Consider the reaction below and its K_{p} value at 350K:
H_{2(g)} + Br_{2(g)} <> 2HBr_{(g)} K_{p} = 3.5 x 10^{4}
If the partial pressures of H_{2} = 0.024 atm and HBr = 5.07 atm, what is the equilibrium partial pressure of Br_{2}?
a. 
12 atm 

b. 
4.7 atm 

c. 
0.26 atm 

d. 
0.031 atm 
Solution:
The Equilibrium Reaction given here is:
H2(g) + Br2(g) > 2HBr(g).
Partial Pressure of H2(g) (pH2) = 0.024atm.
Partial Pressure of HBr(g) (pHBr)= 5.07atm.
The Equilibrium Constant of the Reaction (Kp) = 3.4×10^4.
Consider a Reaction; aX + bY > cZ
If pX ,pY and pZ is the Partial Pressures of the Gases and a,b and c is the Number of Moles of the Gases,then;
The Equilibrium Constant of this Reaction is related to the Partial Pressure of the Gases as:
Kp = (pZ)^c ÷ [(pX)^a × (pY)^b].
Then the Equilibrium Constant of the given Reaction is Calculated as follows:
Kp = (pHBr)^2 ÷ [pH2 × pBr2].
Then; 3.4×10^4 = 5.07^2 ÷ (0.024 atm × pBr2).
pBr2 = 25.7 ÷ (0.024× 3.4×10^4).
Therefore, the Equilibrium Partial Pressure of Br2 (pBr2) = 0.031 atm.
So Option d. is the Correct Answer.
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