Question

For the reaction; $N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$,

Given:

$-\frac{d\left[N_2{O}_5\right]}{dt}=K_1\left[N_2{O}_5\right],\frac{d\left[N{O}_2\right]}{dt}=K_2\left[N_2{O}_5\right],\frac{d\left[O_2\right]}{dt}=K_3\left[N_2{O}_5\right]$.
The relation between $K_1,K_2$ and $K_3$ is:

• A. $2K_1=K_2=4K_3$
• B. $K_1=K_2=K_3$
• C. $2K_1=4K_2=K_3$
• D. none of the above

Answer 1

Answer: (A)

$2K_1=K_2=4K_3$

Given that :-

$N_2{O}_5\leftrightarrow 2N{O}_2+\frac{1}{2}{O}_2$.........(i)

$\frac{-d\left[N_2{O}_5\right]}{dt}=K_1\left[N_2{O}_5\right]$........(ii)

$\frac{d\left[N{O}_2\right]}{dt}=K_2\left[N_2{O}_5\right]$.........(iii)

$\frac{d\left[O_2\right]}{dt}=K_3\left[N_2{O}_5\right]$.......(iv)

From equation (1)

$\frac{-d\left[N_2{O}_5\right]}{dt}=\frac{1}{2}\frac{d\left[N{O}_2\right]}{dt}=\frac{2d\left[O_2\right]}{dt}$

From (1), (iii) and (iv)

$K_1\left[N_2{O}_5\right]=\frac{1}{2}{K}_2\left[N_2{O}_5\right]=2K_3\left[N_2{O}_5\right]$

Multiply by 2

$2K_1\left[N_2{O}_5\right]=K_2\left[N_2{O}_5\right]=4K_3\left[N_2{O}_5\right]$

divided by $N_2{O}_5$

$2K_1=K_1=4K_3$.