Question

In the reaction $2N_2{O}_5\to 4N{O}_2+O_2$, $+\frac{d\left[N{O}_2\right]}{dt}$ at any time t was found to be $2.4\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}$ with the constant $4.4\times {10}^{-4}mi{n}^{-1}$. Hence $-\frac{d\left[N_2{O}_5\right]}{dt}$ at the same time t and the corresponding rate constant of the reactions respectively would be

• A. $1.2\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}and2.2\times {10}^{-4}mi{n}^{-1}$
• B. $1.2\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}and8.8\times {10}^{-4}mi{n}^{-1}$
• C. $4.8\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}and2.2\times {10}^{-4}mi{n}^{-1}$
• D. $2.4\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}and4.4\times {10}^{-4}mi{n}^{-1}$

Answer 1

The correct option is A $1.2\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}and2.2\times {10}^{-4}mi{n}^{-1}$

$-\frac{1}{2}\frac{d\left[N_2{O}_5\right]}{dt}=-\frac{1}{4}\frac{d\left[N{O}_2\right]}{dt}$
For 4 moles of $N{O}_2$ formed, 2 moles of $N_2{O}_5$ consumed in a given time interval, so the rate consumption of $N_2{O}_5$ must be half of the rate of formation of $N{O}_2$, Rate constants w.r.t. these two rates will also have the same relationship.