For the gaseous reaction, N₂ + 3H₂ ↔ 2NH₃ the partial pressures of H₂ and N₂ are 0.4 and 0.8 atm respectively. The total pressure of the entire system is 2.8 atm. What will be the value of K_p if all the concentrations are given in atmospheres??

N₂ + 3H₂ ↔ 2NH₃

partial pressures of H₂ and N₂ are 0.4 and 0.8 atm

so partial pressure of 2NH_{3  = 1.6 atm}

K_p= [NH3] ^2 / ( [N2] [ H2]^3 )

= 1.6 ^ 2 / 0.8 * 0.4 ^3

= 2.56 / 0.0512

= 50

Please approve !

if partial pressures tht u hav givn fr h2 nd n2 r at equilibrium thn...

presure of nh3=1.6

so kp=(1.6)^2/0.8*(0.4)^3

N₂      +      3H₂     =>    2NH3

0.4            0.8                  P                 (at equilibrium)

sum of all pressure = P(total) = 1.2 + p = P_T

  P_T  = 2.8 (given)  so

 1.2 + P = 2.8

         p_NH3 = 1.6

now K_p =  ( p_NH3)² /(P_N2)(p_H2)³

           = (1.6)²/(0.4)(0.8)³

          K_p =  12.5 (atm)^-2

thnx to al......