In a reversible reaction, $2 {NO}_{2} ⇌ N_{2} O_{4}$ the rate of disappearance of NO₂is equal to

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$\frac{2 K_{1} {[{NO}_{2}]}^{2}}{K_{2}}$

$(2 K_{1} - K_{2}) [{NO}_{2}]$

$2 K_{1} {[{NO}_{2}]}^{2} - K_{2} [N_{2} O_{4}]$

$K_{1} {[{NO}_{2}]}^{2} - 2 K_{1} [N_{2} O_{4}]$

answer is C.

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Detailed Solution

NO₂ is consumed in forward reaction and is formed in backward reaction

rate of disappearance of NO₂ in forward reaction :

$- \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = K_{1} {[{NO}_{2}]}^{2}$

rate of formation of NO2 in backward reaction

$\frac{d [{NO}_{2}]}{dt} = K_{2} [N_{2} O_{4}]$

net rate of disappearance of NO₂

$- \frac{d [{NO}_{2}]}{dt} = 2 K_{1} {[{NO}_{2}]}^{2} - K_{2} [N_{2} O_{4}]$

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