Download I - Holland Public Schools

UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
1
I. Introduction
* chemical kinetics - the study of rates of chemical reactions
* examples:
1
K2CrO4(aq) +
4
Fe(s) +
3
1
Pb(NO3)2(aq) 
O2(g) 
2
2
KNO3(aq) +
1
PbCrO4(s)
yellow
Fe2O3(s)
* collision theory - idea that in order for a reaction to occur, the molecules must come in physical contact
with one another (they must collide)
*When you look at the equations above on paper, there is no way to tell which one is faster
It can only be determined by experiment
* reaction rate is affected by two factors:
- collision effectiveness – an effective collision is one in which product is formed
- collision frequency- a measure of how often collisions occur between particles
II. Factors that affect reaction rates:
A. Nature of Reactants - deals with collision effectiveness
- strong bonds require lots of energy to break, so fewer collisions are effective
* covalent bonds are strong - slow reactions  many synthesis, decomposition, SR
* most reactions with ionic compounds are in aqueous solution, therefore ionic bonds
are already broken – double-replacement
* surface area – more exposed surface area allows for more collisions
 powdered reactant reacts faster than a condensed chunk
B. Concentration of Reactants - deals with collision frequency
* Generally, as concentration increases, rates increase
* Why?
As concentration increases, the particles are closer together, therefore
the collision frequency is higher, causing the reaction rate to increase.
C. Temperature - deals with collision frequency
* Generally, as temperature increases, rates increase
* Why?
As temperature increases, the particles move faster, therefore the
collision frequency is higher, causing the reaction rate to increase.
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
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D. Catalysts - substances which increase reaction rates without being a reactant or product
* catalysts make collisions more effective – catalysts act as “chemical matchmakers”.
They provide alternate pathways for the reaction that require less activation
energy (see III below), therefore making the reaction rate increase.
2 H2O2(l)  2 H2O(l) + O2(g)
* example:
* enzyme – a catalyst used in biological reactions
enzyme
products
recycled
substrate
of
reaction
enzyme
active site
III. Potential Energy Diagram
2
2
1
P
4
3
Energy
Energy
1
4
R
3
P
R
endothermic(b)
Course of the Reaction
exothermic(a)
Course of the Reaction
* reactants and products – where are they on the graph?
S. Label the following on the two graphs above:
1) activation energy - the amount of energy required to “jump start” a reaction
2) activated complex – a high energy complex formed between reactant and product
3) enthalpy of reaction (ΔH ) - total amount of energy gained or lost in a chemical reaction
a. exothermic – a reaction that releases heat  H is negative
b. endothermic – a reaction that absorbs heat  H is positive
* label which graph represents the endothermic and exothermic processes
4) effect of a catalyst on energy curve – a catalyst lowers the activation energy, but doesn’t affect
the total energy of the reactants and products
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
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IV. Reaction Mechanisms
* look at this reaction:
2
C2H2(g) +
5
O2(g) 
4
CO2(g) +
2
H2O(l)
* What does collision theory say?
In order for a reaction to occur, the reactant particles must physically collide with each other
In this case, 2 C2H2’s and 5 O2’s would need to collide in the same place at the same time
VERY UNLIKELY
* OK, so how does this work then?
The chemical reaction is divided into a series of steps, each of which produces an intermediate,
a product that is used as a reactant in a later step. Each step only contains one or two particles
that need to collide
* reaction mechanism – a series of steps which describe how a chemical reaction occurs
Give the overall reaction for the following reaction mechanisms:
T.
1)
A+B  C
A+C D
--------------------
* reaction intermediate - C is an intermediate in this example, so
when we determine the overall reaction, the
2A + B  D
2)
intermediate gets “cancelled out”
A
+ B  AB
A
+ AB  A2B
A2B + C
 A2 + BC
--------------------------------------
any intermediates?? AB and A2B
* note that A and A2 do not cancel; they are
different substances by definition
2A + B + C  A2 + BC
S.
1)
H2O2(l) + I-1(aq)  H2O(l) + OI-1(aq)
H2O2(l) + OI-1(aq)  H2O(l) + O2(g) + I-1(aq)
-------------------------------------------------------------------------------
2 H2O2
2)
 2 H2O + O2
catalyst – used as a reactant
early, then comes out as
a product later (I- is a
catalyst, OI- is an intermediate)
O2 + O2
 O3 + O
S + O2
 SO2
SO2 + O  SO3
O3 + SO3  O2 + SO4
-----------------------------------------------
S + 2 O2  SO4
* note also that the overall reaction should always come out balanced
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
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V. Reaction Mechanisms and Reaction Rates
- remember, generally as concentration of a reactant increases, rate increases
- in a mechanism with a series of steps, they move at different speeds
- rate-determining step - slowest step; determines rate of reaction
- example:
Step 1: A + B  I1
(fast)
Step 2: A + I1  I2
(slow)
Step 3: C + I2  D
(fast)
---------------------------------------
2A + B + C  D
- What happens if I increase concentration of B?
*step 1 gets faster, but because it is a fast step, it has no effect on the overall reaction rate
- What happens if I increase concentration of C?
* no effect
- What happens if I increase concentration of A?
* because A is in a slow step, increasing the concentration of A will make the slow step
faster, thus increasing the rate of the overall reaction
S.
Step 1: NO(g) + H2(g) 
H2NO(g)
(slow)
Step 2: H2NO(g)

