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 = pBr_{2} * pH_{2} / pHBr^{2}

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} = x^{2} / (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} = x^{2} / 1.43x10^{-5}

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

x = 1.63x10^{-7} atm = pBr_{2} =
pH_{2}

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

Hope this helps

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