( CH3)3CBr + OH- -> (CH3)3COH +Br- The rate law for the above reaction at 55 deg
ID: 677144 • Letter: #
Question
( CH3)3CBr + OH- -> (CH3)3COH +Br- The rate law for the above reaction at 55 degrees C is rate=k[(CH3)3CBr] with a rate constant of 1.0x10-2 s-1. what is the rateif the concentration of (CH3)3CBr is 3.4x10-3 M? How would the ratechange of this reaction change at 55 degrees C if the concentrationof (CH3)3CBr were held constant and the concentration of OH- weredoubled?A. 3.4 x 10-1 M2 s-1 B. 3.4 x 10-1 M s-1 D. 3.4 x 10-5 M s-1 E. 3.4 x 10-3 M s-1 The rate law for the above reaction at 55 degrees C is rate=k[(CH3)3CBr] with a rate constant of 1.0x10-2 s-1. what is the rateif the concentration of (CH3)3CBr is 3.4x10-3 M? How would the ratechange of this reaction change at 55 degrees C if the concentrationof (CH3)3CBr were held constant and the concentration of OH- weredoubled?
A. 3.4 x 10-1 M2 s-1 B. 3.4 x 10-1 M s-1 D. 3.4 x 10-5 M s-1 E. 3.4 x 10-3 M s-1
Explanation / Answer
rate = k[(CH3)3CBr] so with that jsut multiply k by concentration and you get d nothing would change because oh- is not a factor in teh rateexpression so it doesnt matter what you do with it the expressionwould still be at teh same rate
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