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Q91

In the following reaction; xA

\to

{\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010

'A' and 'B' respectively can be :

A n-Butane and Iso-butane

B C2H4 and C4H8

C C2H4 and C6H6

D N2O4 and NO2

Correct Answer

Option B

Solution

xA $\to$ yB

{1 \over x}\left\{ { - {{d\left[ A \right]} \over {dt}}} \right\} = {1 \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}

$\Rightarrow$

- {{d\left[ A \right]} \over {dt}} = {x \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}

Taking log both sides, we get $\Rightarrow$

{\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + {\log _{10}}\left( {{x \over y}} \right)

Given that,

{\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010

So, by comparing both of them we get

{\log _{10}}\left( {{x \over y}} \right)

= 0.3010 =

{\log _{10}}\left( 2 \right)

$\therefore$

{{x \over y}}

= 2 $\Rightarrow$ x = 2y If x = 2 then y = 1 The reaction is of type 2A $\to$ B So possible reaction is 2C2H4 $\to$ C4H8