For the reaction 2N2O5→4NO2+O2,rate and rate constant are 1.02×10-4 mol L-1s-1 and 3.4×10-5 s-1,respectively. The concentration of N2O5 at that time will be

Rate=KN2O51; 1.02×10-4=3.4×10-5×N2O5; N2O5=3

For the reaction 2N2O5→4NO2+O2,rate and rate constant are 1.02×10-4 mol L-1s-1 and 3.4×10-5 s-1,respectively. The concentration of N2O5 at that time will be

Rate=KN2O51; 1.02×10-4=3.4×10-5×N2O5; N2O5=3

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For the reaction $2 N_{2} O_{5} \to 4 {NO}_{2} + O_{2}, rate and rate constant are 1.02 \times 10^{- 4} mol L^{- 1} s^{- 1}$ and $3.4 \times 10^{- 5} s^{- 1}$ ,respectively. The concentration of $N_{2} O_{5}$ at that time will be

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1.73

$3.4 \times 10^{5}$

$1.02 \times 10^{- 4}$

answer is B.

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$Rate = K {[N_{2} O_{5}]}^{1}; 1.02 \times 10^{- 4} = 3.4 \times 10^{- 5} \times [N_{2} O_{5}]; [N_{2} O_{5}] = 3$

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For the reaction 2N2O5→4NO2+O2,rate and rate constant are 1.02×10-4 mol L-1s-1 and 3.4×10-5 s-1,respectively. The concentration of N2O5 at that time will be

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For the reaction $2 N_{2} O_{5} \to 4 {NO}_{2} + O_{2}, rate and rate constant are 1.02 \times 10^{- 4} mol L^{- 1} s^{- 1}$ and $3.4 \times 10^{- 5} s^{- 1}$ ,respectively. The concentration of $N_{2} O_{5}$ at that time will be