Question

For the given reaction $2N_{2}O_{5} \rightarrow 4NO_{2}+O_{2}$ , the rate of reaction and rate constant values are $1.02\times 10^{- 4}molL^{- 1 }\sec^{- 1}$ and $3.4\times 10^{- 5}\sec^{- 1}$ respectively. Then calculate the concentration of $N_{2}O_{5}$ at that time

• A. $1.732\,M$
• B. $3\,M$
• C. $1.02\times 10^{- 4}\,M$
• D. $3.4\times 10^{5}\,M$

Answer 1

Answer: (B)

The unit of rate constant suggest that it is a first order reaction.
Hence the given rate law is
$R=K\left[\right.A\left]\right.$ , $\text{1.02}\times \text{1}\text{0}^{- 4}$
$=\text{3.4}\times \text{1}\text{0}^{- 5}\times \left[\right.N_{2}O_{5}\left]\right.$
or $\left(N_{2} O_{5}\right)=\frac{1 .02 \times \left(10\right)^{- 4}}{3 .4 \times \left(10\right)^{- 5}}=3M$