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What is the rate constant for a reaction of $(CH_3)_3CBr$ with water is represented below :$CH_3)_3CBr+H_2O\rightarrow (CH_3)_3COH+HBr$ where the following data was obtained experimentally:

$\begin{array} {cccc} \text{Experiment} & [A]\; mol\;L^{-1} & [B]\; mol\;L^{-1} & \text{Initial Rate} \; mol\;L^{-1}min^{-1}\\ 1 & 5 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-6}\\ 2 & 5 \times 10^{-2} & 4 \times 10^{-2} & 2 \times 10^{-6}\\ 3 & 1 \times 10^{-1} & 4 \times 10^{-2} & 4 \times 10^{-6}\\ \end{array}$

$\begin{array}{1 1} 2\times 10^{-5} \\ 8\times 10^{-5} \\ 4\times 10^{-5} \\3\times 10^{-5} \end{array}$

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Let, $R=K[H_2O]^x[(CH_3)_3CBr]^y$
Taking $(CH_3)_3CBr$ constant if we double concentration of $H_2O$ than rate of reaction will not change and if we take $[H_2O]$ constant and 5 times $[(CH_3)_3CBr]$ then rate also becomes 5 times.
So $x=0$ & $y=1$
So $R=k^1[(CH_3)_3CBr]^1$
$K^1=\large\frac{2\times 10^{-6}}{5\times 10^{-2}}$
$\;\;\;\;\;=4\times 10^{-5}$
answered Dec 18, 2013 by sreemathi.v
edited Jul 27, 2014 by balaji.thirumalai
 

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