Question

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

Answer 1

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

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

Answer: K_p = 0.27