The following set of data was obtained by the method of initial rates for the re
ID: 628261 • Letter: T
Question
The following set of data was obtained by the method of initial rates for the reaction:(H3C)3CBr + OH- ? (H3C)3COH + Br-
What is the value of the rate constant, k? [(H3C)3CBr], M [OH-] M Initial Rate M/s 0.25 0.25 1.1x10e-4 0.25 0.252.2x10e-4 0.25 0.50 2.2x10e-4 Please show step by step
The following set of data was obtained by the method of initial rates for the reaction:
(H3C)3CBr + OH- ? (H3C)3COH + Br-
What is the value of the rate constant, k? [(H3C)3CBr], M [OH-] M Initial Rate M/s 0.25 0.25 1.1x10e-4 0.25 0.252.2x10e-4 0.25 0.50 2.2x10e-4 Please show step by step
The following set of data was obtained by the method of initial rates for the reaction:
(H3C)3CBr + OH- ? (H3C)3COH + Br-
What is the value of the rate constant, k? [(H3C)3CBr], M [OH-] M Initial Rate M/s 0.25 0.25 1.1x10e-4 0.25 0.252.2x10e-4 0.25 0.50 2.2x10e-4 Please show step by step
The following set of data was obtained by the method of initial rates for the reaction:
(H3C)3CBr + OH- ? (H3C)3COH + Br-
What is the value of the rate constant, k? [(H3C)3CBr], M [OH-] M Initial Rate M/s 0.25 0.25 1.1x10e-4 0.25 0.252.2x10e-4 0.25 0.50 2.2x10e-4 Please show step by step
Explanation / Answer
rate = k*([(H3C)3CBr]^a)*([OH-]^b)
when [(H3C)3CBr is doubled, keeping [OH-] constant, rate gets doubled => rate is directly proportional to [(H3C)3CBr.
Similarly, when [OH-] is doubled, keeping [(H3C)3CBr]] constant, rate gets doubled => rate is directly proportional to [OH-]
=>.a = 1 and b = 1
=> rate =k*([(H3C)3CBr])*([OH-])
Now, we have, rate 1.1*10^-4, [OH-] = 0.25 [(H3C)CBr] = 0.25
=> k = (1.1*10^-4)/(0.25*0.25)
=> k = 0.00176 M^-1s^-1