To find m, ratio law from exp. 1 and 3 where [Cl
2
] is the same
0.189
0.021
=
k (0.300)
m
(0.100)
n
k (0.100)
m
(0.100)
n
9.00 = (3.00)
m
m = 2
2)
To find n, ratio law from exp. 1 and 2
0.042
0.021
=
k (0.100)
2
(0.200)
n
k (0.100)
2
(0.100)
n
2.00 = (2.00)
n
n = 1
1.4
Therefore:
Rate law = k [NO]
2
[Cl
2
]
It just happened that m and n are also the coefFcients in the reaction …
BUT you can’t assume this is always true!
Notice that you may have used another set of coefFcients, e.g.
NO(g) + 1/2 Cl
2
(g)
→
NOCl(g)
)
Rate law is:
Rate = k [NO]
m
[Cl
2
]
n
2 NO
(g) +
Cl
2
(g)
→
2
NOCl (g)
Initial :
[NO]
[Cl
2
]
Initial Rate (M/s = mol/Ls)
Expt 1
0.100
0.100
0.021
2
0.100
0.200
0.042
3
0.300
0.100
0.189
To determine the order of the reaction, run several experiments
under different reactant concentrations, and measure the initial rates.
*** Sometimes you don’t have two experiments with the same concentration for
a species. Put the different values into the rate laws, and solve for the exponent.
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 Spring '09
 DrK
 Reaction, Chemical reaction, Rate equation

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