For the following equilibrium, $ N_{2} O_{4} \rightleftharpoons $ $ 2 N O_{2} $ in gaseous phase, $ N O_{2} $ is $ 50 \% $ of the total volume when equilibrium is set up. Hence, percent of dissociation of $ N_{2} O_{4} $ is:

Question

Accepted Answer

$ 66.66 \% $. If 1 mole of N2O4 dissociates by x, the equilibrium amounts become N2O4 = 1−x and NO2 = 2x, so total moles = 1+x. Since NO2 is 50% of the mixture, 2x/(1+x)=1/2. Solving gives x=2/3, so the dissociation is 66.66%.

For the following equilibrium, \( N_{2} O_{4} \rightleftharpoons \) \( 2 N O_{2} \) in gaseous phase, \( N O_{2} \) is \( 50 \% \) of the total volume when equilibrium is set up. Hence, percent of dissociation of \( N_{2} O_{4} \) is:

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