# How do you calculate the value of kp for the following question?

Hi, I've trying figure this out, but am having no luck whatsoever;

Consider the following reaction: 2NO2(g) ---> N204 (g) Kp = 7.1 at 25 degrees

what is the value of Kp for the following reaction at 25 degrees?

1/2N204 (g)---> NO2 (g)

Thank you so much

You have given that Kp = 7.1 which is also = Pn2o4 / (Pno2)^2

You want to determine the Kp for the reaction 1/2N2O4(g) ---? NO2(g)

The equilibrium expression is: Kp = Pno2 /(Pn2o4)^1/2

The relation ship between the two equations is that the 2nd one is the first one reversed and the amounts halved. The relationship between the two Kp's is that the second one is the reciprocal of the first that has then been raised to the 1/2 power (that's the square root).

So, your new Kp = (1/7.1)^1/2 = (0.14085)^1/2 = .375 = 0.38.

Did that help?

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