H2O(g) + N(g)
(fast)
Step 3: N(g)
+ N(g) 
N2(g)
(fast)
-------------------------------------------------------------------
NO + H2 + N  N2 + H2O
1) What is the overall reaction? see above
2) What is the rate-determining step? Step 1
3) Name any intermediates. H2NO, N
4) What is the effect of:
a) increasing (NO)?
rate will increase (NO is in slow step)
b) increasing (N)?
no effect
c) increasing (H2)?
rate will increase (H2 is in slow step)
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
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VI. Rate Laws
A. Introduction
- The rate of a reaction is a measure of how quickly products are formed
- We learned earlier that the rate is affected by nature of reactants, temperature, concentration
of reactants, and catalysts
- We are going to focus on how concentration affects the reaction rate, so all other factors are
held constant
B. Expressing a Rate Law
For the reaction aA + bB  cC + dD
rate = k [A]x[B]y
* x, y, and k must be determined by experiment, they do not have anything to do with
the coefficients of the balanced equation
* this reaction is x order in A, y order in B, and (x + y) order overall
T. Using the reaction above, find k and rewrite the rate law if x = 1 and y = 2 for [A] = 0.100 M and [B] = 0.100 M
and the rate is 4.02 x 10-3 M/s
* reaction is 3rd order overall, 1st in A and 2nd in B
rate = k(A)(B)2
4.02 x 10-3 M/s = k (0.100M)(0.100M)2
rate = 4.02(A)(B)2
k = 4.02 M-2s-1
What will the rate be if [A] = 0.234M and [B] = 0.117M?
rate = 4.02(0.234)(0.117)2 = 0.0129 M/s
S. Using the reaction above, find k and rewrite the rate law x = 0, y = 1 for [A] = 0.100 M and [B] = 0.100 M and
the rate is 8.12 x 10-5 M/s
* reaction is 1st order overall, 0 in A and 1st in B
rate = k(B)
8.12 x 10-5 M/s = k (0.100M)
rate = 8.12 x 10-4(B)
k = 8.12 x 10-4 s-1
What will the rate be if [A] = 0.300 M and [B] = 0.300 M?
rate = 8.12 x 10-4(0.300) = 2.44 x 10-4 M/s
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
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VII. Determining a Rate Law by Experiment
T.
For the Reaction:
A + 2 B  C, determine the rate law based on the following data:
Trial
1
2
3
4
5
Initial [A]
0.100
0.200
0.400
0.100
0.100
Initial [B]
0.100
0.100
0.100
0.300
0.600
Initial Rate (M/s)
5.50 x 10-6
2.20 x 10-5
8.80 x 10-5
1.65 x 10-5
3.30 x 10-5
* note that a rate law does not necessarily match the coefficients of the balanced equation. To determine the rate
law, you must look at the effect of each reactant (A & B in this case) on the rate when the other is held the same.
general rate law: rate = k(A)x(B)y
* Let’s look at A: I choose trial 1 vs. 2, because A changes, but B stays the same
  A2  
Rate2
   
Rate1
 A1 
x
 0.200
2.20 x10 5

 0.100
5.50 x10 6


x
2x  4
x2
* Let’s look at B: I choose trial 1 vs. 4, because B changes, but A stays the same
 B2  
Rate2
   
