Question

$\begin{array}{ccccc}\text{Experiment No.}& \text{[A]}\left(mold{m}^{-3}\right)& \text{[B]}\left(mold{m}^{-3}\right)& \text{[C]}\left(mold{m}^{-3}\right)& \text{Rate of the reaction}\\ & & & & \left(mold{m}^{-3}{s}^{-1}\right)\\ 1& 0.2& 0.1& 0.1& 6.0\times {10}^{-5}\\ 2& 0.2& 0.2& 0.1& 6.0\times {10}^{-5}\\ 3& 0.2& 0.1& 0.2& 1.2\times {10}^{-4}\\ 4& 0.3& 0.1& 0.1& 9.0\times {10}^{-5}\end{array}$
The rate of the reaction for $\left[A\right]=0.15mold{m}^{-3},\left[B\right]=0.25mold{m}^{-3}and\left[C\right]=0.15mold{m}^{-3}$ is found to be $Y\times {10}^{-5}mold{m}^{-3}{s}^{-1}$.The value of Y is

Answer 1

The rate of the reaction for $\left[A\right]=0.15mold{m}^{-3},\left[B\right]=0.25mold{m}^{-3}and\left[C\right]=0.15mold{m}^{-3}$ is found to be $Y\times {10}^{-5}mold{m}^{-3}{s}^{-1}.$ The value of Y is
Rate = $k\left[A{]}^x\right[B{]}^y[C{]}^z$
From exp. No. 1 & 2, $2^y=1\Rightarrow y=0$
From exp. No. 1 & 3, $2^z=2\Rightarrow z=1$
By exp. No. 1 & 4, ${1.5}^x=1.5\Rightarrow x=1$
Rate $=k\left[A{]}^1\right[B{]}^0[C{]}^1$
From Exp. No.1,
$6\times {10}^{-5}=k\left(0.2\right)\left(0.1\right)$
$\Rightarrow k=3\times {10}^{-3}$
Now for $\left[A\right]=0.15,\left[B\right]=0.25,\left[C\right]=0.15$
$Rate=k\left[A{]}^1\right[B{]}^0[C{]}^1=3\times {10}^{-3}\times 0.15\times 1\times 0.15=3\times 0.0225\times {10}^{-3}=\underset{Y}{6.75\times {10}^{-5}}mol{L}^{-1}se{c}^{-1}$