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In a hypothetical reaction

A(aq)\rightarrow \:\:\:2B(aq)+C(aq)\:\:\:\:\:\:\:\:(1^{st}order\:\:decomposition)

'A' is optically active (dextro-rotatory) while  ' B ' \:and \:' C ' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with H_{2}O_{2}. Hence the progress of reaction can be monitored by measuring rotation of plane of plane polarised light or by measuring volume of H_{2}O_{2} consumed in titration.

   In an experiment the optical rotation was found to be \Theta = 30^{0}  at  t  20\:\:min  and \Theta = 15^{0} at t=50\:min . from start of the reaction. If the progress would have been monitored by titration method, volume of H_{2}O_{2}  consumed at t = 30 min. (from start) is 3d\:\: ml  then volume of H_{2}O_{2}  consumed at t = 90\: min will be (in ml):

Option: 1

60


Option: 2

45


Option: 3

52.5


Option: 4

90


Answers (1)

 

As only A is optically active. So conc. of A\: at\: t = 20 \:min \propto \:30^{0}

                While concentration of A\: at \:t = 50\: min\propto 15^{0}

                So conc. has decreased to half of its value in 30\:min  min, so t_{\frac{1}{2}}=\:30\:min .

                So volume consumed of H_{2}O_{2} at t = 30\: min = t_{\frac{1}{2}}  , is according to 50 \:^{o}/_{o}production of B.

                at t = 90\: min . production of B = 87.5 \:^{o}/_{o}  (Three half lives)

                So\: volume \:consumed = (30\:ml)+\left ( \frac{30}{2}\:ml \right )+\left ( \frac{30}{4} \right )\:ml

               = 52.5\: ml\:  ans .

Posted by

Kshitij

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