Consider the reaction:
2 NO(g) + Br2(g) ↔ 2 NOBr(g) Kp = 28.4 atm-1 at 298 K
In a reaction mixture at equilibrium, the partial pressure of NO is 173 Torr and that of Br2 is 142 Torr. What is the partial pressure (in Torr) of NOBr in this mixture?
PNOBr = ___ Torr
2 NO(g) + Br2(g) ----------------- 2 NOBr(g)
Kp = 28.4 atm-1
1 torr = 0.001316 atm
partail pressure of NO = 173 torr = 0.2277 atm
partial pressure of Br2= 142 torr = 0.1869 atm
Kp = P^2NOBr/P^2NO xPBr2
28.4 = P^2 NOBr / ( 0.2277)^2 x(0.1869)
PNOBr = 0.525 atm
Partial pressur of NOBr = 0.525 atm = 398.94 torr
Partial pressure of NOBr = 398.94 torr.
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