For , it is found that , ; , then

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

For , it is found that , ; , then

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

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For $large {N_2}{O_5} to 2N{O_2} + frac{1}{2}{O_2}$ , it is found that $large - frac{d}{{dt}}left[ {{N_2}{O_5}} right], = ,{k_1}left[ {{N_2}{O_5}} right]$ , $large [frac{d}{{dt}}left[ {N{O_2}} right], = ,{k_2}left[ {{N_2}{O_5}} right]]$ ; $large frac{d}{{dt}}left[ {{O_2}} right], = ,{k_3}left[ {{N_2}{O_5}} right]$ , then

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k₁ = 2k₂ = 3 k₃

2k₁= 4k₂ = k₃

2k₁ = k₂= 4k₃

k₁= k₂= k₃

answer is C.

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Detailed Solution

$N_{2} O_{5} \to 2 {NO}_{2} + \frac{1}{2} O_{2} - \frac{d [N_{2} O_{5}]}{dt} = \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = 2 \frac{d [O_{2}]}{dt} K_{1} [N_{2} O_{5}] = \frac{1}{2} K_{2} [N_{2} O_{5}] = 2 K_{3} [N_{2} O_{5}] K_{1} = \frac{1}{2} K_{2} = 2 K_{3} 2 K_{1} = K_{2} = 4 K_{3}$