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.
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Jespersen Chap. 14 Sec 3
Dr. C. Yau
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!!
Rate= k [reactant]order
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."
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 orderwith respect to H2SeO3,
third orderwith respect to I-(Nothing to do with 6 in eqn)
second orderwith 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?
Do Practice Exercises 7, 8 & 9 on p.648.
Rate changes with concentrations. The rate law allows us to determine the rate for various concentrations of the reactants.
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?
rate = 0.045 M-1 s-1 [0.2M][0.3M]
rate = 0.0027 M/s
Do Practice Exercises 5 & 6 p.646
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
2.3x102 M-2 s-1
2NO(g) + O2(g)→ 2NO2(g)
Select 2 rate laws that vary in concentration for only one of the substances (NO).
Hint: Write the fractions with the larger R on top.
2NO(g) + O2(g)→ 2NO2(g)
Next choose 2 rate laws where the concentration for the other component
x=2, y = 1
Rate = k [NO]2[O2]
Finally we can solve for k. Use any rate law and the orders that we have determined.
rate = k[NO]2[O2]
0.048M/s =k [0.015M]2[0.015M]
1.4×104 M-2s-1 =k
Do Example 14.5, 14.6 p.651, Exercises 10 thru 14p.651+.
Note that changing the concentration of C had no effect on the rate. We say it is “zero order with respect to C.”
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
(See next 2 slides.)
What is the rate law?
Rate = k[A]?[B]?
Rate = k[A]1[B]2
2 NO2(g) + F2(g)→ 2 NO2F(g):
We cannot always determine the rxn order visually.
For example, if we ended with
How do we determine what x is?
In high-level chemistry courses, x might even be a fraction!