Q. In the reaction: $A+2B \rightarrow $ Product
Rate $=K\left[A\right]\left[B\right]^\circ $
The initial conc. of both $A$ and $B$ are $'a'$ molar. The graph of conc. of $B$ vs. time is $\left(\right.t_{1}=t_{1 / 2}$ for $A$ )

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A

B

C

D

Solution:

$ & A & + & 2B & \rightarrow Product \\ t=0 & a & & a & \\ t=2, & \left(a - x\right) & & \left(a - 2 x\right) & $
$\left[A\right]_{t}=ae^{- kt}$
$a-x=ae^{- kt}$
$x=a\left(1 - e^{- kt}\right)$
At $t=t_{1 / 2}\Rightarrow a-x=\frac{a}{2}$
$x=\frac{a}{2}$
$\left[B\right]_{t}=a-2x$
$=a-2a\left(1 - e^{- kt}\right)$
$=a\left(2 e^{- kt} - 1\right)$
At $t=0,\left[B\right]=a$
$t=t_{1 / 2},\left[B\right]_{t}=a-2x$
$=a-a$
$=0$

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