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Gaseous hydrogen iodide is placed in a closed container  at 425^0 C, Where it...

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roshan-rox | Valedictorian

Posted May 22, 2013 at 7:06 AM via web

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Gaseous hydrogen iodide is placed in a closed container  at 425^0 C, Where it partially decomposes to hydrogen and iodine.

2HI(g)------->H_2(g) + I_2(g).

         <-------

found theat [HI] = 3.53*10^-3M,[H_2] = 4.79*10^-4M, and [I_2] = 4.79*10^-4 M.What is the value of K_c at this temperature?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 22, 2013 at 7:22 AM (Answer #1)

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For a chemical reaction depicted by aA + bB <--> cC + dD, the equilibrium constant `K_c` is defined in terms of the concentration of the reactants and products by` K_c = ([C]^c*[D]^d)/([A]^a*[B]^b)` .

When gaseous hydrogen iodide is placed in a closed container at 425 C, the reaction that takes place is represented by the equation `2HI` (g)<-->`H_2` (g) + `I_2` (g), it is found that `[HI]` = 3.53*10^-3 M, [`H_2` ] = 4.79*10^-4 M, and   [`I_2` ] = 4.79*10^-4 M.

The value of `K_c` in this case is `([H_2]*[I_2])/([HI]^2)`

= `(4.79*10^-4*4.79*10^-4)/(3.53*10^-3)^2`

= 0.0184

The equilibrium constant is approximately 0.0184

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