Transcript Chapter 1: Matter and Measurement
General Chemistry
M. R. Naimi-Jamal
Faculty of Chemistry Iran University of Science & Technology
: مهدراهچ لصف
ییایمیش یاهشنکاو کیتنیس
Contents 14-1 14-2 14-3 14-4 14-5 14-6 14-7 The Rate of a Chemical Reaction Measuring Reaction Rates Effect of Concentration on Reaction Rates: The Rate Law Zero-Order Reactions First-Order Reactions Second-Order Reactions Reaction Kinetics: A Summary
Contents 14-8 14-9 Theoretical Models for Chemical Kinetics The Effect of Temperature on Reaction Rates 14-10 Reaction Mechanisms 14-11 Catalysis
Focus On Combustion and Explosions
همدقم .
اهنآ تعرس لرتنک یاههار و ییایمیش یاهشنکاو تعرس هعلاطم ینعی کیتنیس لص ف رد .
دنوش یم یدنب هقبط نگمهان و نگمه تروص هب ییایمیش یاهشنکاو نگمهان یاهشنکاو و دنریگ یم تروص زاف کی رد اهنت نگمه یاهشنکاو .
اهزاف کرتشم • • •
H
(
aq
) 2
NO
(
g
)
OH O
2 (
g
( )
aq
)
H
2
O
(
l
) 2
NO
2 (
g
) 2
Mg
(
s
)
Zn
(
s
) 2
O
2 (
g
)
H
(
aq
) 2
MgO
(
s
)
Zn
2 (
aq
)
H
2 (
g
)
تعرس هلداعم
.
تسا طبترم نآ داوم تظلغ اب شنکاو تعرس ای هیلوا داوم نتفر نیب زا تعرس ، شنکاو تعرس یضایر ظاحل هب .
تسا نامز دحاو رد لصاح داوم دیلوت تعرس .
دنهد یم شیامن [ ] اب ار رلاوم تظلغ و
R
اب ار شنکاو تعرس : تشون ناوت یم لااب فیرعت قباطم • • • •
A
B R
d
[
A
]
dt
d
[
B
]
dt
14-1 The Rate of a Chemical Reaction • Rate of change of concentration with time.
2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) t = 38.5 s Δt = 38.5 s [Fe 2+ ] = 0.0010 M Δ[Fe 2+ ] = (0.0010 – 0) M Δ[Fe 2+ ] Rate of formation of Fe 2+ = = Δt 0.0010 M 38.5 s = 2.6 x 10 -5 M s -1
Rates of Chemical Reaction 2 Fe 3+ (aq) + Sn 2+ → 2 Fe 2+ (aq) + Sn 4+ (aq) Δ[Sn 4+ ] Δt = 1 2 Δ[Fe 2+ ] = Δt 1 2 Δ[Fe 3+ ] Δt
General Rate of Reaction
a
A +
b
B →
c
C +
d
D Rate of reaction = rate of disappearance of reactants = 1
a
Δ[A] Δt = 1
b
Δ[B] Δt = rate of appearance of products = 1
c
Δ[C] Δt = 1
d
Δ[D] Δt
14-2 Measuring Reaction Rates H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) 2 MnO 4 (aq) + 5 H 2 O 2 (aq) + 6 H + → 2 Mn 2+ + 8 H 2 O(l) + 5 O 2 (g) Experimental set-up for determining the rate of decomposition of H 2 O 2 . Oxygen gas given off by the reaction mixture is trapped, and its volume is measured in the gas buret. The amount of H 2 O 2 remaining concentration of H 2 O 2 consumed and the can be calculated from the measured volume of O 2 (g).
Example:
Determining and Using an Initial Rate of Reaction.
H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) Rate = -Δ[H 2 O 2 ] Δt Initial rate: -(-2.32 M / 1360 s) = 1.7 x 10 -3 M s -1
Example:
What is the concentration at 100s?
[H 2 O 2 ] i = 2.32 M Rate = 1.7 x 10 -3 M s -1 = - Δ[H 2 O 2 ] Δt -Δ[H 2 O 2 ] = -([H 2 O 2 ] f - [H 2 O 2 ] i ) = 1.7 x 10 -3 M s -1 x Δt [H 2 O 2 ] 100 s – 2.32 M = -1.7 x 10 -3 M s -1 x 100 s [H 2 O 2 ] 100 s = 2.32 M - 0.17 M = 2.17 M
14-3 Effect of Concentration on Reaction Rates: The Rate Law
a
A +
b
B …. →
g
G +
h
H ….
Rate of reaction = k [A]
m
[B]
n
….
Rate constant = k Overall order of reaction =
m
+
n
+ ….
شنکاو هبترم
A B
A
B R
d
[
A
]
k dt R
d
[
A
]
k
[
A
]
dt R
d
[
A
]
k
[
A
] 2
dt R
d
[
A
]
k
[
A
] 3
dt
رفص هبترم یاهشنکاو لوا هبترم یاهشنکاو مود هبترم یاهشنکاو موس هبترم یاهشنکاو • • • •
لوا هبترم یاهشنکاو
d
[ [
A
]
A
]
k
[
A
]
dt
d
[
A
]
kdt
[ [
A
]
A
] 0
d
[
A
] [
A
] 0
t
kdt
ln [
A
] 0 [
A
]
kt
بسح رب اهتظلغ تبسن یمتیراگل رادومن .
تسا یطخ نامز
Example:
Establishing the Order of a reaction by the Method of Initial Rates.
Use the data provided establish the order of the reaction with respect to HgCl reaction.
