Question

The rate constant for the reaction, $2N_2{O}_5\to 4N{O}_2+O_2$, is $3.0\times {10}^{-5}se{c}^{-1}$. If the rate is $2.40\times {10}^{-5}$mol $litr{e}^{-1}se{c}^{-1}$, then the concentration of $N_2{O}_5$ ( in mol $litr{e}^{-1}$) is:

• A. $1.4$
• B. $1.1$
• C. $0.04$
• D. $0.8$

Answer 1

Answer: (D)

$0.8$

The given reaction is:

$2N_2{O}_5⟶4N{O}_2+O_2$

The unit of rate constant indicates its first-order reaction.

$Rate=K\left[N_2{O}_5\right]$ (Rate Law)

where $K=$ Rate constant

Given that,

$K=3\times {10}^{-5}se{c}^{-1}$

$Rate=2.4\times {10}^{-5}mol{L}^{-1}{s}^{-1}$

Putting the given values in rate law:-

$2.4\times {10}^{-5}=3.0\times {10}^{-5}\left[N_2{O}_5\right]$

$\Rightarrow \frac{8}{10}=\left[N_2{O}_5\right]$

$\Rightarrow \left[N_2{O}_5\right]=0.8mol{L}^{-1}$