Find K_cKc 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?


Sorry if this has been asked before!

1 Answer
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