Consider the following reaction and associated equalibrium constant: N2 (g) + O2

Question

Consider the following reaction and associated equalibrium constant:

N2 (g) + O2 (g) <--> 2NO (g) Kp= 4

Container A has initial pressures: N2 = 2 atm, O2 = 3atm, NO = 0 atm

Contain B: N2 = 1 atm, O2 = 2atm, NO = 2 atm

Container C: N2 = 1 atm, O2 = 1atm, NO = 2 atm

1. While the partial pressures in container A be identical to the partial pressures in container B at equilibrium (The answer is Yes, but why?)

2. While the partial pressures in container A be identical to the partial pressures in container C at equilibrium (Answer is No)

3. While the partial pressures in container B be identical to the partial pressures in container C at equilibrium (Answer is No)

Thank you!

Explanation / Answer

I think, in this case, Kp=Kc..
Kp=Kc(RT)^n, where n is the change in number of moles from reactant to product

Here, it is 2 moles of reactants forming 2 moles of products.

So, RT^0=1

Therefore, Kp=Kc X 1

Kp=4.10 x 10^-31

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