Posted by Local on Friday, March 30, 2012 at 2:24pm.
Look at the three experiments. Note that A and B have the same concn bromobutane but OH is twice as much in B as in A. Note the rate change is by a factor of two also. 2^x = 2. Of course x = 1 so the reaction is first order with respect to OH^-. Do the same for bromobenzene. Then the overall order is the sum of the two orders.
The rate equation is
rate = k[bromomenzene][OH^-]
why do the same for bromomenzene? it sayes to do so for bromobutane?
My typo. My brain got bromobutane and bromobenzen mixed up and bromobutane got lost.
You have the order for OH. Find the order for bromobutane the same way. Then
rate = k[bromobutane]^x[OH^-]^y
where x and y are the orders. I've already shown y is 1. You need to do x. The overall order is x+y.
Thank you soo much! Really appreciate it. Could you also help with iv please? Thank you.
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