Gaseous BrCl is placed in a closed container at 2054 oC, where it partially decomposes to Br2 and Cl2:
2 BrCl(g) 
1 Br2(g) + 1 Cl2(g)
At equilibrium it is found that p(BrCl) = 0.004430 atm,
p(Br2) = 0.008780 atm, and p(Cl2) = 0.002480
atm. What is the value of KP at this
temperature?
Homework Answers
From the given data,
The equilibrium pressures are as,
PBrCl = 0.004430 atm
PBr2 = 0.008780 atm
PCl2 = 0.002480 atm
Hence , we know that,
Kp = (PBr2).(PCl2)/(PBrCl)^2
Kp =(0.008780)(0.002480)/(0.004430)^2
Kp = 1.109
Hence, equilibrium constant Kp = 1.11
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