Question

In the reaction of $2NO+O_2\to 2{NO}_2$, if the rate of disappearance of $O_2$ is $16gm$ ${min}^{-1}$, then the rate of apperance of ${NO}_2$ is:

• A. $90gm{min}^{-1}$
• B. $46gm{min}^{-1}$
• C. $28gm{min}^{-1}$
• D. $32gm{min}^{-1}$

Answer 1

The correct option is A $90gm{min}^{-1}$

$2NO+O_2\to 2N{O}_2$

$-\frac{1}{2}\frac{d\left[NO\right]}{dt}=\frac{-d\left[O_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}$

$\frac{-d\left[O_2\right]}{dt}=16ℊ{min}^{-1}=\frac{16}{16}mol{min}^{-1}=1mol{min}^{-1}$

$\frac{d\left[N{O}_2\right]}{dt}=2\times \frac{-d\left[O_2\right]}{dt}=2\times 1=2mol/min$

No. of moles$=\frac{weight}{mol.wt}$

$2\times$Mol. wt of $N{O}_2=$wt

$2\times 46=$ wt $\Rightarrow$Weight of $N{O}_2=92ℊ$

Rate of appearance of $N{O}_2=92ℊ/min\approx 90ℊ/min$

Therefore, option $A$ is correct.