#80 Notes
Unit 10: Kinetics, Equilibrium, Energy and Electrochemistry
Ch. Rates, Equilibriums, Energies
Ch.
I. Reaction Rates
NO2(g) + CO(g) → NO(g) + CO2(g)
Rate is defined in terms of the rate of disappearance of one of the reactants, but it can also be
defined by the rate of appearance of one of the products.
Rate = -∆[NO2] = -∆[CO] = ∆[NO] = ∆[CO2]
∆t
∆t
∆t
∆t
- rate can be measured for a specific time
(instantaneous rate = slope)
or over a time interval (average rate)
II. Integrated Rate Laws
Rate = k [reactant 1]x [reactant 2]y
-x,y are usually integers
↑
↑
↑ concentrations (molarity)
rate constant, (k), depends on size, speed, kind of molecule, temperature, etc.
Ex. 1) A + B → C
row 1
row 2
row 3
initial [A]
0.100 M
0.100 M
0.200 M
initial [B]
0.100 M
0.200 M
0.100 M
initial rate
rate = -∆[B]
-5
4.0 x 10
∆t
-5
4.0 x 10
1.6 x 10-4 = 16 x 10-5
Compare two rows where only one concentration is changing (only “A” or only “B”).
comparing row 1 & 2: A remains the same, B is doubling and the rate is the same.
(going down the rows) rate = k [A] [B]
1X
2X
20 equals 1, so [B]0
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comparing row 1 & 3: A is doubling, B is the same and the rate is four times bigger.
(going down the rows) rate = k [A] [B]
4X
2X
22 equals 4, so [A]2
Rate = k[A]2[B]0 or Rate = k[A]2
(2nd order, since powers add up to “2”)
Find rate constant:
(A is 2nd order, B is zero order)
Rate = k[A]2
4.0 x 10-5 = k (0.100 M)2
plugging in values from row #1.
-3
4.0 x 10 = k ← Book does all 3 rows and averages (just do one row).
Ex. 2) (CH3)3CBr + OH- → (CH3)3COH + Brrate = +∆[Br-]
[(CH3)3CBr]0
[OH-]0
∆t c
row 1
0.10 M
0.18 M
1.0 x 10-3
row 2
0.20 M
0.10 M
2.0 x 10-3
row 3
0.10 M
0.10 M
1.0 x 10-3
comparing row 1 & 3: (CH3)3CBr is the same,
(going up the row)
OH- is basically doubling and the rate is the same.
rate = k [(CH3)3CBr] [OH-]
1X
2X
20 equals 1, so [OH-]0
comparing row 2 & 3: (CH3)3CBr is doubling,
(going up the row)
OH- is the same and the rate is doubling.
rate = k [(CH3)3CBr] [OH-]
2X
2X
21 equals 2, so [(CH3)3CBr]1
rate = k [(CH3)3CBr]1 [OH-]0 = k [(CH3)3CBr]1 (1st order overall)
Find the rate constant:
rate = k [(CH3)3CBr]1
(1.0 X10-3) = k (0.10 M) using row #1
1.0 X10-2 = k
What would the rate be, if [(CH3)3CBr] = 0.26 M & [OH-] = 0.45 M?
rate = k [(CH3)3CBr]1
rate = (1.0 X10-2) (0.26 M)
rate = 2.6 X10-3
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#81 Notes
III. Reaction Mechanisms
-are the series of steps or reactions necessary to achieve an overall process.
*The slowest step always determines the rate law.
2 NO2 → NO3 + NO
NO3 + CO → NO2 + CO2
NO2 + CO → NO + CO2
Rate = k [NO2]2
slow
fast
*NO3 is an intermediate . (It’s first a product,
then a reactant, which cancel.)
rate law from the slowest step!
Ex.1a) A→ products
(slow)
rate = k [A]1
b) 2A + B → X
(slow)
rate = k [A]2 [B]1
X+B→ Y
(fast)
Y + Z → products
(fast)
*Find the slow reaction or if only one reaction, assume it is slow! Use the slow reaction.
