What are the partial pressures of H2, Br2, and HBr at equilibrium?

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SUMMARY

The equilibrium partial pressures of H2, Br2, and HBr at 1000K for the reaction H2(g) + Br2(g) <--> 2HBr are calculated using the equilibrium constant Kp of 1.2 x 10^6. After adding 0.952 mol of Br2 to a 1.00L vessel containing 1.25 mol of H2, the calculated partial pressures are Pf(H2) = 24.5 atm, Pf(Br2) = 4.75 x 10^-4 atm, and Pf(HBr) = 156 atm. The calculations utilize the ideal gas law and the relationship between Kp and Kc, confirming the equilibrium state of the system.

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mike1967
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Homework Statement


at 1000K, Kp=1.2*10^6 and Delta H = -101.7 kJ for the reaction H2(g)+ Br2(g) <-->2HBr.

A 0.952 mol quantity of Br2 is added to a 1.00L reaction vessel that contains 1.25 mol of H2 gas at 1000K . What are the partial pressures of H2 ,Br2 ,HBr and at equilibrium?

Homework Equations


PV=nRT
Kc=Kp(RT)^n
Kc=products over reactants reased to power of stoichiometric coefficients

The Attempt at a Solution


P=nRT/v
Pi(H2)=(1.25)(.08206)(1000)=78.121
Pi(Br2)=(.952)(.08206)(1000)=102.575

2.1*10^6=((2x)^2)/(78.121-x)(102.575-x))

x=78.121

Pf(H2)=102.575-x=25.45
Pf(Br2)=78.121-x=0
Pf(HBr2)=2x=156.24

Mastering Chemistry rejected
 
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I just skimmed so it is possible I missed something, but at first sight nothing cries "wrong!", apart from the significant digits in the final answer.

I wonder why they give delta H? It is not clear if the final temperature is still 1000K.
 
It ends up they wanted
Pf(H2) as 24.5
Pf(Br2) as 4.75*10^-4
Pf(HBr) as 156
 

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