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  • For a given chemical reaction <img alt="\gamma_{1} \mathrm{~A}+\gamma_{2} \mathrm{~B} \rightarrow \gamma_{3} \mathrm{C}+\gamma_{4} \mathrm{D}" src="https://entrancecorner.oncodecogs.com/g

For a given chemical reaction

\gamma_{1} \mathrm{~A}+\gamma_{2} \mathrm{~B} \rightarrow \gamma_{3} \mathrm{C}+\gamma_{4} \mathrm{D}

Concentration of C changes from \mathrm{10 \; mmol\; dm^{-3} } to \mathrm{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 \mathrm{9\; mmol\; dm^{-3}\; s^{-1}}. Therefore the rate of reaction is _________\mathrm{mmol} \; \mathrm{d \textrm {m } ^ { - 3 }\; \mathrm { s } ^ { - 1 }}.

(Nearest Integer)

Option: 1

1


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\frac{\Delta[C]}{\Delta t}=1 \mathrm{mmol}\; \mathrm{dm}^{-3} \mathrm{~s}^{-1}

Also,

\begin{aligned} & \frac{\Delta[D]}{\Delta t}=-1.5 \frac{\Delta[B]}{\Delta t} \\ \\&\text { and } \frac{\Delta[B]}{\Delta t}=-2 \frac{\Delta[A]}{\Delta t} \\ \\&\text { and } \frac{\Delta[D]}{\Delta t}=9 \frac{\Delta[C]}{\Delta t} \end{aligned}

\begin{aligned} &\therefore-3 \frac{\Delta[A]}{\Delta t}=-1.5 \frac{\Delta[B]}{\Delta t}=9 \frac{\Delta[C]}{\Delta t}=\frac{\Delta[D]}{\Delta t} \\ \\&\Rightarrow \frac{-\Delta[A]}{3 \Delta t}=-\frac{\Delta[B]}{6 \Delta t}=\frac{\Delta[C]}{\Delta t}=\frac{\Delta[D]}{9 \Delta t} \end{aligned}

Thus, the reaction is

\begin{aligned} &\mathrm{3 A+6 B \rightarrow C+9 D}\\ &\therefore \text { Rate }=\frac{\Delta[C]}{\Delta t}=1 \end{aligned}

Hence, the answer is 1

Posted by

Kshitij

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