for the equation 2NOBr(g)<---> 2NO(g) + Br2(g)
a 1.0L vessel was initislly filled with pure NOBr at a pressure of 4.0 atm at 300K after equilibrium was established the partial pressure of NoBr was 2.5 atm. What is Kp for the reaction
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Expert's answer
2013-03-06T07:03:17-0500
According to the ideal gas law pV= nRT In the beginning: n(NOBr) = pV/RT = 4.0 x1.0/(0.082 x 300) = 0.163 mol
Kp = P(Br2) x P(NO)^2 / P (NOBr)^2
P(Br2), P(NO), P (NOBr) – partial pressures
Pertial pressure is P1 = X1 x P(general), whereX1 = n1 / Sn
X(NOBr) = P(NOBr) / P(general) = 2.5/4.0 = 0.625
X(Br2) = (1 - 0.625)/3 = 0.1875
X(NO) = 2 x X(Br2) = 2 x 0.1875 = 0.375
P(Br2) = X(Br2) x P(general) = 0.1875 x 4.0 =0.75 atm
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