For the following reaction $2X+Y\stackrel{k}{\to }P$ the rate of reaction is $\frac{d\left[P\right]}{dt}-k\left[X\right]$. Two moles of X are mixed with one mole of Y to make 1.0L of solution. At 50 s, 0.2 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is (are)

(Use: ln 2=0.693) 

Rate = $\frac{dp}{dt}=k{\left[X\right]}^1$

$2X+Y\to P$

t=0,        2 mol    1mol

t=50s,    1 mol      0.5 mol 0.5 mol

=$-\frac{1}{2}\frac{d\left[X\right]}{dt}=\frac{d\left[P\right]}{dt}=k{\left[X\right]}^1$

=$-\frac{d\left[X\right]}{dt}=2k{\left[X\right]}^1$

$k=\frac{\ln 2}{100}=6.93\times {10}^{-3}{s}^{-}$

$t_{1/2}=\frac{ln2\times 50}{ln2}=50\sec$

At 50 sec $\frac{-d\left[X\right]}{dt}=2k\times {\left(1\right)}^1=\frac{ln2}{50}$

=$13.86\times {10}^{-3}mol{L}^{-1}{s}^{-1}$

At 100 sec $-\frac{1}{2}\frac{d\left[X\right]}{dt}=\frac{-d\left[Y\right]}{dt}$

$\Rightarrow -\frac{d\left[Y\right]}{dt}=\frac{ln2}{100}\times \frac{1}{2}\left\{\frac{-d\left[Y\right]}{dt}=K{\left[X\right]}^1\right\}$

$\frac{d\left[Y\right]}{dt}=3.46\times {10}^{-3}mol{L}^{-1}{s}^{-1}$