IV. Chemical Kinetics/ Collision Model
NaNO3 + KCl → NaCl + KNO3 Double Displacement
Ionic compounds in reactions do not need to rearrange their electrons, since the ions just
change partners. These reactions occur quickly.
Covalent compounds in reactions must change their electron clouds. The energy to do this
comes from collisions! *The rate can be increased by increasing the temperature, since this
increases the velocity of the particles (KE = ½ mv2). The number and power of collisions
increase.
But the rate does not go up by as much as it should.
Arrhenius proposed the existence of a threshold or activation energy that must be overcome in
order to produce a reaction.
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Ea = activation energy needed for reaction to occur.
∆E = overall energy change for the reaction, above it is (-) so exothermic.
The kinetic energy of the moving molecules is changed into potential energy as the bonds are
broken and formed.
**Only collisions with enough energy (activation energy) will be able to form the activated
complex and then the products.
The reaction rate is still smaller, than the rate of collisions with enough energy to form the
activated complex, because of molecular orientations.
The OH- must hit from the opposite side to eject the Br-.
*Catalysts: lower the Ea, allowing reactions to go faster.
*Inhibitors: slow the reaction.
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#82 Notes
Ch. Chemical Equilibrium
I. Equilibrium Equations
N2(g) + 3H2(g) ↔ 2NH3(g)
at equilibrium:
Forward Rate = Reverse Rate
kf[N2]1[H2]3 = kr[NH3]2
kf = [NH3]2 s
kr [N2]1[H2]3
K=
[NH3]2 s
[N2]1[H2]3 ← Molarity
← products
← reactants
*Exception: Pure solids and pure liquids are not included (“aq” and “g” are included).
{H2O(l) not included, H2O(g)included}
Kp is for gases only!
Kp = PNH32 d
PN2 PH23
Kp = K(RT)∆n
**P in atmospheres
R = 0.08206 L ∙ atm
mol ∙ K
∆n = mols products – mols reactants
Ex. 1) Write the K equation and is K = Kp?
a) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
K = [MgCl2]1 [H2]1
[HCl]2
**(Solids and liquids do not count
in either equation.)
Kp = K (RT)∆n=2 mols product – 2 mols reactant
Kp = K (RT)0
Kp = K
b) Fe2O3(s) + 3 CO(g) ↔ 2 Fe(s) + 3 CO2(g)
K = [CO2]3
[CO]3
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Kp = K (RT) 3 mols product – 3 mols reactant
Kp = K (RT)0 so Kp = K
c) 4 Fe(s) + 3 O2(g) ↔ 2 Fe2O3(s)
Kp = K (RT)0 mols product – 3mols reactant
Kp = K (RT)-3 so Kp ≠ K
K= 1
[O2]3
Ex.2) For N2(g) + 3 H2(g) ↔ 2 NH3(g) : What is the concentration of NH3 at equilibrium, if at
equilibrium there is 0.0402 mol N2 and 0.1207 mol H2 in 2.00 L? K = 0.105
K = [NH3]2__
[N2] [H2]3
MN2 = 0.0402 mol = 0.0201 M
2.00 L
0.105 = [NH3]2______________
(0.0201M)1 (0.06035 M)3
4.64 X10-7 = [NH3]2
6.81 X10-4 M = [NH3]
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MH2 = 0.1207 mol = 0.06035 M
2.00 L
#83 Notes II. Le Chatelier’s Principle
If a system is in equilibrium and a condition is changed, then the system will shift toward
restoring the equilibrium.
