For the given reaction 2N2O5 arrow 4NO2+O2 , the rate of reaction and rate constant values are 1.02× 10- 4molL- 1 sec- 1 and 3.4× 10- 5 sec- 1 respectively. Then calculate the concentration of N2O5 at that time

Question

For the given reaction 2N2O5 arrow 4NO2+O2 , the rate of reaction and rate constant values are 1.02× 10- 4molL- 1 sec- 1 and 3.4× 10- 5 sec- 1 respectively. Then calculate the concentration of N2O5 at that time (A) 1.732 M (B) 3 M (C) 1.02× 10- 4 M (D) 3.4× 105 M

Tardigrade Admin · Accepted Answer

The unit of rate constant suggest that it is a first order reaction.
Hence the given rate law is

R = K [A]

,

1.02 \times 1 0^{- 4}

= 3.4 \times 1 0^{- 5} \times [N_{2} O_{5}]

or

(N_{2} O_{5}) = \frac{1.02 \times ( 10 ) ^{- 4}}{3.4 \times ( 10 ) ^{- 5}} = 3 M

Q. For the given reaction $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ , the rate of reaction and rate constant values are $1.02 \times 1 0^{- 4} m o l L^{- 1} sec^{- 1}$ and $3.4 \times 1 0^{- 5} sec^{- 1}$ respectively. Then calculate the concentration of $N_{2} O_{5}$ at that time

$1.732 M$

$3 M$

$1.02 \times 1 0^{- 4} M$

$3.4 \times 1 0^{5} M$

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

Q. For the given reaction 2N2​O5​→4NO2​+O2​ , the rate of reaction and rate constant values are 1.02×10−4molL−1sec−1 and 3.4×10−5sec−1 respectively. Then calculate the concentration of N2​O5​ at that time

1.732M

3M

1.02×10−4M

3.4×105M