Question

For the reaction, $2N_2{O}_5\to 4N{O}_2+O_2$, the instantaneous rate of reaction can be expressed in terms of time and concentrations by the expression:

• A. $Rate=\frac{d\left[N_2{O}_5\right]}{dt}=-\frac{1}{4}\frac{d\left[N{O}_2\right]}{dt}=\frac{1}{2}\frac{d\left[O_2\right]}{dt}$
• B. $Rate=-\frac{1}{2}\frac{d\left[N_2{O}_5\right]}{dt}=\frac{1}{4}\frac{d\left[N{O}_2\right]}{dt}=\frac{d\left[O_2\right]}{dt}$
• C. $Rate=-\frac{1}{4}\frac{d\left[N_2{O}_5\right]}{dt}=\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=\frac{d\left[O_2\right]}{dt}$
• D. $Rate=-\frac{1}{2}\frac{d\left[N_2{O}_5\right]}{dt}=\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=\frac{1}{2}\frac{d\left[O_2\right]}{dt}$

Answer 1

The correct option is B $Rate=-\frac{1}{2}\frac{d\left[N_2{O}_5\right]}{dt}=\frac{1}{4}\frac{d\left[N{O}_2\right]}{dt}=\frac{d\left[O_2\right]}{dt}$

$2N_2{O}_5\to 4N{O}_2+O_2$
Rate of disappearance = Rate of appearance
$Rate=-\frac{1}{2}\frac{d\left[N_2{O}_5\right]}{dt}=+\frac{1}{4}\frac{d\left[N{O}_2\right]}{dt}=+\frac{d\left[O_2\right]}{dt}$
Option B is the correct answer.