Question

at 400 degrees celsius and 350 bar, a 1:3 mixture of nitrogen and hydrogen gases react to form an equillibrium mixture containing ammonia at a mole fraction of 0.5. Assuming perfect gas behavior, calculate the equillibrium constnat, K, for:

N₂(g) + 3H₂ = 2NH₃

Answer 1

the reaction is N2(g)+3H2(g) <---->2NH3(g)

molar ratio of N2 and H2= 1:3

let moles of N2= 1, moles of H2= 3

let x= drop in moles of N2 to reach equilibrium

at Equilibrium [N2]= 1-x, [H2]= 3-3x, [NH3]=2x

total moles at equilibrium = moles of N2+moles of H2+moles of NH3=1-x+3-3x+2x =4-2x

mole fraction of NH3= 2x/(4-2x)=0.5

x/(2-x)=0.5

x=0.5*(2-x)

x=1-0.5x

1.5x=1

x=1/1.5 = 2/3=0.67

[N2]= 1-0.67=0.33 , [H2]=3-3*0.67= 0.99 and [NH3]= 2*0.67= 1.34

K = equilibrium constant = [NH3]²/ [N2] [H2]³= (1.34)²/(0.33*(0.99)²= 5.55