For the reaction N2O5→2NO2+12O2 Given −dN2O5dt=K1N2O5,dNO2dt=K2N2O5and dO2dt=K3N2O5.The relation in between K1,K2,K3is : -

N2O5→ 2NO2+12O2−dN2O5dt=12dNO2dt=2dO2dt⇒K1N2O5=12K2N2O5=2K3N2O5⇒2K1=K2=4K3

For the reaction N2O5→2NO2+12O2 Given −dN2O5dt=K1N2O5,dNO2dt=K2N2O5and dO2dt=K3N2O5.The relation in between K1,K2,K3is : -

N2O5→ 2NO2+12O2−dN2O5dt=12dNO2dt=2dO2dt⇒K1N2O5=12K2N2O5=2K3N2O5⇒2K1=K2=4K3

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For the reaction $N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}$ Given $\frac{- d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}], \frac{d [N O_{2}]}{d t} = K_{2} [N_{2} O_{5}]$ and $\frac{d [O_{2}]}{d t} = K_{3} [N_{2} O_{5}] .$ The relation in between $K_{1}, K_{2}, K_{3}$ is : -

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$2 K_{1} = K_{2} = 4 K_{3}$

$K_{1} = K_{2} = K_{3}$

$2 K_{1} = 4 K_{2} = K_{3}$

$K_{1} = 2 K_{2} = K_{3}$

answer is A.

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Detailed Solution

$N_{2} O_{5} \overset{}{\to} 2 N O_{2} + \frac{1}{2} O_{2}$
$\begin{array}{l} \frac{- d [N_{2} O_{5}]}{d t} = \frac{1}{2} \frac{d [N O_{2}]}{d t} = 2 \frac{d [O_{2}]}{d t} \\ \Rightarrow K_{1} [N_{2} O_{5}] = \frac{1}{2} K_{2} [N_{2} O_{5}] = 2 K_{3} [N_{2} O_{5}] \\ \Rightarrow 2 K_{1} = K_{2} = 4 K_{3} \end{array}$