Question

For the reaction, $2N_2{O}_5\to 4N{O}_2+O_2$, 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

$+\frac{d\left[N{O}_2\right]}{dt}=k^{\prime }\left[N_2{O}_5\right]$.

Here, $k$ and $k^{\prime }$ are related as:

• A. $k$ $=$ $4k^{\prime }$
• B. $k=2k^{\prime }$
• C. $2k=k^{\prime }$
• D. $k=k^{\prime }$

Answer 1

The correct option is C $2k=k^{\prime }$

From the given reaction, $2N_2{O}_5\to 4N{O}_2+O_2$
$-\frac{d\left[N_2{O}_5\right]}{2dt}=+\frac{d\left[N{O}_2\right]}{4dt}$........(1)
$-\frac{d\left[N_2{O}_5\right]}{dt}=k\left[N_2{O}_5\right]$........(2)
$+\frac{d\left[N{O}_2\right]}{dt}=k^{\prime }\left[N_2{O}_5\right]$........(3)
From eq (1), (2) and (3), we get,
$-\frac{d\left[N_2{O}_5\right]}{2dt}=\frac{k\left[N_2{O}_5\right]}{2}=+\frac{d\left[N{O}_2\right]}{4dt}=\frac{k^{\prime }\left[N_2{O}_5\right]}{4}$
$=>\frac{k}{2}=\frac{k^{\prime }}{4}$
$=>2k=k^{\prime }$