2 NO(g) + O2 (g) <==> 2 NO2 (g)
Given that Kp=7.96×1012 for the reaction above and the following starting conditions:
Initial Concentrations
p(NO)=0.775 atm
p(NO2)=0.000 atm
p(O2)=0.789 atm
Determine the equilibrium concentration of NO2.
........2NO + O2 ==> 2NO2
I.....0.775.0.789....0
C.......2x....-x.....2x
E..0.775-2x.0.789-x..2x
Substitute the E line into Kp expression and solve for x then evaluate the individual pressures.
I should point out that I did NOT work this problem as concentrations since you gave the pressures in atm for initial conditions but called them concns. I think you must have meant pressures. And you gave a Kp also. With no T listed you can't evaluate Kc.
To determine the equilibrium concentration of NO2, we will first need to calculate the partial pressure of NO2 at equilibrium using the given equilibrium constant (Kp) and the initial concentrations of the reactants.
Step 1: Write the expression for Kp using the given balanced chemical equation:
Kp = (P(NO2))^2 / (P(NO))^2 * (P(O2))
Step 2: Substitute the given values into the expression:
Kp = (P(NO2))^2 / (P(NO))^2 * (P(O2))
7.96×1012 = (P(NO2))^2 / (0.775 atm)^2 * (0.789 atm)
Step 3: Rearrange the equation to solve for (P(NO2))^2:
(P(NO2))^2 = Kp * (P(NO))^2 * (P(O2)) / 1
Step 4: Substitute the given values into the equation and solve for (P(NO2))^2:
(P(NO2))^2 = 7.96×1012 * (0.775 atm)^2 * (0.789 atm)
(P(NO2))^2 = 4.519013 atm^2
Step 5: Take the square root of both sides to find the equilibrium partial pressure of NO2:
P(NO2) = √(4.519013 atm^2)
P(NO2) ≈ 2.125 atm
Therefore, the equilibrium partial pressure of NO2 is approximately 2.125 atm.