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Chem 1120

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The KC for the reaction
I2 --> 2I
is 3.8 x 10 ^ -5 at 727 degrees C.
Calculate Kc and Kp for the equilibrium of
2I -> I2
at the same conditions.

I don't understand how to set up or solve this problem. I know Kc=[I]^2 / [I2] = 3.8x10^-5 for the first equation but I don't understand how to relate it to the second equation or how to use the temperature in the process?

  • Chem 1120 - ,

    Kp = Kc(RT)^delta n.
    BTW, isn't I2 a solid. Then does I2 go in the denominator.

  • Chem 1120 - ,

    Oops, my problem states I2 is a gas. I forgot to include it. so does this change the equation?

  • Chem 1120 - ,

    Yes and no.
    You have Kc given in the problem. All you need to do is to substitute Kc into the Kp = Kc*(RT)^delta n and solve for Kp and you're done. The only thing changing I2 from solid to gas changes is delta n. For I2(g) ==>2I^-(g) delta n will be 2-1 = 1

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