For the reaction, $\mathrm{A}+3 \mathrm{~B} \longrightarrow 2 \mathrm{C}$ rate of consumption of A is $1.4…

For the reaction, $\mathrm{A}+3 \mathrm{~B} \longrightarrow 2 \mathrm{C}$ rate of consumption of A is $1.4 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$. Calculate rate of formation of C ?
  1. $0.07 \mathrm{moldm}^{-3} \mathrm{sec}^{-1}$
  2. $1.4 \mathrm{moldm}^{-3} \mathrm{sec}^{-1}$
  3. $2.8 \mathrm{moldm}^{-3} \mathrm{sec}^{-1}$
  4. $3.5 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{sec}^{-1}$

Solution

For the reaction, $A+3 B \rightarrow 2 C$
Rate of reaction $=\frac{-\mathrm{d}[\mathrm{~A}]}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{~d}[\mathrm{~B}]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{~d}[\mathrm{C}]}{\mathrm{dt}}$
Hence, $\begin{aligned} \frac{-\mathrm{d}[\mathrm{~A}]}{\mathrm{dt}} & =\frac{1 \mathrm{~d}[\mathrm{C}]}{2 \mathrm{dt}} \\ \therefore \quad \frac{\mathrm{~d}[\mathrm{C}]}{\mathrm{dt}} & =2 \times 1.4 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1} \\ & =2.8 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1} \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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