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Consider the reversible reaction, $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g);$ At equilibrium, number of moles of reactants is equal to that of products. Degree of decomposition of $N_{2} O_{4}$ is

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0.5

0.4

0.33

0.66

answer is C.

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Detailed Solution

$N_{2} O_{4} ⇌ 2 {NO}_{2} 1 - 1 - x 2 x Given, 1 - x = 2 x x = 0.33$

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