If you have difficulty, please review the book (free at openstaweollege.org) or

Question

If you have difficulty, please review the book (free at openstaweollege.org) or the textbook for this course. There are also video tutorials online at https://www.khanacademy.org/science/chemistry/che.m-kinetics Rate Law and Half-time Consider the following rate data for the reaction below at a particular temperature. 2A + 3B rightarrow Products Experiment Initial [A] Initial [B] Initial Rate of loss of A 0.10 M 0.30 M 7.20 times 10^-5 M middot s^-1 0.10 M 0.60 M 1.44 times 10^4 M middot s^-1 020 M 0.90 M 8.64 times 10^-4 M middot s^-1 The reaction is _____ order in A _____ and order in B.

Explanation / Answer

From 1 to 2, keeping concentration of A constant, concentration of B is doubled which increased rate of reaction by 2 times

So, order of reaction is 1 with respect to B

from 1 to 3, [A] is doubled, [B] is thripled , then rate is increase by 12 times

Let rate law be rate = k [A]m [B]n

Here, n=1

So, rate =k [A]m [B]

so, because of B, rate increrase by 3 times. So, m=2 as doubling [A] should increase the rate by 4 times

Therefore ,Rate = k [A]2[B]

The reaction is second order in A and first order in B

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