If four moles of NH3 react to form two moles of N2 and six moles of H2 , calculate Kp at the same temperature.

In the gas phase one mole of N2 and three moles of H2 react to form two moles of NH3 At T = 298 Kelvin , Kp = 560000 atm-2





If four moles of NH3 react to form two moles of N2 and six moles of H2 , calculate Kp at the same temperture.

If four moles of NH3 react to form two moles of N2 and six moles of H2 , calculate Kp at the same temperature.
Lancenigo di Villorba (TV), Italy





The reaction described is usually written as follows :





N2(g) + 3 H2(g) %26lt;---%26gt; 2 NH3(g)





You gave me several information about it.


At 25°C, the synthetic mixture overcame to chemical equilibrium when the Partial Pressures of reaction's actors obey to :





5.6E+5 (atm)^2 = Kp' = (pNH3)^2 / ((pN2) * (pH2)^3)





If you refer to another writing, as the following :





2 N2(g) + 6 H2(g) %26lt;---%26gt; 4 NH3(g)





then you modify the "equilibrium equation"





(5.6+5)^2 (atm)^4 = (Kp')^2 = (NH3)^4 / ((pN2)^2 * (pH2)^6)





Hence, you must calculate Kp'' = (Kp')^2 = (5.6+5)^2 =


= 3.14E+11 (atm)^4





I hope this helps you.