Rate1
 B1 
y
 0.300
1.65 x10 5
 0.100  5.50 x10 6


y
3y  3
y 1
nd
* now rewrite the rate law with the appropriate powers (2 order in A, 1st order in B) and solve for k
 I used the concentrations of A & B from trial 1, but any of the 5 trials will work
5.50 x 10-6 M/s = k(0.100M)2(0.100M)
k = 5.5 x 10-3 M-2s-1
rate = 5.5 x 10-3(A)2(B)
3rd
* This reaction is
S.
order overall,
2nd
in A and
1st
in B
For the reaction 4 D + 2 C  D2 + 2 DC, determine the rate law based on the following data:
Trial
1
2
3
4
5
6
Initial [D]
0.100
0.200
0.400
0.100
0.100
0.100
Initial [C]
0.100
0.100
0.100
0.200
0.300
0.400
Initial Rate (M/s)
4.530 x 10-4
1.812 x 10-3
7.248 x 10-3
4.530 x 10-4
4.530 x 10-4
4.530 x 10-4
general rate law: rate = k(D)x(C)y
* for D (1 vs. 2)
 0.200
1.812 x10 3

 0.100
4.530 x10  4


x
* for C (1 vs. 4)
y
x2
4.530 x 10-4 M/s = k(0.100M)2
* This reaction is
 0.200
4.530 x10 4

 0.100
4.530 x10 4


2x  4
k = 0.04530 M-1s-1
2nd
order overall,
2nd
in D and
2y 1
y0
rate = 0.04530(D)2
0
in C
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
7
VIII. Using Reaction Order to Make Predictions
t
* First Order Reactions
0.693
t1
A  Ao (0.5) 2
k
t1 
2
ln
A
  kt
Ao
* half-life - the time required for one half of the reactant to convert to product
T. The decomposition of N2O5 is first order with k = 5.1 x 10-4 s-1 at 45oC
2 N2O5(g)  4 NO2(g) + O2(g)
1. If [N2O5] = 0.25 M, what will the concentration be after 3.2 minutes? note 3.2 min = 192s
t1
2
0.693
0.693


 1359 sec
k
5.1x10 4 s 1
A  Ao (0.5)
t
t1
2
 (0.25)(0.5)
192s
1359s
 0.23M
2. How long will it take for the concentration of N2O5 to decrease from 0.25 M to 0.15 M?


 0.15 
4
ln 
   5.1x10 t
 0.25 
ln
A
  kt
Ao
t
 0.511
 1002 sec
 5.1x10  4


 0.511   5.1x104 t
3. How long will it take to convert 62% of the starting material to product?
*note that if 62% of initial stuff is converted to product, 38% will remain (that is A)
ln


 0.38 
4
ln 
   5.1x10 t
 1 
A
  kt
Ao


 0.968   5.1x104 t
t  1897 sec
4. What is the half-life of this reaction?
see #1
S. The decomposition of peroxide (H2O2) is first order with k = 8.3 x 10-4 s-1 at 273 K
2 H2O2(l)  2 H2O(l) + O2(g)
1. If the initial concentration of peroxide is 0.200 M, what will the concentration be after 5.00 minutes?
t1
2
0.693
0.693


 835 sec
k
8.3 x10 4 s 1
A  (0.200)(0.5)
300s
835s
 0.16M
2. How long will it take for the concentration of peroxide to drop from 0.300 M to 0.100 M?


 0.100 
4
ln 
   8.3x10 t
 0.300 


 1.10   8.3x104 t
t  1324 sec
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
8
3. How much time will it take for there to be exactly 1/4 of the original concentration of peroxide?


