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The equilibrium pressure of the reaction NH4CN(s) ---------- NH3(g) + HCN(g) is 0.298 atm. If it is allowed to decompose in the presence of NH3 at 0.25 atm calculate y where partial pressure of HCN is y × 10^(-2)

Siri , 5 Years ago
Grade 12
anser 1 Answers
Khimraj

Last Activity: 5 Years ago

Partial Pressure at equilibrium = 2p = 0.298
p = 1.49
Kp = 0.149 x 0.149 = 0.0222
Kp = P(P+0.25)
0.0222 = P2 + 0.25P
P = 0.0694 atm

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