A) Concentration
N2 + 3 H2 ↔ 2 NH3
Ex. 1) increase N2:
increases collisions, so it speeds the reaction going to the right
(products)
shifts right (too much N2, so it will shift to the other side (products))
Ex. 2) increase NH3: shifts left (too much NH3, shifts to the other side (reactants))
Ex. 3) decrease H2: shifts left (too little H2, so it must shift toward H2 to increase it
(reactants))
B) Pressure (only affects gases)
Ex. 1) increase P, by decreasing volume (if decreasing volume, the reaction will shift toward
the side that is the most compact (smallest volume))
N2 + 3 H2 ↔ 2 NH3
4 mol ↔ 2 mol
x 22.4 L x 22.4 L
88 L
44 L ←smallest volume (= least number of mols)
the right side is more compact
shifts right
(if increase volume, shifts to the left, larger volume side)
Ex. 2) increase P, by adding Ne,
no change (Ne is not in the reaction, other gases will
not change the volume, since gases are “point masses” (ideal))
C) Temperature
Ex. 1) Girls + Boys ↔ Dancing + heat (exothermic, heat produced, ΔH = (-)
energy on product side)
Increase Temperature, Add heat, less dancing, shifts left
Ex. 2) Decrease T, for exothermic reaction (ΔH = (-), energy on product side)
A + B ↔ C + energy
decrease T, too little energy, shifts right
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Ex. 3) Increase T for endothermic reaction (ΔH = (+),energy on reactant side)
energy + A ↔ C
increase T, too much energy, shifts right
Ch. Energies
I. ∆H, ∆G, ∆S
∆H = Enthalpy (heat energy)
(+)∆H = endothermic
(-)∆H = exothermic
∆G = Gibb’s Free Energy
(chemical potential energy)
∆S = Entropy
(energy of disorder)
(+)∆G = nonspontaneous
(-)∆G = spontaneous
(+)∆S = increasing
(-)∆S = decreasing
ΔHreaction = Σ n ΔHproducts(final) – Σ n ΔHreactants(initial)
ΔGreaction = Σ n ΔGproducts(final) – Σ n ΔGreactants(initial)
ΔSreaction = Σ n ΔSproducts(final) – Σ n ΔSreactants(initial)
Ex. 1a) Calculate ΔGreaction for: 2 HF(g) → H2(g) + F2(g) Is it spontaneous?
ΔGreaction = Σ n ΔGproducts(final) – Σ n ΔGreactants(initial)
ΔGreaction = [(1 mol H2) ( 0 kJ/mol) + (1 mol F2) ( 0 kJ/mol) ]
– [( 2 mol HF) ( -273.2 kJ/mol)]
ΔGreaction = 0 + 0 + 546.4 kJ = +546.4 kJ nonspontaneous
Ex. 1b) Calculate ΔSreaction
ΔSreaction = Σ n ΔSproducts(final) – Σ n ΔSreactants(initial)
ΔSreaction = [(1 mol H2) ( 130.684 J/(mol·K)) + (1 mol F2) ( 202.78 J/(mol·K)) ]
– [( 2 mol HF) ( 173.779 J/(mol·K))]
ΔSreaction = 130.684 J/K + 202.78 J/K – 347.558 J/K = -14.094 J/K decreasing
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#84 Notes
II. Entropy (S)
-is the disorder in a system.
In nature systems get more messed up (gain entropy).
Ex. 1) Is entropy increasing or decreasing?
a) melting Al
Al(s) → Al(l)
↑
↑
More ordered
More messy
(lower entropy)
(higher entropy)
I2(s) → I2(g)
↑
↑
More ordered
More messy
(lower entropy)
(higher entropy)
ΔS = (+) increasing
b) sublimating I2
ΔS = (+) increasing
c) 2 CO(g) + O2(g) → 2 CO2(g)
all gases, so look at amounts
3mols
2 mols
Three objects can get more disorganized than 2 objects.
High entropy
Low entropy
ΔS = (-) decreasing
d) Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)
↑
↑ aqueous to aqueous is no change
↑
↑
Solid
Gas
More messy, higher entropy
ΔS = (+) increasing
Ex. 2) Which has more entropy?
a) C6H12O6(s) or H2O(s)
C6H12O6(s) is more complex
b) He at 0 oC or He at 200 oC
200 oC is a higher temperature, more motion
c) He at 1 atm or He at 2 atm
1 atm has lower pressure, less organized
** The higher the temperature and the lower the pressure, the higher the entropy!
STP = 273 K & 1 atm
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III. Relationship between ∆H, ∆G, & ∆S
ΔG = ΔH – TΔS
ΔG in kJ/mol, ΔH in kJ/mol, T in Kelvin,
ΔS in J/(mol·K) **so change to kJ/(mol·K)
Ex.1) Will a reaction be spontaneous at 25 oC, if ΔH = 542 kJ/mol and ΔS = -14 J/(mol·K)?
