Quiz 2 #3


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Kevin Pham 2G
Posts: 12
Joined: Fri Jul 15, 2016 3:00 am

Quiz 2 #3

Postby Kevin Pham 2G » Sun Mar 12, 2017 8:44 pm

#3. "Consider the following reaction mechanism
Step 1: (CH3)3C-Br ->k1 <-k1' (CH3)3C+ +Br-
Step 2: (CH3)3C+ + OH- ->k2 (CH3)3C-OH
Given that k1 << k2"

I know that Step 1 is the rate limiting step b/c the rate constant of Step 1 is much smaller than that of Step 2, hence Step 1 is the slow step. How do you approach (c) Write the rate law based on the above information if Step 1 reaction has ->k1 and <-k1'?

Milan Hirpara 3K
Posts: 20
Joined: Sat Jul 09, 2016 3:00 am

Re: Quiz 2 #3

Postby Milan Hirpara 3K » Sun Mar 12, 2017 9:12 pm

The rate law is just RATE= k[(CH3)3C-Br]. Even though the equation is in equilibrium, you still approach it the same way as you always would.


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