ΔH, ΔE, q and w
ΔE = q + w
ΔE = ΔH - P ΔV
(P ΔV is the amount of work done by expanding gases, but in most reactions, there is a
very small volume change, so ΔE ~ ΔH in many cases.)
q , ΔH
q , ΔH
w
w
ΔE
ΔE
Sign
+
+
+
-
Condition
Heat transferred to system
Heat transferred from system
Work done by surroundings
Work done by system
q > 0 and w > 0
q < 0 and w < 0
Mullis
3
Internal Energy



Internal energy = ΔH – PΔV = ΔH – ΔnRT
If number of moles of gas change in the
reaction, include the # moles term.
The combustion of C3H8(g) to produce gaseous
products has ΔHr = -2044.5 kJ-mol-1 at 298 K.
Find the change in internal energy for this
reaction.
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (g)
Δn = 7-6 = 1
ΔnRT = 1 mol(8.314 J mol-1K-1)(298K) = 2478J
Int. Energy = -2044.5 kJ -2.478 kJ = -2047 kJ
Mullis
4
Calorimetry





Calorimeter = Measures temp change in a process
“Bomb” calorimeter = constant volume
Under constant pressure, heat transferred = enthalpy
change ( q = ΔE )
Heat capacity = amount of heat to raise temp by 1K (or
1º C)
Specific heat = heat capacity for 1 g of a substance
(Symbol for specific heat is usually C)

Amount of heat absorbed by a substance calculated
using mass, specific heat and temperature change:
q = ΔE = mc Δ T
Mullis
5
Heat Measurements Using Calorimeter
50.0 mL of 0.400 M CuSO4 at 23.35 ºC is mixed with 50.0 mL of
0.600 M NaOH at 23.35 ºC in a coffee-cup calorimeter with heat
capacity of 24.0 J/ ºC. After reaction, the temp is 25.23 ºC. The
density of the final solution is 1.02 g/mL. Calculate the amount
of heat evolved. Specific heat of water is 4.184J/g ºC .
q = ΔE = mc Δ T
Mass = (50 mL+50 mL)(1.02g/mL) = 102 g
Δ T = 25.23-23.35 = 1.88 ºC
Heat absorbed by solution = (102g)(4.18J)(1.88 ºC) = 801J
g ºC
Add this to heat absorbed by calorimeter = (24.0J)(1.88 ºC) = 45.1J
g ºC
Total heat liberated by this reaction = 846 J
Mullis
6
Bomb Calorimeter
1.00 g ethanol, C2H5OH, is burned in a bomb calorimeter with heat
capacity of 2.71 kJ/g ºC. The temp of 3000 g H2O rose from 24.28 to
26.22 ºC. Determine the ΔE for the reaction in kJ/g of ethanol and
then in kJ/mol ethanol. Specific heat of water is 4.184J/g ºC .
q = ΔE = mc Δ T
Δ T = 26.22 - 24.28- = 1.94 ºC
Heat to warm H2O = (3000g)(4.18J)(1.94 ºC) = 24.3 x 103J = 24.3 kJ
g ºC
Add this to heat to warm calorimeter = (2.71kJ)(1.94 ºC) = 5.26 kJ
g ºC
Total heat absorbed by calorimeter and water = 29.6 kJ = 29.6 x 103J
Since this is combustion, heat is liberated so change sign for heat of
reaction: - 29.6kJ/g ethanol
-29.6kJ | 46.1 g = -1360 kJ ethanol
g | mol
mol
Mullis
7
Hess’s Law



Law of heat summation
The enthalpy change for a reaction is the same
whether it occurs by one step or by any series of
steps.
A state function such as enthalpy does not depend on
the steps



State functions are analogous to your bank account: Many
combinations of deposits and withdrawals result in what you
watch: The account balance.
ΔHºrxn = ΔHºa + ΔHºb + ΔHºc +…
Application is an accounting/algebra exercise to get
resulting reaction ΔHºrxn by adding reactions of known
ΔH.
Mullis
8
Hess’s Law




Write resulting equation and arrange the
step equations to get products in step
reactions on same side of products in the
final reaction.
If reaction is reversed, change sign of ΔH.
Multiply step equations as needed to get
the resulting equation. Fractions are
allowed.
Recall that coefficients correspond to
moles.
Mullis
9
Hess’s Law Example
C2H5OH + 3O2  2CO2 + 3H2O ΔH = -1367 kJ/mol
C2H4 + 3O2  2 CO2 + 2H2O ΔH = -1411 kJ/mol
Find ΔH for C2H4 + H2O  C2H5OH
2CO2 + 3H2O  C2H5OH + 3O2 ΔH = 1367 kJ/mol
C2H4 + 3O2  2 CO2 + 2H2O ΔH = -1411 kJ/mol
___________________________________________
C2H4 + H2O  C2H5OH
ΔH = -44 kJ/mol
Mullis
10