Question

For a given chemical reaction
${\gamma }_{1}A+{\gamma }_{2}B\to {\gamma }_{3}C+{\gamma }_{4}D$
Concentration of $C$ changes from $10mmol$ $d{m}^{-3}$ to $20mmold{m}^{-3}$ in $10$ seconds. Rate of appearance of $D$ is $1.5$ times the rate of disappearance of $B$ which is twice the rate of disappearance $A$. The rate of appearance of $D$ has been experimentally determined to be $9mmol$ $d{m}^{-3}{s}^{-1}$. Therefore the rate of reaction is _____mmol $d{m}^{-3}{s}^{-1}$. (Nearest Integer)

Answer 1

Answer: 1

${\gamma }_{1}A+{\gamma }_{2}B⟶{\gamma }_{3}C+{\gamma }_{4}D$
Given: $+\frac{d[D]}{dt}=\frac{-3}{2}\frac{d[B]}{dt}$
$\Rightarrow \frac{-1}{2}\frac{d[B]}{dt}=\frac{+1}{3}\frac{d[D]}{dt}$
$-\frac{d[B]}{dt}=-2\frac{d[A]}{dt}\Rightarrow -\frac{1}{2}\frac{d[B]}{dt}=\frac{-d(A)}{dt}$
$+\frac{d[B]}{dt}=9mmold{m}^{-3}{s}^{-1}$
$\frac{+d[C]}{dt}=\frac{20-10}{10}=1mmold{m}^{-3}{s}^{-1}$
$\frac{+d[C]}{dt}=\frac{1}{9}\times \frac{+d[D]}{dt}$
$\Rightarrow 3A+6B⟶+\frac{1}{3}C+3D$
Rate of reaction $=\frac{+d[C]}{dt}=1mmold{m}^{-3}{s}^{-1}$