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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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.

Serenity Allen · Accepted Answer

Remember to include the coefficients in the change in concentratio, as well as the exponents. I get #K_c = 1.69#. If #K_p = K_c(RT)^(Deltan_"gas")#, what would #K_p# be if this reaction has been occurring at #"298 K"#? #R = "0.08206 L"cdot"atm/mol"cdot"K"#. Highlight below to see. #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 #K_c# is in terms of concentrations. #"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 #"NH"_3# having the initial concentration of #"2.0 M"# and final of #"1.0 M"#. Therefore, we should be able to find #x# already, but let's assume we don't know #x# yet. #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# Recall that the exponents and concentration change with the coefficients. Now, the #K_c# would be: #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 #x#. This is because #2.0 - 2x = "1.0 M"#. Consequently: #x = "0.5 M"#. Consequently, #color(blue)(K_c) = (27(0.5)^4)/(2.0 - 2(0.5))^2# #= color(blue)(1.69)#

Cooper Adams · Answer

undefined The equilibrium constant $ K_c $ is equal to 1.5.

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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