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

AI Thread Summary
At 1000K, the equilibrium constant Kp for the reaction H2(g) + Br2(g) <--> 2HBr is 1.2*10^6. After adding 0.952 mol of Br2 to a 1.00L vessel containing 1.25 mol of H2, calculations for partial pressures were attempted using the ideal gas law and equilibrium expressions. The initial partial pressures were calculated as approximately 78.121 for H2 and 102.575 for Br2. However, the final equilibrium partial pressures were found to be 24.5 for H2, 4.75*10^-4 for Br2, and 156 for HBr, indicating discrepancies in the initial calculations. The discussion also questioned the relevance of the given ΔH value and whether the temperature remained constant at 1000K.
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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