The rate constant for the reaction; 2N2O5→4NO2+O2 is 3.0×10−5sec−1. If the rate is 2.40×10−5M sec−1, then the concentration of N2O5 (in M) is

r=K[N2O5] for I order as unit of K=sec-12.40×10−5=3.0×10−5[N2O5][N2O5]=2.403.0=0.8M

The rate constant for the reaction; 2N2O5→4NO2+O2 is 3.0×10−5sec−1. If the rate is 2.40×10−5M sec−1, then the concentration of N2O5 (in M) is

r=K[N2O5] for I order as unit of K=sec-12.40×10−5=3.0×10−5[N2O5][N2O5]=2.403.0=0.8M

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The rate constant for the reaction; $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ is $3.0 \times 10^{- 5} s e c^{- 1} .$ If the rate is $2.40 \times 10^{- 5} M s e c^{- 1},$ then the concentration of $N_{2} O_{5}$ (in M) is

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$r = K [N_{2} O_{5}]$ for I order as unit of K=sec^-1
$\begin{array}{l} 2.40 \times 10^{- 5} = 3.0 \times 10^{- 5} [N_{2} O_{5}] \\ [N_{2} O_{5}] = \frac{2.40}{3.0} = 0.8 M \end{array}$

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The rate constant for the reaction; 2N2O5→4NO2+O2 is 3.0×10−5sec−1. If the rate is 2.40×10−5M sec−1, then the concentration of N2O5 (in M) is

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The rate constant for the reaction; $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$ is $3.0 \times 10^{- 5} s e c^{- 1} .$ If the rate is $2.40 \times 10^{- 5} M s e c^{- 1},$ then the concentration of $N_{2} O_{5}$ (in M) is