For a given chemical reaction γ _ 1 A + γ _ 2 B → γ _ 3 C + γ _ 4 D...

\dfrac{d\left[c\right]}{dt} = \left(\dfrac{20 - 10}{10}\right) = 1milimole / {dm}^{3}sec$</p><p>${\gamma}_{1}A + {\gamma}_{2}B \to {\gamma}_{3}C + {\gamma}_{4}D$</p><p>$\dfrac{d\left[D\right]}{dt} = 1 . 5\left(- \dfrac{d}{dt}\left[B\right]\right) = 9milimole / {dm}^{3}sec$</p><p>\left(- \dfrac{d}{dt}\left[B\right]\right) = 2\left(- \dfrac{d}{dt}\left[A\right]\right)$

Rate $= - \dfrac{1}{{\gamma}_{1}}\dfrac{d\left[A\right]}{dt} = - \dfrac{1}{{\gamma}_{2}}\dfrac{d\left[B\right]}{dt} = \dfrac{1}{{\gamma}_{3}}\dfrac{d\left[C\right]}{dt} = \dfrac{1}{{\gamma}_{4}}\dfrac{d\left[D\right]}{dt}$

(i) $\dfrac{d\left[D\right]}{dt} = \dfrac{{\gamma}_{4}}{{\gamma}_{2}}\left[- \dfrac{d\left[B\right]}{dt}\right] = 1 . 5\left(- \dfrac{d}{dt}\left[B\right]\right)$

$\dfrac{{\gamma}_{4}}{{\gamma}_{2}} = 1 . 5$

(ii) $\dfrac{{\gamma}_{2}}{{\gamma}_{1}}\left[- \dfrac{d\left[A\right]}{dt}\right] = \left(- \dfrac{d}{dt}\left[B\right]\right) = 2\left(- \dfrac{d}{dt}\left[A\right]\right)$

$\dfrac{{\gamma}_{2}}{{\gamma}_{1}} = 2$

(iii) $\dfrac{d\left[C\right]}{dt} = \dfrac{{\gamma}_{3}}{{\gamma}_{4}}\left[\dfrac{d\left[D\right]}{dt}\right]$

$1 = \dfrac{{\gamma}_{3}}{{\gamma}_{4}} \times 9$

$\dfrac{{\gamma}_{4}}{{\gamma}_{3}} = 9$

So, if

${\gamma}_{3} = 1than{\gamma}_{4} = 9{\gamma}_{2} = 6and{\gamma}_{1} = 3Rateofreaction = \dfrac{1}{{\gamma}_{3}}\dfrac{d [ C ]}{dt} = 1mmol{dm}^{- 3}{s}^{- 1}$