Q. For the reaction: $A+3B \rightarrow 2C+D$ The data is given below :

$5\times 10^{- 5}$

The rate law for the above reaction is :

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A

$r=5\times 10^{- 3}L\cdot mol^{- 1}s^{- 1}\left[A\right]^{2}$

B

$r=5\times 10^{- 2}L^{2}mol^{- 2}s^{- 1}\left[A\right]^{2}\left[B\right]$

C

$r=5\times 10^{- 4}s^{- 1}\left[A\right]$

D

$r=5\times 10^{- 3}L^{- 1}mol^{- 1}s^{- 1}\left[\right.A\left]\right.\left[\right.B\left]\right.$

Solution:

Rate $\left(r\right)=k\left(\left[A\right]\right)^{x}\left(\left[B\right]\right)^{y}$ ......(1)
$\left(\right.1\left.\right)\Rightarrow 1.25\times \left(10\right)^{- 5}=k\left(0 . 050\right)^{x}\left(0 . 025\right)^{y}......\left(2\right)$
$\left(\right.1\left.\right)\Rightarrow 5\times \left(10\right)^{- 5}=k\left(\right.0.100\left(\left.\right)^{x}\left(\right.0.050\left(\left.\right)^{y}......\left(3\right)$
$\left(\right.1\left.\right)\Rightarrow 5\times \left(10\right)^{- 5}=k\left(\right.0.100\left(\left.\right)^{x}\left(\right.0.100\left(\left.\right)^{y} . . . . . . \left(4\right)$
$\left(\right.4\left.\right)\div\left(\right.3\left.\right) \rightarrow 1=\left(2\right)^{y}\therefore y=0$
$\left(3\right)\div\left(2\right) \rightarrow 4=\left(2\right)^{^{x}}\therefore x=2$
$\therefore $ eqn. $\left(4\right)\Rightarrow k=\frac{5 \times \left(10\right)^{- 5} \left(molL\right)^{- 1} s^{- 1}}{\left(10\right)^{- 2} \left(\left(moll\right)^{- 1} s^{- 1}\right)^{2}}=5\times \left(10\right)^{- 3}L\left(mol\right)^{- 1}s^{- 1}$

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