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

Question

#\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?

Sorry if this has been asked before!

Caleb Adams · Accepted Answer

#k=3xx10^-4#

The given reversible gaseous reaction's balanced equation

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

Molar masses of the products and reactants

#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#

Serenity Allen · Answer

undefined To find Kc for a reaction, you need to write the balanced chemical equation and determine the concentrations of the reactants and products at equilibrium. Then, plug these values into the expression for Kc, which is the equilibrium constant for concentrations.