Homework Help: Chemistry

Posted by Tracy on Monday, March 5, 2012 at 8:35pm.

At 389K, this reaction has a Kc value of 0.0682. 2X(g) + 2Y(g) <-> Z(g) .. Calculate Kp at 389K. Kp=

this is what I did.. can somebody confirm or find any mistakes in my thought process?
Kp = Kc(R)(T)^Delta n
Delta n = (1)-(2+2) = -3
Kp = (0.0682)(0.08206)(398K)^-3 = 0.09678

i'm confused because I'm not sure if delta n is allowed to be negative. please help!

Chemistry - DrBob222, Monday, March 5, 2012 at 9:12pm
Yes, delta n may be negative and in this instance you are right that it is -3. However, you have parentheses in the wrong place. The formula is
Kp = Kc (RT)^delta n.
Kp = 0.0628(0.08206*398)^-3
Chemistry - Tracy, Monday, March 5, 2012 at 9:46pm
Thank you for proofreading!

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