Kp = Kc(RT)^delta n.
BTW, isn't I2 a solid. Then does I2 go in the denominator.
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?
3 answers
Oops, my problem states I2 is a gas. I forgot to include it. so does this change the equation?
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
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
1 answer