Question

For $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2,$ it is found that $\frac{-d}{dt}\left[N_2{O}_5\right]=k_1\left[N_2{O}_5\right],\frac{d}{dt}\left[N{O}_2\right]=k_2\left[N_2{O}_5\right]and$ $\frac{d}{dt}\left[O_2\right]=k_3\left[N_2{O}_3\right],$ then:

• A. $k_1=2k_2=3k_3$
• B. $2k_1=4k_2=k_3$
• C. $2k_1=k_2=4k_3$
• D. $k_1=k_2=k_3$

Answer 1

Answer: (C)

$2k_1=k_2=4k_3$

$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$

By rate law:

Rate of reaction $\Rightarrow \frac{-1d\left(N_2{O}_3\right)}{dt}=\frac{+1}{2}\frac{d\left(N{O}_2\right)}{dt}-\frac{+1}{\left(\frac{1}{2}\right)}\frac{d\left(O_2\right)}{dt}....\left(i\right)$

Now given that

$\frac{-d}{dt}\left[N_2{O}_5\right]=k_1\left[N_2{O}_5\right],\frac{d}{dt}\left[N{O}_2\right]=k_2\left[N_2{O}_5\right]and$ $\frac{d}{dt}\left[O_2\right]=k_3\left[N_2{O}_3\right].$

Put all the values in equation (i)

We get,

$\frac{-d\left(N_2{O}_3\right)}{dt}=\frac{+1}{2}\frac{d\left(N{O}_2\right)}{dt}=2\frac{d\left(O_2\right)}{dt}$

$k_1\left[N_2{O}_5\right]=\frac{1}{2}{k}_2\left[N_2{O}_5\right]=2\times k_2\left[N_2{O}_5\right]$

$k_1=\frac{1}{2}{k}_2=2k_3$

Multiply by 2

$2k_1=k_2=4k_3$