Solving Partial Pressures for H2 & Cl2 given K & PHCl

  • Thread starter Thread starter jtruth914
  • Start date Start date
  • Tags Tags
    Partial
Click For Summary
SUMMARY

The equilibrium constant K for the reaction H2(g) + Cl2(g) <--> 2HCl(g) at 298K is calculated to be 3.2 x 10^-34 using the Gibbs free energy equation dG* = -RTlnK. Given that the equilibrium partial pressure of HCl is 1 bar, the equilibrium partial pressures of H2 and Cl2 can be determined using the relationship K = (PHCl)^2 / (PH2)(PCl2). By applying stoichiometric principles and the assumption that H2 and Cl2 are products of dissociation, the partial pressures can be accurately calculated.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Familiarity with the Gibbs free energy equation
  • Knowledge of stoichiometry in chemical reactions
  • Ability to manipulate equilibrium constant expressions
NEXT STEPS
  • Study the derivation of the Gibbs free energy equation in chemical thermodynamics
  • Learn how to calculate equilibrium constants for various reactions
  • Explore the concept of partial pressures in gas-phase reactions
  • Practice solving equilibrium problems involving multiple species
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical engineering or thermodynamics who are looking to deepen their understanding of equilibrium calculations and gas behavior.

jtruth914
Messages
20
Reaction score
0
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?
 
Physics news on Phys.org
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
 

Similar threads

Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
7K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K