 0.25 
4
ln 
   8.3x10 t
 1 


 1.39   8.3x104 t
4. What is the half-life of this reaction?
t  1670 sec
see #1
IX. Relating Rate Laws to Mechanisms
* The rate law expression comes from the concentrations of the reactants in the slow step of a reaction
mechanism
* For example:
If the slow step is: A + B  C the rate law is: Rate = k[A][B]
and the reaction is first order in A, first order in B and 2nd order overall
If the slow step is: A  B the rate law is: Rate = k[A], and the reaction is first order in A
If the slow step is: A + A  A2, the rate law is: Rate = k[A]2, and the reaction is second
order in A
S.
1. The mechanism for the decomposition of peroxide is:
H2O2  2 OH
H2O2 + OH  H2O + HO2
HO2 + OH  H2O + O2
------------------------------------------------
2 H2O2  2 H2O + O2
a. Give the overall reaction
b. Name any intermediates: OH and HO2
c. Given that the rate law is Rate = k [H2O2] which step is the rate-determining step? 1st step
* first order in H2O2 matches with the one H2O2 on the reactant side of step 1
2. Given the mechanism:
NO + Cl2  NOCl2 (slow)
NOCl2 + NO  2 NOCl (fast)
-----------------------------------------------
2 NO + Cl2  2 NOCl
a. Give the overall reaction
b. Name any intermediates: NOCl2
c. Give a rate law which is consistent with this mechanism
rate = k(NO)(Cl2)
* slow step contains one NO and one Cl2 on reactant side, so 1st order in each
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
9
X. Introduction to Thermodynamics
* enthalpy(H) - a measure of the amount of heat absorbed or released by a chemical change
- heat of formation (Hfo) - the energy change associated with the formation of a compound
from its elements at normal conditions (25oC and 1 atm)
example: H2(g) + ½ O2(g)  H2O(l)
Hfo = -286 kJ/mol
S.
1) How much heat is lost when 3.00 moles of water is formed from its elements?
3.00 mol H 2 O 
- 286 kJ
 858kJ
1 mol H 2 O
2) How much heat is lost when 15.0g of water is formed from its elements?
15.0 g H 2 O 
1 mol H 2 O - 286 kJ

