Solving Partial Pressures for H2 & Cl2 given K & PHCl

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The equilibrium constant K for the reaction H2(g) + Cl2(g) <--> 2HCl(g) is calculated to be 3.2 x 10^-34 at 298K. Given that the equilibrium partial pressure of HCl is 1 bar, the partial pressures of H2 and Cl2 can be determined using the equation K = (PHCl)^2 / (PH2)(PCl2). By applying stoichiometry and the assumption that all H2 and Cl2 are products of dissociation, the necessary calculations can be performed. The solution involves using the total pressure and the relationship between the partial pressures to find the unknowns. The problem was successfully resolved with this approach.
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a) Determine K at 298K for the reaction H2(g) + Cl2(g) <---> 2HCl(g)
b) The equilibrium partial pressure of HCl is 1 bar. Determine the equilibrium partial pressures of H2 and Cl2.


This question was on my test. I got K= 3.2 x 10^-34 using dG*=-RTlnK when dG=0 at equilibrium. I'm having trouble determining the partial pressures. I know that Pj=(xj)(Ptot). I also know that K= (PHCl)^2/(PH2)(PCl2), but I still don't know how to determine the partial pressure just based off of PHCl=1 bar?
 
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You are most likely to assume you started with HCl and all H2 and Cl2 are products of dissociation. This, plus stoichiometry, gives you all information you need to solve the problem.
 
I figured it out. thanks
 
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