Answer to Question #25674 in Inorganic Chemistry for austinrich

Question #25674
for the equation 2NOBr(g)&lt;---&gt; 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
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) &ndash; 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

P(NO) = X(NO) x P(general) = 0.375 x 4.0 = 1.5atm

Kp = 0.75 x 1.5^2 / 2.5^2 = 0.27

Answer: Kp = 0.27

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