Using an ICE Table, how do we calculate #K_c# for the following reaction?

Question

At #"373 K"#, #"0.1 mols"# of #N_2O_4# is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is #"0.12 mols"#.

#N_2O_4 rightleftharpoons 2NO_2#

Naomi Aronson · Accepted Answer

#K_c = 0.36# Firstly, we assume ideal gases, which are both gases at room temperature, of course. Next, we review what a two-component equilibrium's concentration equilibrium constant is defined as: #K_c = ([B]^(nu_B))/([A]^(nu_A))#, for the equilibrium #nu_A A rightleftharpoons nu_B B#.  Even though we are looking at two gases, we are given the #bb"mol"#s and the total vessel's volume, so we can still get initial and equilibrium concentrations as usual.  However, since the vessel is not going to change size, and ideal gases are assumed to fill the vessel completely and evenly, let us simply work in #"mol"#s for the ICE table, and divide by #"1 L"# later. Usually, this is what we would write: #N_2O_4(g) rightleftharpoons 2NO_2(g)# #"I"" "" "0.1" "" "" "0#
#"C"" "" "- x" "+2x#
#"E"" " 0.1 - x" "" "2x#  noting that the consumption of #x# #"mol"#s of #"N"_2"O"_4# yields #2x# #"mol"#s of #"NO"_2(g)#, and not just #x#. But we know what #x# is. Since at equilibrium, #n_(NO_2) = "0.12 mols"#, we have that #2x = "0.12 mol"#s, or #x = "0.06 mols"#. Therefore, we can already calculate #K_c#: #K_c = ([NO_2]^(2))/([N_2O_4])# #= ((2x)^2" mols"^cancel(2)"/" cancel("1") "L"^cancel(2))/((0.1 - x)cancel"mols""/" cancel"1 L")# #= (4x^2)/(0.1 - x)# #"M"# #= (4(0.06)^2)/(0.1 - 0.06)# #"M"# However, #K_c# is conventionally reported without units, so: #color(blue)(K_c = 0.36)#

Avery Adams · Answer

undefined To calculate $ K_c $ for a reaction using an ICE table, follow these steps:

1. Write the balanced chemical equation.
2. Set up an ICE table with initial, change, and equilibrium concentrations.
3. Use stoichiometry to determine changes in concentration.
4. Substitute equilibrium concentrations into the expression for $ K_c $.
5. Solve for $ K_c $ using the equilibrium concentrations.

Using an ICE Table, how do we calculate #K_c# for the following reaction?

At #"373 K"#, #"0.1 mols"# of #N_2O_4# is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is #"0.12 mols"#. #N_2O_4 rightleftharpoons 2NO_2#

At #"373 K"#, #"0.1 mols"# of #N_2O_4# is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is #"0.12 mols"#.

#N_2O_4 rightleftharpoons 2NO_2#