# Laws Order

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Laws Order document sample

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6/30/2011
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```							Slide 1      Section 12.2
Rate Laws & Reaction Order

For this reaction:
aA + bB  Products
• A and B are chemical formulas for the
reactants
• a and b are the coefficients in the balanced
equation

Rate Law expression= k[A]m [B]n
How to determine
Slide 2        m and n
• For simple reactions (the ones found in this
class) an exponent of 1 means that the rate
depends linearly on the concentration of the
corresponding reactant.
• For example if m=1 and the [A] is doubled, the
rate doubles.
• If m = 2 and the [A] is doubled, [A]2 quadruples
and the rate increases by a factor of 4.
• If m=0, the rate is independent of the [A]
Slide 3

The m and n that you put as exponents are
the rate orders for the reactants.

For the given reaction:
Rate = k [ A]2 [B]

The reaction is in the 2nd order for A, first
order for B and third order for the whole
reaction
Slide 5

Section 12.3
Experimental Determination
Of Rate Law
Slide 6
Let’s look at the following reaction:

2NO(g) + O2(g)  2NO2(g)

And the following experimental data
Experiment   Initial [NO] Initial [O2]   Initial Rate
M/s
1      0.015         0.015          0.024
2      0.030         0.015         0.096
3      0.015         0.030         0.048
4      0.030         0.030         0.192
Slide 7

Analyzing the data you see that when the
[NO] is doubled (comparing experiment 1 to
2) and the [O2] stays the same, the rate
increases by a factor of 4, telling you that
[NO] is in the second order.

And when the [O2] is doubled (comparing
experiment 1 to 3) the rate doubled, from
the initial rate, telling you that [O2] is in the
first order.
Slide 4
Units for the reaction:

Rate Law           Overall Rate     Units for k
Order

Rate = k[B]        First order      1/s

Rate = k [A]2      Second order     1/M(s)

Rate = k[A]2[B]    Third order      1/M2(s)

```
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