For a given chemical reaction $γ_{1} A + γ_{2} B \to γ_{3} C + γ_{4} D$
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)

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Detailed Solution

$γ_{1} A + γ_{2} B ⟶ γ_{3} C + γ_{4} D$

$Given : + \frac{d [D]}{dt} = \frac{- 3}{2} \frac{d [B]}{dt}$

$\begin{aligned} \Rightarrow \frac{- 1}{2} \frac{d [B]}{dt} = \frac{+ 1}{3} \frac{d [D]}{dt} \\ - \frac{d [B]}{dt} = - 2 \frac{d [A]}{dt} \Rightarrow - \frac{1}{2} \frac{d [B]}{dt} = \frac{- d (A)}{dt} \end{aligned}$

$+ \frac{d [B]}{dt} = 9 {mmoldm}^{- 3} s^{- 1}$

$\frac{+ d [C]}{dt} = \frac{20 - 10}{10} = 1 {mmoldm}^{- 3} s^{- 1}$

$\frac{+ d [C]}{dt} = \frac{1}{9} \times \frac{+ d [D]}{dt}$

$\begin{aligned} 1 A + 2 B ⟶ \frac{1}{3} C + 3 D \\ 3 A + 6 B ⟶ C + 9 D \end{aligned}$

$Rate of reaction = \frac{+ d [C]}{dt} = 1 {mmoldm}^{- 3} s^{- 1}$

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