Question

The rate of the reaction $2N_2{O}_5\to 4N{O}_2+O_2$ can be written in three ways:

$\frac{-d\left[N_2{O}_5\right]}{dt}=k\left[N_2{O}_5\right]$

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

$\frac{d\left[O_2\right]}{dt}=k^{\prime \prime }\left[N_2{O}_5\right]$

The relationship between $k$ and $k^{\prime }$ are:

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

Answer 1

Answer: (C)

$k^{\prime }=2k,k^{\prime \prime }=k/2$

${2N}_2{O}_5⟶{4NO}_2+O_2$

Applying rate law.

$\frac{-1}{2}\frac{d\left[N_2{O}_5\right]}{dt}=\frac{1}{4}\frac{d\left[{NO}_2\right]}{dt}=\frac{d\left[O_2\right]}{dt}=$ Rate of Reaction.

Using rate law.

$\left(1\right)\frac{1}{2}K\left[N_2{O}_5\right]=\frac{1}{4}{K}^{\prime }\left[N_2{O}_5\right]$

$\Rightarrow K^{\prime }=2K$

$\left(2\right)\frac{1}{2}K\left[N_2{O}_5\right]=K^{\prime \prime }\left[N_2{O}_5\right]$

$\Rightarrow K^{\prime \prime }=\frac{K}{2}$