Question

For a chemical reaction, $2X+Y\to Z$, the rate of appearance of $Z$ is $0.05mol{L}^{-1}$ per min. Find the rate of disappearance of $x$ in $mol{L}^{-1}mi{n}^{-1}$

Answer 1

Answer: 0.1

Rate of reaction
$=-\frac{1}{2}\frac{d[X]}{dt}=-\frac{d[Y]}{dt}=+\frac{d[Z]}{dt}$ ____ (i)
$\therefore$ Rate of appearance of
$z=\frac{d[Z]}{dt}=0.05mol{L}^{-1}$ per min
$\therefore$ Rate of disappearance of $x=-\frac{d[X]}{dt}$
From equation (i), we get
$-\frac{1}{2}\frac{d[X]}{dt}=\frac{s[Z]}{dt}$
$\therefore -\frac{d[X]}{dt}=2\times \frac{d[Z]}{dt}$
$=2\times \frac{d[Z]}{dt}=2\times 0.05=0.1mol{L}^{-1}mi{n}^{-1}$