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First Order Reaction

For a reversi...

Question

For a reversible first - order reaction $A ⇌ B K_{1} = 10^{- 2} s^{- 1}$ and $[B]_{e q} = 4$ . If $[A]_{0} = 0.01$ mole $L^{- 1}$ and $[B]_{0} = 0$ , what will be the concentration of B after 30 s?

A

0.0025 m

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B

0.0013 m

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C

0.0026 m

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D

0.0030 m

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Solution

The correct option is A 0.0025 m
As we know,
$k_{1} + k_{2} = (1 / t) l o g (x_{e} - x_{0}) / (x_{e} - x)$ and
$k_{e} = k_{1} / k_{2}$
And from given data, conc. of B after 30 sec.
$k_{1} + k_{2} = 1 / 30) l o g (4 - 0 / 4 - x)$
$x = 0.0025$ m

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