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
$K_c=([I_2(g)][Cl_2(g)])/([ICl (g)]$
$=(x/V*x/V)/((\alpha-2x)/V)$
$=x^2/(V(\alpha-2x))$
$=(1.5xx10^-4)^2/(0.625(4.2xx10^(-4)-2*1.5xx10^-4))$
$=(2.25xx10^-8)/(0.625xx1.2xx10^-4)=3xx10^-4$
