Question

At 284 oC the equilibrium constant for the reaction: 2 HBr(g) H2(g) + Br2(g) is KP = 1.85e-09. If the initial pressure of HBr is 0.00378 atm, what are the equilibrium partial pressures of HBr, H2, and Br2? p(HBr) = .

p(H2) = .

p(Br2) = .

Answer 1

I'll assume the reaction is as follow: 2HBr_(g) <--------> H_2(g) + Br_2(g)

Kp = pBr₂ * pH₂ / pHBr²

2HBr_(g) <--------> H_2(g) + Br_2(g)   doing a ICE chart:

i. 0.00378 0 0

e. 0.00378-x x x

1.85x10^-9 = x² / (0.00378-x)² Kp is a very low value, so we can assume that x is really low too so, we can assume too that 0.00378-x = 0.00378

1.85x10^-9 = x² / 1.43x10^-5

x^{2 =} 1.85x10^-9 * 1.43x10^-5

x = 1.63x10^-7 atm = pBr₂ = pH₂

pHBr = 0.00378 - 1.63x10^-7 = 0.0037789 atm

Hope this helps