A) 4
B) 5
C) 3
D) 4
Correct Answer: B
Solution :
[b] \[\frac{-d[RX]}{dt}={{k}_{2}}[RX]\,[\overset{\bigcirc -}{\mathop{O}}\,H]\] (by \[{{S}_{N}}2\]path way) \[{{k}_{2}}\]= rate constant of \[{{S}_{N}}2\] reaction \[\frac{-d[RX]}{dt}={{k}_{1}}[RX]\] (by \[{{S}_{N}}1\] path way) \[{{k}_{1}}=\] rate constant of \[{{S}_{N}}1\] reaction \[\frac{-d[RX]}{dt}={{k}_{2}}[RX]\,\,[\overset{\bigcirc -}{\mathop{O}}\,H]+{{k}_{1}}[RX]\] \[-\frac{1}{[RX]}\frac{d[RX]}{dt}={{k}_{2}}[\overset{\bigcirc -}{\mathop{O}}\,H]+{{k}_{1}}\] This is the equation of a straight line for \[-\frac{1}{[RX]}=\frac{d[RX]}{dt}\] vs \[[\overset{\bigcirc -}{\mathop{O}}\,H]\] plot with slope equal to \[{{k}_{2}}\] and intercept equal to \[{{k}_{1}},\] From question: \[{{k}_{2}}=2\times {{10}^{3}}mo{{l}^{-1}}L\,h{{r}^{-1}},\,{{k}_{1}}=1\times {{10}^{2}}h{{r}^{-1}}\] \[[RX]=1.0M\] and \[[\overset{\bigcirc -}{\mathop{O}}\,H]=0.1M\] Hence, \[\frac{-d[RX]}{dt}=2\times {{10}^{3}}\times 1\times 0.1+1\times {{10}^{2}}\times 1\] \[=300\,mol\,{{L}^{-1}}h{{r}^{-1}}\] \[=5\,mol\,{{L}^{-1}}\,{{\min }^{-1}}\]
You need to login to perform this action.
You will be redirected in
3 sec
