Find #K_c# for the reaction below (see details)?

#\sf{2ICl(g)\harrI_2(g)+Cl_2(g)#

A 0.0682-gram sample of #"ICl (g)"# is placed in a 625-mL reaction vessel at 628 K. When equilibrium is reached between the #"ICl (g)"# and #I_2" (g)"# formed by dissociation, 0.0383-grams of #I_2# are present.
What is #K_c# for this reaction?

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1 Answer
P dilip_k · seol
Aug 5, 2018

#k=3xx10^-4#

Explanation:

The balanced equation of the given reversible gaseous reaction

#\sf{2ICl(g)rightleftharpoonsI_2(g)+Cl_2(g)#

Molar masses of reactant and products

#ICl->162.5" "g*mol^-1#

#I_2->254" "g*mol^-1#

#Cl_2->71" "g*mol^-1#

Volume of reaction vessel #V=625mL=0.625L#

ICE Table

#" "" "\sf{2ICl(g)" "" "rightleftharpoons" "I_2(g)" "+" "Cl_2(g)#

#I" " " "\alpha" "mol" "" "" "" "0" "mol" "" "0" "mol#

#C" "-2x" "mol" "" "" "x" "mol" "" "x" "mol#

#E" "\alpha-2x" "mol" "" "" "x" "mol" "" "x" "mol#

By the problem initial amount of #ICl(g)#

#\alpha=(0.0682" g "ICl)/(162.5" g/mol "ICl)~~4.2xx10^(-4)mol#

Amount of #I_2(g)# as well as #Cl_2(g)# in equilibrium mixture

#x=(0.0383" g")/(254" g/mol")~~1.5xx10^-4mol#

The equilibrium constant #K_c# of the reaction

#K_c=([I_2(g)][Cl_2(g)])/([ICl (g)]#

#=(x/V*x/V)/((\alpha-2x)/V)#

#=x^2/(V(\alpha-2x))#

#=(1.5xx10^-4)^2/(0.625(4.2xx10^(-4)-2*1.5xx10^-4))#

#=(2.25xx10^-8)/(0.625xx1.2xx10^-4)=3xx10^-4#

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