13.8g of N2O4 was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium N2O4(g) ≤==≥ 2NO2(g).The total pressure at equilibrium was found to be 9.15 bar.Calculate Kc,Kp and partial pressure at equilibrium.

13.8g of N2O4 was placed in a 1L reaction vessel at 400K and allowed t

1.	13.8g of N2O4 was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium N2O4(g) ≤==≥ 2NO2(g).The total pressure at equilibrium was found to be 9.15 bar.Calculate Kc,Kp and partial pressure at equilibrium.
Answer» - 0.025 atmSolution:- The given balanced equilibrium EQUATION is:2NO_2(g)\leftrightarrow 2NO(g)+O_2(g)2NO2(g)↔2NO(g)+O2(g)From the equation, there is 2:1 mole ratio between NO and O_2O2 .Their equilibrium partial pressure would be in the ratio. Equilibrium partial pressure of oxygen is given as 0.25 ATM. So, the equilibrium partial pressure of NO = 2(0.25 atm) = 0.50 atmEquilibrium expression for the given equation is written as:Let's plug in the VALUES in it and solve it for equilibrium partial pressure of NO_2NO2 . Let's SAY the equilibrium partial pressure of nitrogen dioxide is p.100=\frac{(0.50)^2(0.25))}{p^2}100=p2(0.50)2(0.25))100=\frac{0.0625}{p^2}100=p20.0625On rearranging the above equation:p^2=\frac{0.0625}{100}p2=1000.0625On taking square root to both sides:p=\frac{0.25}{10}p=100.25p = 0.025 atmSo, the equilibrium partial pressure of NO_2NO2 is 0.025 atm.