Question

For a given chemical reaction
$\gamma_{1} A +\gamma_{2} B \rightarrow \gamma_{3} C +\gamma_{4} D$
Concentration of $C$ changes from $10\, mmol$ $dm ^{-3}$ to $20\, mmol dm ^{-3}$ 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 $9 mmol$ $dm ^{-3} s ^{-1}$. Therefore the rate of reaction is _____mmol $dm ^{-3} s ^{-1}$. (Nearest Integer)

Answer 1

$\gamma_{1} A +\gamma_{2} B \longrightarrow \gamma_{3} C +\gamma_{4} D$
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 } $
$+\frac{ d [ B ]}{ dt }=9\, m\,mol\,dm ^{-3} s ^{-1}$
$\frac{+ d [ C ]}{ dt }=\frac{20-10}{10}=1 \,m\,mol \,dm ^{-3} s ^{-1} $
$\frac{+ d [ C ]}{ dt }=\frac{1}{9} \times \frac{+ d [ D ]}{ dt }$
$\Rightarrow 3 A +6 B \longrightarrow +\frac{1}{3} C +3 D$
Rate of reaction $=\frac{+ d [ C ]}{ dt }=1 \,m\,mol \,dm ^{-3} s ^{-1}$