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with k = 6.73 10 −6 L/mol/s
Trial | [NO] (mol/L) | [Cl 2 ] (mol/L) | |
---|---|---|---|
1 | 0.10 | 0.10 | 0.00300 |
2 | 0.10 | 0.15 | 0.00450 |
3 | 0.15 | 0.10 | 0.00675 |
As in [link] , we can approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k . In this example, however, we will use a different approach to determine the values of m and n :
Determine the value of m
from the data in which [NO] varies and [Cl
2 ] is constant. We can write the ratios with the subscripts
x and
y to indicate data from two different trials:
Using the third trial and the first trial, in which [Cl
2 ] does not vary, gives:
After canceling equivalent terms in the numerator and denominator, we are left with:
which simplifies to:
We can use natural logs to determine the value of the exponent
m :
We can confirm the result easily, since:
Determine the value of n
from data in which [Cl
2 ] varies and [NO]is constant.
Cancelation gives:
which simplifies to:
Thus
n must be 1, and the form of the rate law is:
Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol 3 /L 3 . The units for k should be mol −2 L 2 /s so that the rate is in terms of mol/L/s.
To determine the value of
k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for
k :
Trial | [OCl − ] (mol/L) | [I − ] (mol/L) | Initial Rate (mol/L/s) |
---|---|---|---|
1 | 0.0040 | 0.0020 | 0.00184 |
2 | 0.0020 | 0.0040 | 0.00092 |
3 | 0.0020 | 0.0020 | 0.00046 |
Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.
2.00 = 2.00
y
y = 1
Substituting the concentration data from trial 1 and solving for
k yields:
In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.
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