for the equilibrium: 2 no (g) < > n2(g) + o2 (g), kp=2400. if initially, only no is present at a partial pressure of 37.30 atm, what will the partial pressures of n2 and o2 be at equilibrium? 1827 atm 38.08 atm 1.725 atm 36.55 atm

Explanation:

Hello,

In this case, for the given chemical reaction, the law of mass action turns out:

Whereas the equilibrium pressures, based on the stoichiometry and the change , changes to:

Solving for via quadratic equation, one obtains:

In such a way, the feasible solution is 18.27 atm since the other pressure lead to a negative pressure of NO at the equilibrium, therefore, the equilibrium pressures of nitrogen and oxygen that are equal, result in:

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