# At a certain temperature, 4.0 mol #NH_3# is introduced into a 2.0 L container, and the #NH_3# partially dissociates to #2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)#. At equilibrium, 2.0 mol #NH_3# remains. What is the value of #K_c#?

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**A clearer version:**

At a certain temperature, 4.0 mol #NH_3# is introduced into a 2.0 L container, and the #NH_3# partially dissociates to #2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)# .

At equilibrium, 2.0 mol #NH_3# remains. What is the value of #K_c# ?

**Note:** I started an ICE table but I don't know what to put in the #N_2# and #3H_2# columns.

**A clearer version:**

At a certain temperature, 4.0 mol

At equilibrium, 2.0 mol

**Note:** I started an ICE table but I don't know what to put in the

##### 1 Answer

Remember to include the coefficients in the change in concentratio, as well as the exponents. I get

If

#color(white)("From the reaction, "Deltan_"gas" = (1+3) - (2) = 2,)#

#color(white)("since there are 3 mols H"_2, "1 mol N"_2, "and 2 mols NH"_3.)#

#color(white)("So,")#

#color(white)(K_p = 1.69 cdot ("0.08206 L"cdot"atm/mol"cdot"K" cdot "298 K")^((1+3)-(2)))#

#color(white)(~~ 1010" in implied units of atm")#

Let's first find the concentrations, because

#"4.0 mols NH"_3/("2.0 L") = "2.0 M"#

#"2.0 mols NH"_3/("2.0 L") = "1.0 M"#

The ICE table uses

#color(red)(2)"NH"_3(g) rightleftharpoons "N"_2(g) + color(red)(3)"H"_2(g)#

#"I"" "2.0" "" "" "" "0" "" "" "0#

#"C"" "-color(red)(2)x" "" "+x" "" "+color(red)(3)x#

#"E"" "2.0-color(red)(2)x" "" "x" "" "color(red)(3)x#

*Remember that the coefficients go into the change in concentration and exponents.*

Now, the

#K_c = (x(color(red)(3)x)^color(red)(3))/(2.0 - color(red)(2)x)^color(red)(2)#

#= (27x^4)/(2.0 - 2x)^2#

But like mentioned, we know

#2.0 - 2x = "1.0 M"# .

Therefore:

#x = "0.5 M"# .

As a result,

#color(blue)(K_c) = (27(0.5)^4)/(2.0 - 2(0.5))^2#

#= color(blue)(1.69)#