Question

the decomposition of amonia is shown below. if kp is 1.5x10³ at 400 C, what is the partial pressure of amonia when the partial pressure of H₂ is 0.15 atm and N₂ is 0.10 atm at equilibrium?

2 NH3(g)<____> N2 (g) + 3 H2 (g)

The correct answer is 4.47x10^-4 atm please and thank you

Answer 1

                               2 NH_3(g) N_{2 (g)} + 3 H_{2 (g)}

Given that the equilibrium partial pressure of H₂ is, p_H2(g) = 0.15 atm

                 the equilibrium partial pressure of N₂ is, p_N2(g) = 0.10 atm

                   Equilibrium constant , K_p = 1.5 x 10 ³

For the equation Equilibrium constant , K_p = ( p_N2(g) x p³_H2(g) ) / p²_NH3(g)

p²_NH3(g) =( p_N2(g) x p³_H2(g) ) / K_p

                                                                   = (0.10 x 0.15³ ) / ( 1.5 x 10 ³ )

                                                                   = 2.25x10 ^-7

                                                   p_NH3(g) = (2.25x10 ^-7 )

                                                                 = 4.74x10 ^-4 atm

So partial pressure of ammonia is p _NH3(g) = 4.74x10 ^-4 atm