Question

For the reaction: $3A_{\left(\right. g \left.\right)} \rightarrow 2B_{\left(\right. g \left.\right)}$ , the rate of formation of $B$ at $298K$ is represented as $ln\left(\frac{d \left[B\right]}{dt}\right)= \, -4.606+2ln \, \left[A\right] \, $ . The order of reaction is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$ln\left(\frac{d \left[B\right]}{d t}\right)=-4.606+2ln \left[\right.A\left]\right. \, $
$2.303log\left(\frac{d \left[B\right]}{dt}\right)=-4.606+2\times 2.303 \, log \, \left[A\right] \, $
$log\left(\frac{d \left[B\right]}{dt}\right)=\frac{- 4 .606}{2 .303}+2log\left[A\right]$
$log\left(\frac{d \left[B\right]}{dt}\right)=-2+2 \, log\left[A\right]$
$log\left(\frac{d \left[B\right]}{dt}\right)=log \, \left(10\right)^{- 2}+log\left(\left[A\right]\right)^{2}$
$log\left(\frac{d \left[B\right]}{dt}\right)=log\left(\left(10\right)^{- 2} \left(\left[A\right]\right)^{2}\right)$
$\left(\frac{d \left[B\right]}{d t}\right)=\left(1 0^{- 2} \left(\left[A\right]\right)^{2}\right)$
Compare with $\left(\frac{d \left[B\right]}{d t}\right)=k\left(\left[A\right]\right)^{m}$
$\left(\frac{d \left[B\right]}{d t}\right)=\left(1 0^{- 2} \left(\left[A\right]\right)^{2}\right)$
Order = 2