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Kinetics Part II: Rate Laws & Order of Reaction. Jespersen Chap. 14 Sec 3. Dr. C. Yau Spring 2014. 1. Rate of Rxn vs . Rate Law. Rate of reaction is based on one component (reactant or product) of the reaction: disappearance of a reactant or formation of a product. - PowerPoint PPT Presentation
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Kinetics Part II:Rate Laws & Order of Reaction
Dr. C. Yau
Spring 201411
Jespersen Chap. 14 Sec 3
Rate of Rxn vs. Rate Law
22
Rate of reaction is based on one component (reactant or product) of the reaction: disappearance of a reactant or formation of a product.
Rate law is a rate expression that includes all reactants.
LEARN THESE TERMS SO YOU KNOW WHAT IS BEING ASKED FOR!!
The Rate Law Depends On The Concentrations Used
Rate= k [reactant]order
•k is a reaction rate constant, a measure of time efficiency (not to be confused with “Rate”).
•High values of k mean high efficiency.(Reaction goes fast.)•k must be determined experimentally.•Each experiment has its own rate law.•Rate law must be determined experimentally.
3
A + B products
Rate = k [A]m[B]n
where m and n are the "orders of reaction" and are found by experiment, NOT based on the coefficients of the chemical equation,
and k is the "rate constant."
This expression is called the "rate law."
4
H2SeO3 + 6I- +4H+ Se + 2I3- + 3H2O
Rate = 5.0x105 L5mol-5 s-1 [H2SeO3] [I-]3[H+]2
5.0x105 mol-5 s-1 is the rate constant (k).
We speak of the reaction as being…first order with respect to H2SeO3,
third order with respect to I-(Nothing to do with 6 in eqn)
second order with respect to H+, and
the overall order of reaction is 6 (sum of all the orders).
Learn this terminology!
What is the unit of Rate in the equation shown above?
5Do Practice Exercises 7, 8 & 9 on p.648.
Example:
The rate law for the reaction 2A +B→3C is
Rate = 0.045M-1s-1 [A][B]
If the concentration of A is 0.2M and that of B is 0.3M, what will be the reaction rate?
6rate = 0.0027 M/srate = 0.045 M-1 s-1 [0.2M][0.3M]
What is a rate law used for?Rate changes with concentrations. The rate law allows us to determine the rate for various concentrations of the reactants.
Do Practice Exercises 5 & 6 p.646
Chlorine dioxide, ClO2, is a reddish-yellow gas that is soluble in water. In basic solution it gives ClO3
- and ClO2
- ions.
ClO2(aq) + OH(aq) ClO3(aq) + ClO2
(aq) + H2O (l)
The rate law is Rate=k[ClO2]2[OH-]. What is the value of the rate constant given that when [ClO2]=0.060M, [OH-] = 0.030M, the reaction rate is 0.0248 M/s
A. 0.02 M-1 /s
B. 0.02 M/s
C. 0.02 s-
D. None of these
7
2.3x102 M-2 s-1
Orders…
• are indicated for each reactant,
• the overall reaction order is the sum of individual reactant orders,
• may be zero, negative, fractional or integers, but in this course we will usually encounter positive integers, and
• must be determined from experimental data.
8
Determining The Rate Law:
• Run reaction under the same conditions, varying only the concentrations of reactants (not the temperature).
• A ratio of rate laws for each experiment allows us to determine the order of each reactant.
• The rate law is unique to temperature and concentration conditions. Therefore, when a rate law is stated, it must include the temperature at which it is determined.
9
Use Rate Laws To Determine Orders :
Select 2 rate laws that vary in concentration for only one of the substances (NO).
10
exp [NO] [O2] RNO2 M/s
1 .015 .015 .048
2 .030 .015 .192
3 .015 .030 .096
2NO(g) + O2(g) → 2NO2(g)
2. bemust x
2.0 .004
[0.015MNO]
]MNO030.0[ 4.00
[0.015MNO]
]MNO030.0[ 4.00
]MO015.0[]k[0.015MNO
]MO015.0[]MNO030.0[k
s/M048.0
s/M192.0
x
x
x
x
y2
x
y2
x
Hint: Write the fractions with the larger R on top.
Use Rate Laws To Determine Orders :
Next choose 2 rate laws where the concentration for the other component
(O2) changes.
11
exp [NO] [O2] RNO2 M/s
1 .015 .015 .048
2 .030 .015 .192
3 .015 .030 .096
x=2, y = 1
so…..
Rate = k [NO]2[O2]
2NO(g) + O2(g) → 2NO2(g)
1. bemust y
2.0 .002
][0.015MO
]MO030.0[ 2.00
]MO015.0[
]MO030.0[ 2.00
]MO015.0[]k[0.015MNO
]MO030.0[]MNO015.0[k
s/M048.0
s/M096.0
y
y
2
2
y2
y2
y2
x
y2
x
Determining The Value Of k
Finally we can solve for k. Use any rate law and the orders that we have determined.
exp [NO] [O2] RNO2
M/s
1 .015 .015 .048
2 .030 .015 .192
3 .015 .030 .096
12
rate = k[NO]2[O2]
0.048M/s =k [0.015M]2[0.015M]
1.4×104 M-2s-1 =kDo Example 14.5, 14.6 p.651, Exercises 10 thru 14p.651+.
Determine The Rate Law From Given Data
[A] [B] [C]Rate
M/s
0.01 0.02 0.15 0.0002
0.02 0.02 0.15 0.0004
0.01 0.01 0.15 0.0001
0.01 0.02 0.3 0.0002
13
z=0x=1 y=1
rate=k[A][B]
Note that changing the concentration of C had no effect on the rate. We say it is “zero order with respect to C.”
]15.0[]02[.]M01[.k
]15.0[]02[.]M02[.k
s/M0002.0
s/M0004.0zyx
zyx
]15.0[]01[.]M01[.k
]15.0[]02[.]M01[.k
s/M0001.0
s/M0002.0zyx
zyx
]3.0[]02[.]M01[.k
]15.0[]02[.]M01[.k
s/M0002.0
s/M0002.0zyx
zyx
Effect of Order of Rxn on Rate
Consider Rate = k[A]n
If n = 0, change in conc has no effect on rate.
If n = 1, Rate = k[A]1 and when conc is 2x,
rate is 2x.
If n = 2, Rate = k[A]2 and when conc is 2x,
rate is 4x
If n = 2, when conc is tripled, rate is …?
rate is 9x
If n = 3, and conc is doubled, rate is…?
rate is 8x 14
Visual Determination of Reaction Order
• Once you understand how you can predict effect of a change in concentration on rates (as in the previous slide), you can often determine the rxn order visually without doing complicated calculations.
• HOWEVER, that is only if the conc were neatly doubled or tripled, etc.
(See next 2 slides.)
15
What is the rate law?
Rate = k[A]?[B]?
Rate = k[A]1[B]2 16
p. 648
For the following data, determine the order of NO2 in the reaction at 25°C
2 NO2(g) + F2(g)→ 2 NO2F(g):
Exp [NO2] [F2] Rate NO2 disappearance (M/s)
1 0.001 0.005 2 x10-4
2 0.002 0.005 4 x10-4
3 0.006 0.002 4.8 x10-4
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A. 0 B. 1 C. 2 D. 3 E. not enough information given
When Visual Determination Fails...We cannot always determine the rxn order
visually.
For example, if we ended with
32.1=3.18x
How do we determine what x is?
In high-level chemistry courses, x might even be a fraction!
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