## Quiz 2 #3

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

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

### Quiz 2 #3

#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

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.