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Gaseous NOBr is placed in a closed container at -165 oC, where it partially decomposes to...

Gaseous NOBr is placed in a closed container at -165 oC, where it partially decomposes to NO and Br2:

2 NOBr(g) revrxnarrow.gif 2 NO(g) + 1 Br2(g)

At equilibrium it is found that p(NOBr) = 0.003400 atm, p(NO) = 0.005450 atm, and p(Br2) = 0.007050 atm. What is the value of KP at this temperature?


KP = ______

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Answer #1

2 NORG) = 2 Norg) + 1 emzeg) Kp = (Pro)? ( Pean) 1 (PROBA)? Kp = (0.00545012(0.007050) (0.002400) Kp= 0.018

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