Transcript Metathesis Problems (and Some Solutions) Identified

Chapter 19: Thermodynamics and
Equilibrium
Chemistry 1062: Principles of Chemistry II
Andy Aspaas, Instructor
First Law of Thermodynamics
• First Law: conservation of energy applied to
thermodynamic systems
• Internal Energy (U): sum of kinetic and potential energies in
a system (energy of motion, and energy contained in
chemical bonds and intermolecular forces)
• U is a state function
– Changing from one state to another will give you the
change in internal energy, and this is independent of
path (ΔU = Uf – Ui is independent of path)
• Changes of U are caused by exchanges of energy between
system and surroundings
– As either heat (due to temp change) or work (due to a
force moving an object a certain distance)
Enthalpy
• Change in enthalpy (ΔH) = heat of reaction at
constant pressure (qp)
• More specifically, H = U + PV
• ΔH for a reaction can be calculated by summing
standard ΔHf values for products and subtracting
summed ΔHf values for reactants
Entropy
• Entropy (S): measure of how dispersed the energy
in a system is among all possible ways a system
can contain energy
– As a coffee cup cools, it heats the surroundings
and disperses its energy. S increases
– If a gas is allowed to enter an empty chamber,
the kinetic energy of the gas molecules
distributes through the whole volume: S
increases.
Entropy changes
• ΔS = Sf – Si
• For H2O(s)  H2O(l), ΔS = 22 J/K
• Second Law of Thermodynamics
– Total entropy for a system and surroundings increases in
a spontaneous process
• Entropy associated with heat flow = q / T
• Total entropy change for the system = entropy created plus
entropy associated with heat flow (ΔS = Screated + q / T)
• Since entropy must be created in a spontaneous process,
if ΔS > q / T the process must be spontaneous
• At equilibrium, no new entropy is being created, so any
entropy change is just due to heat flow
ΔS = q / T in a system at equilibrium
Enthalpy, entropy, and spontaneity
• Since ΔH = qp, ΔS > ΔH / T for any spontaneous
process (at constant temp and pressure)
• ΔH – TΔS < 0 for a spontaneous process
> 0 for a nonspontaneous process
= 0 for a system at equilibrium
3rd Law of Thermodynamics
• 3rd Law: a perfectly crystalline solid at 0 K has
entropy of zero
• When heat is transferred, ΔS = q / T
• So = Standard entropy
– Calculated by changing the temperature slightly
and measuring the heat absorbed
Predicting changes in entropy
• Three occurrences will generally increase entropy
– A single molecule is broken into 2 or more
molecules
– The number of moles of gas increases
– Phase changes: (s)  (l) or (g); or (l)  (g)
• An increase in entropy means the sign of ΔS is (+)
• The opposite of any of the above corresponds with a
decrease in entropy, ΔS = (-)
Predicting the sign of ΔS
• CS2 (l)  CS2 (g)
• 2Hg (l) + O2 (g)  2HgO (s)
• CaCO3 (s)  CaO (s) + CO2 (g)
Calculating ΔSo
ΔSo = nSo(products) − mSo(reactants)
• So CS2 (l) = 151.3 J/K
• So CS2 (g) = 237.9 J/K
Free energy
• Gibbs Free Energy, G = H − TS
• Free energy change, ΔG = ΔH − TΔS
ΔG = (−) for a spontaneous process
ΔG = (+) for a nonspontaneous process
ΔG = 0 for a process at equilibrium
Standard free energy change
• ΔGo = ΔHo − TΔSo
• Calculate ΔGo for the following reaction, and predict
its spontaneity
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
ΔHfo -74.87
0
-393.5
-241.8 (kJ)
So
186.1
205.0 213.7
188.7 (J/K)
Standard free energies of formation
• ΔGfo = ΔG for production of 1 mol of a compound
under standard conditions from elements in their
standard forms
ΔGo =  n ΔGfo (products) −  m ΔGfo (reactants)
More precise predictions of spontaneity
• If ΔGo < -10 kJ, the reaction is spontaneous as
written, and reactants transform nearly entirely to
products when equilibrium is reached
• If ΔGo > +10 kJ, the reaction is nonspontaneous as
written, and the reactants will not give a significant
amount of products at equilibrium
• If ΔGo is between -10 kJ and +10 kJ, an equilibrium
mixture of reactants and products will be obtained,
with significant amounts of each
Relating ΔGo to the equilibrium constant
• ΔGo can be converted to ΔG (for nonstandard
temperatures)
ΔG = ΔGo + RT ln Q
(where Q is the reaction quotient)
• At equilibrium, ΔG = 0, and Q = K
ΔGo = -RT ln K
– (this equation relates standard free energy
change for a reaction with that reaction’s
equilibrium constant)
Calculating K from ΔGo
• 2NH3(g) + CO2(g) = NH2CONH2(aq) + H2O(l)
ΔGo = -13.6 kJ
R = 8.31 J/(K mol)
ΔGo = -RT ln K
• When K > 1, ΔGo < 0
• When K < 1, ΔGo > 0
Change of free energy with temperature
• ΔGTo = ΔHo −TΔSo (assuming ΔHo and ΔSo are constant
with respect to temperature)
ΔHo
ΔSo
ΔGo
Description
-
+
-
Spont. at all T
+
-
+
Nonspont at all T
-
-
+ or -
Spont at low T, nonspont at high T
+
+
+ or -
Nonspont at low T, spont at high T
Example
Ba(OH)2 · 8H2O(s) + 2NH4NO3(s)  Ba(NO3)2(aq) +
2NH3(g) + 10H2O(l)
• Predict sign of ΔS
• ΔHo =
• ΔSo =
• ΔGo at room temperature =
• At what temperature does this reaction switch from
being spontaneous to nonspontaneous?