An equilibrium mixture in an 8.00 L vessel contains 1.60 mol NH3, 0.800 mol N2, and 1.20 mol H2. What is the value of Kc?

Consider the equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g) at a certain temperature.

An equilibrium mixture in an 8.00 L vessel contains 1.60 mol NH3, 0.800 mol N2, and 1.20 mol H2. What is the value of Kc?

A)1.85
B)29.6
C)37.4
D)75.8
E)119

1 Answer
Apr 23, 2018

#"E) 119 L"^2"/mol"^2#

Explanation:

#"Concentration" = "Moles of solute"/"Volume of solution"#

At equilibrium concentrations are

  • #["N"_2] = "0.8 mol"/"8.00 L" = "0.1 mol/L"#
  • #["H"_2] = "1.2 mol"/"8.00 L" = "0.15 mol/L"#
  • #["NH"_3] = "1.6 mol"/"8.00 L" = "0.2 mol/L"#

#"N"_2 + "3H"_2 rightleftharpoons "2NH"_3#

Equilibrium constant #"K"_"c"# for above reaction is given by

#"K"_"c" = ["NH"_3]^2 / (["N"_2]["H"_2]^3)#

Where #["NH"_3], ["H"_2]# and #["N"_2]# are concentrations at equilibrium.

#"K"_"c" = ("0.2 mol/L")^2/(("0.1 mol/L") × ("0.15 mol/L")^3)#

#"K"_"c" = ("0.04 mol"^2"/L"^2)/("0.1 mol/L" × "0.003375 mol"^3"/L"^3)#

#"K"_"c" = "118.52 L"^2"/mol"^2 ≈ "119 L"^2"/mol"^2#