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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)

S solutionqc

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

M Make It Possible

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