Question

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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Sorry if this has been asked before!

Answer 1

`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`