What volume flask should be used to get 0.37 moles #I# present for every mole #I_2# at equilibrium?

Question

For the dissociation of #I_2"(g)"# at 1200°C, #K_c=0.011#. What volume flask should we use if we want 0.37 moles of #I# to be present for every mole of #I_2# present at equilibrium?

#I_2"(g)"\rightleftharpoons2I"(g)"#

My work:

(can't see? it sets equilibrium pressures as stated below:)
#P_I=\frac{0.370.0821(1200+273)}{V}=\frac{0.37121}{V}" atm"#
#P_(I_2)=\frac{1.000.0821(1200+273)}{V}=\frac{1.00121}{V}=121/V" atm"#

after using those pressures for #K_p# in the formula #\color(red)(K_c=K_p(RT)^(\Deltan))#, I got #V=0.0307/0.011\approx0.279" L"#

Jeremiah Adams · Accepted Answer

Are we missing information? We need to be able to figure out the exact number of mols of #"I"# or #"I"_2#... no ratios, an actual number. The ICE table is where we start: #"I"_2(g) " "rightleftharpoons" " 2"I"(g)# #"I"" "["I"_2]_i" "" "" "" "0#
#"C"" "-x" "" "" "+2x#
#"E"" "["I"_2]_i-x" "" "2x# In order to obtain:, we write the mass action expression. #K_c = 0.011 = (["I"]^2)/(["I"_2])# #= (2x)^2/(["I"_2]_i - x)# It seems that what we desire is #(2x)/(["I"_2]_i - x) = 0.37 = (["I"])/(["I"_2]) = ("mols I"//cancel"L")/("mols I"_2//cancel"L")# Now, we possess a set of equations: #(2x)^2/(["I"_2]_i - x) = 0.011# #" "" "bb((1))#
(Note that #0.011# is in implied units of #"M"#.) #(2x)/(["I"_2]_i - x) = 0.37# #" "" "bb((2))#
(Note that #0.37# is unitless.) Upon examination, we have: #(2x)^2/(["I"_2]_i - x) = 0.011 = 0.37(2x) = 0.74x# Thus, #x = ("0.011 M")/0.74 = ul"0.0149 M"# As a result, take #(2)# to get #(2 cdot "0.0149 M")/(["I"_2]_i - "0.0149 M") = 0.37# #2 cdot "0.0149 M" = 0.37(["I"_2]_i - "0.0149 M")# #"0.0297 M" = 0.37["I"_2]_i - "0.0055 M"# Therefore, the initial concentration of #"I"_2# is: #color(blue)(["I"_2]_i) = ("0.0352 M")/(0.37) = color(blue)("0.0952 M")# and the percentage of detachment is #alpha = x/(["I"_2]_i) = 0.1565# Now we just need to know the ACTUAL mols of either #"I"# or #"I"_2#... the ratio desired can be ANY NUMBER of combinations of quantities...  The fraction of dissociation #alpha# and the concentration lost #x# also vary with concentration #["I"_2]_i#, so we cannot use those to determine the volume.

Jose Ayala · Answer

undefined A 2-liter flask should be used to get 0.37 moles of I present for every mole of I2 at equilibrium.