# Rates of Reaction - Download Now PowerPoint

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```							Rates of Reaction
Chapter 6
Factors Affecting Reaction Rate

1.   Concentration
2.   Surface Area            p. 276 Summary

3.   Temperature
4.   Catalyst
Know the effects of these factors on reaction
rate! (Recall demos!)
Be able to explain them! (We will get into
further detail with Collision Theory!!)
Rate of Reaction
For compound A in a reaction,

Rate of Reaction = ∆ Amount of A
∆time

Need to know:
- Difference between average and instantaneous rate
- Secant and Tangent lines
- Concentration vs Time graphs

- Do Thought Lab on p. 270 (refer Figure 6.2 p. 269)
Expressing Rates of Reaction

2N2O5(g)  4NO2(g) + O2(g)

2 ways to express rate of reaction:
1. Rate of disappearance of N2O5
2. Rate of formation of NO2 or O2
Expressing Rates of Reaction

2N2O5(g)  4NO2(g) + O2(g)

For every 1 mol of O2, how many moles of NO2 are
produced?      4 mol of NO2 are produced.

Therefore,
Rate of production of O2 is ¼ rate of production of NO2
Δ[O2] =    1 Δ[NO2]
Δt        4 Δt
Expressing Rates of Reaction

2N2O5(g)  4NO2(g) + O2(g)

For every 1 mol of O2 produced, how many moles
of N2O5 are consumed? 2 mol of N2O5 are consumed.

Therefore,
Rate of production of O2 is ½ rate of disappearance
of N2O5
Δ[O2] = -1 Δ[N2O5]
Δt          2 Δt
Example 2
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

At a specific time in the reaction, ammonia is
disappearing at a rate of 0.068mol/(L·s)

What is the corresponding rate of production of
water?
Since 6H2O produced for every 4NH3,
The rate of production of H2O is 6/4 the rate of reaction of NH3
Rate of appearance of H2O = 6/4 x (rate of disappearance of NH3)
= 3/2 x 0.068 mol/(L·s)
= 0.10 mol/(L·s)
Do Practice Problems #1, 3, 4 on p. 272
Methods for Measuring Reaction Rates

Various techniques to measure
concentration of a reactant or product:
Mass, pH, and Conductivity
Pressure
Colour
Volume

Read p. 272-273 for descriptions
The Rate Law
Reactant Concentration and Rate

Consider, aA + bB  products
We know effect of concentration on
reaction rate???
But what is the quantitative
relationship???
The Rate Law Equation!
Rate = k[A]m[B]n
Reactant Concentration and Rate

The Rate Law Equation!
Rate = k[A]m[B]n

Rate Constant:            Rate Law Exponents:
-Depends on temperature   -Do not change with temperature
-Determined experimentally
Reactant Concentration and Rate

Rate Constant, k:
Different rate constant at different temperature
Small k means slow reaction
Large k means fast reaction
Exponents, m and n:
Determined experimentally
Usually integers. 1 and 2 most common.
Values of 0, 3 and fractions occur as well.
m and n values DO NOT correspond to
stoichiometric coefficients!!!
Reaction Order

Overall Order of a Reaction
Sum of the individual exponents of each
reactant in the rate law

Example 1:
2N2O5(g)  4NO2(g) + 5O2(g)
Rate = k[N2O5]1

The overall order is 1.
This is a first-order reaction.
Reaction Order
Example 2:
(CH3)3CBr(aq) + H2O(l)  (CH3)3OH(aq) + H+(aq) + Br-(aq)

Rate = k[(CH3)3CBr(aq)]1[H2O(l)]0
The overall order is 1 + 0 = 1.
This is a first-order reaction.
Example 3:
NO(g) + O3(g)  NO2(g) + O2(g)

Rate = k[NO]1[O3]1

The overall order is 1 + 1 = 2.
This is a second-order reaction.
Reaction Order
Example 4:
W(s)
2NH3(q)  N2(q) + 3H2(g)

Rate = k[NH3]0 = k

The overall order is 0. This is a zero-order reaction.
The rate of the reaction is a constant.

The rate of decomposition of ammonia is independent
of the concentration of ammonia!
Determining Rate Law Values!
2 N2O5 (g)  2NO2(g) + O2(g)
Experiment        [N2O5] (mol/L)     Rate (mol/L·s)
1                 0.010              4.8x10-6

2                 0.020              9.6x10-6

3                 0.030              1.5x10-5

a. What is the rate law equation?
b. What is k?
Determining Rate Law Values!
 Solution:

 Rate = k[N2O5]
 K = 4.8x10-4 s-1
Determining Rate Law Values!
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)

Experiment       [NO] (mol/L)      [H2] (mol/L)    Rate (mol/L•s)

1            0.210              0.122            0.0339

2            0.420              0.122            0.136

3            0.210              0.244            0.0678

a.   What is the rate law equation? Using variables.
b.   What are the rate law exponents?
c.   What is the value of k?
d.   Express rate law equation with values obtained.
Determining Rate Law Values!
Solutions:

a.   Rate = k[NO]a[H2]b
b.   a = 2, b = 1
c.   k = 6.3
d.   Rate = 6.3[NO]2[H2]

e. Concept check! What are the units for k?
Note: units for k change for different orders of
reaction!!
Practice, Practice, Practice!!!

Sample Problem p. 282
Practice Problems #5-8 p. 284
Section Review #1-6
Thought Lab: Rate of Formation of Oxygen Gas

0.0090                                                           Tangent at 4800s
slope = 4.0E-07
0.0080
Concentration of Oxygen (mol/L)

0.0070

0.0060
Tangent at
1200s
0.0050
slope = 2.1E-06

0.0040
Secant
0.0030                                         slope = 1.6E-06

0.0020

0.0010

0.0000
0      600     1200   1800   2400     3000     3600     4200   4800        5400   6000
Time (s)

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