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Chemistry Question on Rate of a Chemical Reaction

For the reaction 2N2O5→4NO2+O22N_2O_5 \to 4NO_2 + O_2, rate and rate constant are 1.02×10−41.02 \times 10^{-4} mol lir−1lir^{-1} sec−1sec^{-1} and 3.4×10−53.4 \times 10^{-5} sec−1sec^{-1} respectively then concentration of N2O5N_2O_5 at that time will be

A

1.732 M

B

3 M

C

3.4×10−5M3.4\times 10^{-5}M

D

1.02×10−4M1.02 \times 10^{-4}M

Answer

3 M

Explanation

Solution

2N2O5→4NO2+O22N_2O_5 \to 4NO_2 + O_2
from the imit of rate constant it is clear that the reaction follow first order kinetics.
Hence
by rate law equation, r=k[N2O5]r = k [N_2O_5]
wherer =1.02×10−4,k=3.4×10−5= 1.02 \times 10^{-4}, k= 3.4 \times 10^{-5}
1.02×10−4=3.4×[N2O5]1.02 \times 10^{-4} = 3.4 \times \left[N_{2}O_{5}\right]
[N2O5]=3M\left[N_{2}O_{5}\right] = 3M