Question

For the reaction, $2N_{2}O_{5} \to 4NO_{2} + O_{2}$, the rate equation can be expressed in two ways $- \frac{d\left[N_{2}O_{5}\right]}{dt} = k \left[N_{2}O_{5}\right]$ and $+\frac{d\left[NO_{2}\right]}{dt} = k' \left[N_{2}O_{5}\right]$
k and k' are related as :

• A. k = k'
• B. 2k = k'
• C. k = 2k'
• D. k = 4k'

Answer 1

Answer: (B)

$2N_{2}O_{5} \to4NO_{2}+O_{2}$

$-\frac{d}{d t} \left[N_{2}O_{5}\right]=k\left[N_{2}O_{5}\right]$

Now

$\Rightarrow \quad-\frac{1}{2} \frac{d}{d t} \left[N_{2}O_{5}\right]=\frac{1}{4} \times K^{1} \left[N_{2}O_{5}\right]$

$\Rightarrow 2k=k'$