Hi, I've been OK, still doin' this chemistry stuff! In your experiment, you doubled one reactant concentration while not changing the other, the usual procedure, in order to determine the orders in the rate law for the reaction,
Rate = k[ClO2]x[OH�?/SUP>]y
In exps 1 and 2, [ClO2] was doubled and [OH�?/SUP>] was not changed. The rate of the reaction increased by a factor of 0.230 Ms�? / 0.0575 Ms�? = 4.00. Thus, 4.00 = 2x, x = 2.
More fully,
Rate2 = 0.230 Ms�? = 4.00 = k(0.100 M)x(0.100 M)y
Rate1 = 0.0575 Ms�? k(0.050 M)x(0.100 M)y
k and (0.100 M)y cancel out, leaving
4.00 = (0.100 M)x = 2x
(0.050 M)x
4.00 = 2x, log (4.00) = x log (2), x = log (4.00) / log (2) = 2.
With "exact" numbers like these, we can tell that x = 2 without using logarithms, but that is the general method. 
Calculate Y in the rate law the same way, using exps 3 and 2. When [OH�?/SUP>] is doubled from 0.050 M to 0.100 M, the rate increases by a factor of 0.230 Ms�? / 0.115 Ms�? = 2.00. 2.00 = 2y, y = ____.
Take any of the experiments and plug the numbers into the rate law (the concentrations and the values of x and y) and solve for k. Be sure to write the correct units for k!
Steve
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