Where is uv light found?
1 answer:
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Answer:
0.364
Explanation:
Let's do an equilibrium chart for the reaction of combustion of ammonia:
2NH₃(g) + (3/2)O₂(g) ⇄ N₂(g) + 3H₂O(g)
4.8atm 1.9atm 0 0 Initial
-2x -(3/2)x +x +3x Reacts (stoichiometry is 2:3/2:1:3)
4.8-2x 1.9-(3/2)x x 3x Equilibrium
At equilibrium the velocity of formation of the products is equal to the velocity of the formation of the reactants, thus the partial pressures remain constant.
If pN₂ = 0.63 atm, x = 0.63 atm, thus, at equilibrium
pNH₃ = 4.8 - 2*0.63 = 3.54 atm
pO₂ = 1.9 -(3/2)*0.63 = 0.955 atm
pH₂O = 3*0.63 = 1.89 atm
The pressure equilibrium constant (Kp) is calculated with the partial pressure of the gases substances:
Kp = [(pN₂)*(pH₂O)³]/[(pNH₃)²*]
Kp = [0.63*(1.89)³]/[(3.54)²*]
Kp = 4.2533/11.6953
Kp = 0.364