Explanation:
[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
c) Concentration of [tex]H_2[/tex] in this mixture
[tex]K_p=0.0489[/tex]
[tex]\Delta n_g[/tex]:
=(gaseous mole on product side)-(gaseous mole on reactant side)=2-4=-2
R = 8.314 J/mol K
T = 256 °C = 256+273 = 529 K
[tex]K_p=0.0489=K_c(RT)^{\Delta n_g}[/tex]
[tex]K_c=940,524.93[/tex]
[tex]K_c=940,524.93=\frac{[NH_3]^2}{[N_2][H_2]^3}=\frac{(0.0300 mol/L)^3}{0.0100 mol/L\times [H_2]^3}[/tex]
[tex][H_2]^3=10,450,277[/tex]
[tex][H_2]=218.62 mol/L[/tex]
The concentration of hydrogen gas is 218.62 mol/L
d)A fourth equilibrium mixture contains equal concentrations of all three chemicals
[tex][N_2]=[H_2]=[NH_3]=x mol/L[/tex]
[tex]K_c=940,524.93=\frac{[NH_3]^2}{[N_2][H_2]^3}=\frac{x^2}{x\times x^3}=\frac{1}{x^2}[/tex]
x =[tex]0.001031 mol/L=1.031\times 10^{-3}mol/L[/tex]
According ideal gas equation:
PV=nRT
[tex]P=\frac{n}{V}RT=\text{(total concentration of gases)}RT[/tex]
[tex]=3\times 1.031\times 10^{-3}mol/L\times 0.0821 L atm/mol K\times 529K=0.1343 atm[/tex]
Total pressure,P = 0.1343 atm
Partial pressure of nitrogen gas[tex]p_{N_2}=P\times \chi_{N_2}=0.1343\times\frac{1}{3}=0.0447 atm[/tex]
Partial pressure of hydrogen gas[tex]p_{H_2}=P\times \chi_{H_2}=0.1343\times\frac{1}{3}=0.0447 atm[/tex]
Partial pressure of ammonia gas[tex]p_{NH_3}=P\times \chi_{NH_3}=0.1343\times\frac{1}{3}=0.0447 atm[/tex]
