The equilibrium constant for the reaction N_{2}O_{4} <-> 2NO_{2} is 0.113. If one were to start with 1 mole of N_{2}O_{4} at 1 atm pressure and 298 K, what fraction of it would decompose if pressure and temperature are both maintained at 1 atm and 298 K?

By doing a simple equilibrium calculation:

0.113 = (P_{NO2})^{2} / (P_{N2O4})

= (2x)^{2} / (1 - x)

I get a value of 0.155 for x.

However, the solution my instructor gives:

0.113 = (P_{NO2})^{2} / (P_{N2O4})

= (2x / 1 + x)^{2} / ( (1 - x) / (1 + x))

Where (1 + x) represents the total pressure of the system at equilibirum, results in an x value of 0.166. Which solution is correct and why?