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 rate 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 will be:

• A. $1.2\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}$ and $4.4\times {10}^{4}mi{n}^{-1}$
• B. $1.2\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}$ and $8.8\times {10}^{4}mi{n}^{-1}$
• C. $4.8\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}$ and $2.2\times {10}^{4}mi{n}^{-1}$
• D. $2.4\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}$ and $4.4\times {10}^{4}mi{n}^{-1}$

Answer 1

Answer: (A)

$1.2\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}$ and $4.4\times {10}^{4}mi{n}^{-1}$

2 moles of dinitrogen pentoxide consume to form 4 moles of nitrogen dioxide.
$-\frac{1}{2}\frac{d\left[N_2{O}_5\right]}{dt}=-\frac{1}{4}\frac{d\left[N{O}_2\right]}{dt}$
Thus, the rate of consumption of dinitrogen pentoxide is half of the rate of formation of nitrogen dioxide.
Rate constant only depend upon temperature, so it will remain same.
Thus, the rate of consumption of dinitrogen pentoxide $-\frac{d\left[N_2{O}_5\right]}{dt}=\frac{2.4\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}}{2}=1.2\times {10}^{-4}mole{L}^{-1}mi{n}^{-1}$
Also, the rate constant for consumption of dinitrogen pentoxide $=4.4\times {10}^{4}mi{n}^{-1}$