-14 J │1 kJ
= -0.014 kJ/(mol·K)
3
mol•K│1 X10 J
ΔG = ΔH – TΔS = (542 kJ/mol) – [(298 K) ( -0.014 kJ/(mol·K))]
ΔG = 542 kJ/mol + 4.172 kJ/mol = 546 kJ/mol (+) so nonspontaneous
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#85 Notes
IV. Free Energy and Equilibrium
∆G = ∆Go + RT ln Qp
∆G is at non-equilibrium conditions
∆Go is at equilibrium, calculated by
∆Greaction = ∑ n∆Gproducts - ∑ n∆Greactants
R = 8.31 J/mol·K
Qp is Kp at non-equilibrium conditions
Remember: Kp = K (RT) ∆n
At equilibrium ∆G = 0, so
0 = ∆Go + RT ln Kp
∆Go = -RT ln Kp
Ex. 1) Find ∆G for N2(g) + 3 H2(g) ↔ 2 NH3(g) at 25 oC, if PN2 = 0.050 atm,
PH2 = 0.0010 atm, and PNH3 = 0.012 atm.
∆Goreaction = ∑ n∆Gproducts - ∑ n∆Greactants (old ΔG energy table)
∆Go = [ (2 mol NH3) ( -16.45 kJ/mol) ] –
[ (1 mol N2) (0 kJ/mol) + (3 mol H2) (0 kJ/mol) ]
o
∆G = -32.9 kJ
Qp = PNH32
PN21 PH23
= (0.012 atm)2
(0.050 atm)1 (0.0010 atm)3
= 2.88 X106
∆G = ∆Go + RT ln Qp
∆G = (-32.9 kJ │1 X103 J ) + (8.31 J/mol∙K) (298 K) ln (2.88 X106)
│ 1 kJ
(14.87)
∆G = -32900 J + 36832 J = 3932 = 3.9 X103 J or 3.9 kJ
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#86 Notes
Ch. Electrochemistry
I. Galvanic Cells
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Zn(s) → Zn2+(aq) + 2 eCu2+(aq) + 2 e- → Cu(s)
oxidizing
reducing
Oxidized
Reduced
Anode (-)
Cathode (+)
The salt bridge adds ions to keep the solutions balanced:
(-) ions are added to
(+) ions are added to
balance the arriving
replace the departing
2+
Zn (+) ions.
Cu2+ (+) ions.
** or cells separated by porous disk
Cell potential = electromotive force (emf) = ε cell
From Reduction Potential Table:
Zn2+(aq) + 2 e- → Zn(s) -0.76 v
Cu2+(aq) + 2 e- → Cu(s) 0.34 v
Zn(s) → Zn2+(aq) + 2 eCu2+(aq) + 2 e- → Cu(s)
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
+0.76 v
0.34 v
1.10 v
(+) is spontaneous
** If given 2 reduction potentials, make the reaction spontaneous. (flip one of them)
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Ex.1a) Find the cell potential for:
Al3+ + 3 e- → Al
-1.676 v
2+
Ni + 2 e → Ni
-0.257 v
Reaction must be spontaneous! Which should we flip?
2X
3X
Al → Al3+ + 3eNi2+ + 2e- → Ni
3 Ni2+ + 2 Al→ 2 Al3+ + 3 Ni
+1.676 v **Don’t multiply voltages!
-0.257 v
1.419v (spontaneous)
b) Find cell potential and sketch the galvanic cell.
Ag+ + e- → Ag
0.7991 v
2+
Co + 2 e → Co
-0.277 v
2X
Ag+ + e- → Ag
Co → Co2+ + 2 e2 Ag+ + Co → 2Ag + Co2+
0.7991 v
+0.277 v
1.076 v
Reducing
Oxidizing
Cathode (+)
Anode (-)
The salt bridge adds ions to keep the solutions balanced:
(+) ions are added to
(-) ions are added to
replace the departing
balance the arriving
+
Ag (+) ions.
Co2+ (+) ions.
*End of Notes*
(Assignments #87-88 are Review Assignments. There are no notes for these assignments.)
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