Equilibrium Constant Keq: 2H2S(g) <-> H2(g) + S2(g) at 830°C

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Discussion Overview

The discussion revolves around the equilibrium constant (Keq) for the reaction 2H2S(g) <-> H2(g) + S2(g) at 830°C. Participants explore the implications of the given Keq value, the calculation of equilibrium concentrations, and the interpretation of results in the context of chemical equilibrium.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states the value of Keq as 4.20x10^-6 and questions whether reactants or products are favored in the reaction.
  • Another participant suggests writing out the expression for Keq using the law of mass action and proposes an ICE table for the reaction.
  • A participant claims to have solved for x but encounters a negative concentration for H2S, indicating a potential error in calculations.
  • Further posts express confusion about solving the resulting polynomial equation and question the presence of terms like 4x^3.
  • Some participants assert that products are favored based on the Keq value but express uncertainty about solving the equation.

Areas of Agreement / Disagreement

Participants generally agree that products are favored based on the Keq value, but there is no consensus on how to solve the resulting equations or the implications of their calculations.

Contextual Notes

Participants have not resolved the mathematical steps necessary to find the equilibrium concentrations, and there are indications of errors in earlier calculations that remain unaddressed.

Fusilli_Jerry89
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The value of Keq for the reaction: 2H2S(g) <-> 2H2(g) + S2(g) is 4.20x10^-6 at 830 degrees celsius.
a) What is favoured in this reaction, reactants or products?
b) What concentration of S2 can be expected at equilibrium after 0.200 mol of H2S is injected into an empty 1.00 L flask?
 
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a) Do you know how to write out Keq? (Hint: Use the law of mass action).
b) Make an ice box
2H2S(g) <--> 2H2(g) + S2 (g)
i .200 0 0
c -2x +2x +x
e .200 - 2x 2x x

Keq = [((2x)^2)(x)] / [(.200 - 2x)^2]
 
Last edited:
i solved x to equal 1.431 but when u plug it back into the original question, H2S comes out to a negative number
 
oh nm i made a mistake
 
k got down to 4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0 now what
 
Fusilli_Jerry89 said:
k got down to 4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0 now what
What does that mean?

Look to your rules of equilibrium to determine which side of the reaction is favoured.

As for x, you can solve for it.
 
i put the products are favoured, but i do not know how to solve this equation because it has an 4^3 x^2 and an x
 
Fusilli_Jerry89 said:
i put the products are favoured, but i do not know how to solve this equation because it has an 4^3 x^2 and an x
Why do you have [itex]4x^{3}[/itex] in your equation? Solve for x.
 

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