2 and C 2 O 2 2 and also the overall order of the
Example: Notice that concentration changes between reactions are by a factor of 2.
Write and take ratios of rate laws taking this into account.
Example: R 3 = k [HgCl 2 ] 3 m [C 2 O 4 2 ] 3 n R 2 = k[HgCl 2 ] 2 m [C 2 O 4 2 ] 2 n R 2 R 3 = k (0.105) m [C 2 O 4 2 ] 2 n k (0.052) m [C 2 O 4 2 ] 3 n R 2 R 3 = 2 m = 7.1 x 10 -5 3.5 x 10 -5 2 m = 2.0 therefore m = 1.0
Example: R 2 = k[HgCl 2 ] 2 1 [C 2 O 4 2 ] 2 n = k(0.105)( 0.30
) n R 1 = k[HgCl 2 ] 1 1 [C 2 O 4 2 ] 1 n = k(0.105)( 0.15
) n R 2 R 1 = k(0.105)(0.30) k(0.105)(0.15) n n R 2 R 1 = (0.30) n (0.15) n = 2 n = 7.1
x 10 -5 1.8
x 10 -5 = 3.94 2 n = 3.98 therefore n = 2.0
Example: First order R = k [HgCl 2 ] [C 2 O 4 2 ] 2
15-4 Zero-Order Reactions A → products R rxn = k [A] 0 R rxn = k [k] = mol L -1 s -1
Integrated Rate Law -Δ[A] = k Δt Move to the infinitesimal -d[A] = k dt And integrate from 0 to time t [A] t -∫ [A] 0 d[A] 0 t dt -[A] t + [A] 0 = kt [A] t = [A] 0 - kt
15-5 First-Order Reactions H 2 O 2 (aq) → H 2 O(l) + ½ O 2 (g) d[H 2 O 2 ] dt = -k [H 2 O 2 ] ; ∫ [A] t [A] 0 d[H 2 O 2 ] [H 2 O 2 ] 0 t [k] = s -1 ln [A] t [A] 0 = -kt ln[A] t = -kt + ln[A] 0
First-Order Reactions
Half-Life • t ½ is the time taken for one-half of a reactant to be consumed.
For a first order reaction: ln [A] t [A] 0 = -kt ln ½[A] 0 [A] 0 = -kt ½ ln 2 = kt ½ t ½ = ln 2 k = 0.693
k
Half-Life Bu t OOBu t (g) → 2 CH 3 CO(g) + C 2 H 4 (g)
Some Typical First-Order Processes Some typical first-order processes
15-6 Second-Order Reactions • Rate law where sum of exponents m + n + … = 2 A → products d[A] = -k[A] 2 ; dt [A] t ∫ [A] 0 d[A] [A] 2 t ∫ 0 dt 1 [A] t 1 = kt + [A] 0 [k] = M -1 s -1 = L mol -1 s -1
Second-Order Reaction 1 [A] t 1 = kt + [A] 0
Pseudo First-Order Reactions • • Simplify the kinetics of complex reactions Rate laws become easier to work with CH 3 CO 2 C 2 H 5 + H 2 O → CH 3 CO 2 H + C 2 H 5 OH • • If the concentration of water does not change appreciably during the reaction.
– Rate law appears to be first order Typically hold one or more reactants constant by using high concentrations and low concentrations of the reactants under study.
Testing for a Rate Law Plot [A] vs t.
Plot ln[A] vs t.
Plot 1/[A] vs t. 2nd order
15-7 Reaction Kinetics: A Summary • Calculate the rate of a reaction from a known rate law using: Rate of reaction = k [A]
m
[B]
n
….
• Determine the instantaneous rate of the reaction by: Finding the slope of the tangent line of [A] vs t or, Evaluate –Δ[A]/Δt, with a short Δt interval.
Summary of Kinetics • Determine the order of reaction by: Using the method of initial rates Find the graph that yields a straight line Test for the half-life to find first order reactions Substitute data into integrated rate laws to find the rate law that gives a consistent value of k.
Summary of Kinetics • Find the rate constant k by: Determining the slope of a straight line graph.
Evaluating k with the integrated rate law.
Measuring the half life of first-order reactions.
• Find reactant concentrations or times for certain conditions using the integrated rate law after determining k.
Activation Energy • For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).
• Activation Energy is: – The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.
Activation Energy
Kinetic Energy
Collision Theory • If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower.
• As temperature increases, reaction rate increases.
• Orientation of molecules may be important.
Collision Theory
Transition State Theory • The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.
15-9 Effect of Temperature on Reaction Rates • Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation: k = A
e -Ea/RT
-
Ea
R 1 ln k = + ln A T
Arrhenius Plot N 2 O 5 (CCl 4 ) → N 2 O 4 (CCl 4 ) + ½ O 2 (g) -
E a
= -1.2
x 10 4 K R
E a
= 1.0
x 10 2 kJ mol -1
Arrhenius Equation k = A
e -Ea/RT
-
E a
R 1 ln k = + ln A T ln k 2 – ln k 1 -
E a
= + ln A R 1 T 2 -
E a
R 1 T 1 - ln A k k 2
E
1 ln = 1 R T 1 1 T 2 k k 2
E
1 log = 1 2.3 R T 1 1 T 2
A Rate Determining Step
11-5 Catalysis • • • Alternative reaction pathway of lower energy.
Homogeneous catalysis.
– All species in the reaction are in solution. Heterogeneous catalysis.
– – The catalyst is in the solid state.
Reactants from gas or solution phase are adsorbed.
– Active sites on the catalytic surface are important .
11-5 Catalysis