 238kJ
18 g H 2 O 1 mol H 2 O
Thermodynamic Information for Selected Compounds
Substance
Al(s)
AlCl3(s)
Al2O3(s)
Br2(l)
C(s)
CH4(g)
C2H2(g)
C2H6(g)
C3H8(g)
C6H6(l)
C2H5OH(l)
CCl4(l)
CO(g)
CO2(g)
Cl2(g)
Cu(s)
Cu(NO3)2(aq)
CuO(s)
Fe(s)
Fe2O3(s)
H2(g)
Hfo (kJ/mol)
0
-704
-1676
0
0
-75.0
+226.6
-84.7
-103.8
+49.04
-277
-139.5
-110
-394
0
0
-350
-155
0
-822.2
0
So (J/K)
28.3
111
51.0
152.0
5.7
186.2
200.8
229.49
259.9
124.5
161.0
214.4
197.9
213.6
222.9
33.1
193.0
43.5
27.2
90.0
130.6
Substance
HBr(g)
HCl(g)
HCl(aq)
HNO3(aq)
H2O(g)
H2O(l)
KBr(s)
KCl(s)
KClO3(s)
MgCO3(s)
MgO(s)
MgCl2(aq)
N2(g)
NH3(g)
NH4ClO4(aq)
NO(g)
NO2(g)
N2O5(g)
NaCl(s)
O2(g)
SO2(g)
Hfo (kJ/mol)
-53.4
-92.3
-167.2
-207
-242
-286
-392
-435.9
-391.4
-1113
-602
-801.2
0
-46.0
-295
+90.4
+33.1
+11.3
-411
0
-297
So (J/K)
199.0
187.0
56.5
53.3
188.7
69.9
96.4
82.7
143.0
66
26.8
-25.1
191.5
192.5
186
210.6
240.5
356.0
72.4
205.0
248.5
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
10
XI. Hess’ Law of Constant Heat Summation - a method used to find energy changes in chemical reactions
* actual definition: If you can express a chemical reaction as a series of formation reactions, the total heat
of the overall reaction is equal to the sum of the heats of formation for the added reactions.
* that is:
CuO(s) + H2(g)
 Cu(s) + H2O(g)
H = ????
CuO(s)
 Cu(s) + ½ O2(g)
H1 = +155 kJ/mol
[switch signs]
½ O2(g) + H2(g)
 H2O(g)
H2 = -242 kJ/mol
--------------------------------------------------------H =
-87 kJ
* there is a shorter way:
H =
 H
f
’s for products -
o
 H
Calculate H for the following reactions:
* Also, label each reaction as
exothermic or endothermic
T.
1)
f
’s for reactants
o
ROUND ALL ANSWERS TO THE NEAREST
“ONES” DIGIT
CuO(s) + H2(g)  Cu(s) + H2O(g)
ΔH = (Cu) + (H2O) – (CuO) – (H2) = (0) + (-242) – (-155) – (0) = -87 kJ
*exothermic
2)
2 CO(g) + O2(g)  2 CO2(g)
ΔH = 2(CO2) – 2(CO) – (O2) = 2 (-394) – 2 (-110) – (0) = -568 kJ
*exothermic
S.
1)
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
ΔH = 2 (-394) + 3 (-286) – (-277) – (0) = -1369 kJ
* exothermic
2)
Cl2(g) + 2 HBr(g)  2 HCl(g) + Br2(l)
ΔH = 2 (-92.3) + (0) – 2 (-53.4) – (0) = -78 kJ
*exothermic
3)
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
ΔH = 2(0) + 3(-394) – (-822) – 3(-110) = -3θ kJ
* exothermic
4)
Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
ΔH = 2(0) + 3 (-242) – (-1676) – (0) = +950 kJ
*endothermic
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
11
XII. Entropy - a measure of the disorder contained within a system; more disorder = higher entropy
* liquids have higher entropies than solids
* gases have higher entropies than liquids
* particles in solution have greater entropies than solids
* two moles of a substance has greater entropy than one mole
* a system at a higher temperature has greater entropy than a system at
a lower temperature
* NOTE:
1)
2)
3)
4)
Round answers to nearest “ones” digit also
So values are in J / K not kJ!!!
So values for pure elements are not 0
So values are positive
Calculate the entropy change (S) for the following reactions:
* label each reaction as favoring less disorder or more disorder
T.
1)
CuO(s) + H2(g)  Cu(s) + H2O(g)
ΔS = (Cu) + (H2O) – (CuO) – (H2) = (33.1) + (188.7) – (43.5) – (130.6) = +48 J/K
*more disorder
2)
2 CO(g) + O2(g)  2 CO2(g)
ΔS = 2(CO2) – 2(CO) – (O2) = 2 (213.6) – 2 (197.9) – (205) = -174 J/K
*less disorder
S.
1)
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
ΔS = 2 (213.6) + 3 (69.9) – (161) – 3(205) = -139 J/K
* less disorder
2)
Cl2(g) + 2 HBr(g)  2 HCl(g) + Br2(l)
ΔS = 2 (187) + (152) – 2 (199) – (222.9) = -95 J/K
*less disorder
3)
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
ΔS = 2(27.2) + 3(213.6) – (90) – 3(197.9) = +12 J/K
* more disorder
4)
Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
ΔS = 2(28.3) + 3 (188.7) – (51) – 3(130.6) = +180 J/K
*more disorder
UNIT XII - CHEMICAL KINETICS & THERMODYNAMICS
12
XIII. Gibbs Free Energy Equation (G); determines whether a reaction is spontaneous (opp. nonspontaneous)
1) a waterfall goes downhill, but never up, spontaneously
2) a lump of sugar spontaneously dissolves in water, but dissolved sugar never
spontaneously forms a sugar cube.
3) Water freezes spontaneously below 0oC and ice melts spontaneously above 0oC.
4) A gas will spontaneously expand from a closed container, but an uncontained
gas will not spontaneously gather into a container.
5) Iron exposed to oxygen will spontaneously rust, but rust will not spontaneously
reform iron and oxygen.
* we know (-)H is favorable and (+)S is favorable, Gibbs relates these two quantities and shows
whether reaction is spontaneous
G = H - TS
* temperature must be in Kelvins – we will assume room temp (298K) for all of these calculations
Calculate the spontenaity (G) for the following reactions:
*label each reaction as spontaneous or nonspontaneous
T.
* note that we will use the ΔH’s from p. 10 and the ΔS’s from p. 11
* note also that ΔS and ΔH must match in units, so I convert ΔS from J  kJ
1)
CuO(s) + H2(g)  Cu(s) + H2O(g)
S  48
J 1kJ
kJ
 3  0.048
K 10 J
K
ΔG = -87 kJ – (298K)(0.048 kJ/K) = -101 kJ
*spontaneous
2)
2 CO(g) + O2(g)  2 CO2(g)
ΔG = -568 kJ – (298K)(-0.174 kJ/K) = -516 kJ
*spontaneous
S.
1)
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
ΔG = -1369 kJ – (298K)(-0.139 kJ/K) = -1328 kJ
*spontaneous
2)
Cl2(g) + 2 HBr(g)  2 HCl(g) + Br2(l)
ΔG = -78 kJ – (298K)(-0.095 kJ/K) = -5θ kJ
*spontaneous
3)
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
ΔG = -3θ kJ – (298K)(0.012 kJ/K) = -34 kJ
*spontaneous
4)
Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g)
ΔG = +950 kJ – (298K)(0.180 kJ/K) = +896 kJ
*nonspontaneous