Consider the reaction of nitrogen monoxide with oxygen at 25 oC to produce nitro

Question

Consider the reaction of nitrogen monoxide with oxygen at 25 oC to produce nitrogen dioxide and its corresponding partial pressure equilibrium constant, Kp to answer the next two questions:2 NO (g) + O2 (g) Û 2 NO2 (g)                  Kp = 2.2 x 1012

A.)Given that Kp = 2.2 x 1012 for the reaction above, what is the concentration equilibrium constant, Kc for this reaction?

B.)Given that the equilibrium partial pressure constant, Kpfor the reaction above is 2.2 x 1012at 25oC, what is the expected partial pressure of NO2when the following equilibrium partial pressures were observed: PNO = 3.0 x 10-4atm and PO2 = 3.5 x 10-4atm?

Explanation / Answer

Answer: According to question ,

A] here we have to use the equation Kp  = Kc * (0.0821 T)n'

here 0.0821 is the value of R , T is temperature in kelvine , n' is equal to the number of moles of gaseos product- moles of gaseous reactant

hence n' =2 -3 = -1 [ now putting this value in equation and using some mathematical step we get :

hence Kc = Kp * 0.0821 *T

here T [ temperature is = 25 + 273 = 298 K

Kc = 2.2 * 1012 * 0.0821 * 298

Kc  = 53.82 * 1012 = 5.382 * 1013

hence it is all about the first part .

B] According to the given informations , here first we have to write the kp expression of the given equation which is writen as :

Kp = [ PNO2]2  / [Po2 ] [P no]2  

Now putting the all given values we get ,

2.2 * 1012   = P2 / [3.5 * 10-4] [ 3.0 * 10-4]2   

2.2 * 1012  * 3.5 * 10-4 * 9 * 10-8 = P2

Now solve for P to get the required answer

Hence it is all about the given question , it helps you to understand the steps .

thanks you

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