Question

For the reaction $3A\left(g\right)\stackrel{k}{\to }B\left(g\right)+C\left(g\right)$, k is ${10}^{-4}Lmo{l}^{-1}mi{n}^{-1}$. If [A] = 0.5 M then the value of $\frac{-d\left[A\right]}{\mathrm{dt}}$ (in $M{s}^{-1}$) is

• A. $7.5\times {10}^{-5}$
• B. $3\times {10}^{-4}$
• C. $2.5\times {10}^{-5}$
• D. $1.25\times {10}^{-6}$

Answer 1

The correct option is D $1.25\times {10}^{-6}$

$\mathrm{Since},k={10}^{-14}L/\mathrm{molmin},\mathrm{the}\mathrm{reaction}\mathrm{will}\mathrm{follow}{2}^{\mathrm{nd}}\mathrm{order}\mathrm{kinetics}.-\frac{1}{3}\frac{d\left[A\right]}{\mathrm{dt}}=k[A{]}^2-\frac{d\left[A\right]}{\mathrm{dt}}=3\times {10}^{-4}\times (0.5{)}^2=7.5\times {10}^{-5}{\mathrm{Mmin}}^{-1}=\frac{7.5\times {10}^{-5}}{60}{\mathrm{Ms}}^{-1}=1.25\times {10}^{-6}{\mathrm{Ms}}^{-1}$