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Solve it, - Chemical kinetics - JEE Main

In the following reaction; xA\rightarrow yB

\log_{10}\left [ -\frac{d\left [ A \right ]}{dt} \right ]=\log_{10}\left [ \frac{d\left [ B \right ]}{dt} \right ]+0.3010

'A' and 'B' respectively can be : 

  • Option 1)

    N_2O_4 and NO_2

  • Option 2)

    C_2H_4 and C_4H_8

  • Option 3)

    C_2H_2 and C_6H_6

  • Option 4)

    n-Butane and Iso-butane

Answers (2)
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First Order Reaction -

The rate of the reaction is proportional to the first power of the concentration of the reactant.

- wherein

Formula:

R    \rightarrow        P

a                 0

a-x             x

rate[r]=K[R]^{1}

\frac{-d(a-x)}{dt}=K(a-x)

\frac{-dx}{dt}=K(a-x)  [differentiate rate law]

ln \:[\frac{a}{a-x}]=kt \:(Integrated\: rate\: law)

Unit of k=sec^{-1}

t_\frac{1}{2}=\frac{0.693}{k}

 

 

 

\log_{10}\left [ -\frac{d\left [ A \right ]}{dt} \right ]=\log_{10}\left [ \frac{d\left [ B \right ]}{dt} \right ]+0.3010

\Rightarrow \log\left [ -\frac{d\left [ A \right ]}{dt} \right ]=\log\left [ \frac{d\left [ B \right ]}{dt} \right ]+\log 2

\Rightarrow \frac{-d\left [ A \right ]}{dt}=2\frac{dB}{dt}

\Rightarrow -\frac{1}{2}\frac{d\left [ A \right ]}{dt}=\frac{d\left [ B \right ]}{st}

\therefore x=2  &  y=1

C_{2}H_{4}  &   C_{4}H_{8}


Option 1)

N_2O_4 and NO_2

Option 2)

C_2H_4 and C_4H_8

Option 3)

C_2H_2 and C_6H_6

Option 4)

n-Butane and Iso-butane

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