Answer:
The numerical value of equilibrium constant is 0.0560
Explanation:
Initial Concentration:
[tex][NH^3]= \frac{3mol}{1l}[/tex] = 3M
[[tex]N_2[/tex]] = 1M
[[tex]H_2[/tex]] = 2 M
[tex]N_2 +3H_2 \rightleftarrows 2NH3[/tex]
at the end, [[tex]NH_3[/tex]] = 1.96 M
Thus the change is 3 - 1.96 = 1.04 M
Thus 1.04 moles of reacted
By stoichiometry, 1.04 moles ([tex]NH_3 \times\frac{1 mol N_2}{2 mol NH_3}[/tex]) = 0.52 mol created (in addition to 1 mol already in vessel)
By similar reasoning = [tex]1.04 \times\frac{3}{2 }[/tex] 1.56 moles created
Final concentrations:
[[tex]NH_3[/tex]] = 1.96 M
[[tex]N_2[/tex]] = 0.52 + 1 = 1.52 M
[[tex]H^2[/tex]] = 1.56 + 2 = 3.56 M
[tex]K_c =[/tex] [tex]\frac{[NH^3]^2}{N^2[H^2]^3}[/tex]
[tex]K_c =[/tex] [tex]\frac{1.96^2}{(1.52 )\times(3.56)^3}[/tex]
[tex]K_c =[/tex] 0.0560
Therefore, the numerical value of equilibrium constant is 0.0560
