Given the data below, what are the orders of reaction with respect to #"[A]"# and #"[B]"#?

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1 Answer
Ernest Z.
Mar 12, 2015

The orders are #m = 1.16; n = "-1.26"#.

Explanation:

A typical rate law is #r = k"[A]"^m"[B]"^n#

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From Expts. 2 and 1: #r_2/r_1 = 9.6/6.4 = (0.00208/0.00146)^m (0.00060/0.00060)^n#

#1.50 = 1.42^m#

#log1.50 = mlog1.42#

#m = "log1.50"/"log1.42" = 0.176/0.152 = 1.16#

From Expts. 2 and 3: #r_2/r_3 = 9.6/4.0 = (0.00208/0.00208)^1.16 (0.00060/0.00120)^n#

#2.40 = 0.50^n#

#n = "log2.40"/"log0.50" = 0.380/("-0.301") = "-1.26"#

Note: The results are not integers. I presume that this is just an exercise to demonstrate how to find orders of reaction by the method of initial rates.

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