Question

For the reaction, $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$, consider the equations, $-\frac{d}{dt}\left[N_2{O}_5\right]=R_1\left[N_2{O}_5\right]$, $\frac{d}{dt}\left[N{O}_2\right]=R_2\left[N_2{O}_5\right]$ and $\frac{d}{dt}\left[O_2\right]=R_3\left[N_2{O}_5\right]$.
The relation between $R_1,R_2$ and $R_3$ is:

• A. $2R_1=R_2=4R_3$
• B. $R_1=R_2=R_3$
• C. $2R_1=4R_2=R_3$
• D. $R_1=4R_2=R_3$

Answer 1

The correct option is A $2R_1=R_2=4R_3$

The expression for the rate of the reaction is $-\frac{d\left[N_2{O}_5\right]}{dt}=\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=2\times \frac{d\left[O_2\right]}{dt}$ ......(1)
But, $-\frac{d\left[N_2{O}_5\right]}{dt}=R_1\left[N_2{O}_5\right]$ .....(2)

$\frac{d\left[N{O}_2\right]}{dt}=R_2\left[N_2{O}_5\right]$ ......(3)
$\frac{d\left[O_2\right]}{dt}=R_3\left[N_2{O}_5\right]$ ......(4)
By substituting values of equations $\left(2\right),\left(3\right)$ and $\left(4\right)$ in equation (1), we get
$R_1\left[N_2{O}_5\right]=\frac{1}{2}{R}_2\left[N_2{O}_5\right]=2\times R_3\left[N_2{O}_5\right]$
$2R_1=R_2=4R_3$