Answer:
The pressure equilibrium constant for the combustion of ammonia is 0.132
Step-by-step explanation:
the balanced chemical reaction of the combustion of ammonia is as follows:
4NH3 + 3O2 = 2N2 + 6H2O
initially we have 4.8 1.9 0 0
in equilibrium 3.54 0.955 0.63 1.89
therefore the value of the equilibrium constant in terms of pressures is equal to:
Keq = ((PN2)^2 * (PH2O)^6)/((PNH3)^4 * (PO2)^3) = ((0.63^2)*(1.89^6))/((3.54^4) * (0.955^3)) = 0.132