Sections 12'1 to 12'3

1
Lecture 5

• Sections 12.1 to 12.3

2
First, the Questions

• What is the rate of a reaction, expressed in
  terms of concentration of substance A and in
  terms of time t?
• On a graph of concentration versus time, what is
  the slope of the line equal to?
• A differential rate law expresses rate as a
  function of what?

3
More Questions

• An integrated rate law expresses concentration as
  a function of what?
• If a reaction is second order with respect to
  reactant A and first order with respect to
  reactant B, what is the overall order of the
  reaction?

4
Now, Your Questions

• ?

5
Section 12.1

• Rate of a reaction, in terms of reactant or
  product A, is
• Rate ?A/?t
• A means concentration of A in mol/L (molarity).
• It may seem strange to talk about
  concentrations of gases, but mixtures of gases
  are solutions.
• PV nRT, so n/V P/RT

6
Rates and Stoichiometry

• In Table 12.2, when average rate of decomposition
  of NO2 is 1.4x10-5 M/s
• What is the rate of formation of NO?
• 1.4x10-5 M/s
• What is the rate of formation of O2?
• 7.0x10-6 M/s
• Note that ?NO/?t 2(?O2/?t )
• Or ½(?NO/?t ) (?O2/?t )

7
A Useful Generality

• For aA bB ? cC dD
• -(1/a)(?A/?t) -(1/b)(?B/?t)
  (1/c)(?C/?t) (1/d)(?D/?t)
• For 4NH3(g) 5O2(g) ? 4NO(g) 6H2O(g)
• If rate of formation of NO is 0.020 M/s, how fast
  is O2 disappearing? How fast is water appearing?
• 0.025 M/s and 0.030 M/s

8
Section 12.2

• Choose conditions in which only the forward
  reaction is important
• Usually right at the beginning of the reaction,
  called the initial rate.
• Not enough products have formed to have an effect
  on the forward rate.
• Under these conditions, the rate depends only on
  concentrations of reactants.

9
The Differential Rate Law

• This law expresses the rate of a reaction in
  terms of concentrations of reactants.
• For aA bB ? products
• Rate kAnBm
• k is called the rate constant.
• Values of n and m are determined by experiment.
• They may be equal to a and b, but they may not be.

10
Examples

• 2NO(g) Br2(g) ? 2NOBr(g)
• Rate kNO2Br2
• This reaction is 2nd order in NO, 1st order in
  Br2, 3rd order overall.
• NO2(g) CO(g) ? NO(g) CO2(g)
• Rate kNO22
• 2nd order in NO2, 0 order in CO, 2nd order
  overall
• You cannot assume that the exponent (the order)
  is equal to the coefficient in the balanced
  equation.

11
Section 12.3

• First, some math review
• R kA (first order)
• Double A, what happens to R?
• It doubles
• Triple A, what happens to R?
• It triples
• R kA2 (second order)
• Double A, what happens? Triple A, what happens?
• R quadruples, R increases by a factor of nine.

12
More Math Review

• R kA3 (third order)
• Double A, what happens? Triple A, what happens?
• R increases by factors of 8 and 27
• R kA2B (third order overall)
• Double A and B. What happens?
• R increases by a factor of 8.
• Cut A in half, double B. What happens?
• R is cut in half.

13
Determining the Rate Law

• When the concentration was cut in half, the rate
  was cut in half. The reaction is
• First order, Rate kN2O5
• If the problems are simple enough, you can figure
  out the exponents just by thinking through it.
• Be prepared to do it mathematically, however.
• Just like Lab 0 question 9!

14
One More Example

• 2A B ? Products
• Exp. A B Rate
• 1 0.130 M 0.180 M 1.45 M/min
• 2 0.210 M 0.140 M 1.42 M/min
• 3 0.130 M 0.230 M 2.37 M/min
• Find n, m, and k in Rate kAnBm
• n 1, m 2, k 344 /(M2min)