Consider the following reaction:
A+B+C→D
The rate law for this reaction is as follows:
Rate=k[A][C]2[B]1/2
Suppose the rate of the reaction at certain initial concentrations
of A, B, and C is 1.19×10−2 mol⋅L−1⋅s−1 .
What is the rate of the reaction if the concentrations of A and C are doubled and the concentration of B is tripled?
Given that
Rate = k[A][C]2[B]1/2 = 1.19 *10-2 mol /L.s
Now, Both A and C are doubled where as B is tripled
SO, New rate = k [2A][3B]1/2[2C]2 = 8(3)1/2 k[A][C]2[B]1/2 = 8(3)1/2 ( 1.19 *10-2 mol /L.s) = 0.165mol/L.s or 1.65 *10-1 mol /L.s
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