For the reaction 2N2O5→4NO2+O2rate and rate constant are 1.02×10−4mollit−1sec−1 and 3.4×10−5sec−1respectively then concentration of N2O5at that time will be

A
1.732 M
B
3M
C
$3 .4 \times 10^{5} M$
D
$1 .02 \times 10^{- 4} M$

Solution:

$2 N_{2} O_{5} \to 4 {NO}_{2} + O_{2}$

From the unit of rate constant it is clear that the reaction follow first order kinetics. Hence by rate law equation, $r=k [N_{2} O_{5}]$
where $r = 1 .02 \times 10^{- 4}, k = 3 .4 \times 10^{- 5}$
${1.02×10}^{- 4} = 3 .4 \times 10^{- 5} [N_{2} O_{5}] [N_{2} O_{5}] = 3 M$

For the reaction $2 N_{2} O_{5} \to 4 {NO}_{2} + O_{2}$ rate and rate constant are ${1.02×10}^{- 4} mol {lit}^{- 1} \sec^{- 1} a n d 3 .4 \times 10^{- 5} \sec^{- 1}$ respectively then concentration of $N_{2} O_{5}$ at that time will be

Solution:

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